diff --git a/EXPLANATIONS.md b/EXPLANATIONS.md
index e859c33..929708a 100644
--- a/EXPLANATIONS.md
+++ b/EXPLANATIONS.md
@@ -176,6 +176,43 @@ band-plan lookup), say that plainly and cite the band plan in
`source` — confidence stays low (3–4) until someone finds a deeper
hook.
+### The Hilfsmittel note
+
+Candidates may use the official exam aid (`Hilfsmittel_12062024.pdf`)
+during the exam. Its complete contents — every formula and table, with
+**printed page numbers** — are catalogued in
+[`references/Hilfsmittel.md`](references/Hilfsmittel.md). When a
+question's answer is a lookup or a direct application of something in
+that sheet, append a short note to the end of the `explanation` body:
+
+```
+... Hilfsmittel:
+```
+
+Rules — this is the part that was historically done badly (generic
+boilerplate stamped on everything), so be strict:
+
+- **Only cite what is actually in the sheet.** Verify against
+ `references/Hilfsmittel.md`. If the fact the question turns on is a
+ memory item — diode forward voltages (~0.6 V / ~0.3 V), `tan δ = 1/Q`,
+ S-meter step = 6 dB, harmonics = n × fundamental, ppm/percent
+ arithmetic, semiconductor behaviour, definitions, antenna
+ length/shortening factors — **do not add a Hilfsmittel note at all.**
+- **Name the specific formula(s) or table, and the page.** Not "the
+ formula is in the Formelsammlung" but e.g. `P = U²/R (Leistung,
+ S.12)` or `the Widerstands-Farbcode table (S.11)`.
+- **For multi-step calculations, give the order:** "first
+ `U_eff = Û/√2` (Wechselspannung, S.12), then `P = U_eff²/R`
+ (Leistung, S.12)". State plainly when a value comes from outside the
+ sheet (e.g. a diode drop) versus from a sheet formula.
+- **For a pure table lookup**, say it is a lookup and cite the table +
+ page (e.g. "a table lookup, not a memory item — the band limits …
+ are in the Frequenzbereichszuweisung (Anlage 1, Tabellarische
+ Übersicht, S. 2–3)").
+- **Page numbering:** the printed page number = PDF page − 2 (the cover
+ and the "Hinweis" page are unnumbered). Always cite the printed
+ number, as `references/Hilfsmittel.md` does.
+
---
## 5. Confidence scale
diff --git a/explanations.json b/explanations.json
index bb65165..a2763aa 100644
--- a/explanations.json
+++ b/explanations.json
@@ -25,73 +25,73 @@
},
"AA105": {
"revision": 3,
- "explanation": "For power ratios, gain in dB is 10 log10(P2/P1); 10 log10(40) is about 16 dB. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "For power ratios, gain in dB is 10 log10(P2/P1); 10 log10(40) is about 16 dB. Hilfsmittel: apply g = 10·log10(P2/P1) (Pegel, S.15); the table has no 16 dB row — combine +10 dB (×10) and +6 dB (×4) → ×40.",
"source": "https://50ohm.de/NEA_dezibel_2.html#AA105",
"confidence": 8
},
"AA106": {
"revision": 3,
- "explanation": "A 16 dB power gain is approximately 40 times; with 1 W drive the output is about 40 W, below the 100 W maximum. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "A 16 dB power gain is approximately 40 times; with 1 W drive the output is about 40 W, below the 100 W maximum. Hilfsmittel: reverse g = 10·log10(P2/P1) (Pegel, S.15); 16 dB is not a table row — it is +10 dB (×10) plus +6 dB (×4) = ×40.",
"source": "https://50ohm.de/NEA_dezibel_2.html#AA106",
"confidence": 8
},
"AA107": {
"revision": 3,
- "explanation": "1 W is 0 dBW, and a 10 dB amplifier raises the power level by 10 dB, giving 10 dBW. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "1 W is 0 dBW, and a 10 dB amplifier raises the power level by 10 dB, giving 10 dBW. Hilfsmittel: apply p = 10·log10(P/1W) dBW (Pegel, S.15); +10 dB simply adds 10 to the dBW level.",
"source": "https://50ohm.de/NEA_dezibel_2.html#AA107",
"confidence": 8
},
"AA108": {
"revision": 2,
- "explanation": "dBW is referenced to 1 W, so 20 dBW means 10^(20/10) W = 100 W = 10^2 W. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "dBW is referenced to 1 W, so 20 dBW means 10^(20/10) W = 100 W = 10^2 W. Hilfsmittel: invert p = 10·log10(P/1W) dBW (Pegel, S.15): P = 1W·10^(p/10).",
"source": "https://50ohm.de/NEA_dezibel_2.html#AA108",
"confidence": 8
},
"AA109": {
"revision": 3,
- "explanation": "The amplifier output is 10 W; in dBm that is 10 W = 10000 mW, and 10 log10(10000) = 40 dBm. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "The amplifier output is 10 W; in dBm that is 10 W = 10000 mW, and 10 log10(10000) = 40 dBm. Hilfsmittel: apply p = 10·log10(P/1mW) dBm (Pegel, S.15) after converting 10 W to 10000 mW.",
"source": "https://50ohm.de/NEA_dezibel_2.html#AA109",
"confidence": 8
},
"AA110": {
"revision": 3,
- "explanation": "dBm is referenced to 1 mW: 0 dBm is 1 mW, +3 dB is about double or 2 mW, and +20 dB is 100 times or 100 mW. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "dBm is referenced to 1 mW: 0 dBm is 1 mW, +3 dB is about double or 2 mW, and +20 dB is 100 times or 100 mW. Hilfsmittel: use p = 10·log10(P/1mW) dBm (Pegel, S.15); the dB↔ratio table (S.15) gives the +3/+20 dB factors.",
"source": "https://50ohm.de/NEA_dezibel_2.html#AA110",
"confidence": 8
},
"AA111": {
"revision": 3,
- "explanation": "For voltage ratios use 20 log10(U2/U1); 20 log10(15) is about 23.5 dB. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "For voltage ratios use 20 log10(U2/U1); 20 log10(15) is about 23.5 dB. Hilfsmittel: apply the voltage form g = 20·log10(U2/U1) (Pegel, S.15).",
"source": "https://50ohm.de/NEA_dezibel_2.html#AA111",
"confidence": 8
},
"AA112": {
"revision": 2,
- "explanation": "120 dB relative to 1 microvolt per meter is a voltage ratio of 10^(120/20) = 10^6, so the field is 1 V/m. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "120 dB relative to 1 microvolt per meter is a voltage ratio of 10^(120/20) = 10^6, so the field is 1 V/m. Hilfsmittel: invert g = 20·log10(U2/U1) (Pegel, S.15): ratio = 10^(120/20) = 10⁶.",
"source": "https://50ohm.de/NEA_dezibel_2.html#AA112",
"confidence": 8
},
"AA113": {
"revision": 3,
- "explanation": "Each S-step is 6 dB; S4 to S7 is three steps, so 3 x 6 dB = 18 dB. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Each S-step is 6 dB; S4 to S7 is three steps, so 3 x 6 dB = 18 dB.",
"source": "https://50ohm.de/NEA_s_meter.html#AA113",
"confidence": 8
},
"AA114": {
"revision": 3,
- "explanation": "From S9+20 dB down to S9 removes 20 dB, and from S9 to S8 removes another 6 dB, totaling 26 dB. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "From S9+20 dB down to S9 removes 20 dB, and from S9 to S8 removes another 6 dB, totaling 26 dB.",
"source": "https://50ohm.de/NEA_s_meter.html#AA114",
"confidence": 8
},
"AA115": {
"revision": 3,
- "explanation": "1 ppm is one part in one million; 435 MHz divided by 10^6 is 435 Hz. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "1 ppm is one part in one million; 435 MHz divided by 10^6 is 435 Hz.",
"source": "https://50ohm.de/NEA_frequenzgenauigkeit.html#AA115",
"confidence": 8
},
"AA116": {
"revision": 3,
- "explanation": "10 ppm at 14.200000 MHz is 14.2 MHz x 10/10^6 = 142 Hz, so the possible frequency is 14.200000 MHz plus or minus 0.000142 MHz. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "10 ppm at 14.200000 MHz is 14.2 MHz x 10/10^6 = 142 Hz, so the possible frequency is 14.200000 MHz plus or minus 0.000142 MHz.",
"source": "https://50ohm.de/NEA_frequenzgenauigkeit.html#AA116",
"confidence": 8
},
@@ -157,19 +157,19 @@
},
"AB202": {
"revision": 2,
- "explanation": "Maximum power transfer occurs when the load resistance equals the source internal resistance. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Maximum power transfer occurs when the load resistance equals the source internal resistance.",
"source": "https://50ohm.de/NEA_innenwiderstand.html#AB202",
"confidence": 8
},
"AB203": {
"revision": 2,
- "explanation": "Voltage matching minimizes voltage drop inside the source, so the load resistance must be much larger than the source internal resistance. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Voltage matching minimizes voltage drop inside the source, so the load resistance must be much larger than the source internal resistance.",
"source": "https://50ohm.de/NEA_innenwiderstand.html#AB203",
"confidence": 8
},
"AB204": {
"revision": 2,
- "explanation": "Current matching uses a source whose internal resistance is much larger than the load, so the load current stays nearly constant. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Current matching uses a source whose internal resistance is much larger than the load, so the load current stays nearly constant.",
"source": "https://50ohm.de/NEA_innenwiderstand.html#AB204",
"confidence": 8
},
@@ -193,13 +193,13 @@
},
"AB208": {
"revision": 2,
- "explanation": "The voltage drop is 0.2 V at 20 A, so $R_i = 0.2 V / 20 A = 0.01 ohm = 10 milliohm$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "The voltage drop is 0.2 V at 20 A, so $R_i = 0.2 V / 20 A = 0.01 ohm = 10 milliohm$. Hilfsmittel: apply R_i = ΔU/ΔI (Innenwiderstand, S.11).",
"source": "https://50ohm.de/NEA_innenwiderstand.html#AB208",
"confidence": 8
},
"AB209": {
"revision": 3,
- "explanation": "The six 2 V cells are in series, so voltages add to 12 V while the ampere-hour capacity remains that of one cell, 10 Ah. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "The six 2 V cells are in series, so voltages add to 12 V while the ampere-hour capacity remains that of one cell, 10 Ah. Hilfsmittel: series voltages add: U_G = U1+U2 (Spannungsteiler, S.12); the Ah capacity stays that of one cell (outside the sheet).",
"source": "https://50ohm.de/NEA_akku.html#AB209",
"confidence": 7
},
@@ -235,13 +235,13 @@
},
"AB301": {
"revision": 2,
- "explanation": "For a sine current, $I_eff = I_max / sqrt(2)$; power is $I_eff^2 R = (0.5/sqrt(2))^2 x 20 = 2.5 W$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "For a sine current, $I_eff = I_max / sqrt(2)$; power is $I_eff^2 R = (0.5/sqrt(2))^2 x 20 = 2.5 W$. Hilfsmittel: first I_eff = Î/√2 (from Û = U_eff·√2, Wechselspannung, S.12), then P = I_eff²·R (Leistung, S.12).",
"source": "https://50ohm.de/NEA_wechselstrom_leistung.html#AB301",
"confidence": 8
},
"AB302": {
"revision": 2,
- "explanation": "Point X3 is three quarters of a cycle after zero, which is 270 degrees or $3 pi / 2$ radians. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Point X3 is three quarters of a cycle after zero, which is 270 degrees or $3 pi / 2$ radians.",
"source": "https://50ohm.de/NEA_phase.html#AB302",
"confidence": 7
},
@@ -307,19 +307,19 @@
},
"AB501": {
"revision": 3,
- "explanation": "Stored energy in watt-hours is voltage times ampere-hour capacity: $12 V x 5 Ah = 60 Wh$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Stored energy in watt-hours is voltage times ampere-hour capacity: $12 V x 5 Ah = 60 Wh$. Hilfsmittel: W = U·I·t, i.e. W = P·t with P = U·I (Arbeit/Leistung, S.12); U·Ah gives Wh.",
"source": "https://50ohm.de/NEA_akku.html#AB501",
"confidence": 8
},
"AB502": {
"revision": 3,
- "explanation": "Power is $230 V x 0.63 A = 144.9 W$; over 7 h this is $144.9 W x 7 h = 1014 Wh$, about 1.01 kWh. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Power is $230 V x 0.63 A = 144.9 W$; over 7 h this is $144.9 W x 7 h = 1014 Wh$, about 1.01 kWh. Hilfsmittel: first P = U·I, then W = P·t (Leistung/Arbeit, S.12).",
"source": "https://50ohm.de/NEA_ladung_energie.html#AB502",
"confidence": 8
},
"AB503": {
"revision": 3,
- "explanation": "The resistor has $P = U^2/R = 10^2/100 = 1 W$; over one hour that is 1 Wh, equal to 3600 J. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "The resistor has $P = U^2/R = 10^2/100 = 1 W$; over one hour that is 1 Wh, equal to 3600 J. Hilfsmittel: first P = U²/R (Leistung, S.12), then W = P·t (S.12); 1 Wh = 3600 J.",
"source": "https://50ohm.de/NEA_ladung_energie.html#AB503",
"confidence": 7
},
@@ -373,7 +373,7 @@
},
"AC108": {
"revision": 3,
- "explanation": "First find reactance from $X_C = U/I = 16 V / 0.032 A = 500 ohm$; then $C = 1/(2 pi f X_C)$ gives about 6.37 microfarad. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "First find reactance from $X_C = U/I = 16 V / 0.032 A = 500 ohm$; then $C = 1/(2 pi f X_C)$ gives about 6.37 microfarad. Hilfsmittel: first X_C = U/I (Ohmsches Gesetz, S.11 for reactance), then C = 1/(2π·f·X_C) (X_C = 1/(ω·C), ω = 2πf, S.13/12).",
"source": "https://50ohm.de/NEA_kondensator_2.html#AC108",
"confidence": 8
},
@@ -385,13 +385,13 @@
},
"AC110": {
"revision": 2,
- "explanation": "Capacitor loss is commonly described by loss factor tan delta; high loss means low quality factor, with tan delta equal to the reciprocal of Q. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Capacitor loss is commonly described by loss factor tan delta; high loss means low quality factor, with tan delta equal to the reciprocal of Q.",
"source": "https://50ohm.de/NEA_kondensator_2.html#AC110",
"confidence": 8
},
"AC111": {
"revision": 2,
- "explanation": "An ideal capacitor draws reactive current but no real power in steady-state AC, so the real power is approximately zero. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "An ideal capacitor draws reactive current but no real power in steady-state AC, so the real power is approximately zero.",
"source": "https://50ohm.de/NEA_kondensator_2.html#AC111",
"confidence": 8
},
@@ -409,7 +409,7 @@
},
"AC203": {
"revision": 2,
- "explanation": "With DC only the winding resistance limits current; with AC the inductive reactance is added, so the total impedance is higher and current is smaller. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "With DC only the winding resistance limits current; with AC the inductive reactance is added, so the total impedance is higher and current is smaller.",
"source": "https://50ohm.de/NEA_spule_2.html#AC203",
"confidence": 8
},
@@ -421,31 +421,31 @@
},
"AC205": {
"revision": 2,
- "explanation": "For a core with AL value, $L = N^2 x AL$; $14^2 x 1.5 nH = 294 nH = 0.294 microhenry$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "For a core with AL value, $L = N^2 x AL$; $14^2 x 1.5 nH = 294 nH = 0.294 microhenry$. Hilfsmittel: apply L = N²·A_L (Ringkernspulen L = N²·A_L, S.13).",
"source": "https://50ohm.de/NEA_spule_2.html#AC205",
"confidence": 8
},
"AC206": {
"revision": 2,
- "explanation": "Use $L = N^2 x AL$; $300^2 x 1250 nH = 112500000 nH = 112.5 mH$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Use $L = N^2 x AL$; $300^2 x 1250 nH = 112500000 nH = 112.5 mH$. Hilfsmittel: apply L = N²·A_L (Ringkernspulen L = N²·A_L, S.13).",
"source": "https://50ohm.de/NEA_spule_2.html#AC206",
"confidence": 8
},
"AC207": {
"revision": 2,
- "explanation": "Rearrange to $N = sqrt(L/AL)$; $sqrt(2 mH / 250 nH) = sqrt(8000)$, about 89 turns. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Rearrange to $N = sqrt(L/AL)$; $sqrt(2 mH / 250 nH) = sqrt(8000)$, about 89 turns. Hilfsmittel: rearrange L = N²·A_L → N = √(L/A_L) (Ringkernspulen L = N²·A_L, S.13).",
"source": "https://50ohm.de/NEA_spule_2.html#AC207",
"confidence": 8
},
"AC208": {
"revision": 2,
- "explanation": "Rearrange to $N = sqrt(L/AL)$; $sqrt(12 microhenry / 30 nH) = sqrt(400) = 20 turns$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Rearrange to $N = sqrt(L/AL)$; $sqrt(12 microhenry / 30 nH) = sqrt(400) = 20 turns$. Hilfsmittel: rearrange L = N²·A_L → N = √(L/A_L) (Ringkernspulen L = N²·A_L, S.13).",
"source": "https://50ohm.de/NEA_spule_2.html#AC208",
"confidence": 8
},
"AC209": {
"revision": 2,
- "explanation": "Coil losses are represented by an equivalent series resistance; the loss factor tan delta is used and equals the reciprocal of the quality factor Q. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Coil losses are represented by an equivalent series resistance; the loss factor tan delta is used and equals the reciprocal of the quality factor Q.",
"source": "https://50ohm.de/NEA_spule_2.html#AC209",
"confidence": 8
},
@@ -469,19 +469,19 @@
},
"AC302": {
"revision": 3,
- "explanation": "With losses neglected, primary and secondary power are equal: $6 V x 1.15 A = 6.9 W$, and $6.9 W / 230 V = 0.030 A$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "With losses neglected, primary and secondary power are equal: $6 V x 1.15 A = 6.9 W$, and $6.9 W / 230 V = 0.030 A$. Hilfsmittel: the transformer relation U_P·I_P = U_S·I_S (from ü = U_P/U_S = I_S/I_P, S.13) gives the primary current; equivalently equal powers P = U·I (S.12).",
"source": "https://50ohm.de/NEA_uebertrager_2.html#AC302",
"confidence": 8
},
"AC303": {
"revision": 2,
- "explanation": "Impedance transforms with the square of the turns ratio; with 1:4, the input sees $16 kOhm / 4^2 = 1 kOhm$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Impedance transforms with the square of the turns ratio; with 1:4, the input sees $16 kOhm / 4^2 = 1 kOhm$. Hilfsmittel: impedance transforms with the turns ratio squared (Übersetzungsverhältnis ü = N_P/N_S = U_P/U_S = I_S/I_P = √(Z_P/Z_S), S.13).",
"source": "https://50ohm.de/NEA_uebertrager_2.html#AC303",
"confidence": 7
},
"AC304": {
"revision": 2,
- "explanation": "The same 1:4 transformer reflects the secondary load by a factor of 16, so $6.4 kOhm / 16 = 0.4 kOhm$ at a-b. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "The same 1:4 transformer reflects the secondary load by a factor of 16, so $6.4 kOhm / 16 = 0.4 kOhm$ at a-b. Hilfsmittel: impedance transforms with the turns ratio squared (Übersetzungsverhältnis ü = N_P/N_S = U_P/U_S = I_S/I_P = √(Z_P/Z_S), S.13).",
"source": "https://50ohm.de/NEA_uebertrager_2.html#AC304",
"confidence": 7
},
@@ -499,7 +499,7 @@
},
"AC307": {
"revision": 3,
- "explanation": "The wire area is $pi d^2/4 = pi x 0.5^2/4 = 0.196 mm2$; at 2.5 A/mm2 the current is about 0.49 A. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "The wire area is $pi d^2/4 = pi x 0.5^2/4 = 0.196 mm2$; at 2.5 A/mm2 the current is about 0.49 A. Hilfsmittel: first A = d²·π/4 (S.11), then I = S·A with S ≈ 2,5 A/mm² (Belastbarkeit von Wicklungen, S.13).",
"source": "https://50ohm.de/NEA_uebertrager_2.html#AC307",
"confidence": 8
},
@@ -529,13 +529,13 @@
},
"AC405": {
"revision": 2,
- "explanation": "Antiparallel silicon diodes clip the waveform when either polarity exceeds about 0.6 V, so the output is the sine wave limited at that threshold. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Antiparallel silicon diodes clip the waveform when either polarity exceeds about 0.6 V, so the output is the sine wave limited at that threshold.",
"source": "https://50ohm.de/NEA_diode_2.html#AC405",
"confidence": 7
},
"AC406": {
"revision": 2,
- "explanation": "Germanium diodes have a lower threshold, about 0.3 V, so the same limiter clips the waveform earlier and more strongly than silicon diodes. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Germanium diodes have a lower threshold, about 0.3 V, so the same limiter clips the waveform earlier and more strongly than silicon diodes.",
"source": "https://50ohm.de/NEA_diode_2.html#AC406",
"confidence": 7
},
@@ -637,55 +637,55 @@
},
"AC515": {
"revision": 2,
- "explanation": "Base current is $5 mA / 298 = 16.8 microampere$; with about 0.6 V base-emitter drop, $R_1 = (12 - 0.6) V / 16.8 microampere$, about 680 kOhm. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Base current is $5 mA / 298 = 16.8 microampere$; with about 0.6 V base-emitter drop, $R_1 = (12 - 0.6) V / 16.8 microampere$, about 680 kOhm. Hilfsmittel: first base current I_B = I_C/B (from B = I_C/I_B, S.14), then R = U/I (Ohmsches Gesetz, S.11); the 0,6 V base drop is added knowledge.",
"source": "https://50ohm.de/NEA_transistor_2.html#AC515",
"confidence": 7
},
"AC516": {
"revision": 2,
- "explanation": "Making the divider current much larger than base current keeps the base voltage mostly set by the divider, so transistor beta and temperature changes disturb the operating point less. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Making the divider current much larger than base current keeps the base voltage mostly set by the divider, so transistor beta and temperature changes disturb the operating point less.",
"source": "https://50ohm.de/NEA_transistor_2.html#AC516",
"confidence": 7
},
"AC517": {
"revision": 2,
- "explanation": "Base current is $2 mA/200 = 10 microampere$; R2 carries ten times that, so R1 carries 110 microampere. With 1 V at the emitter, the base is about 1.6 V, giving $R_1 = 8.4 V/110 microampere = 76.4 kOhm$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Base current is $2 mA/200 = 10 microampere$; R2 carries ten times that, so R1 carries 110 microampere. With 1 V at the emitter, the base is about 1.6 V, giving $R_1 = 8.4 V/110 microampere = 76.4 kOhm$. Hilfsmittel: first base current I_B = I_C/B (B = I_C/I_B, S. 14); R2 carries ten times that and R1 the sum, then R = U/I (Ohmsches Gesetz, S. 11). The 0,6 V base-emitter drop is added knowledge.",
"source": "https://50ohm.de/NEA_transistor_2.html#AC517",
"confidence": 7
},
"AC518": {
"revision": 2,
- "explanation": "Base current is 10 microampere and R2 current is 100 microampere, so R1 current is 110 microampere; with the base near 0.6 V, $R_1 = 9.4 V/110 microampere = 85.5 kOhm$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Base current is 10 microampere and R2 current is 100 microampere, so R1 current is 110 microampere; with the base near 0.6 V, $R_1 = 9.4 V/110 microampere = 85.5 kOhm$. Hilfsmittel: first base current I_B = I_C/B (B = I_C/I_B, S.14), then R = U/I (Ohmsches Gesetz, S.11); the 0,6 V base drop is added knowledge.",
"source": "https://50ohm.de/NEA_transistor_2.html#AC518",
"confidence": 7
},
"AC519": {
"revision": 2,
- "explanation": "If R1 is open, the base receives no forward bias, the transistor switches off, and with no collector current the collector rises to the supply voltage. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "If R1 is open, the base receives no forward bias, the transistor switches off, and with no collector current the collector rises to the supply voltage.",
"source": "https://50ohm.de/NEA_transistor_2.html#AC519",
"confidence": 7
},
"AC520": {
"revision": 2,
- "explanation": "If R2 is open, the base is driven too strongly through R1, so the transistor saturates; collector current is then limited mainly by RC and collector voltage falls near saturation voltage. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "If R2 is open, the base is driven too strongly through R1, so the transistor saturates; collector current is then limited mainly by RC and collector voltage falls near saturation voltage.",
"source": "https://50ohm.de/NEA_transistor_2.html#AC520",
"confidence": 7
},
"AC521": {
"revision": 2,
- "explanation": "The gate draws negligible current, so the divider gives $U_G = 44 V x 1 kOhm/(10 kOhm + 1 kOhm) = 4 V$; with the source at reference, that is $U_GS$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "The gate draws negligible current, so the divider gives $U_G = 44 V x 1 kOhm/(10 kOhm + 1 kOhm) = 4 V$; with the source at reference, that is $U_GS$. Hilfsmittel: apply the divider U_G = U·R/(R1+R2) (Spannungsteiler, S.12).",
"source": "https://50ohm.de/NEA_transistor_2.html#AC521",
"confidence": 7
},
"AC522": {
"revision": 2,
- "explanation": "For a divider, $R_2 = R_1 U_G/(U_B - U_G)$; $10 kOhm x 2.8/(44 - 2.8)$ gives about 680 ohm. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "For a divider, $R_2 = R_1 U_G/(U_B - U_G)$; $10 kOhm x 2.8/(44 - 2.8)$ gives about 680 ohm. Hilfsmittel: rearrange the divider for R2 (Spannungsteiler, S.12).",
"source": "https://50ohm.de/NEA_transistor_2.html#AC522",
"confidence": 7
},
"AC523": {
"revision": 2,
- "explanation": "Conduction loss is $P = I^2 R$; $25^2 x 0.004 ohm = 2.5 W$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Conduction loss is $P = I^2 R$; $25^2 x 0.004 ohm = 2.5 W$. Hilfsmittel: apply P = I²·R (Leistung, S.12).",
"source": "https://50ohm.de/NEA_transistor_2.html#AC523",
"confidence": 8
},
@@ -721,109 +721,109 @@
},
"AD101": {
"revision": 2,
- "explanation": "Series capacitors add by reciprocals: $1/C = 1/100 pF + 1/47 pF + 1/22 pF$, giving about 13.0 pF. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Series capacitors add by reciprocals: $1/C = 1/100 pF + 1/47 pF + 1/22 pF$, giving about 13.0 pF. Hilfsmittel: series caps via 1/C_G = Σ1/Ci (Kapazität, S.13).",
"source": "https://50ohm.de/NEA_reihe_parallel_gemischt.html#AD101",
"confidence": 8
},
"AD102": {
"revision": 3,
- "explanation": "Series inductances add directly: 2200 nH is 2.2 microhenry, 0.033 mH is 33 microhenry, so the sum is 2.2 + 33 + 150 = 185.2 microhenry. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Series inductances add directly: 2200 nH is 2.2 microhenry, 0.033 mH is 33 microhenry, so the sum is 2.2 + 33 + 150 = 185.2 microhenry. Hilfsmittel: inductances in series add: L_G = ΣLi (Induktivität, S.13).",
"source": "https://50ohm.de/NEA_reihenschaltung_spule.html#AD102",
"confidence": 8
},
"AD103": {
"revision": 2,
- "explanation": "The shown capacitances are effectively parallel, so they add: 100 pF + 1500 pF + 220 pF + 1 pF = 1821 pF. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "The shown capacitances are effectively parallel, so they add: 100 pF + 1500 pF + 220 pF + 1 pF = 1821 pF. Hilfsmittel: parallel caps add: C_G = ΣCi (Kapazität, S.13).",
"source": "https://50ohm.de/NEA_reihe_parallel_gemischt.html#AD103",
"confidence": 7
},
"AD104": {
"revision": 2,
- "explanation": "At 1 MHz and 1 nF, $X_C$ is about 159 ohm; the series impedance magnitude is $sqrt(100^2 + 159^2)$, about 188 ohm. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "At 1 MHz and 1 nF, $X_C$ is about 159 ohm; the series impedance magnitude is $sqrt(100^2 + 159^2)$, about 188 ohm. Hilfsmittel: first X_C = 1/(ω·C) (X_C = 1/(ω·C) bzw. X_L = ω·L, S.13), then |Z| = √(R²+X²) (Z = √(R²+X²), S.12).",
"source": "https://50ohm.de/NEA_reihe_parallel_gemischt.html#AD104",
"confidence": 8
},
"AD105": {
"revision": 2,
- "explanation": "At 1 MHz and 100 microhenry, $X_L$ is about 628 ohm; $|Z| = sqrt(100^2 + 628^2)$, about 636 ohm. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "At 1 MHz and 100 microhenry, $X_L$ is about 628 ohm; $|Z| = sqrt(100^2 + 628^2)$, about 636 ohm. Hilfsmittel: first X_L = ω·L (X_C = 1/(ω·C) bzw. X_L = ω·L, S.13), then |Z| = √(R²+X²) (Z = √(R²+X²), S.12).",
"source": "https://50ohm.de/NEA_reihe_parallel_gemischt.html#AD105",
"confidence": 8
},
"AD106": {
"revision": 2,
- "explanation": "If 1 mA flows through R3, the parallel section has 10 V across it; R2 draws another 1 mA, so 2 mA through R1 drops 20 V, making the total 30 V. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "If 1 mA flows through R3, the parallel section has 10 V across it; R2 draws another 1 mA, so 2 mA through R1 drops 20 V, making the total 30 V. Hilfsmittel: resistors add in series / combine in parallel (Reihen-/Parallelschaltung, S.12); voltages via U = R·I (Ohmsches Gesetz, S.11).",
"source": "https://50ohm.de/NEA_reihe_parallel_widerstandsnetz_2.html#AD106",
"confidence": 7
},
"AD107": {
"revision": 2,
- "explanation": "R2 and R3 in parallel give 5 kOhm, in series with R1 gives 15 kOhm; 15 V / 15 kOhm is 1 mA total, split equally so R3 has 0.5 mA. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "R2 and R3 in parallel give 5 kOhm, in series with R1 gives 15 kOhm; 15 V / 15 kOhm is 1 mA total, split equally so R3 has 0.5 mA. Hilfsmittel: parallel then series (Reihen-/Parallelschaltung, S.12); current via I = U/R (Ohmsches Gesetz, S.11).",
"source": "https://50ohm.de/NEA_reihe_parallel_widerstandsnetz_2.html#AD107",
"confidence": 7
},
"AD108": {
"revision": 2,
- "explanation": "The total current is 1 mA, so the parallel section has 5 V across it; R2 power is $5^2/10000 = 0.0025 W = 2.5 mW$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "The total current is 1 mA, so the parallel section has 5 V across it; R2 power is $5^2/10000 = 0.0025 W = 2.5 mW$. Hilfsmittel: reduce the network (Reihen-/Parallelschaltung, S.12), then P = U²/R (Leistung, S.12).",
"source": "https://50ohm.de/NEA_reihe_parallel_widerstandsnetz_2.html#AD108",
"confidence": 7
},
"AD109": {
"revision": 2,
- "explanation": "The input is 200 ohm plus 100 ohm in parallel with 200 ohm + R; at R = 0 this is about 267 ohm, and at R = 1 kOhm it is about 292 ohm. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "The input is 200 ohm plus 100 ohm in parallel with 200 ohm + R; at R = 0 this is about 267 ohm, and at R = 1 kOhm it is about 292 ohm. Hilfsmittel: series adds, parallel = R1·R2/(R1+R2) (Reihen-/Parallelschaltung, S.12).",
"source": "https://50ohm.de/NEA_reihe_parallel_widerstandsnetz_2.html#AD109",
"confidence": 7
},
"AD110": {
"revision": 2,
- "explanation": "Each side branch is 2.2 kOhm + 220 ohm = 2420 ohm, and two equal branches in parallel give half that value, 1210 ohm. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Each side branch is 2.2 kOhm + 220 ohm = 2420 ohm, and two equal branches in parallel give half that value, 1210 ohm. Hilfsmittel: series adds, then two equal branches in parallel halve (Reihen-/Parallelschaltung, S.12).",
"source": "https://50ohm.de/NEA_reihe_parallel_widerstandsnetz_2.html#AD110",
"confidence": 7
},
"AD111": {
"revision": 2,
- "explanation": "A bridge has zero branch voltage when the two divider ratios are equal, which gives $R1/R2 = R3/R4$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "A bridge has zero branch voltage when the two divider ratios are equal, which gives $R1/R2 = R3/R4$. Hilfsmittel: the bridge balances when the two divider ratios match: U1/U2 = R1/R2 (Spannungsteiler, S.12).",
"source": "https://50ohm.de/NEA_brueckenschaltung.html#AD111",
"confidence": 7
},
"AD112": {
"revision": 3,
- "explanation": "All four resistors are equal, so the two divider midpoints sit at the same potential; the bridge voltage from A to B is 0 V. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "All four resistors are equal, so the two divider midpoints sit at the same potential; the bridge voltage from A to B is 0 V. Hilfsmittel: equal divider ratios give equal midpoints (Spannungsteiler, S.12); the bridge voltage is 0.",
"source": "https://50ohm.de/NEA_brueckenschaltung.html#AD112",
"confidence": 7
},
"AD113": {
"revision": 2,
- "explanation": "The left divider gives point A at 10 V and the right divider gives point B at 1 V, so measured from A to B the bridge voltage is +9 V. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "The left divider gives point A at 10 V and the right divider gives point B at 1 V, so measured from A to B the bridge voltage is +9 V. Hilfsmittel: each side is a voltage divider (Spannungsteiler, S.12); subtract the two midpoint voltages.",
"source": "https://50ohm.de/NEA_brueckenschaltung.html#AD113",
"confidence": 7
},
"AD114": {
"revision": 2,
- "explanation": "The load is parallel to R2: $2.2 kOhm || 8.2 kOhm$ is about 1.73 kOhm. The divider output is $12 V x 1.73/(10 + 1.73)$, about 1.8 V. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "The load is parallel to R2: $2.2 kOhm || 8.2 kOhm$ is about 1.73 kOhm. The divider output is $12 V x 1.73/(10 + 1.73)$, about 1.8 V. Hilfsmittel: first the load in parallel with R2 (Reihen-/Parallelschaltung, S.12), then the divider (Spannungsteiler, S.12).",
"source": "https://50ohm.de/NEA_spannungsteiler_2.html#AD114",
"confidence": 7
},
"AD115": {
"revision": 2,
- "explanation": "Adding the load lowers the effective lower resistance of the divider, increasing the supply current through R1; with higher current, R1 dissipates more heat. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Adding the load lowers the effective lower resistance of the divider, increasing the supply current through R1; with higher current, R1 dissipates more heat.",
"source": "https://50ohm.de/NEA_spannungsteiler_2.html#AD115",
"confidence": 7
},
"AD201": {
"revision": 3,
- "explanation": "An RC high-pass cutoff is $f_g = 1/(2 pi R C)$; with 4.7 kOhm and 2.2 nF this is about 15.4 kHz. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "An RC high-pass cutoff is $f_g = 1/(2 pi R C)$; with 4.7 kOhm and 2.2 nF this is about 15.4 kHz. Hilfsmittel: apply f_g = 1/(2π·R·C) (RC-Hochpass, S.14).",
"source": "https://50ohm.de/NEA_schwingkreis_2.html#AD201",
"confidence": 7
},
"AD202": {
"revision": 3,
- "explanation": "An RC low-pass has the same cutoff formula, $f_g = 1/(2 pi R C)$; with 10 kOhm and 47 nF this is about 339 Hz. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "An RC low-pass has the same cutoff formula, $f_g = 1/(2 pi R C)$; with 10 kOhm and 47 nF this is about 339 Hz. Hilfsmittel: apply f_g = 1/(2π·R·C) (RC-Tiefpass, S.14).",
"source": "https://50ohm.de/NEA_schwingkreis_2.html#AD202",
"confidence": 7
},
"AD203": {
"revision": 2,
- "explanation": "The relevant low-pass is R1 with C1; C2 is supply decoupling and the amplifier input is very high impedance. $1/(2 pi x 4.7 kOhm x 6.8 nF)$ is about 5 kHz. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "The relevant low-pass is R1 with C1; C2 is supply decoupling and the amplifier input is very high impedance. $1/(2 pi x 4.7 kOhm x 6.8 nF)$ is about 5 kHz. Hilfsmittel: apply f_g = 1/(2π·R·C) (RC-Tiefpass, S.14).",
"source": "https://50ohm.de/NEA_schwingkreis_2.html#AD203",
"confidence": 7
},
@@ -847,37 +847,37 @@
},
"AD207": {
"revision": 2,
- "explanation": "In a series resonant circuit the L and C reactances cancel, leaving only the real series resistance R as the impedance. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "In a series resonant circuit the L and C reactances cancel, leaving only the real series resistance R as the impedance. Hilfsmittel: at resonance X_C = X_L cancel, leaving R (Schwingkreis, S.14).",
"source": "https://50ohm.de/NEA_schwingkreis_2.html#AD207",
"confidence": 7
},
"AD208": {
"revision": 2,
- "explanation": "Use Thomson's formula $f = 1/(2 pi sqrt(L C))$; with 1.2 microhenry and 6.8 pF the result is about 55.7 MHz. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Use Thomson's formula $f = 1/(2 pi sqrt(L C))$; with 1.2 microhenry and 6.8 pF the result is about 55.7 MHz. Hilfsmittel: apply f₀ = 1/(2π·√(L·C)), S.14.",
"source": "https://50ohm.de/NEA_schwingkreis_2.html#AD208",
"confidence": 7
},
"AD209": {
"revision": 2,
- "explanation": "The resistor does not set the ideal resonant frequency; $1/(2 pi sqrt(10 microhenry x 1 nF))$ is about 1.592 MHz. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "The resistor does not set the ideal resonant frequency; $1/(2 pi sqrt(10 microhenry x 1 nF))$ is about 1.592 MHz. Hilfsmittel: apply f₀ = 1/(2π·√(L·C)), S.14 (R does not set f₀).",
"source": "https://50ohm.de/NEA_schwingkreis_2.html#AD209",
"confidence": 7
},
"AD210": {
"revision": 2,
- "explanation": "Using $f = 1/(2 pi sqrt(L C))$ with 100 microhenry and 0.01 microfarad gives about 159 kHz. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Using $f = 1/(2 pi sqrt(L C))$ with 100 microhenry and 0.01 microfarad gives about 159 kHz. Hilfsmittel: apply f₀ = 1/(2π·√(L·C)), S.14.",
"source": "https://50ohm.de/NEA_schwingkreis_2.html#AD210",
"confidence": 7
},
"AD211": {
"revision": 2,
- "explanation": "For the parallel resonant circuit, $f = 1/(2 pi sqrt(2.2 microhenry x 56 pF))$, giving about 14.34 MHz. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "For the parallel resonant circuit, $f = 1/(2 pi sqrt(2.2 microhenry x 56 pF))$, giving about 14.34 MHz. Hilfsmittel: apply f₀ = 1/(2π·√(L·C)), S.14.",
"source": "https://50ohm.de/NEA_schwingkreis_2.html#AD211",
"confidence": 7
},
"AD212": {
"revision": 2,
- "explanation": "The parallel capacitances add to about 1.82 nF; with 1.2 mH, $1/(2 pi sqrt(L C))$ gives about 107.7 kHz. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "The parallel capacitances add to about 1.82 nF; with 1.2 mH, $1/(2 pi sqrt(L C))$ gives about 107.7 kHz. Hilfsmittel: first add the parallel caps (S.13), then f₀ = 1/(2π·√(L·C)), S.14.",
"source": "https://50ohm.de/NEA_schwingkreis_2.html#AD212",
"confidence": 7
},
@@ -919,13 +919,13 @@
},
"AD219": {
"revision": 3,
- "explanation": "Bandwidth is read as the frequency difference between the two points on the curve at the specified level; at -60 dB the marked span is about 4 kHz. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Bandwidth is read as the frequency difference between the two points on the curve at the specified level; at -60 dB the marked span is about 4 kHz.",
"source": "https://50ohm.de/NEA_schwingkreis_2.html#AD219",
"confidence": 7
},
"AD220": {
"revision": 2,
- "explanation": "Filter bandwidth is the difference between the two frequencies where the voltage has fallen to 0.7 of the resonant maximum, the -3 dB points. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Filter bandwidth is the difference between the two frequencies where the voltage has fallen to 0.7 of the resonant maximum, the -3 dB points.",
"source": "https://50ohm.de/NEA_schwingkreis_2.html#AD220",
"confidence": 8
},
@@ -943,25 +943,25 @@
},
"AD223": {
"revision": 2,
- "explanation": "For a series resonant circuit, bandwidth is $B = R/(2 pi L)$; $10/(2 pi x 100 microhenry)$ is about 15.9 kHz. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "For a series resonant circuit, bandwidth is $B = R/(2 pi L)$; $10/(2 pi x 100 microhenry)$ is about 15.9 kHz. Hilfsmittel: apply B = R/(2π·L) (Reihenschwingkreis, S.14).",
"source": "https://50ohm.de/NEA_schwingkreis_2.html#AD223",
"confidence": 8
},
"AD224": {
"revision": 2,
- "explanation": "For the parallel case, $B = 1/(2 pi R C)$; with 1 kOhm and 56 pF this is about 2.84 MHz. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "For the parallel case, $B = 1/(2 pi R C)$; with 1 kOhm and 56 pF this is about 2.84 MHz. Hilfsmittel: apply B = 1/(2π·R·C) (Parallelschwingkreis, S.14).",
"source": "https://50ohm.de/NEA_schwingkreis_2.html#AD224",
"confidence": 8
},
"AD225": {
"revision": 2,
- "explanation": "For the series circuit, Q is resonant frequency divided by bandwidth; about 159 kHz / 15.9 kHz = 10. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "For the series circuit, Q is resonant frequency divided by bandwidth; about 159 kHz / 15.9 kHz = 10. Hilfsmittel: apply Q = f₀/B (Schwingkreis, S.14).",
"source": "https://50ohm.de/NEA_schwingkreis_2.html#AD225",
"confidence": 8
},
"AD226": {
"revision": 2,
- "explanation": "For the parallel circuit, Q is resonant frequency divided by bandwidth; about 14.34 MHz / 2.84 MHz is about 5. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "For the parallel circuit, Q is resonant frequency divided by bandwidth; about 14.34 MHz / 2.84 MHz is about 5. Hilfsmittel: apply Q = f₀/B (Schwingkreis, S.14).",
"source": "https://50ohm.de/NEA_schwingkreis_2.html#AD226",
"confidence": 8
},
@@ -985,19 +985,19 @@
},
"AD301": {
"revision": 3,
- "explanation": "In each series string the cell voltages add, giving $30 x 0.6 V = 18 V$; four identical strings in parallel add their short-circuit currents to 4 A. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "In each series string the cell voltages add, giving $30 x 0.6 V = 18 V$; four identical strings in parallel add their short-circuit currents to 4 A. Hilfsmittel: series voltages add (U_G = U1+U2) and parallel currents add (I_G = I1+I2), both S.12; the 0,6 V cell voltage is given.",
"source": "https://50ohm.de/NEA_photovoltaik.html#AD301",
"confidence": 7
},
"AD302": {
"revision": 3,
- "explanation": "The unloaded smoothing capacitor charges close to the peak of the secondary AC voltage; about 15 V RMS times $sqrt(2)$ gives roughly 21 V. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "The unloaded smoothing capacitor charges close to the peak of the secondary AC voltage; about 15 V RMS times $sqrt(2)$ gives roughly 21 V. Hilfsmittel: the cap charges to the peak: Û = U_eff·√2 (Wechselspannung, S.12).",
"source": "https://50ohm.de/NEA_gleichrichter_2.html#AD302",
"confidence": 7
},
"AD303": {
"revision": 3,
- "explanation": "A 20:1 transformer gives 230 V / 20 = 11.5 V RMS; the peak is about 16.3 V, and adding 50 percent safety gives about 24.4 V, so choose at least 25 V. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "A 20:1 transformer gives 230 V / 20 = 11.5 V RMS; the peak is about 16.3 V, and adding 50 percent safety gives about 24.4 V, so choose at least 25 V. Hilfsmittel: first U_S = U_P·N_S/N_P (Übersetzungsverhältnis ü = N_P/N_S = U_P/U_S = I_S/I_P = √(Z_P/Z_S), S.13), then Û = U_eff·√2 (Wechselspannung, S.12).",
"source": "https://50ohm.de/NEA_gleichrichter_2.html#AD303",
"confidence": 7
},
@@ -1015,7 +1015,7 @@
},
"AD306": {
"revision": 2,
- "explanation": "The secondary peak is the mains peak divided by 8: $230 V x 1.414 / 8$ is about 40.6 V, which is the unloaded capacitor voltage. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "The secondary peak is the mains peak divided by 8: $230 V x 1.414 / 8$ is about 40.6 V, which is the unloaded capacitor voltage. Hilfsmittel: first scale by the turns ratio (Übersetzungsverhältnis ü = N_P/N_S = U_P/U_S = I_S/I_P = √(Z_P/Z_S), S.13), then Û = U_eff·√2 (Wechselspannung, S.12).",
"source": "https://50ohm.de/NEA_brueckengleichrichter.html#AD306",
"confidence": 7
},
@@ -1033,13 +1033,13 @@
},
"AD309": {
"revision": 3,
- "explanation": "The ripple span is the difference between the high and low points, 3 V, and a full-wave rectifier on 50 Hz mains produces ripple at 100 Hz. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "The ripple span is the difference between the high and low points, 3 V, and a full-wave rectifier on 50 Hz mains produces ripple at 100 Hz.",
"source": "https://50ohm.de/NEA_restwelligkeit.html#AD309",
"confidence": 7
},
"AD310": {
"revision": 3,
- "explanation": "A full-wave rectifier produces one output pulse for each half-cycle, so 50 Hz mains becomes 100 Hz ripple frequency. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "A full-wave rectifier produces one output pulse for each half-cycle, so 50 Hz mains becomes 100 Hz ripple frequency.",
"source": "https://50ohm.de/NEA_restwelligkeit.html#AD310",
"confidence": 8
},
@@ -1069,7 +1069,7 @@
},
"AD315": {
"revision": 3,
- "explanation": "The Z-diode regulator clamps the output near the Zener voltage, so the output between A and B is approximately 5 V despite the varying input. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "The Z-diode regulator clamps the output near the Zener voltage, so the output between A and B is approximately 5 V despite the varying input.",
"source": "https://50ohm.de/NEA_spannungsstabilisierung.html#AD315",
"confidence": 7
},
@@ -1087,13 +1087,13 @@
},
"AD318": {
"revision": 3,
- "explanation": "The load current is $5 V / 10 ohm = 0.5 A$ and the regulator drops $13.8 V - 5 V = 8.8 V$; loss is $8.8 V x 0.5 A = 4.4 W$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "The load current is $5 V / 10 ohm = 0.5 A$ and the regulator drops $13.8 V - 5 V = 8.8 V$; loss is $8.8 V x 0.5 A = 4.4 W$. Hilfsmittel: first I = U/R (Ohmsches Gesetz, S.11), then the drop·I via P = U·I (Leistung, S.12).",
"source": "https://50ohm.de/NEA_spannungsstabilisierung.html#AD318",
"confidence": 7
},
"AD319": {
"revision": 3,
- "explanation": "A linear regulator dissipates the voltage drop times current: $(13.8 V - 9 V) x 0.9 A = 4.32 W$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "A linear regulator dissipates the voltage drop times current: $(13.8 V - 9 V) x 0.9 A = 4.32 W$. Hilfsmittel: apply P = (U_in − U_out)·I, i.e. P = U·I (Leistung, S.12).",
"source": "https://50ohm.de/NEA_spannungsstabilisierung.html#AD319",
"confidence": 8
},
@@ -1105,7 +1105,7 @@
},
"AD321": {
"revision": 2,
- "explanation": "The load power is $4.7 V x 10 mA = 47 mW$; input power is $13.8 V x (10 + 15) mA = 345 mW$, so efficiency is about 0.14. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "The load power is $4.7 V x 10 mA = 47 mW$; input power is $13.8 V x (10 + 15) mA = 345 mW$, so efficiency is about 0.14. Hilfsmittel: apply η = P_ab/P_zu (Wirkungsgrad, S.12) with P = U·I for each side.",
"source": "https://50ohm.de/NEA_spannungsstabilisierung.html#AD321",
"confidence": 7
},
@@ -1123,7 +1123,7 @@
},
"AD324": {
"revision": 2,
- "explanation": "C1 is the RF coupling capacitor toward the receiver; it passes RF but blocks the DC supply from reaching the receiver input. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "C1 is the RF coupling capacitor toward the receiver; it passes RF but blocks the DC supply from reaching the receiver input.",
"source": "https://50ohm.de/NEA_fernspeiseweiche.html#AD324",
"confidence": 7
},
@@ -1141,7 +1141,7 @@
},
"AD402": {
"revision": 2,
- "explanation": "An emitter follower has voltage gain just below unity because the emitter follows the base voltage, and it is non-inverting, so the phase shift is 0 degrees. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "An emitter follower has voltage gain just below unity because the emitter follows the base voltage, and it is non-inverting, so the phase shift is 0 degrees.",
"source": "https://50ohm.de/NEA_kollektorschaltung.html#AD402",
"confidence": 7
},
@@ -1165,7 +1165,7 @@
},
"AD406": {
"revision": 2,
- "explanation": "Without DC bias the transistor conducts only when the base-emitter voltage exceeds about 0.6 V; the collector voltage then dips, so the output is a clipped, inverted pulse-like waveform. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Without DC bias the transistor conducts only when the base-emitter voltage exceeds about 0.6 V; the collector voltage then dips, so the output is a clipped, inverted pulse-like waveform.",
"source": "https://50ohm.de/NEA_emitterschaltung.html#AD406",
"confidence": 7
},
@@ -1177,7 +1177,7 @@
},
"AD408": {
"revision": 2,
- "explanation": "The emitter-stage output is taken at the collector, so the collector waveform is inverted relative to the input while the bias and coupling points keep their shown DC roles. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "The emitter-stage output is taken at the collector, so the collector waveform is inverted relative to the input while the bias and coupling points keep their shown DC roles.",
"source": "https://50ohm.de/NEA_emitterschaltung.html#AD408",
"confidence": 7
},
@@ -1189,31 +1189,31 @@
},
"AD410": {
"revision": 2,
- "explanation": "A bypassed emitter stage can provide large voltage gain, and the collector output is inverted relative to the base input, so the phase shift is 180 degrees. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "A bypassed emitter stage can provide large voltage gain, and the collector output is inverted relative to the base input, so the phase shift is 180 degrees.",
"source": "https://50ohm.de/NEA_emitterschaltung.html#AD410",
"confidence": 7
},
"AD411": {
"revision": 2,
- "explanation": "R1 and R2 form a voltage divider feeding the base, setting the transistor's DC bias point before the AC signal is applied. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "R1 and R2 form a voltage divider feeding the base, setting the transistor's DC bias point before the AC signal is applied.",
"source": "https://50ohm.de/NEA_emitterschaltung.html#AD411",
"confidence": 7
},
"AD412": {
"revision": 2,
- "explanation": "The coupling capacitors pass the AC signal into and out of the stage while blocking the DC bias voltages from adjacent circuits. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "The coupling capacitors pass the AC signal into and out of the stage while blocking the DC bias voltages from adjacent circuits.",
"source": "https://50ohm.de/NEA_emitterschaltung.html#AD412",
"confidence": 7
},
"AD413": {
"revision": 2,
- "explanation": "The emitter bypass capacitor shorts the emitter resistor for AC, reducing emitter degeneration and therefore maximizing AC voltage gain. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "The emitter bypass capacitor shorts the emitter resistor for AC, reducing emitter degeneration and therefore maximizing AC voltage gain.",
"source": "https://50ohm.de/NEA_emitterschaltung.html#AD413",
"confidence": 7
},
"AD414": {
"revision": 2,
- "explanation": "Removing the emitter bypass capacitor leaves the emitter resistor active for AC feedback, so emitter degeneration lowers the voltage gain. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Removing the emitter bypass capacitor leaves the emitter resistor active for AC feedback, so emitter degeneration lowers the voltage gain.",
"source": "https://50ohm.de/NEA_emitterschaltung.html#AD414",
"confidence": 7
},
@@ -1225,7 +1225,7 @@
},
"AD416": {
"revision": 2,
- "explanation": "Moving the bias point upward increases the conduction angle: C is below cutoff most of the cycle, B sits at cutoff, AB is slightly above it, and A conducts for the whole cycle. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Moving the bias point upward increases the conduction angle: C is below cutoff most of the cycle, B sits at cutoff, AB is slightly above it, and A conducts for the whole cycle.",
"source": "https://50ohm.de/NEA_verstaerker_klasse.html#AD416",
"confidence": 7
},
@@ -1273,31 +1273,31 @@
},
"AD424": {
"revision": 2,
- "explanation": "The DC input power is $50 V x 2 A = 100 W$; class A efficiency is about 40%, so expected RF output is about $0.4 x 100 W = 40 W$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "The DC input power is $50 V x 2 A = 100 W$; class A efficiency is about 40%, so expected RF output is about $0.4 x 100 W = 40 W$. Hilfsmittel: P_zu = U·I (Leistung, S.12), then P_ab = η·P_zu (Wirkungsgrad, S.12); the ~40 % class-A efficiency is outside knowledge.",
"source": "https://50ohm.de/NEA_verstaerker_klasse.html#AD424",
"confidence": 8
},
"AD425": {
"revision": 2,
- "explanation": "The DC input power is $50 V x 2 A = 100 W$; using about 85% efficiency for class C gives about $0.85 x 100 W = 85 W$ RF output. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "The DC input power is $50 V x 2 A = 100 W$; using about 85% efficiency for class C gives about $0.85 x 100 W = 85 W$ RF output. Hilfsmittel: P_zu = U·I (Leistung, S.12), then P_ab = η·P_zu (Wirkungsgrad, S.12); the ~85 % class-C efficiency is outside knowledge.",
"source": "https://50ohm.de/NEA_verstaerker_klasse.html#AD425",
"confidence": 8
},
"AD426": {
"revision": 3,
- "explanation": "A 16 dB power gain is a ratio of $10^(16/10) = 39.8$, so 1 W input becomes about 40 W output. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "A 16 dB power gain is a ratio of $10^(16/10) = 39.8$, so 1 W input becomes about 40 W output. Hilfsmittel: convert via G = 10^(g/10dB), the inverse of g = 10·log10(P2/P1) (Pegel, S.15); 16 dB = +10 dB (×10) + +6 dB (×4) → ≈×40 (no direct 16 dB table row).",
"source": "https://50ohm.de/NEA_verstaerkungsleistung.html#AD426",
"confidence": 8
},
"AD427": {
"revision": 3,
- "explanation": "For equal impedances, voltage gain in dB is $20 log10(U2/U1)$; $20 log10(4 mV / 1 mV) = 20 log10(4) = 12 dB$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "For equal impedances, voltage gain in dB is $20 log10(U2/U1)$; $20 log10(4 mV / 1 mV) = 20 log10(4) = 12 dB$. Hilfsmittel: apply the voltage form g = 20·log10(U2/U1) (Pegel, S.15).",
"source": "https://50ohm.de/NEA_verstaerkungsleistung.html#AD427",
"confidence": 8
},
"AD428": {
"revision": 3,
- "explanation": "Power gain in dB is $10 log10(P2/P1)$; $10 log10(38 W / 2.5 W) = 10 log10(15.2) = 11.8 dB$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Power gain in dB is $10 log10(P2/P1)$; $10 log10(38 W / 2.5 W) = 10 log10(15.2) = 11.8 dB$. Hilfsmittel: apply g = 10·log10(P2/P1) (Pegel, S.15).",
"source": "https://50ohm.de/NEA_verstaerkungsleistung.html#AD428",
"confidence": 8
},
@@ -1483,7 +1483,7 @@
},
"AD616": {
"revision": 2,
- "explanation": "C1 and C2 form the capacitive voltage divider of the Colpitts oscillator; a fraction of the output is fed back to sustain oscillation. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "C1 and C2 form the capacitive voltage divider of the Colpitts oscillator; a fraction of the output is fed back to sustain oscillation.",
"source": "https://50ohm.de/NEA_oszillator_schaltungen.html#AD616",
"confidence": 7
},
@@ -1525,7 +1525,7 @@
},
"AD703": {
"revision": 3,
- "explanation": "In this integer-N PLL, the smallest output step equals the reference frequency at the phase detector, so a 12.5 kHz channel spacing requires 12.5 kHz at point A. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "In this integer-N PLL, the smallest output step equals the reference frequency at the phase detector, so a 12.5 kHz channel spacing requires 12.5 kHz at point A.",
"source": "https://50ohm.de/NEA_oszillator_pll.html#AD703",
"confidence": 7
},
@@ -1555,25 +1555,25 @@
},
"AD803": {
"revision": 2,
- "explanation": "For power ratios, 20 dB corresponds to $10^(20/10) = 100$, so input power is 100 times the load power. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "For power ratios, 20 dB corresponds to $10^(20/10) = 100$, so input power is 100 times the load power. Hilfsmittel: convert dB→power ratio via the inverse of g = 10·log10(P2/P1) (Pegel, S.15); the table (S.15) gives 20 dB → ×100.",
"source": "https://50ohm.de/NEA_daempfungsglieder.html#AD803",
"confidence": 7
},
"AD804": {
"revision": 2,
- "explanation": "For power ratios, 6 dB corresponds approximately to $10^(6/10) = 3.98$, so the practical ratio is 4. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "For power ratios, 6 dB corresponds approximately to $10^(6/10) = 3.98$, so the practical ratio is 4. Hilfsmittel: convert dB→power ratio (inverse of g = 10·log10(P2/P1), Pegel, S.15); table (S.15): 6 dB → ×4.",
"source": "https://50ohm.de/NEA_daempfungsglieder.html#AD804",
"confidence": 7
},
"AD805": {
"revision": 2,
- "explanation": "A symmetrical attenuator designed for a 50 ohm system presents 50 ohm at its input when its output is terminated with the matching 50 ohm load. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "A symmetrical attenuator designed for a 50 ohm system presents 50 ohm at its input when its output is terminated with the matching 50 ohm load.",
"source": "https://50ohm.de/NEA_daempfungsglieder.html#AD805",
"confidence": 7
},
"AD806": {
"revision": 2,
- "explanation": "A 20 dB attenuator reduces power by a factor of 100, so 100 W input leaves 1 W at the matched load; the remaining 99 W is dissipated as heat in the pad. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "A 20 dB attenuator reduces power by a factor of 100, so 100 W input leaves 1 W at the matched load; the remaining 99 W is dissipated as heat in the pad. Hilfsmittel: 20 dB → ×100 via the dB↔ratio table (Pegel, S.15); the rest is power bookkeeping.",
"source": "https://50ohm.de/NEA_daempfungsglieder.html#AD806",
"confidence": 7
},
@@ -1633,7 +1633,7 @@
},
"AE209": {
"revision": 3,
- "explanation": "SSB phone is normally limited to about 2.7 kHz, so about 3 kHz spacing gives a small guard margin between adjacent SSB signals. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "SSB phone is normally limited to about 2.7 kHz, so about 3 kHz spacing gives a small guard margin between adjacent SSB signals.",
"source": "https://50ohm.de/NEA_ssb_3.html#AE209",
"confidence": 8
},
@@ -1711,31 +1711,31 @@
},
"AE308": {
"revision": 4,
- "explanation": "Carson's rule gives $B \\approx 2(\\Delta f + f_{mod,max})$, where $\\Delta f$ is FM deviation, German Hub/Frequenzhub. With $\\Delta f=2.5 kHz$ and $f_{mod,max}=2.7 kHz$, $B=2(2.5+2.7)=10.4 kHz$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Carson's rule gives $B \\approx 2(\\Delta f + f_{mod,max})$, where $\\Delta f$ is FM deviation, German Hub/Frequenzhub. With $\\Delta f=2.5 kHz$ and $f_{mod,max}=2.7 kHz$, $B=2(2.5+2.7)=10.4 kHz$. Hilfsmittel: apply Carson B ≈ 2·(Δf + f_mod,max) (FM, S.16).",
"source": "https://50ohm.de/NEA_fm_3.html#AE308",
"confidence": 8
},
"AE309": {
"revision": 3,
- "explanation": "Using Carson's rule, $B = 2 x (1.8 kHz + 2.0 kHz) = 7.6 kHz$; the 145 MHz carrier frequency does not enter this bandwidth estimate. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Using Carson's rule, $B = 2 x (1.8 kHz + 2.0 kHz) = 7.6 kHz$; the 145 MHz carrier frequency does not enter this bandwidth estimate. Hilfsmittel: apply Carson B ≈ 2·(Δf + f_mod,max) (FM, S.16).",
"source": "https://50ohm.de/NEA_fm_3.html#AE309",
"confidence": 8
},
"AE310": {
"revision": 4,
- "explanation": "Narrowband FM in a 12.5 kHz channel uses a typical peak deviation, German Hub/Frequenzhub, of about 2.5 kHz. That leaves room for the modulation sidebands inside the channel spacing. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Narrowband FM in a 12.5 kHz channel uses a typical peak deviation, German Hub/Frequenzhub, of about 2.5 kHz. That leaves room for the modulation sidebands inside the channel spacing.",
"source": "https://50ohm.de/NEA_fm_3.html#AE310",
"confidence": 8
},
"AE311": {
"revision": 4,
- "explanation": "Rearrange Carson's rule $B \\approx 2(\\Delta f + f_{mod})$. Here $\\Delta f$ is deviation, German Hub/Frequenzhub, so $f_{mod}=B/2-\\Delta f=10 kHz/2-2.5 kHz=2.5 kHz$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Rearrange Carson's rule $B \\approx 2(\\Delta f + f_{mod})$. Here $\\Delta f$ is deviation, German Hub/Frequenzhub, so $f_{mod}=B/2-\\Delta f=10 kHz/2-2.5 kHz=2.5 kHz$. Hilfsmittel: rearrange Carson B ≈ 2·(Δf + f_mod) for f_mod (FM, S.16).",
"source": "https://50ohm.de/NEA_fm_3.html#AE311",
"confidence": 8
},
"AE312": {
"revision": 4,
- "explanation": "Rearrange Carson's rule $B \\approx 2(\\Delta f + f_{mod})$. The deviation $\\Delta f$, German Hub/Frequenzhub, is $B/2-f_{mod}=10 kHz/2-2.7 kHz=2.3 kHz$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Rearrange Carson's rule $B \\approx 2(\\Delta f + f_{mod})$. The deviation $\\Delta f$, German Hub/Frequenzhub, is $B/2-f_{mod}=10 kHz/2-2.7 kHz=2.3 kHz$. Hilfsmittel: rearrange Carson B ≈ 2·(Δf + f_mod) for Δf (FM, S.16).",
"source": "https://50ohm.de/NEA_fm_3.html#AE312",
"confidence": 8
},
@@ -1891,7 +1891,7 @@
},
"AF103": {
"revision": 3,
- "explanation": "A tenfold power increase is +10 dB. From S8, +6 dB reaches S9 and the remaining +4 dB gives S9+4 dB. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "A tenfold power increase is +10 dB. From S8, +6 dB reaches S9 and the remaining +4 dB gives S9+4 dB.",
"source": "https://50ohm.de/NEA_s_meter.html#AF103",
"confidence": 8
},
@@ -1915,13 +1915,13 @@
},
"AF107": {
"revision": 3,
- "explanation": "The IF is $24.94 MHz - 14.24 MHz = 10.70 MHz$; the image on the other side of the oscillator is $24.94 MHz + 10.70 MHz = 35.64 MHz$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "The IF is $24.94 MHz - 14.24 MHz = 10.70 MHz$; the image on the other side of the oscillator is $24.94 MHz + 10.70 MHz = 35.64 MHz$. Hilfsmittel: first the IF f_ZF = |f_E − f_OSZ| (Zwischenfrequenz f_ZF = |f_E − f_OSZ| (S.14)), then the image Spiegelfrequenz f_S = 2·f_OSZ − f_E (S.14).",
"source": "https://50ohm.de/NEA_spiegelfrequenzen.html#AF107",
"confidence": 7
},
"AF108": {
"revision": 3,
- "explanation": "With high-side injection the oscillator is $28.5 MHz + 10.7 MHz = 39.2 MHz$; the image is another IF above that, $39.2 MHz + 10.7 MHz = 49.9 MHz$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "With high-side injection the oscillator is $28.5 MHz + 10.7 MHz = 39.2 MHz$; the image is another IF above that, $39.2 MHz + 10.7 MHz = 49.9 MHz$. Hilfsmittel: first f_OSZ from f_ZF = |f_E − f_OSZ| (Zwischenfrequenz f_ZF = |f_E − f_OSZ| (S.14)), then the image Spiegelfrequenz f_S = 2·f_OSZ − f_E (S.14).",
"source": "https://50ohm.de/NEA_spiegelfrequenzen.html#AF108",
"confidence": 7
},
@@ -2005,13 +2005,13 @@
},
"AF202": {
"revision": 3,
- "explanation": "The IF is $145.6 MHz - 134.9 MHz = 10.7 MHz$; the image is the other signal 10.7 MHz from the oscillator, $134.9 MHz - 10.7 MHz = 124.2 MHz$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "The IF is $145.6 MHz - 134.9 MHz = 10.7 MHz$; the image is the other signal 10.7 MHz from the oscillator, $134.9 MHz - 10.7 MHz = 124.2 MHz$. Hilfsmittel: first f_ZF = |f_E − f_OSZ| (Zwischenfrequenz f_ZF = |f_E − f_OSZ| (S.14)), then the image mirrored about f_OSZ (Spiegelfrequenz f_S = 2·f_OSZ − f_E (S.14)).",
"source": "https://50ohm.de/NEA_spiegelfrequenzen.html#AF202",
"confidence": 7
},
"AF203": {
"revision": 3,
- "explanation": "The image is mirrored around the oscillator frequency: $2 x 39 MHz - 28.3 MHz = 49.7 MHz$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "The image is mirrored around the oscillator frequency: $2 x 39 MHz - 28.3 MHz = 49.7 MHz$. Hilfsmittel: the image is mirrored about f_OSZ: Spiegelfrequenz f_S = 2·f_OSZ − f_E (S.14).",
"source": "https://50ohm.de/NEA_spiegelfrequenzen.html#AF203",
"confidence": 8
},
@@ -2053,13 +2053,13 @@
},
"AF210": {
"revision": 3,
- "explanation": "For a 50 MHz first IF and a 3 to 30 MHz receive range, the VFO can use either difference mixing $50 - f_rx$ = 47 to 20 MHz or sum mixing $50 + f_rx$ = 53 to 80 MHz. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "For a 50 MHz first IF and a 3 to 30 MHz receive range, the VFO can use either difference mixing $50 - f_rx$ = 47 to 20 MHz or sum mixing $50 + f_rx$ = 53 to 80 MHz. Hilfsmittel: the VFO is f_OSZ = f_IF ± f_rx (from f_ZF = |f_E − f_OSZ|, Zwischenfrequenz f_ZF = |f_E − f_OSZ| (S.14)).",
"source": "https://50ohm.de/NEA_doppelueberlagerungsempfaenger_doppelsuper.html#AF210",
"confidence": 7
},
"AF211": {
"revision": 3,
- "explanation": "For CW, the BFO is offset from the last IF by an audible tone frequency; about 800 Hz gives a comfortable received beat note. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "For CW, the BFO is offset from the last IF by an audible tone frequency; about 800 Hz gives a comfortable received beat note.",
"source": "https://50ohm.de/NEA_bfo_2.html#AF211",
"confidence": 8
},
@@ -2119,7 +2119,7 @@
},
"AF221": {
"revision": 2,
- "explanation": "IP3 is a measure of third-order intermodulation behavior, so it indicates how well the receiver handles large signals without generating distortion products. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "IP3 is a measure of third-order intermodulation behavior, so it indicates how well the receiver handles large signals without generating distortion products.",
"source": "https://50ohm.de/NEA_intermodulation_kreuzmodulation.html#AF221",
"confidence": 8
},
@@ -2209,7 +2209,7 @@
},
"AF305": {
"revision": 2,
- "explanation": "After the balanced modulator has made DSB, the marked filter must be a narrow bandpass, commonly a crystal filter, that selects the desired sideband. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "After the balanced modulator has made DSB, the marked filter must be a narrow bandpass, commonly a crystal filter, that selects the desired sideband.",
"source": "https://50ohm.de/NEA_modulatoren.html#AF305",
"confidence": 7
},
@@ -2221,7 +2221,7 @@
},
"AF307": {
"revision": 3,
- "explanation": "The USB carrier frequency is symmetric to the LSB carrier around the 9 MHz filter center: $9.0000 MHz - (9.0015 - 9.0000) MHz = 8.9985 MHz$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "The USB carrier frequency is symmetric to the LSB carrier around the 9 MHz filter center: $9.0000 MHz - (9.0015 - 9.0000) MHz = 8.9985 MHz$.",
"source": "https://50ohm.de/NEA_modulatoren.html#AF307",
"confidence": 7
},
@@ -2233,7 +2233,7 @@
},
"AF309": {
"revision": 2,
- "explanation": "The balancing network trims amplitude and phase so the carrier components cancel as well as possible in the balanced modulator. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "The balancing network trims amplitude and phase so the carrier components cancel as well as possible in the balanced modulator.",
"source": "https://50ohm.de/NEA_modulatoren.html#AF309",
"confidence": 7
},
@@ -2263,7 +2263,7 @@
},
"AF314": {
"revision": 2,
- "explanation": "Only the sequence $12 MHz x 2 x 2 x 3 x 3$ passes through 144 MHz as an intermediate result: 24 MHz, 48 MHz, 144 MHz, then 432 MHz. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Only the sequence $12 MHz x 2 x 2 x 3 x 3$ passes through 144 MHz as an intermediate result: 24 MHz, 48 MHz, 144 MHz, then 432 MHz.",
"source": "https://50ohm.de/NEA_frequenzvervielfacher_2.html#AF314",
"confidence": 8
},
@@ -2323,7 +2323,7 @@
},
"AF410": {
"revision": 2,
- "explanation": "C1 and C2 are part of the matching network, setting the impedance transformation between the transistor stage and the connected circuit. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "C1 and C2 are part of the matching network, setting the impedance transformation between the transistor stage and the connected circuit.",
"source": "https://50ohm.de/NEA_leistungsvertaerker.html#AF410",
"confidence": 7
},
@@ -2347,49 +2347,49 @@
},
"AF414": {
"revision": 2,
- "explanation": "The transformer couples stages while transforming the output impedance of one emitter stage to the input impedance of the following emitter stage. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "The transformer couples stages while transforming the output impedance of one emitter stage to the input impedance of the following emitter stage.",
"source": "https://50ohm.de/NEA_leistungsvertaerker.html#AF414",
"confidence": 7
},
"AF415": {
"revision": 2,
- "explanation": "Large capacitors are effective at low frequencies but poorer at very high RF; small capacitors keep low impedance at high frequencies, so the parallel pair decouples over a wider range. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Large capacitors are effective at low frequencies but poorer at very high RF; small capacitors keep low impedance at high frequencies, so the parallel pair decouples over a wider range.",
"source": "https://50ohm.de/NEA_leistungsvertaerker.html#AF415",
"confidence": 7
},
"AF416": {
"revision": 2,
- "explanation": "The resistor damps the transformer winding, reducing excessive Q and helping prevent parasitic oscillations. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "The resistor damps the transformer winding, reducing excessive Q and helping prevent parasitic oscillations.",
"source": "https://50ohm.de/NEA_parasitaere_schwingungen.html#AF416",
"confidence": 7
},
"AF417": {
"revision": 2,
- "explanation": "The transformers provide broadband impedance transformation between the 50 ohm system and the low transistor input and output impedances. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "The transformers provide broadband impedance transformation between the 50 ohm system and the low transistor input and output impedances.",
"source": "https://50ohm.de/NEA_leistungsvertaerker.html#AF417",
"confidence": 7
},
"AF418": {
"revision": 2,
- "explanation": "An inductor in series with shunt capacitors forms an LC low-pass section: it passes DC/low-frequency supply current but diverts RF to ground. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "An inductor in series with shunt capacitors forms an LC low-pass section: it passes DC/low-frequency supply current but diverts RF to ground.",
"source": "https://50ohm.de/NEA_leistungsvertaerker.html#AF418",
"confidence": 7
},
"AF419": {
"revision": 2,
- "explanation": "The choke and bypass capacitors form supply-line filtering, reducing RF components on the DC supply line rather than filtering the transmitted RF path itself. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "The choke and bypass capacitors form supply-line filtering, reducing RF components on the DC supply line rather than filtering the transmitted RF path itself.",
"source": "https://50ohm.de/NEA_leistungsvertaerker.html#AF419",
"confidence": 7
},
"AF420": {
"revision": 2,
- "explanation": "Moving R3 toward position 3 lowers the gate bias for both LDMOS devices in the DC equivalent circuit, so both drain currents decrease. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Moving R3 toward position 3 lowers the gate bias for both LDMOS devices in the DC equivalent circuit, so both drain currents decrease.",
"source": "https://50ohm.de/NEA_leistungsvertaerker.html#AF420",
"confidence": 7
},
"AF421": {
"revision": 2,
- "explanation": "For DC bias, the gates draw negligible current and the resistor network acts as a voltage divider; at stop 1 the divider sets the gate-source voltage to 3.5 V. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "For DC bias, the gates draw negligible current and the resistor network acts as a voltage divider; at stop 1 the divider sets the gate-source voltage to 3.5 V. Hilfsmittel: the FET gate draws negligible current (device knowledge), so the resistor chain is an unloaded voltage divider U_G = U_B · R/(ΣR) (Spannungsteiler, S. 12).",
"source": "https://50ohm.de/NEA_leistungsvertaerker.html#AF421",
"confidence": 7
},
@@ -2401,49 +2401,49 @@
},
"AF423": {
"revision": 2,
- "explanation": "Increasing LDMOS quiescent current means raising both gate-bias voltages, so both bias controls are moved toward UBIAS. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Increasing LDMOS quiescent current means raising both gate-bias voltages, so both bias controls are moved toward UBIAS.",
"source": "https://50ohm.de/NEA_leistungsvertaerker.html#AF423",
"confidence": 7
},
"AF424": {
"revision": 2,
- "explanation": "R4 affects only the bias path for transistor 1 in the shown circuit; moving its wiper toward UBIAS raises that gate bias and drain current, while transistor 2 is unchanged. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "R4 affects only the bias path for transistor 1 in the shown circuit; moving its wiper toward UBIAS raises that gate bias and drain current, while transistor 2 is unchanged.",
"source": "https://50ohm.de/NEA_leistungsvertaerker.html#AF424",
"confidence": 7
},
"AF425": {
"revision": 2,
- "explanation": "The resistor must drop $13.5 V - 4 V = 9.5 V$ at 10 mA, so $R = 9.5 V / 0.010 A = 950 ohm$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "The resistor must drop $13.5 V - 4 V = 9.5 V$ at 10 mA, so $R = 9.5 V / 0.010 A = 950 ohm$. Hilfsmittel: apply R = U/I (Ohmsches Gesetz, S.11) to the resistor's drop and current.",
"source": "https://50ohm.de/NEA_integrierte_schaltkreise.html#AF425",
"confidence": 7
},
"AF426": {
"revision": 2,
- "explanation": "The bias resistor drops $13.8 V - 4 V = 9.8 V$ at 15 mA; $9.8 V / 0.015 A = 653 ohm$, so the nearest listed standard value is 680 ohm. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "The bias resistor drops $13.8 V - 4 V = 9.8 V$ at 15 mA; $9.8 V / 0.015 A = 653 ohm$, so the nearest listed standard value is 680 ohm. Hilfsmittel: apply R = U/I (Ohmsches Gesetz, S.11); then pick the nearest E-series value (S.12).",
"source": "https://50ohm.de/NEA_integrierte_schaltkreise.html#AF426",
"confidence": 7
},
"AF427": {
"revision": 2,
- "explanation": "With a 4 V MMIC drop, the resistor current is $(9 V - 4 V) / 470 ohm = 10.6 mA$; MMIC heat is $4 V x 10.6 mA = 42.6 mW$, about 43 mW. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "With a 4 V MMIC drop, the resistor current is $(9 V - 4 V) / 470 ohm = 10.6 mA$; MMIC heat is $4 V x 10.6 mA = 42.6 mW$, about 43 mW. Hilfsmittel: first I = U/R (Ohmsches Gesetz, S.11), then heat P = U·I (Leistung, S.12).",
"source": "https://50ohm.de/NEA_integrierte_schaltkreise.html#AF427",
"confidence": 7
},
"AF428": {
"revision": 3,
- "explanation": "Overall gain in dB is the output level minus the input level in dBm; the diagram's level difference is 48 dB when cable losses are ignored. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Overall gain in dB is the output level minus the input level in dBm; the diagram's level difference is 48 dB when cable losses are ignored. Hilfsmittel: overall gain in dB is the level difference (out − in), i.e. g = 10·log10(P2/P1) (Pegel, S.15).",
"source": "https://50ohm.de/NEA_leistungsvertaerker.html#AF428",
"confidence": 7
},
"AF501": {
"revision": 2,
- "explanation": "The converter uses the 9th harmonic of the crystal oscillator and maps the 436 to 440 MHz range to 28 to 30 MHz, so $f_Q = (f_rx - f_IF) / 9$ gives 45.333 MHz and 45.556 MHz. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "The converter uses the 9th harmonic of the crystal oscillator and maps the 436 to 440 MHz range to 28 to 30 MHz, so $f_Q = (f_rx - f_IF) / 9$ gives 45.333 MHz and 45.556 MHz. Hilfsmittel: the down-conversion uses f_Q = (f_rx − f_IF)/9, i.e. a difference (f_ZF = |f_E − f_OSZ|, Zwischenfrequenz f_ZF = |f_E − f_OSZ| (S.14)) divided by the 9th harmonic.",
"source": "https://50ohm.de/NEA_transverter_2.html#AF501",
"confidence": 7
},
"AF502": {
"revision": 2,
- "explanation": "For the lower 70 cm segment, the same conversion maps 430 to 434 MHz down to 28 to 30 MHz using the 9th harmonic, so $(430 - 28) / 9 = 44.667 MHz$ and $(434 - 30) / 9 = 44.889 MHz$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "For the lower 70 cm segment, the same conversion maps 430 to 434 MHz down to 28 to 30 MHz using the 9th harmonic, so $(430 - 28) / 9 = 44.667 MHz$ and $(434 - 30) / 9 = 44.889 MHz$. Hilfsmittel: the down-conversion uses f_Q = (f_rx − f_IF)/9, i.e. a difference (f_ZF = |f_E − f_OSZ|, Zwischenfrequenz f_ZF = |f_E − f_OSZ| (S.14)) divided by the 9th harmonic.",
"source": "https://50ohm.de/NEA_transverter_2.html#AF502",
"confidence": 7
},
@@ -2502,14 +2502,14 @@
"confidence": 8
},
"AF610": {
- "revision": 3,
- "explanation": "Eight bits give 256 steps across 1 V; $1 V / 256$ is about 3.9 mV. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "revision": 4,
+ "explanation": "An $n$-bit converter has $2^n$ output levels (codes); 8 bits give $2^8 = 256$. The step size (LSB) is the full-scale range divided by the number of steps: by the usual convention $\\text{LSB} = 1\\,\\text{V}/2^8 = 1/256 \\approx 3.9\\,\\text{mV}$. (Taking it as $1\\,\\text{V}/(2^8-1) = 1/255 \\approx 3.92\\,\\text{mV}$ gives the same to this precision.) So about 4 mV. Hilfsmittel: the Zweierpotenzen table (S.11) gives 2⁸ = 256; divide the 1 V range by that.",
"source": "https://50ohm.de/NEA_digital_analog_umsetzer.html#AF610",
"confidence": 8
},
"AF611": {
- "revision": 3,
- "explanation": "Ten bits give 1024 steps across 1 V; $1 V / 1024$ is about 0.98 mV. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "revision": 4,
+ "explanation": "An $n$-bit converter has $2^n$ output levels (codes); 10 bits give $2^{10} = 1024$. The step size (LSB) is the full-scale range over the number of steps: $\\text{LSB} = 1\\,\\text{V}/2^{10} = 1/1024 \\approx 0.98\\,\\text{mV}$. (Using $1\\,\\text{V}/1023$ gives the same to this precision.) So about 1 mV. Hilfsmittel: the Zweierpotenzen table (S.11) gives 2¹⁰ = 1024; divide the 1 V range by that.",
"source": "https://50ohm.de/NEA_digital_analog_umsetzer.html#AF611",
"confidence": 8
},
@@ -2551,7 +2551,7 @@
},
"AF618": {
"revision": 2,
- "explanation": "Nyquist requires a sampling rate greater than twice the highest signal frequency, so the smallest safe rate is just above $2 f_{max}$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Nyquist requires a sampling rate greater than twice the highest signal frequency, so the smallest safe rate is just above $2 f_{max}$. Hilfsmittel: apply the Abtasttheorem f_abtast > 2·f_max (S.18).",
"source": "https://50ohm.de/NEA_abtasttheorem.html#AF618",
"confidence": 8
},
@@ -2581,7 +2581,7 @@
},
"AF623": {
"revision": 2,
- "explanation": "For an 8 ksample/s speech ADC, the useful passband must end below the 4 kHz Nyquist limit and then attenuate sharply. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "For an 8 ksample/s speech ADC, the useful passband must end below the 4 kHz Nyquist limit and then attenuate sharply. Hilfsmittel: the passband must stay below f_abtast/2 (Abtasttheorem, S.18).",
"source": "https://50ohm.de/NEA_anti_alias_filter.html#AF623",
"confidence": 8
},
@@ -2593,7 +2593,7 @@
},
"AF625": {
"revision": 2,
- "explanation": "The reconstruction filter should pass the wanted speech band but reject images above the 4 kHz Nyquist frequency for an 8 ksample/s stream. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "The reconstruction filter should pass the wanted speech band but reject images above the 4 kHz Nyquist frequency for an 8 ksample/s stream. Hilfsmittel: reject everything above f_abtast/2 (Abtasttheorem, S.18).",
"source": "https://50ohm.de/NEA_rekonstruktionsfilter.html#AF625",
"confidence": 8
},
@@ -2635,7 +2635,7 @@
},
"AF632": {
"revision": 2,
- "explanation": "Quadrature modulation needs two carriers in quadrature; quadrature means a 90 degree phase difference. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Quadrature modulation needs two carriers in quadrature; quadrature means a 90 degree phase difference.",
"source": "https://50ohm.de/NEA_iq_verfahren.html#AF632",
"confidence": 8
},
@@ -2647,19 +2647,19 @@
},
"AF634": {
"revision": 3,
- "explanation": "Complex I/Q sampling represents positive and negative baseband frequencies around the carrier, so 48 ksample/s covers -24 kHz to +24 kHz. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Complex I/Q sampling represents positive and negative baseband frequencies around the carrier, so 48 ksample/s covers -24 kHz to +24 kHz. Hilfsmittel: each sampled I and Q channel obeys the Nyquist bound f_max < f_abtast/2 (Abtasttheorem, S. 18); mapping that magnitude limit to the symmetric −f_s/2 … +f_s/2 complex-I/Q span is outside knowledge.",
"source": "https://50ohm.de/NEA_iq_verfahren.html#AF634",
"confidence": 8
},
"AF635": {
"revision": 3,
- "explanation": "For complex I/Q data, the displayed baseband span equals the sample rate, split symmetrically around zero: 96 ksample/s gives +/-48 kHz. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "For complex I/Q data, the displayed baseband span equals the sample rate, split symmetrically around zero: 96 ksample/s gives +/-48 kHz. Hilfsmittel: each sampled I and Q channel obeys the Nyquist bound f_max < f_abtast/2 (Abtasttheorem, S. 18); mapping that magnitude limit to the symmetric −f_s/2 … +f_s/2 complex-I/Q span is outside knowledge.",
"source": "https://50ohm.de/NEA_iq_verfahren.html#AF635",
"confidence": 8
},
"AF636": {
"revision": 3,
- "explanation": "With I and Q each sampled at 10 Msample/s, the complex baseband span is 10 MHz total, i.e. -5 MHz to +5 MHz. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "With I and Q each sampled at 10 Msample/s, the complex baseband span is 10 MHz total, i.e. -5 MHz to +5 MHz. Hilfsmittel: each sampled I and Q channel obeys the Nyquist bound f_max < f_abtast/2 (Abtasttheorem, S. 18); mapping that magnitude limit to the symmetric −f_s/2 … +f_s/2 complex-I/Q span is outside knowledge.",
"source": "https://50ohm.de/NEA_iq_verfahren.html#AF636",
"confidence": 8
},
@@ -2731,31 +2731,31 @@
},
"AG101": {
"revision": 2,
- "explanation": "A half-wave dipole has total length $0.95 c/(2f)$; at 14.2 MHz that is about 10.04 m total, or 5.02 m per side. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "A half-wave dipole has total length $0.95 c/(2f)$; at 14.2 MHz that is about 10.04 m total, or 5.02 m per side.",
"source": "https://50ohm.de/NEA_verkuerzungsfaktor_2.html#AG101",
"confidence": 8
},
"AG102": {
"revision": 2,
- "explanation": "Using $0.95 c/(2f)$ at 7.1 MHz gives about 20.08 m total dipole length, so each half is about 10.04 m. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Using $0.95 c/(2f)$ at 7.1 MHz gives about 20.08 m total dipole length, so each half is about 10.04 m.",
"source": "https://50ohm.de/NEA_verkuerzungsfaktor_2.html#AG102",
"confidence": 8
},
"AG103": {
"revision": 3,
- "explanation": "For a shortened half-wave dipole, $f = 0.95 c/(2l)$; with 20 m total length this is about 7.125 MHz. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "For a shortened half-wave dipole, $f = 0.95 c/(2l)$; with 20 m total length this is about 7.125 MHz.",
"source": "https://50ohm.de/NEA_verkuerzungsfaktor_2.html#AG103",
"confidence": 8
},
"AG104": {
"revision": 2,
- "explanation": "A quarter-wave groundplane uses quarter-wave radiator and radials; $0.95 c/(4 \\cdot 7.1 MHz)$ is about 10.04 m. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "A quarter-wave groundplane uses quarter-wave radiator and radials; $0.95 c/(4 \\cdot 7.1 MHz)$ is about 10.04 m.",
"source": "https://50ohm.de/NEA_verkuerzungsfaktor_2.html#AG104",
"confidence": 8
},
"AG105": {
"revision": 2,
- "explanation": "A 5/8-wave vertical length is $0.97 \\cdot 5/8 \\cdot c/f$; at 14.2 MHz this is about 12.8 m. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "A 5/8-wave vertical length is $0.97 \\cdot 5/8 \\cdot c/f$; at 14.2 MHz this is about 12.8 m.",
"source": "https://50ohm.de/NEA_verkuerzungsfaktor_2.html#AG105",
"confidence": 8
},
@@ -2809,19 +2809,19 @@
},
"AG114": {
"revision": 3,
- "explanation": "In a 20/15/10 m trap dipole, the inner trap pair must stop the 15 m current, so it is tuned near 21.2 MHz. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "In a 20/15/10 m trap dipole, the inner trap pair must stop the 15 m current, so it is tuned near 21.2 MHz.",
"source": "https://50ohm.de/NEA_traps.html#AG114",
"confidence": 8
},
"AG115": {
"revision": 3,
- "explanation": "The outer trap pair separates the 10 m section, so it is tuned near the 10 m operating frequency around 29 MHz. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "The outer trap pair separates the 10 m section, so it is tuned near the 10 m operating frequency around 29 MHz.",
"source": "https://50ohm.de/NEA_traps.html#AG115",
"confidence": 8
},
"AG116": {
"revision": 2,
- "explanation": "For 80 m a half-wave section is roughly 40 m, and traps for a 160/80 m antenna are tuned near the 80 m band around 3.65 MHz. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "For 80 m a half-wave section is roughly 40 m, and traps for a 160/80 m antenna are tuned near the 80 m band around 3.65 MHz.",
"source": "https://50ohm.de/NEA_traps.html#AG116",
"confidence": 8
},
@@ -2881,7 +2881,7 @@
},
"AG126": {
"revision": 2,
- "explanation": "Circular polarization requires two perpendicular linear fields with equal amplitude and 90 degree phase difference, so one Yagi must be delayed by a quarter wavelength. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Circular polarization requires two perpendicular linear fields with equal amplitude and 90 degree phase difference, so one Yagi must be delayed by a quarter wavelength.",
"source": "https://50ohm.de/NEA_yagi_uda_3.html#AG126",
"confidence": 8
},
@@ -2929,31 +2929,31 @@
},
"AG207": {
"revision": 2,
- "explanation": "A center-fed half-wave dipole and its odd harmonics have current maximum at the feed point, giving series resonance and low feed impedance. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "A center-fed half-wave dipole and its odd harmonics have current maximum at the feed point, giving series resonance and low feed impedance.",
"source": "https://50ohm.de/NEA_strom_spannung_speisung_2.html#AG207",
"confidence": 8
},
"AG208": {
"revision": 2,
- "explanation": "At even harmonics the center feed lies at a voltage maximum/current minimum, so the dipole is voltage-fed and high impedance. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "At even harmonics the center feed lies at a voltage maximum/current minimum, so the dipole is voltage-fed and high impedance.",
"source": "https://50ohm.de/NEA_strom_spannung_speisung_2.html#AG208",
"confidence": 8
},
"AG209": {
"revision": 2,
- "explanation": "At resonance the reactive parts cancel, leaving the feedpoint impedance mainly resistive. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "At resonance the reactive parts cancel, leaving the feedpoint impedance mainly resistive.",
"source": "https://50ohm.de/NEA_fusspunktimpedanz_2.html#AG209",
"confidence": 8
},
"AG210": {
"revision": 2,
- "explanation": "Below resonance a dipole is electrically too short and capacitive; above resonance it is electrically too long and inductive. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Below resonance a dipole is electrically too short and capacitive; above resonance it is electrically too long and inductive.",
"source": "https://50ohm.de/NEA_fusspunktimpedanz_2.html#AG210",
"confidence": 8
},
"AG211": {
"revision": 2,
- "explanation": "A half-wave dipole high enough above ground has a feed resistance close to the textbook free-space value, roughly 65 to 75 ohms. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "A half-wave dipole high enough above ground has a feed resistance close to the textbook free-space value, roughly 65 to 75 ohms.",
"source": "https://50ohm.de/NEA_fusspunktimpedanz_2.html#AG211",
"confidence": 8
},
@@ -2965,49 +2965,49 @@
},
"AG213": {
"revision": 2,
- "explanation": "Antenna gain over a dipole compares the power needed by the reference dipole with the power needed by the directional antenna for the same field strength. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Antenna gain over a dipole compares the power needed by the reference dipole with the power needed by the directional antenna for the same field strength.",
"source": "https://50ohm.de/NEA_vor_rueck_verhaeltnis.html#AG213",
"confidence": 8
},
"AG214": {
"revision": 2,
- "explanation": "Front-to-back ratio compares radiation in the main direction with radiation in the opposite rear direction. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Front-to-back ratio compares radiation in the main direction with radiation in the opposite rear direction.",
"source": "https://50ohm.de/NEA_vor_rueck_verhaeltnis.html#AG214",
"confidence": 8
},
"AG215": {
"revision": 3,
- "explanation": "The rear ERP is transmit power times forward gain divided by front-to-back ratio: 100 W x 10 / 100 = 10 W. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "The rear ERP is transmit power times forward gain divided by front-to-back ratio: 100 W x 10 / 100 = 10 W. Hilfsmittel: convert the dB figures via the dB↔ratio table (S. 15): 10 dB → ×10, 20 dB → ×100; the rear-ERP = P · gain ÷ front-to-back relationship is antenna knowledge, outside the sheet.",
"source": "https://50ohm.de/NEA_vor_rueck_verhaeltnis.html#AG215",
"confidence": 8
},
"AG216": {
"revision": 3,
- "explanation": "15 dBd is a factor 31.6 and 25 dB front-to-back is a factor 316; $6 W \\cdot 31.6 / 316$ is about 0.6 W rear ERP. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "15 dBd is a factor 31.6 and 25 dB front-to-back is a factor 316; $6 W \\cdot 31.6 / 316$ is about 0.6 W rear ERP. Hilfsmittel: first convert the dBd/dB figures to factors with the inverse of g = 10·log10(P2/P1) (Pegel, S.15), then take the power ratio.",
"source": "https://50ohm.de/NEA_vor_rueck_verhaeltnis.html#AG216",
"confidence": 8
},
"AG217": {
"revision": 3,
- "explanation": "Front-to-back ratio is a power ratio: $10 log10(15/0.6)$ is about 14 dB. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Front-to-back ratio is a power ratio: $10 log10(15/0.6)$ is about 14 dB. Hilfsmittel: apply g = 10·log10(P2/P1) (Pegel, S.15) to the forward/back power ratio.",
"source": "https://50ohm.de/NEA_vor_rueck_verhaeltnis.html#AG217",
"confidence": 8
},
"AG218": {
"revision": 3,
- "explanation": "For field strengths, use 20 log10 of the voltage ratio: 300/128 gives 7.4 dBd, and 300/20 gives 23.5 dB front-to-back. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "For field strengths, use 20 log10 of the voltage ratio: 300/128 gives 7.4 dBd, and 300/20 gives 23.5 dB front-to-back. Hilfsmittel: use the voltage form g = 20·log10(U2/U1) (Pegel, S.15) for both ratios.",
"source": "https://50ohm.de/NEA_vor_rueck_verhaeltnis.html#AG218",
"confidence": 8
},
"AG219": {
"revision": 2,
- "explanation": "Half-power is 3 dB down; field strength is proportional to the square root of power, so the boundary is $1/sqrt(2) = 0.707$ of maximum field. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Half-power is 3 dB down; field strength is proportional to the square root of power, so the boundary is $1/sqrt(2) = 0.707$ of maximum field. Hilfsmittel: −3 dB is half power, a 0,71 voltage ratio (dB↔ratio table, S.15); the half-power-beamwidth definition itself is antenna knowledge.",
"source": "https://50ohm.de/NEA_halbwertsbreite.html#AG219",
"confidence": 8
},
"AG220": {
"revision": 2,
- "explanation": "The half-power beamwidth is read at the 0.707 relative-field circle, which is the point marked c in the diagram. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "The half-power beamwidth is read at the 0.707 relative-field circle, which is the point marked c in the diagram.",
"source": "https://50ohm.de/NEA_halbwertsbreite.html#AG220",
"confidence": 8
},
@@ -3025,7 +3025,7 @@
},
"AG223": {
"revision": 2,
- "explanation": "A 5/8-wave vertical over ground has a low elevation-angle maximum, useful for flat long-distance HF radiation. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "A 5/8-wave vertical over ground has a low elevation-angle maximum, useful for flat long-distance HF radiation.",
"source": "https://50ohm.de/NEA_antennenformen_3.html#AG223",
"confidence": 8
},
@@ -3042,32 +3042,32 @@
"confidence": 8
},
"AG226": {
- "revision": 2,
- "explanation": "Parabolic gain rises with aperture diameter over wavelength: with 30 cm at 5.7 GHz and ideal efficiency, $G=(pi D/lambda)^2$ gives about 25 dBi. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "revision": 3,
+ "explanation": "First the wavelength: $\\lambda = c/f = (3\\cdot10^8)/(5.7\\cdot10^9) \\approx 0.0526\\,\\text{m}$. Then $\\pi D/\\lambda = \\pi \\cdot 0.30/0.0526 \\approx 17.9$, so the linear gain is $(\\pi D/\\lambda)^2 \\cdot \\eta \\approx 17.9^2 \\approx 321$, and in decibels $g_i = 10\\log_{10}(321) \\approx 25.1\\,\\text{dBi}$. (Note $(\\pi D/\\lambda)^2$ is the linear gain, not dBi; take $10\\log_{10}$ of it to get dBi.) Hilfsmittel: first λ = c/f (S.17), then g_i = 10·log10[(π·d/λ)²·η] (Parabolspiegel, S.15).",
"source": "https://50ohm.de/NEA_parbolspiegel_2.html#AG226",
"confidence": 8
},
"AG227": {
- "revision": 2,
- "explanation": "At the same frequency, increasing dish diameter to 80 cm increases aperture area strongly; the ideal-gain formula gives about 33.6 dBi. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "revision": 3,
+ "explanation": "$\\lambda = c/f = (3\\cdot10^8)/(5.7\\cdot10^9) \\approx 0.0526\\,\\text{m}$; $\\pi D/\\lambda = \\pi \\cdot 0.80/0.0526 \\approx 47.7$; linear gain $(\\pi D/\\lambda)^2 \\approx 47.7^2 \\approx 2280$; $g_i = 10\\log_{10}(2280) \\approx 33.6\\,\\text{dBi}$. Hilfsmittel: first λ = c/f (S.17), then g_i = 10·log10[(π·d/λ)²·η] (Parabolspiegel, S.15).",
"source": "https://50ohm.de/NEA_parbolspiegel_2.html#AG227",
"confidence": 8
},
"AG228": {
- "revision": 2,
- "explanation": "For a fixed 80 cm dish, the shorter 10.4 GHz wavelength increases aperture gain to about 38.8 dBi. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "revision": 3,
+ "explanation": "$\\lambda = c/f = (3\\cdot10^8)/(10.4\\cdot10^9) \\approx 0.0288\\,\\text{m}$; $\\pi D/\\lambda = \\pi \\cdot 0.80/0.0288 \\approx 87.1$; linear gain $\\approx 87.1^2 \\approx 7590$; $g_i = 10\\log_{10}(7590) \\approx 38.8\\,\\text{dBi}$. Hilfsmittel: first λ = c/f (S.17), then g_i = 10·log10[(π·d/λ)²·η] (Parabolspiegel, S.15).",
"source": "https://50ohm.de/NEA_parbolspiegel_2.html#AG228",
"confidence": 8
},
"AG229": {
- "revision": 2,
- "explanation": "A 1.2 m dish at 10.4 GHz has a large diameter-to-wavelength ratio, so the ideal parabolic-gain formula gives about 42.3 dBi. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "revision": 3,
+ "explanation": "$\\lambda = c/f = (3\\cdot10^8)/(10.4\\cdot10^9) \\approx 0.0288\\,\\text{m}$; $\\pi D/\\lambda = \\pi \\cdot 1.20/0.0288 \\approx 130.7$; linear gain $\\approx 130.7^2 \\approx 17080$; $g_i = 10\\log_{10}(17080) \\approx 42.3\\,\\text{dBi}$. Hilfsmittel: first λ = c/f (S.17), then g_i = 10·log10[(π·d/λ)²·η] (Parabolspiegel, S.15).",
"source": "https://50ohm.de/NEA_parbolspiegel_2.html#AG229",
"confidence": 8
},
"AG301": {
"revision": 2,
- "explanation": "A shielded feed line keeps high RF fields inside the cable and reduces coupling into building wiring and equipment. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "A shielded feed line keeps high RF fields inside the cable and reduces coupling into building wiring and equipment.",
"source": "https://50ohm.de/NEA_uebertragungsleitungen_3.html#AG301",
"confidence": 8
},
@@ -3091,13 +3091,13 @@
},
"AG305": {
"revision": 2,
- "explanation": "For open wire line, $Z_0$ rises with conductor spacing and falls with conductor diameter; the given 20 cm spacing and 2 mm wire give about 635 ohms. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "For open wire line, $Z_0$ rises with conductor spacing and falls with conductor diameter; the given 20 cm spacing and 2 mm wire give about 635 ohms. Hilfsmittel: apply the two-wire line Z = 120Ω/√εr·ln(2a/d) (Wellenwiderstand, S.17).",
"source": "https://50ohm.de/NEA_wellenwiderstand.html#AG305",
"confidence": 8
},
"AG306": {
"revision": 2,
- "explanation": "For air coax, $Z_0 = 60 ln(D/d)$; with 5 mm over 1 mm this is about 97 ohms. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "For air coax, $Z_0 = 60 ln(D/d)$; with 5 mm over 1 mm this is about 97 ohms. Hilfsmittel: apply the coax line Z = 60Ω/√εr·ln(D/d) (Wellenwiderstand, S.17).",
"source": "https://50ohm.de/NEA_wellenwiderstand.html#AG306",
"confidence": 8
},
@@ -3163,7 +3163,7 @@
},
"AG317": {
"revision": 2,
- "explanation": "A quarter wave at 145 MHz is about 0.517 m in free space; with velocity factor 0.66 it is about 0.342 m. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "A quarter wave at 145 MHz is about 0.517 m in free space; with velocity factor 0.66 it is about 0.342 m. Hilfsmittel: first λ = c/f (c = f·λ (λ[m] ≈ 300/f[MHz]), S.17), take a quarter, then apply the Verkürzungsfaktor k_v = 1/√εr (S.17).",
"source": "https://50ohm.de/NEA_uebertragungsleitungen_3.html#AG317",
"confidence": 8
},
@@ -3193,7 +3193,7 @@
},
"AG402": {
"revision": 3,
- "explanation": "A 3 dB line loss halves the power on the way to the far end; with open or short circuit, that 50 W reaching the end is reflected. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "A 3 dB line loss halves the power on the way to the far end; with open or short circuit, that 50 W reaching the end is reflected. Hilfsmittel: −3 dB → ×0,5 power via the dB↔ratio table (Pegel, S.15); a half-power loss reflects on open/short.",
"source": "https://50ohm.de/NEA_swr_3.html#AG402",
"confidence": 8
},
@@ -3223,25 +3223,25 @@
},
"AG407": {
"revision": 2,
- "explanation": "A quarter wavelength is one quarter of a full 360 degree RF cycle, so the phase shift is 90 degrees. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "A quarter wavelength is one quarter of a full 360 degree RF cycle, so the phase shift is 90 degrees.",
"source": "https://50ohm.de/NEA_leitung_phasenverschiebung.html#AG407",
"confidence": 8
},
"AG408": {
"revision": 2,
- "explanation": "One full wavelength produces a 360 degree phase shift, which is equivalent to 0 degrees at the input reference. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "One full wavelength produces a 360 degree phase shift, which is equivalent to 0 degrees at the input reference.",
"source": "https://50ohm.de/NEA_leitung_phasenverschiebung.html#AG408",
"confidence": 8
},
"AG409": {
"revision": 2,
- "explanation": "A quarter-wave line inverts impedance: a short circuit at one end appears as very high impedance at the other. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "A quarter-wave line inverts impedance: a short circuit at one end appears as very high impedance at the other.",
"source": "https://50ohm.de/NEA_lecherleitung.html#AG409",
"confidence": 8
},
"AG410": {
"revision": 2,
- "explanation": "The quarter-wave section transforms the far-end condition so point X is at a current maximum and nearly zero impedance. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "The quarter-wave section transforms the far-end condition so point X is at a current maximum and nearly zero impedance.",
"source": "https://50ohm.de/NEA_lecherleitung.html#AG410",
"confidence": 8
},
@@ -3259,49 +3259,49 @@
},
"AG413": {
"revision": 2,
- "explanation": "A half-wave dipole is low impedance at its feed, and a half-wave feed line repeats that low impedance at its input. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "A half-wave dipole is low impedance at its feed, and a half-wave feed line repeats that low impedance at its input.",
"source": "https://50ohm.de/NEA_impedanztransformation.html#AG413",
"confidence": 8
},
"AG414": {
"revision": 2,
- "explanation": "A full-wave dipole is voltage-fed and high impedance at the feed point, and a half-wave line repeats that high impedance. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "A full-wave dipole is voltage-fed and high impedance at the feed point, and a half-wave line repeats that high impedance.",
"source": "https://50ohm.de/NEA_impedanztransformation.html#AG414",
"confidence": 8
},
"AG415": {
"revision": 2,
- "explanation": "A quarter-wave line transforms impedance, so a high-impedance full-wave dipole becomes low impedance at the line input. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "A quarter-wave line transforms impedance, so a high-impedance full-wave dipole becomes low impedance at the line input.",
"source": "https://50ohm.de/NEA_impedanztransformation.html#AG415",
"confidence": 8
},
"AG416": {
"revision": 2,
- "explanation": "A half-wave transmission line repeats the load impedance independent of its own characteristic impedance, so the input remains 70 ohms. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "A half-wave transmission line repeats the load impedance independent of its own characteristic impedance, so the input remains 70 ohms.",
"source": "https://50ohm.de/NEA_impedanztransformation.html#AG416",
"confidence": 8
},
"AG417": {
"revision": 2,
- "explanation": "For a quarter-wave transformer, $Z_t = sqrt(Z_1 Z_2)$; $sqrt(60 \\cdot 240)$ is 120 ohms. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "For a quarter-wave transformer, $Z_t = sqrt(Z_1 Z_2)$; $sqrt(60 \\cdot 240)$ is 120 ohms. Hilfsmittel: apply Z = √(Z_E·Z_A) (Viertelwellentransformator, S.17).",
"source": "https://50ohm.de/NEA_impedanztransformation.html#AG417",
"confidence": 8
},
"AG418": {
"revision": 2,
- "explanation": "The quarter-wave transformer impedance is $sqrt(240 \\cdot 600)$, which is about 380 ohms. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "The quarter-wave transformer impedance is $sqrt(240 \\cdot 600)$, which is about 380 ohms. Hilfsmittel: apply Z = √(Z_E·Z_A) (Viertelwellentransformator, S.17).",
"source": "https://50ohm.de/NEA_impedanztransformation.html#AG418",
"confidence": 8
},
"AG419": {
"revision": 2,
- "explanation": "An end-fed resonant wire has a high feed impedance when its length is a half wavelength or an integer multiple of that. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "An end-fed resonant wire has a high feed impedance when its length is a half wavelength or an integer multiple of that.",
"source": "https://50ohm.de/NEA_antennenformen_3.html#AG419",
"confidence": 8
},
"AG420": {
"revision": 2,
- "explanation": "A coax feed is unbalanced while a dipole is balanced, so a balun or equivalent phasing line provides the symmetry transition. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "A coax feed is unbalanced while a dipole is balanced, so a balun or equivalent phasing line provides the symmetry transition.",
"source": "https://50ohm.de/NEA_umwegleitung.html#AG420",
"confidence": 8
},
@@ -3319,13 +3319,13 @@
},
"AG423": {
"revision": 2,
- "explanation": "The half-wave bypass line provides both impedance transformation and phase reversal, matching the folded dipole's high balanced impedance to lower unbalanced coax. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "The half-wave bypass line provides both impedance transformation and phase reversal, matching the folded dipole's high balanced impedance to lower unbalanced coax.",
"source": "https://50ohm.de/NEA_umwegleitung.html#AG423",
"confidence": 8
},
"AG424": {
"revision": 2,
- "explanation": "Each folded-dipole terminal is about 120 ohms to ground; the half-wave detour preserves magnitude but reverses phase, so the two 120 ohm paths combine to 60 ohms. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Each folded-dipole terminal is about 120 ohms to ground; the half-wave detour preserves magnitude but reverses phase, so the two 120 ohm paths combine to 60 ohms.",
"source": "https://50ohm.de/NEA_umwegleitung.html#AG424",
"confidence": 8
},
@@ -3343,13 +3343,13 @@
},
"AG427": {
"revision": 2,
- "explanation": "Common-mode current arises when the antenna/feed system is unbalanced or nearby objects couple RF onto the outside of the coax shield. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Common-mode current arises when the antenna/feed system is unbalanced or nearby objects couple RF onto the outside of the coax shield.",
"source": "https://50ohm.de/NEA_mantelwellen_2.html#AG427",
"confidence": 8
},
"AG428": {
"revision": 2,
- "explanation": "A common-mode choke or suitable balun raises the impedance for the unwanted shield current while leaving the wanted differential feed current mostly unaffected. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "A common-mode choke or suitable balun raises the impedance for the unwanted shield current while leaving the wanted differential feed current mostly unaffected.",
"source": "https://50ohm.de/NEA_mantelwellen_2.html#AG428",
"confidence": 8
},
@@ -3367,13 +3367,13 @@
},
"AG502": {
"revision": 2,
- "explanation": "First subtract feed-line loss from transmitter power to get antenna input power; multiplying by antenna gain relative to a dipole gives ERP. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "First subtract feed-line loss from transmitter power to get antenna input power; multiplying by antenna gain relative to a dipole gives ERP. Hilfsmittel: apply P_ERP = P_S·10^((g_d−a)/10dB) (ERP, S.15); loss and gain combine in dB in the exponent — you cannot subtract dB from watts.",
"source": "https://50ohm.de/NEA_effektive_strahlungsleistung_erp_2.html#AG502",
"confidence": 8
},
"AG503": {
"revision": 3,
- "explanation": "A 20 dB loss is a factor of 100; 50 W divided by 100 gives 0.5 W ERP. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "A 20 dB loss is a factor of 100; 50 W divided by 100 gives 0.5 W ERP. Hilfsmittel: 20 dB → ×100 via the dB↔ratio table (Pegel, S.15); divide power by that factor.",
"source": "https://50ohm.de/NEA_effektive_strahlungsleistung_erp_2.html#AG503",
"confidence": 8
},
@@ -3433,7 +3433,7 @@
},
"AH202": {
"revision": 3,
- "explanation": "In a sunspot minimum the higher HF bands open less reliably, while 20 m often remains the best daily long-distance band. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "In a sunspot minimum the higher HF bands open less reliably, while 20 m often remains the best daily long-distance band.",
"source": "https://50ohm.de/NEA_ionosphaere_3.html#AH202",
"confidence": 8
},
@@ -3457,7 +3457,7 @@
},
"AH206": {
"revision": 2,
- "explanation": "MUF literally means maximum usable frequency: the highest frequency that can still support the specified ionospheric path. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "MUF literally means maximum usable frequency: the highest frequency that can still support the specified ionospheric path.",
"source": "https://50ohm.de/NEA_muf_luf_2.html#AH206",
"confidence": 8
},
@@ -3475,7 +3475,7 @@
},
"AH209": {
"revision": 2,
- "explanation": "Using $MUF = f_k/sin(45°)$ gives about 4.2 MHz; the optimum working frequency is about 85% of MUF, or 3.6 MHz. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Using $MUF = f_k/sin(45°)$ gives about 4.2 MHz; the optimum working frequency is about 85% of MUF, or 3.6 MHz. Hilfsmittel: apply MUF ≈ f_c/sin(α) then f_opt = MUF·0,85 (S.18).",
"source": "https://50ohm.de/NEA_muf_luf_2.html#AH209",
"confidence": 8
},
@@ -3649,7 +3649,7 @@
},
"AI104": {
"revision": 3,
- "explanation": "The meter input current is $I = U/R = 0.5 V / 10 MOhm = 50 nA$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "The meter input current is $I = U/R = 0.5 V / 10 MOhm = 50 nA$. Hilfsmittel: apply I = U/R (Ohmsches Gesetz, S.11).",
"source": "https://50ohm.de/NEA_strom_spannung_messung_3.html#AI104",
"confidence": 8
},
@@ -3733,19 +3733,19 @@
},
"AI305": {
"revision": 3,
- "explanation": "From the trace the peak voltage is 100 V; $P_{PEP} = (100/sqrt(2))^2 / 50$ gives 100 W. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "From the trace the peak voltage is 100 V; $P_{PEP} = (100/sqrt(2))^2 / 50$ gives 100 W. Hilfsmittel: first U_eff = Û/√2 (from Û = U_eff·√2, Wechselspannung, S.12), then P = U_eff²/R (Leistung, S.12). Both formulas are on S.12; there is no ready-made PEP formula.",
"source": "https://50ohm.de/NEA_ssb_3.html#AI305",
"confidence": 8
},
"AI306": {
"revision": 3,
- "explanation": "A 10:1 probe means the real peak voltage is ten times the displayed value; $P_{PEP} = (60/sqrt(2))^2 / 50$ gives 36 W. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "A 10:1 probe means the real peak voltage is ten times the displayed value; $P_{PEP} = (60/sqrt(2))^2 / 50$ gives 36 W. Hilfsmittel: correct the 10:1 probe (×10), then U_eff = Û/√2 (Wechselspannung, S.12) and P = U_eff²/R (Leistung, S.12).",
"source": "https://50ohm.de/NEA_ssb_3.html#AI306",
"confidence": 8
},
"AI401": {
"revision": 2,
- "explanation": "An SWR meter uses directional couplers to sample forward and reflected line voltages and compares them. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "An SWR meter uses directional couplers to sample forward and reflected line voltages and compares them.",
"source": "https://50ohm.de/NEA_swr_meter_2.html#AI401",
"confidence": 8
},
@@ -3793,31 +3793,31 @@
},
"AI506": {
"revision": 3,
- "explanation": "0.01% is $10^{-4}$; $29 MHz \\cdot 10^{-4} = 2900 Hz$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "0.01% is $10^{-4}$; $29 MHz \\cdot 10^{-4} = 2900 Hz$.",
"source": "https://50ohm.de/NEA_frequenzgenauigkeit.html#AI506",
"confidence": 8
},
"AI507": {
"revision": 2,
- "explanation": "0.00001% is $10^{-7}$; $14100 kHz \\cdot 10^{-7}$ gives a maximum error of 1.410 Hz. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "0.00001% is $10^{-7}$; $14100 kHz \\cdot 10^{-7}$ gives a maximum error of 1.410 Hz.",
"source": "https://50ohm.de/NEA_frequenzgenauigkeit.html#AI507",
"confidence": 8
},
"AI508": {
"revision": 2,
- "explanation": "One ppm is one part in a million; at 100 MHz that is 100 Hz. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "One ppm is one part in a million; at 100 MHz that is 100 Hz.",
"source": "https://50ohm.de/NEA_frequenzgenauigkeit.html#AI508",
"confidence": 8
},
"AI509": {
"revision": 3,
- "explanation": "10 ppm of 145 MHz is 1450 Hz, so the counter may read 145 MHz plus or minus 0.00145 MHz. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "10 ppm of 145 MHz is 1450 Hz, so the counter may read 145 MHz plus or minus 0.00145 MHz.",
"source": "https://50ohm.de/NEA_frequenzgenauigkeit.html#AI509",
"confidence": 8
},
"AI510": {
"revision": 3,
- "explanation": "Protecting the 144.400 MHz beacon edge needs room for 2.7 kHz USB audio plus 1 ppm frequency error of 144 Hz, totaling 2.844 kHz below the edge. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Protecting the 144.400 MHz beacon edge needs room for 2.7 kHz USB audio plus 1 ppm frequency error of 144 Hz, totaling 2.844 kHz below the edge.",
"source": "https://50ohm.de/NEA_frequenzgenauigkeit.html#AI510",
"confidence": 8
},
@@ -3829,7 +3829,7 @@
},
"AI601": {
"revision": 3,
- "explanation": "The resistor network uses 48 one-watt resistors to obtain about 50 ohms while sharing dissipation, so the continuous rating is 48 W. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "The resistor network uses 48 one-watt resistors to obtain about 50 ohms while sharing dissipation, so the continuous rating is 48 W.",
"source": "https://50ohm.de/NEA_dummy_load_2.html#AI601",
"confidence": 8
},
@@ -3858,14 +3858,14 @@
"confidence": 8
},
"AI606": {
- "revision": 2,
- "explanation": "Correcting the 15.3 V reading for the Schottky drop gives the sampled RF peak; scaling from the 5 ohm tap to the 50 ohm load yields about 60 W. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "revision": 3,
+ "explanation": "The two Schottky diodes form a voltage doubler, whose DC output is $U_{DC} \\approx 2\\hat{U} - 2U_F$. So the RF peak at the 5 Ω tap is $\\hat{U}_{tap} = U_{DC}/2 + U_F = 15.3/2 + 0.23 = 7.88\\,\\text{V}$. The tap sits at the 5 Ω point of the 50 Ω dummy load and carries the full load current, so the peak across the whole 50 Ω is ten times larger, $\\hat{U} = 78.8\\,\\text{V}$. Then $U_{eff} = \\hat{U}/\\sqrt{2} = 55.7\\,\\text{V}$ and $P = U_{eff}^2/R = 55.7^2/50 \\approx 62\\,\\text{W} \\approx 60\\,\\text{W}$. (Treating the circuit as a plain peak detector — forgetting the doubler's factor of 2 — would wrongly give about 240 W.) Hilfsmittel: Û = U_eff·√2 and P = U_eff²/R (both Leistung/Wechselspannung, S.12); the voltage-doubler behaviour, the Schottky drop and the 5 Ω-tap ×10 factor are added knowledge.",
"source": "https://50ohm.de/NEA_sender_messungen.html#AI606",
"confidence": 8
},
"AI607": {
- "revision": 2,
- "explanation": "The detector voltage plus Schottky drop gives the RF peak at the measurement point; applying $P = U_{rms}^2/R$ gives roughly 600 mW. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "revision": 3,
+ "explanation": "The two Schottky diodes form a voltage doubler, whose DC output is $U_{DC} \\approx 2\\hat{U} - 2U_F$, so the RF peak at the input is $\\hat{U} = U_{DC}/2 + U_F = 15.3/2 + 0.23 = 7.88\\,\\text{V}$, i.e. $U_{eff} = \\hat{U}/\\sqrt{2} = 5.57\\,\\text{V}$. The circuit input impedance is $56\\,\\Omega \\parallel 470\\,\\Omega \\approx 50\\,\\Omega$, so $P = U_{eff}^2/R = 5.57^2/50 \\approx 0.62\\,\\text{W} \\approx 600\\,\\text{mW}$. (Forgetting the doubler's factor of 2 would wrongly give about 2.4 W.) Hilfsmittel: Û = U_eff·√2 and P = U_eff²/R (both Leistung/Wechselspannung, S.12); the voltage-doubler behaviour, the Schottky drop and the 56 Ω∥470 Ω ≈ 50 Ω input are added knowledge.",
"source": "https://50ohm.de/NEA_sender_messungen.html#AI607",
"confidence": 8
},
@@ -3877,25 +3877,25 @@
},
"AI609": {
"revision": 3,
- "explanation": "The measuring head is not rated for the full expected 15 W directly; a 20 dB, 20 W attenuator reduces both level and risk. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "The measuring head is not rated for the full expected 15 W directly; a 20 dB, 20 W attenuator reduces both level and risk.",
"source": "https://50ohm.de/NEA_sender_messungen.html#AI609",
"confidence": 8
},
"AI610": {
"revision": 2,
- "explanation": "For 1 W into 50 ohms, $U_{rms}=sqrt(50)$ and peak voltage is about 10 V; the divider and diode drop leave about 4.8 V DC at the output. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "For 1 W into 50 ohms, $U_{rms}=sqrt(50)$ and peak voltage is about 10 V; the divider and diode drop leave about 4.8 V DC at the output. Hilfsmittel: for 1 W into 50 Ω, U_eff = √(P·R) (Leistung, S.12) and Û = U_eff·√2 (Wechselspannung, S.12); the divider/diode drop is added knowledge.",
"source": "https://50ohm.de/NEA_sender_messungen.html#AI610",
"confidence": 8
},
"AI611": {
"revision": 2,
- "explanation": "Add the silicon diode drop to the DC reading to recover peak RF voltage, convert to RMS, then use $P=U^2/R$ with 54.1 ohms to get about 9.7 W. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Add the silicon diode drop to the DC reading to recover peak RF voltage, convert to RMS, then use $P=U^2/R$ with 54.1 ohms to get about 9.7 W. Hilfsmittel: add the diode drop (knowledge), convert peak→RMS (Û = U_eff·√2, Wechselspannung, S.12), then P = U_eff²/R (Leistung, S.12).",
"source": "https://50ohm.de/NEA_sender_messungen.html#AI611",
"confidence": 8
},
"AI612": {
"revision": 2,
- "explanation": "Detector diodes, resistors and layout introduce systematic errors, so accurate RF power readings require calibration correction values. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Detector diodes, resistors and layout introduce systematic errors, so accurate RF power readings require calibration correction values.",
"source": "https://50ohm.de/NEA_sender_messungen.html#AI612",
"confidence": 8
},
@@ -4033,19 +4033,19 @@
},
"AJ201": {
"revision": 3,
- "explanation": "The second harmonic is twice the fundamental: $2 \\cdot 3.730 MHz = 7.460 MHz$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "The second harmonic is twice the fundamental: $2 \\cdot 3.730 MHz = 7.460 MHz$.",
"source": "https://50ohm.de/NEA_nicht_sinus_signale.html#AJ201",
"confidence": 8
},
"AJ202": {
"revision": 3,
- "explanation": "The third harmonic is three times the fundamental: $3 \\cdot 7.050 MHz = 21.150 MHz$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "The third harmonic is three times the fundamental: $3 \\cdot 7.050 MHz = 21.150 MHz$.",
"source": "https://50ohm.de/NEA_nicht_sinus_signale.html#AJ202",
"confidence": 8
},
"AJ203": {
"revision": 3,
- "explanation": "The third overtone is the fourth harmonic, so $4 \\cdot 7.20 MHz = 28.80 MHz$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "The third overtone is the fourth harmonic, so $4 \\cdot 7.20 MHz = 28.80 MHz$.",
"source": "https://50ohm.de/NEA_unerwuenschte_aussendungen_3.html#AJ203",
"confidence": 8
},
@@ -4057,13 +4057,13 @@
},
"AJ205": {
"revision": 3,
- "explanation": "Odd harmonics are 1st, 3rd, 5th, ...; the second odd harmonic is the 3rd, so $3 \\cdot 144.690 MHz = 434.070 MHz$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Odd harmonics are 1st, 3rd, 5th, ...; the second odd harmonic is the 3rd, so $3 \\cdot 144.690 MHz = 434.070 MHz$.",
"source": "https://50ohm.de/NEA_nicht_sinus_signale.html#AJ205",
"confidence": 8
},
"AJ206": {
"revision": 4,
- "explanation": "Of the harmonics of 144.300 MHz, the 3rd at 432.900 MHz lands in the 70 cm amateur band and the 9th at 1298.700 MHz lands in the 23 cm band; other harmonics fall outside amateur allocations and don't disturb amateur operation. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Of the harmonics of 144.300 MHz, the 3rd at 432.900 MHz lands in the 70 cm amateur band and the 9th at 1298.700 MHz lands in the 23 cm band; other harmonics fall outside amateur allocations and don't disturb amateur operation.",
"source": "https://50ohm.de/NEA_nicht_sinus_signale.html#AJ206",
"confidence": 8
},
@@ -4195,13 +4195,13 @@
},
"AK103": {
"revision": 2,
- "explanation": "The simple distance formula assumes far-field behaviour; below roughly $lambda/(2 pi)$ or for electrically small antennas, measurement or near-field modelling is needed. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "The simple distance formula assumes far-field behaviour; below roughly $lambda/(2 pi)$ or for electrically small antennas, measurement or near-field modelling is needed. Hilfsmittel: the simple field/power relation only holds in the far field, d > λ/(2π) (E = √(30Ω·P_EIRP)/d (Feldstärke im Fernfeld, S.15)).",
"source": "https://50ohm.de/NEA_nahfeld.html#AK103",
"confidence": 8
},
"AK104": {
"revision": 2,
- "explanation": "Feed-line loss reduces transmitter power before it reaches the antenna, so antenna input power is transmitter power multiplied by the loss factor. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Feed-line loss reduces transmitter power before it reaches the antenna, so antenna input power is transmitter power multiplied by the loss factor. Hilfsmittel: feed-line loss in dB subtracts from transmitter power before the antenna; a = 10·log10(P_in/P_out) (Pegel, S.15).",
"source": "https://50ohm.de/NEA_personenschutzabstand_3.html#AK104",
"confidence": 8
},
@@ -4218,8 +4218,8 @@
"confidence": 8
},
"AK107": {
- "revision": 3,
- "explanation": "Rearranging the field-strength formula for power with 5 m distance, 28 V/m limit and 6 dBd gain gives roughly 100 W transmitter output. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "revision": 4,
+ "explanation": "Rearrange $E = \\sqrt{30\\,\\Omega \\cdot P_{EIRP}}/d$ for the radiated power: $P_{EIRP} = (E \\cdot d)^2/30 = (28 \\cdot 5)^2/30 = 140^2/30 \\approx 653\\,\\text{W}$ EIRP. Convert the antenna gain to an isotropic factor: $6\\,\\text{dBd} = 8.15\\,\\text{dBi} \\to G = 10^{8.15/10} \\approx 6.53$. The transmitter output is then $P = P_{EIRP}/G = 653/6.53 \\approx 100\\,\\text{W}$. Hilfsmittel: rearrange E = √(30Ω·P_EIRP)/d (Feldstärke im Fernfeld, S.15) for power, using the gain factor.",
"source": "https://50ohm.de/NEA_personenschutzabstand_3.html#AK107",
"confidence": 8
},
@@ -4236,8 +4236,8 @@
"confidence": 8
},
"AK110": {
- "revision": 2,
- "explanation": "Convert 11.5 dBd gain and 1.5 dB cable loss to linear factors, then apply $d = sqrt(30 P_{EIRP})/E$ to get about 6.86 m. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "revision": 3,
+ "explanation": "Net isotropic gain is $11.5\\,\\text{dBd} + 2.15 - 1.5\\,\\text{dB} = 12.15\\,\\text{dBi}$, a factor $G = 10^{12.15/10} \\approx 16.4$. So $P_{EIRP} = 75\\,\\text{W} \\cdot 16.4 \\approx 1231\\,\\text{W}$. Then $d = \\sqrt{30 \\cdot P_{EIRP}}/E = \\sqrt{30 \\cdot 1231}/28 \\approx 192/28 \\approx 6.86\\,\\text{m}$. Hilfsmittel: rearrange E = √(30Ω·P_EIRP)/d (Feldstärke im Fernfeld, S.15) for distance: d = √(30·P_EIRP)/E.",
"source": "https://50ohm.de/NEA_naeherungsformel_2.html#AK110",
"confidence": 8
},
@@ -4255,19 +4255,19 @@
},
"AK113": {
"revision": 3,
- "explanation": "12.15 dBi is a linear gain of about 16.4; $sqrt(30 \\cdot 250 W \\cdot 16.4)/30 m$ is about 11.7 V/m. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "12.15 dBi is a linear gain of about 16.4; $sqrt(30 \\cdot 250 W \\cdot 16.4)/30 m$ is about 11.7 V/m. Hilfsmittel: apply E = √(30Ω·P_EIRP)/d (Feldstärke im Fernfeld, S.15) after converting 12,15 dBi to a factor (G = 10^(g/10dB), S.15).",
"source": "https://50ohm.de/NEA_personenschutzabstand_3.html#AK113",
"confidence": 8
},
"AK114": {
"revision": 3,
- "explanation": "A vertical dipole has about 2.15 dBi gain; applying the free-space field formula at 10 W and 10 m gives roughly 2.2 V/m. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "A vertical dipole has about 2.15 dBi gain; applying the free-space field formula at 10 W and 10 m gives roughly 2.2 V/m. Hilfsmittel: apply E = √(30Ω·P_EIRP)/d (Feldstärke im Fernfeld, S.15) with the dipole gain (g_i = 2,15 dBi, S.15).",
"source": "https://50ohm.de/NEA_personenschutzabstand_3.html#AK114",
"confidence": 8
},
"AK115": {
"revision": 3,
- "explanation": "Convert 100 W ERP to about 164 W EIRP, then $sqrt(30 \\cdot 164 W)/100 m$ gives about 0.7 V/m. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Convert 100 W ERP to about 164 W EIRP, then $sqrt(30 \\cdot 164 W)/100 m$ gives about 0.7 V/m. Hilfsmittel: first ERP→EIRP (×1,64), then apply E = √(30Ω·P_EIRP)/d (Feldstärke im Fernfeld, S.15).",
"source": "https://50ohm.de/NEA_personenschutzabstand_3.html#AK115",
"confidence": 8
},
@@ -4453,307 +4453,307 @@
},
"BC101": {
"revision": 2,
- "explanation": "The 10 m amateur band is around 28 MHz, which lies in the 3-30 MHz HF shortwave range. Hilfsmittel: the frequency-allocation table in the official exam aids (Hilfsmittel) lists these band limits and usage parameters directly, so in the exam this is a table lookup rather than a memory item.",
+ "explanation": "The 10 m amateur band is around 28 MHz, which lies in the 3-30 MHz HF shortwave range.",
"source": "https://50ohm.de/NEA_frequenzspektrum.html#BC101",
"confidence": 9
},
"BC102": {
"revision": 2,
- "explanation": "The 2 m band is around 144-146 MHz, which lies in the 30-300 MHz VHF range. Hilfsmittel: the frequency-allocation table in the official exam aids (Hilfsmittel) lists these band limits and usage parameters directly, so in the exam this is a table lookup rather than a memory item.",
+ "explanation": "The 2 m band is around 144-146 MHz, which lies in the 30-300 MHz VHF range.",
"source": "https://50ohm.de/NEA_frequenzspektrum.html#BC102",
"confidence": 9
},
"BC103": {
"revision": 2,
- "explanation": "The 70 cm band is around 430-440 MHz, which lies in the 300-3000 MHz UHF range. Hilfsmittel: the frequency-allocation table in the official exam aids (Hilfsmittel) lists these band limits and usage parameters directly, so in the exam this is a table lookup rather than a memory item.",
+ "explanation": "The 70 cm band is around 430-440 MHz, which lies in the 300-3000 MHz UHF range.",
"source": "https://50ohm.de/NEA_frequenzspektrum.html#BC103",
"confidence": 9
},
"BC104": {
"revision": 2,
- "explanation": "HF is defined as 3-30 MHz; those wavelengths are roughly 100 m to 10 m, hence shortwave/KW. Hilfsmittel: the frequency-allocation table in the official exam aids (Hilfsmittel) lists these band limits and usage parameters directly, so in the exam this is a table lookup rather than a memory item.",
+ "explanation": "HF is defined as 3-30 MHz; those wavelengths are roughly 100 m to 10 m, hence shortwave/KW.",
"source": "https://50ohm.de/NEA_frequenzspektrum.html#BC104",
"confidence": 9
},
"BC105": {
"revision": 2,
- "explanation": "VHF is defined as 30-300 MHz; in German amateur practice this is the UKW range. Hilfsmittel: the frequency-allocation table in the official exam aids (Hilfsmittel) lists these band limits and usage parameters directly, so in the exam this is a table lookup rather than a memory item.",
+ "explanation": "VHF is defined as 30-300 MHz; in German amateur practice this is the UKW range.",
"source": "https://50ohm.de/NEA_frequenzspektrum.html#BC105",
"confidence": 9
},
"BC106": {
"revision": 2,
- "explanation": "UHF is defined as 300-3000 MHz; its wavelengths are in the decimetre range. Hilfsmittel: the frequency-allocation table in the official exam aids (Hilfsmittel) lists these band limits and usage parameters directly, so in the exam this is a table lookup rather than a memory item.",
+ "explanation": "UHF is defined as 300-3000 MHz; its wavelengths are in the decimetre range.",
"source": "https://50ohm.de/NEA_frequenzspektrum.html#BC106",
"confidence": 9
},
"BC201": {
"revision": 2,
- "explanation": "IARU band plans are coordination recommendations, not law, but following them prevents incompatible modes from crowding each other. Hilfsmittel: the IARU band plan reproduced in the official exam aids (Hilfsmittel) gives this directly, so it can be looked up in the exam rather than memorized.",
+ "explanation": "IARU band plans are coordination recommendations, not law, but following them prevents incompatible modes from crowding each other.",
"source": "https://50ohm.de/NEA_iaru_bandplan.html#BC201",
"confidence": 7
},
"BC202": {
"revision": 2,
- "explanation": "The HF band-plan convention uses lower sideband below 10 MHz, so 80 m normally uses LSB. Hilfsmittel: the IARU band plan reproduced in the official exam aids (Hilfsmittel) gives this directly, so it can be looked up in the exam rather than memorized.",
+ "explanation": "The HF band-plan convention uses lower sideband below 10 MHz, so 80 m normally uses LSB.",
"source": "https://50ohm.de/NEA_trxmodulation.html#BC202",
"confidence": 7
},
"BC203": {
"revision": 2,
- "explanation": "The HF band-plan convention uses upper sideband above 10 MHz, so 20 m normally uses USB. Hilfsmittel: the IARU band plan reproduced in the official exam aids (Hilfsmittel) gives this directly, so it can be looked up in the exam rather than memorized.",
+ "explanation": "The HF band-plan convention uses upper sideband above 10 MHz, so 20 m normally uses USB.",
"source": "https://50ohm.de/NEA_trxmodulation.html#BC203",
"confidence": 7
},
"BC204": {
"revision": 2,
- "explanation": "Band plans put narrow Morse activity at the lower edge of most bands, leaving wider modes farther up the band. Hilfsmittel: the IARU band plan reproduced in the official exam aids (Hilfsmittel) gives this directly, so it can be looked up in the exam rather than memorized.",
+ "explanation": "Band plans put narrow Morse activity at the lower edge of most bands, leaving wider modes farther up the band.",
"source": "https://50ohm.de/NEA_iaru_bandplan.html#BC204",
"confidence": 7
},
"BC205": {
"revision": 2,
- "explanation": "The Region 1 VHF plan marks 145.500 MHz as the 2 m FM calling frequency, so it is used for general FM calls. Hilfsmittel: the IARU band plan reproduced in the official exam aids (Hilfsmittel) gives this directly, so it can be looked up in the exam rather than memorized.",
+ "explanation": "The Region 1 VHF plan marks 145.500 MHz as the 2 m FM calling frequency, so it is used for general FM calls. Hilfsmittel: a lookup in the IARU-Bandplan 2 m (S. 9).",
"source": "https://50ohm.de/NEA_iaru_bandplan_2m.html#BC205",
"confidence": 7
},
"BC206": {
"revision": 2,
- "explanation": "The Region 1 UHF plan marks 433.500 MHz as the 70 cm FM calling frequency, so it is used for general FM calls. Hilfsmittel: the IARU band plan reproduced in the official exam aids (Hilfsmittel) gives this directly, so it can be looked up in the exam rather than memorized.",
+ "explanation": "The Region 1 UHF plan marks 433.500 MHz as the 70 cm FM calling frequency, so it is used for general FM calls. Hilfsmittel: a lookup in the IARU-Bandplan 70 cm (S. 10).",
"source": "https://50ohm.de/NEA_iaru_bandplan_70cm.html#BC206",
"confidence": 7
},
"BC207": {
"revision": 2,
- "explanation": "The 2 m band plan lists 145.375 MHz for digital voice calling, separating it from analogue FM calling traffic. Hilfsmittel: the IARU band plan reproduced in the official exam aids (Hilfsmittel) gives this directly, so it can be looked up in the exam rather than memorized.",
+ "explanation": "The 2 m band plan lists 145.375 MHz for digital voice calling, separating it from analogue FM calling traffic. Hilfsmittel: a lookup in the IARU-Bandplan 2 m (S. 9).",
"source": "https://50ohm.de/NEA_iaru_bandplan_2m.html#BC207",
"confidence": 7
},
"BC208": {
"revision": 2,
- "explanation": "The 70 cm band plan lists 433.450 MHz for digital voice calling, separating it from analogue FM calling traffic. Hilfsmittel: the IARU band plan reproduced in the official exam aids (Hilfsmittel) gives this directly, so it can be looked up in the exam rather than memorized.",
+ "explanation": "The 70 cm band plan lists 433.450 MHz for digital voice calling, separating it from analogue FM calling traffic. Hilfsmittel: a lookup in the IARU-Bandplan 70 cm (S. 10).",
"source": "https://50ohm.de/NEA_iaru_bandplan_70cm.html#BC208",
"confidence": 7
},
"BC209": {
"revision": 2,
- "explanation": "145.450 MHz falls in the 2 m FM simplex channel area, so it is suitable for an FM voice contact under the band plan. Hilfsmittel: the IARU band plan reproduced in the official exam aids (Hilfsmittel) gives this directly, so it can be looked up in the exam rather than memorized.",
+ "explanation": "145.450 MHz falls in the 2 m FM simplex channel area, so it is suitable for an FM voice contact under the band plan. Hilfsmittel: a lookup in the IARU-Bandplan 2 m (S. 9).",
"source": "https://50ohm.de/NEA_iaru_bandplan_2m.html#BC209",
"confidence": 7
},
"BC210": {
"revision": 2,
- "explanation": "144.310 MHz sits near the 144.300 MHz SSB centre of activity, so it is appropriate for 2 m SSB voice. Hilfsmittel: the IARU band plan reproduced in the official exam aids (Hilfsmittel) gives this directly, so it can be looked up in the exam rather than memorized.",
+ "explanation": "144.310 MHz sits near the 144.300 MHz SSB centre of activity, so it is appropriate for 2 m SSB voice. Hilfsmittel: a lookup in the IARU-Bandplan 2 m (S. 9).",
"source": "https://50ohm.de/NEA_iaru_bandplan_2m.html#BC210",
"confidence": 7
},
"BC211": {
"revision": 2,
- "explanation": "The 2 m band plan uses 144.300 MHz as the SSB centre of activity. Hilfsmittel: the IARU band plan reproduced in the official exam aids (Hilfsmittel) gives this directly, so it can be looked up in the exam rather than memorized.",
+ "explanation": "The 2 m band plan uses 144.300 MHz as the SSB centre of activity. Hilfsmittel: a lookup in the IARU-Bandplan 2 m (S. 9).",
"source": "https://50ohm.de/NEA_iaru_bandplan_2m.html#BC211",
"confidence": 7
},
"BC212": {
"revision": 2,
- "explanation": "The 70 cm band plan uses 432.200 MHz as the SSB centre of activity. Hilfsmittel: the IARU band plan reproduced in the official exam aids (Hilfsmittel) gives this directly, so it can be looked up in the exam rather than memorized.",
+ "explanation": "The 70 cm band plan uses 432.200 MHz as the SSB centre of activity. Hilfsmittel: a lookup in the IARU-Bandplan 70 cm (S. 10).",
"source": "https://50ohm.de/NEA_iaru_bandplan_70cm.html#BC212",
"confidence": 7
},
"BC213": {
"revision": 2,
- "explanation": "144.075 MHz lies in the narrow Morse-preferred segment, so wider or keyboard digital modes should use their own segments. Hilfsmittel: the IARU band plan reproduced in the official exam aids (Hilfsmittel) gives this directly, so it can be looked up in the exam rather than memorized.",
+ "explanation": "144.075 MHz lies in the narrow Morse-preferred segment, so wider or keyboard digital modes should use their own segments. Hilfsmittel: a lookup in the IARU-Bandplan 2 m (S. 9).",
"source": "https://50ohm.de/NEA_iaru_bandplan_2m.html#BC213",
"confidence": 7
},
"BC214": {
"revision": 2,
- "explanation": "Around 144.125 MHz the 2 m band plan is for Morse and narrow digital work, not local FM voice. Hilfsmittel: the IARU band plan reproduced in the official exam aids (Hilfsmittel) gives this directly, so it can be looked up in the exam rather than memorized.",
+ "explanation": "Around 144.125 MHz the 2 m band plan is for Morse and narrow digital work, not local FM voice. Hilfsmittel: a lookup in the IARU-Bandplan 2 m (S. 9).",
"source": "https://50ohm.de/NEA_iaru_bandplan_2m.html#BC214",
"confidence": 7
},
"BC215": {
"revision": 2,
- "explanation": "Around 144.450 MHz the 2 m band plan reserves beacon use, so an ordinary local FM QSO would occupy the wrong segment. Hilfsmittel: the IARU band plan reproduced in the official exam aids (Hilfsmittel) gives this directly, so it can be looked up in the exam rather than memorized.",
+ "explanation": "Around 144.450 MHz the 2 m band plan reserves beacon use, so an ordinary local FM QSO would occupy the wrong segment. Hilfsmittel: a lookup in the IARU-Bandplan 2 m (S. 9).",
"source": "https://50ohm.de/NEA_iaru_bandplan_2m.html#BC215",
"confidence": 7
},
"BC216": {
"revision": 2,
- "explanation": "The 145.500-145.5625 MHz FM simplex area is channelised for narrow FM, so keeping to about 12 kHz avoids adjacent-channel interference. Hilfsmittel: the IARU band plan reproduced in the official exam aids (Hilfsmittel) gives this directly, so it can be looked up in the exam rather than memorized.",
+ "explanation": "The 145.500-145.5625 MHz FM simplex area is channelised for narrow FM, so keeping to about 12 kHz avoids adjacent-channel interference. Hilfsmittel: a lookup in the IARU-Bandplan 2 m (S. 9).",
"source": "https://50ohm.de/NEA_fm.html#BC216",
"confidence": 7
},
"BC217": {
"revision": 2,
- "explanation": "145.600 MHz is in the 2 m repeater output area, so a direct local FM contact would interfere with repeater operation. Hilfsmittel: the IARU band plan reproduced in the official exam aids (Hilfsmittel) gives this directly, so it can be looked up in the exam rather than memorized.",
+ "explanation": "145.600 MHz is in the 2 m repeater output area, so a direct local FM contact would interfere with repeater operation. Hilfsmittel: a lookup in the IARU-Bandplan 2 m (S. 9).",
"source": "https://50ohm.de/NEA_iaru_bandplan_2m.html#BC217",
"confidence": 7
},
"BC218": {
"revision": 2,
- "explanation": "145.800 MHz belongs to the 2 m space-communication segment, so it should be kept clear for satellite and other space contacts. Hilfsmittel: the IARU band plan reproduced in the official exam aids (Hilfsmittel) gives this directly, so it can be looked up in the exam rather than memorized.",
+ "explanation": "145.800 MHz belongs to the 2 m space-communication segment, so it should be kept clear for satellite and other space contacts. Hilfsmittel: a lookup in the IARU-Bandplan 2 m (S. 9).",
"source": "https://50ohm.de/NEA_iaru_bandplan_2m.html#BC218",
"confidence": 7
},
"BC219": {
"revision": 2,
- "explanation": "432.040 MHz lies in the 70 cm Morse/narrow digital segment, so local FM voice would be the wrong bandwidth and mode there. Hilfsmittel: the IARU band plan reproduced in the official exam aids (Hilfsmittel) gives this directly, so it can be looked up in the exam rather than memorized.",
+ "explanation": "432.040 MHz lies in the 70 cm Morse/narrow digital segment, so local FM voice would be the wrong bandwidth and mode there. Hilfsmittel: a lookup in the IARU-Bandplan 70 cm (S. 10).",
"source": "https://50ohm.de/NEA_iaru_bandplan_70cm.html#BC219",
"confidence": 7
},
"BC220": {
"revision": 2,
- "explanation": "432.450 MHz is assigned to beacon activity in the 70 cm plan, so it should not be used for an ordinary local FM contact. Hilfsmittel: the IARU band plan reproduced in the official exam aids (Hilfsmittel) gives this directly, so it can be looked up in the exam rather than memorized.",
+ "explanation": "432.450 MHz is assigned to beacon activity in the 70 cm plan, so it should not be used for an ordinary local FM contact. Hilfsmittel: a lookup in the IARU-Bandplan 70 cm (S. 10).",
"source": "https://50ohm.de/NEA_iaru_bandplan_70cm.html#BC220",
"confidence": 7
},
"BC221": {
"revision": 2,
- "explanation": "435.500 MHz lies in the 70 cm satellite segment, so terrestrial local FM would risk interfering with satellite operation. Hilfsmittel: the IARU band plan reproduced in the official exam aids (Hilfsmittel) gives this directly, so it can be looked up in the exam rather than memorized.",
+ "explanation": "435.500 MHz lies in the 70 cm satellite segment, so terrestrial local FM would risk interfering with satellite operation. Hilfsmittel: a lookup in the IARU-Bandplan 70 cm (S. 10).",
"source": "https://50ohm.de/NEA_iaru_bandplan_70cm.html#BC221",
"confidence": 7
},
"BC222": {
"revision": 2,
- "explanation": "439.200 MHz is in the 70 cm repeater output area, so a direct local FM contact would occupy repeater spectrum. Hilfsmittel: the IARU band plan reproduced in the official exam aids (Hilfsmittel) gives this directly, so it can be looked up in the exam rather than memorized.",
+ "explanation": "439.200 MHz is in the 70 cm repeater output area, so a direct local FM contact would occupy repeater spectrum. Hilfsmittel: a lookup in the IARU-Bandplan 70 cm (S. 10).",
"source": "https://50ohm.de/NEA_iaru_bandplan_70cm.html#BC222",
"confidence": 7
},
"BD101": {
"revision": 3,
- "explanation": "German club-station call signs use zero in the numeral position; DA0ABC therefore identifies a club station. Hilfsmittel: the German call-sign plan (Rufzeichenplan) in the official exam aids (Hilfsmittel) lists which call-sign series belong to which class and use, so this is a lookup in the exam rather than a memory item.",
+ "explanation": "German club-station call signs use zero in the numeral position; DA0ABC therefore identifies a club station. Hilfsmittel: a table lookup — the call-sign-series → class/use table is the Rufzeichenplan, Suffixe (S. 5): DA0 = Klubstation.",
"source": "https://50ohm.de/NEA_klubstationen.html#BD101",
"confidence": 9
},
"BD102": {
"revision": 3,
- "explanation": "AFuV §16 allows BNetzA to permit special experimental or technical-scientific studies and to make that dependent on assigning another call sign. Hilfsmittel: the German call-sign plan (Rufzeichenplan) in the official exam aids (Hilfsmittel) lists which call-sign series belong to which class and use, so this is a lookup in the exam rather than a memory item.",
+ "explanation": "AFuV §16 allows BNetzA to permit special experimental or technical-scientific studies and to make that dependent on assigning another call sign. Hilfsmittel: a table lookup — Rufzeichenplan, Suffixe (S. 5): DA5 = „SZ“, defined just below the table as Rufzeichen für besondere experimentelle Studien nach § 16 Abs. 2 AFuV.",
"source": "https://50ohm.de/NEA_experimentelle_studien.html#BD102",
"confidence": 10
},
"BD103": {
"revision": 3,
- "explanation": "DL0 is in the German club-station pattern for class A, and the zero distinguishes it from person-bound DL1-DL9 calls. Hilfsmittel: the German call-sign plan (Rufzeichenplan) in the official exam aids (Hilfsmittel) lists which call-sign series belong to which class and use, so this is a lookup in the exam rather than a memory item.",
+ "explanation": "DL0 is in the German club-station pattern for class A, and the zero distinguishes it from person-bound DL1-DL9 calls. Hilfsmittel: a table lookup — Rufzeichenplan, Suffixe (S. 5): DL0 = club station, class A.",
"source": "https://50ohm.de/NEA_klubstationen.html#BD103",
"confidence": 9
},
"BD104": {
"revision": 3,
- "explanation": "In the German call-sign plan, DL1-DL9 with normal two- or three-letter suffixes are person-bound class A call signs. Hilfsmittel: the German call-sign plan (Rufzeichenplan) in the official exam aids (Hilfsmittel) lists which call-sign series belong to which class and use, so this is a lookup in the exam rather than a memory item.",
+ "explanation": "In the German call-sign plan, DL1-DL9 with normal two- or three-letter suffixes are person-bound class A call signs. Hilfsmittel: a table lookup — Rufzeichenplan, Suffixe (S. 5): DL1–DL9 = person-bound class A.",
"source": "https://50ohm.de/NEA_persoenliche_rufzeichen.html#BD104",
"confidence": 9
},
"BD105": {
"revision": 3,
- "explanation": "The German call-sign plan assigns DN9 to person-bound class N call signs. Hilfsmittel: the German call-sign plan (Rufzeichenplan) in the official exam aids (Hilfsmittel) lists which call-sign series belong to which class and use, so this is a lookup in the exam rather than a memory item.",
+ "explanation": "The German call-sign plan assigns DN9 to person-bound class N call signs. Hilfsmittel: a table lookup — Rufzeichenplan, Suffixe (S. 5): DN9 = person-bound class N.",
"source": "https://50ohm.de/NEA_persoenliche_rufzeichen.html#BD105",
"confidence": 9
},
"BD106": {
"revision": 3,
- "explanation": "The German call-sign plan assigns DO1-DO9 with normal suffixes to person-bound class E call signs. Hilfsmittel: the German call-sign plan (Rufzeichenplan) in the official exam aids (Hilfsmittel) lists which call-sign series belong to which class and use, so this is a lookup in the exam rather than a memory item.",
+ "explanation": "The German call-sign plan assigns DO1-DO9 with normal suffixes to person-bound class E call signs. Hilfsmittel: a table lookup — Rufzeichenplan, Suffixe (S. 5): DO1–DO9 = person-bound class E.",
"source": "https://50ohm.de/NEA_persoenliche_rufzeichen.html#BD106",
"confidence": 9
},
"BD107": {
"revision": 4,
- "explanation": "DP0GVN is one of the German exterritorial class A station patterns; DP0 is used for special locations outside ordinary German territory. Hilfsmittel: the German call-sign plan (Rufzeichenplan) in the official exam aids (Hilfsmittel) lists which call-sign series belong to which class and use, so this is a lookup in the exam rather than a memory item.",
+ "explanation": "DP0GVN is one of the German exterritorial class A station patterns; DP0 is used for special locations outside ordinary German territory. Hilfsmittel: a table lookup — Rufzeichenplan, Suffixe (S. 5): DP0–DP1 = exterritorial class A.",
"source": "https://50ohm.de/NEA_exterritoriale_stationen.html#BD107",
"confidence": 8
},
"BD108": {
"revision": 4,
- "explanation": "DP0POL follows the same exterritorial class A pattern as other German Antarctic or special-location stations. Hilfsmittel: the German call-sign plan (Rufzeichenplan) in the official exam aids (Hilfsmittel) lists which call-sign series belong to which class and use, so this is a lookup in the exam rather than a memory item.",
+ "explanation": "DP0POL follows the same exterritorial class A pattern as other German Antarctic or special-location stations. Hilfsmittel: a table lookup — Rufzeichenplan, Suffixe (S. 5): DP0–DP1 = exterritorial class A.",
"source": "https://50ohm.de/NEA_exterritoriale_stationen.html#BD108",
"confidence": 8
},
"BD109": {
"revision": 3,
- "explanation": "Low-power transmitters for direction finding may identify with short MO-series markers instead of a normal amateur call sign. Hilfsmittel: the German call-sign plan (Rufzeichenplan) in the official exam aids (Hilfsmittel) lists which call-sign series belong to which class and use, so this is a lookup in the exam rather than a memory item.",
+ "explanation": "Low-power transmitters for direction finding may identify with short MO-series markers instead of a normal amateur call sign. Hilfsmittel: a lookup — the beacon identifiers MO, MOE, MOI, MOS, MOH, MO5 are in the Rufzeichenplan § 6 (Kennungen leistungsschwacher Sender, S. 7).",
"source": "https://50ohm.de/NEA_ardf.html#BD109",
"confidence": 9
},
"BD201": {
"revision": 3,
- "explanation": "The suffix /am means aeronautical mobile: the station is operating from an aircraft. Hilfsmittel: the list of internationally used call-sign suffixes in the official exam aids (Hilfsmittel, Rufzeichenplan) gives this directly — a lookup in the exam, not a memory item.",
+ "explanation": "The suffix /am means aeronautical mobile: the station is operating from an aircraft. Hilfsmittel: a lookup — see “International gebräuchliche Rufzeichenzusätze” in the Rufzeichenplan (S. 8).",
"source": "https://50ohm.de/NEA_rufzeichenzusaetze.html#BD201",
"confidence": 8
},
"BD202": {
"revision": 2,
- "explanation": "VE is a Canadian call-sign series, and /am adds that the station is being operated from an aircraft. Hilfsmittel: the list of internationally used call-sign suffixes in the official exam aids (Hilfsmittel, Rufzeichenplan) gives this directly — a lookup in the exam, not a memory item.",
+ "explanation": "VE is a Canadian call-sign series, and /am adds that the station is being operated from an aircraft. Hilfsmittel: the suffix /am (aboard an aircraft) is in „International gebräuchliche Rufzeichenzusätze“ (S. 8); that VE is a Canadian prefix is outside the sheet (Landeskenner are not in the aid).",
"source": "https://50ohm.de/NEA_rufzeichenzusaetze.html#BD202",
"confidence": 9
},
"BD203": {
"revision": 3,
- "explanation": "The suffix /m means mobile; for amateur operation that includes a station moving in a land vehicle. Hilfsmittel: the list of internationally used call-sign suffixes in the official exam aids (Hilfsmittel, Rufzeichenplan) gives this directly — a lookup in the exam, not a memory item.",
+ "explanation": "The suffix /m means mobile; for amateur operation that includes a station moving in a land vehicle. Hilfsmittel: a lookup — see “International gebräuchliche Rufzeichenzusätze” in the Rufzeichenplan (S. 8).",
"source": "https://50ohm.de/NEA_rufzeichenzusaetze.html#BD203",
"confidence": 8
},
"BD204": {
"revision": 3,
- "explanation": "The suffix /m can also mark mobile operation on inland waterways, distinct from /mm on the high seas. Hilfsmittel: the list of internationally used call-sign suffixes in the official exam aids (Hilfsmittel, Rufzeichenplan) gives this directly — a lookup in the exam, not a memory item.",
+ "explanation": "The suffix /m can also mark mobile operation on inland waterways, distinct from /mm on the high seas. Hilfsmittel: a lookup — see “International gebräuchliche Rufzeichenzusätze” in the Rufzeichenplan (S. 8).",
"source": "https://50ohm.de/NEA_rufzeichenzusaetze.html#BD204",
"confidence": 8
},
"BD205": {
"revision": 3,
- "explanation": "The suffix /mm means maritime mobile, so the station is aboard a vessel at sea rather than on land or inland water. Hilfsmittel: the list of internationally used call-sign suffixes in the official exam aids (Hilfsmittel, Rufzeichenplan) gives this directly — a lookup in the exam, not a memory item.",
+ "explanation": "The suffix /mm means maritime mobile, so the station is aboard a vessel at sea rather than on land or inland water. Hilfsmittel: a lookup — see “International gebräuchliche Rufzeichenzusätze” in the Rufzeichenplan (S. 8).",
"source": "https://50ohm.de/NEA_rufzeichenzusaetze.html#BD205",
"confidence": 8
},
"BD206": {
"revision": 2,
- "explanation": "The suffix /p is used as extra information for portable or temporarily fixed operation. Hilfsmittel: the list of internationally used call-sign suffixes in the official exam aids (Hilfsmittel, Rufzeichenplan) gives this directly — a lookup in the exam, not a memory item.",
+ "explanation": "The suffix /p is used as extra information for portable or temporarily fixed operation. Hilfsmittel: a lookup — see “International gebräuchliche Rufzeichenzusätze” in the Rufzeichenplan (S. 8).",
"source": "https://50ohm.de/NEA_rufzeichenzusaetze.html#BD206",
"confidence": 9
},
"BD207": {
"revision": 2,
- "explanation": "AFuV allows internationally customary suffixes but does not require /p for portable or temporary fixed operation in Germany. Hilfsmittel: the list of internationally used call-sign suffixes in the official exam aids (Hilfsmittel, Rufzeichenplan) gives this directly — a lookup in the exam, not a memory item.",
+ "explanation": "AFuV allows internationally customary suffixes but does not require /p for portable or temporary fixed operation in Germany. Hilfsmittel: a lookup — see “International gebräuchliche Rufzeichenzusätze” in the Rufzeichenplan (S. 8).",
"source": "https://50ohm.de/NEA_rufzeichenzusaetze.html#BD207",
"confidence": 9
},
"BD208": {
"revision": 2,
- "explanation": "AFuV §11 names Remote for speech and /R for telegraphy or digital modes when marking remote operation. Hilfsmittel: the list of internationally used call-sign suffixes in the official exam aids (Hilfsmittel, Rufzeichenplan) gives this directly — a lookup in the exam, not a memory item.",
+ "explanation": "AFuV §11 names Remote for speech and /R for telegraphy or digital modes when marking remote operation. Hilfsmittel: a lookup — see the Remotebetrieb rules in the Rufzeichenplan (§ 11, S. 8).",
"source": "https://50ohm.de/NEA_rufzeichenzusaetze.html#BD208",
"confidence": 10
},
"BD209": {
"revision": 2,
- "explanation": "For training operation, AFuV §11 requires /Trainee in speech, so the trainee uses the instructor's call sign plus that suffix. Hilfsmittel: the list of internationally used call-sign suffixes in the official exam aids (Hilfsmittel, Rufzeichenplan) gives this directly — a lookup in the exam, not a memory item.",
+ "explanation": "For training operation, AFuV §11 requires /Trainee in speech, so the trainee uses the instructor's call sign plus that suffix. Hilfsmittel: a lookup — see the Ausbildungsfunkbetrieb rules in the Rufzeichenplan (§ 10, S. 8).",
"source": "https://50ohm.de/NEA_ausbildungsfunk.html#BD209",
"confidence": 10
},
"BD210": {
"revision": 2,
- "explanation": "Training operation may use the club-station call sign, but AFuV §11 requires the training suffix /Trainee for speech or /T for telegraphy/digital modes. Hilfsmittel: the list of internationally used call-sign suffixes in the official exam aids (Hilfsmittel, Rufzeichenplan) gives this directly — a lookup in the exam, not a memory item.",
+ "explanation": "Training operation may use the club-station call sign, but AFuV §11 requires the training suffix /Trainee for speech or /T for telegraphy/digital modes. Hilfsmittel: a lookup — see the Ausbildungsfunkbetrieb rules in the Rufzeichenplan (§ 10, S. 8).",
"source": "https://50ohm.de/NEA_ausbildungsrufzeichen.html#BD210",
"confidence": 10
},
"BD211": {
"revision": 2,
- "explanation": "For training in Morse or digital modes, AFuV §11 requires the short /T suffix on the instructor's call sign. Hilfsmittel: the list of internationally used call-sign suffixes in the official exam aids (Hilfsmittel, Rufzeichenplan) gives this directly — a lookup in the exam, not a memory item.",
+ "explanation": "For training in Morse or digital modes, AFuV §11 requires the short /T suffix on the instructor's call sign. Hilfsmittel: a lookup — see the Ausbildungsfunkbetrieb rules in the Rufzeichenplan (§ 10, S. 8).",
"source": "https://50ohm.de/NEA_ausbildungsrufzeichen.html#BD211",
"confidence": 10
},
"BD212": {
"revision": 2,
- "explanation": "CEPT guest operation uses the visited country's prefix before the home call sign, so a UK G3MM station temporarily in Germany signs with DL/. Hilfsmittel: the list of internationally used call-sign suffixes in the official exam aids (Hilfsmittel, Rufzeichenplan) gives this directly — a lookup in the exam, not a memory item.",
+ "explanation": "CEPT guest operation uses the visited country's prefix before the home call sign, so a UK G3MM station temporarily in Germany signs with DL/. Hilfsmittel: the prefix syntax DL/ (class A) or DO/ (class E) for a foreign short-term licensee is in Rufzeichenplan § 5 (S. 7); that CEPT permits the visit at all is outside the sheet.",
"source": "https://50ohm.de/NEA_funken_im_ausland.html#BD212",
"confidence": 9
},
"BD213": {
"revision": 2,
- "explanation": "CEPT Novice guest operation in Switzerland uses the Swiss novice visitor prefix HB3 before the German class E call sign. Hilfsmittel: the list of internationally used call-sign suffixes in the official exam aids (Hilfsmittel, Rufzeichenplan) gives this directly — a lookup in the exam, not a memory item.",
+ "explanation": "CEPT Novice guest operation in Switzerland uses the Swiss novice visitor prefix HB3 before the German class E call sign.",
"source": "https://50ohm.de/NEA_funken_im_ausland.html#BD213",
"confidence": 9
},
"BD214": {
"revision": 2,
- "explanation": "CEPT guest operation in Switzerland uses the Swiss HB9 prefix before the German class A call sign. Hilfsmittel: the list of internationally used call-sign suffixes in the official exam aids (Hilfsmittel, Rufzeichenplan) gives this directly — a lookup in the exam, not a memory item.",
+ "explanation": "CEPT guest operation in Switzerland uses the Swiss HB9 prefix before the German class A call sign.",
"source": "https://50ohm.de/NEA_funken_im_ausland.html#BD214",
"confidence": 9
},
@@ -5329,13 +5329,13 @@
},
"EA101": {
"revision": 3,
- "explanation": "Capacitance is the charge stored per volt, $C = Q/U$, so its named SI-derived unit is the farad (F) — one coulomb per volt. Ohm ($\\Omega$) is resistance, henry (H) is inductance, and ampere-hours measure charge, so none of those fit capacitance. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Capacitance is the charge stored per volt, $C = Q/U$, so its named SI-derived unit is the farad (F) — one coulomb per volt. Ohm ($\\Omega$) is resistance, henry (H) is inductance, and ampere-hours measure charge, so none of those fit capacitance.",
"source": "https://50ohm.de/NEA_kondensator_1.html#EA101",
"confidence": 8
},
"EA102": {
"revision": 3,
- "explanation": "Inductance describes how much magnetic flux linkage a conductor or coil produces per ampere: $L = \\Psi/I$. A larger inductance stores more magnetic-field energy for the same current, and its named SI-derived unit is the henry (H). Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Inductance describes how much magnetic flux linkage a conductor or coil produces per ampere: $L = \\Psi/I$. A larger inductance stores more magnetic-field energy for the same current, and its named SI-derived unit is the henry (H).",
"source": "https://50ohm.de/NEA_spule_1.html#EA102",
"confidence": 8
},
@@ -5365,61 +5365,61 @@
},
"EA107": {
"revision": 4,
- "explanation": "A power level in decibels is $L = 10\\log_{10}(P_2/P_1)$. Doubling power gives $10\\log_{10}(2) = 10 \\cdot 0.301 \\approx 3$ dB. Worth memorising: $\\times 2$ power $= +3$ dB and $\\times 10$ power $= +10$ dB. (Doubling a voltage is $+6$ dB, because voltage ratios use $20\\log_{10}$.) Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "A power level in decibels is $L = 10\\log_{10}(P_2/P_1)$. Doubling power gives $10\\log_{10}(2) = 10 \\cdot 0.301 \\approx 3$ dB. Worth memorising: $\\times 2$ power $= +3$ dB and $\\times 10$ power $= +10$ dB. (Doubling a voltage is $+6$ dB, because voltage ratios use $20\\log_{10}$.) Hilfsmittel: apply g = 10·log10(P2/P1) (Pegel, S.15); the table (S.15) gives ×2 → 3 dB, ×10 → 10 dB.",
"source": "https://50ohm.de/NEA_dezibel_1.html#EA107",
"confidence": 8
},
"EA108": {
"revision": 3,
- "explanation": "Micro ($\\mu$) means $10^{-6}$. Expressing $0.00042$ A in microamperes shifts the decimal six places: $0.00042$ A $= 420 \\cdot 10^{-6}$ A $= 420\\ \\mu$A. Option D, $42 \\cdot 10^{-6}$ A, is ten times too small. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Micro ($\\mu$) means $10^{-6}$. Expressing $0.00042$ A in microamperes shifts the decimal six places: $0.00042$ A $= 420 \\cdot 10^{-6}$ A $= 420\\ \\mu$A. Option D, $42 \\cdot 10^{-6}$ A, is ten times too small. Hilfsmittel: use the Zehnerpotenzen/SI-Präfix table (S.11): micro = 10⁻⁶.",
"source": "https://50ohm.de/NEA_zehnerpotenzen.html#EA108",
"confidence": 8
},
"EA109": {
"revision": 3,
- "explanation": "Milli means $10^{-3}$. Move from amperes to milliamperes by multiplying by 1000: $0.042 A = 42 mA = 42 \\cdot 10^{-3} A$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Milli means $10^{-3}$. Move from amperes to milliamperes by multiplying by 1000: $0.042 A = 42 mA = 42 \\cdot 10^{-3} A$. Hilfsmittel: use the Zehnerpotenzen/SI-Präfix table (S.11): milli = 10⁻³.",
"source": "https://50ohm.de/NEA_zehnerpotenzen.html#EA109",
"confidence": 8
},
"EA110": {
"revision": 3,
- "explanation": "Scientific notation keeps one non-zero digit before the decimal point. Moving the decimal in $4,200,000$ six places gives $4.2 \\cdot 10^6 Hz$, i.e. 4.2 MHz. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Scientific notation keeps one non-zero digit before the decimal point. Moving the decimal in $4,200,000$ six places gives $4.2 \\cdot 10^6 Hz$, i.e. 4.2 MHz. Hilfsmittel: use the Zehnerpotenzen/SI-Präfix table (S.11) for the powers of ten (scientific notation).",
"source": "https://50ohm.de/NEA_zehnerpotenzen.html#EA110",
"confidence": 8
},
"EA111": {
"revision": 3,
- "explanation": "A millivolt is $10^{-3} V$ and a microvolt is $10^{-6} V$. So $0.01 mV = 0.01 \\cdot 10^{-3} V = 10 \\cdot 10^{-6} V = 10 µV$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "A millivolt is $10^{-3} V$ and a microvolt is $10^{-6} V$. So $0.01 mV = 0.01 \\cdot 10^{-3} V = 10 \\cdot 10^{-6} V = 10 µV$. Hilfsmittel: use the Zehnerpotenzen/SI-Präfix table (S.11): milli = 10⁻³, micro = 10⁻⁶.",
"source": "https://50ohm.de/NEA_zehnerpotenzen.html#EA111",
"confidence": 8
},
"EA112": {
"revision": 3,
- "explanation": "Mega means $10^6$. Therefore $0.002 MOhm = 0.002 \\cdot 10^6 Ohm = 2000 Ohm = 2 kOhm$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Mega means $10^6$. Therefore $0.002 MOhm = 0.002 \\cdot 10^6 Ohm = 2000 Ohm = 2 kOhm$. Hilfsmittel: use the Zehnerpotenzen/SI-Präfix table (S.11): mega = 10⁶.",
"source": "https://50ohm.de/NEA_zehnerpotenzen.html#EA112",
"confidence": 8
},
"EA113": {
"revision": 3,
- "explanation": "A microwatt is $10^{-6} W$. To convert watts to microwatts, divide by $10^{-6}$: $2 \\cdot 10^{-7} W / 10^{-6} = 0.2 µW$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "A microwatt is $10^{-6} W$. To convert watts to microwatts, divide by $10^{-6}$: $2 \\cdot 10^{-7} W / 10^{-6} = 0.2 µW$. Hilfsmittel: use the Zehnerpotenzen/SI-Präfix table (S.11): micro = 10⁻⁶.",
"source": "https://50ohm.de/NEA_zehnerpotenzen.html#EA113",
"confidence": 8
},
"EA114": {
"revision": 3,
- "explanation": "$5 \\cdot 10^{-1} W$ is $0.5 W$. Since $1 W = 1000 mW$, $0.5 W = 500 mW$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "$5 \\cdot 10^{-1} W$ is $0.5 W$. Since $1 W = 1000 mW$, $0.5 W = 500 mW$. Hilfsmittel: use the Zehnerpotenzen/SI-Präfix table (S.11): 10⁻¹ and milli = 10⁻³.",
"source": "https://50ohm.de/NEA_zehnerpotenzen.html#EA114",
"confidence": 8
},
"EA115": {
"revision": 4,
- "explanation": "Micro is $10^{-6}$ and nano is $10^{-9}$, so one microfarad is 1000 nanofarads. Thus $0.22 µF = 0.22 \\cdot 1000 nF = 220 nF$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Micro is $10^{-6}$ and nano is $10^{-9}$, so one microfarad is 1000 nanofarads. Thus $0.22 µF = 0.22 \\cdot 1000 nF = 220 nF$. Hilfsmittel: use the Zehnerpotenzen/SI-Präfix table (S.11): micro = 10⁻⁶, nano = 10⁻⁹.",
"source": "https://50ohm.de/NEA_zehnerpotenzen.html#EA115",
"confidence": 8
},
"EA116": {
"revision": 4,
- "explanation": "Kilo is $10^3$ and mega is $10^6$, so converting kHz to MHz divides by 1000. $3750 kHz = 3.750 MHz$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Kilo is $10^3$ and mega is $10^6$, so converting kHz to MHz divides by 1000. $3750 kHz = 3.750 MHz$. Hilfsmittel: use the Zehnerpotenzen/SI-Präfix table (S.11): kilo = 10³, mega = 10⁶.",
"source": "https://50ohm.de/NEA_zehnerpotenzen.html#EA116",
"confidence": 8
},
@@ -5479,19 +5479,19 @@
},
"EB102": {
"revision": 5,
- "explanation": "In the uniform field of a plate capacitor, field strength is voltage over gap: $E = U/d$. Convert the spacing, $0.6\\ \\text{cm} = 0.006$ m, then $E = 9\\ \\text{V} / 0.006\\ \\text{m} = 1500$ V/m. The unit V/m comes straight out of the formula. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "In the uniform field of a plate capacitor, field strength is voltage over gap: $E = U/d$. Convert the spacing, $0.6\\ \\text{cm} = 0.006$ m, then $E = 9\\ \\text{V} / 0.006\\ \\text{m} = 1500$ V/m. The unit V/m comes straight out of the formula. Hilfsmittel: apply E = U/d (E-Feld im homogenen Feld, S.13); convert the gap to metres.",
"source": "https://50ohm.de/NEA_e_feld.html#EB102",
"confidence": 8
},
"EB103": {
"revision": 5,
- "explanation": "Use $E = U/d$ with the dielectric thickness as the gap: $0.15\\ \\text{mm} = 1.5 \\cdot 10^{-4}$ m. Then $E = 300\\ \\text{V} / (1.5 \\cdot 10^{-4}\\ \\text{m}) = 2.0 \\cdot 10^6$ V/m $= 2000$ kV/m. Thin dielectrics produce enormous field strengths even at modest voltages. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Use $E = U/d$ with the dielectric thickness as the gap: $0.15\\ \\text{mm} = 1.5 \\cdot 10^{-4}$ m. Then $E = 300\\ \\text{V} / (1.5 \\cdot 10^{-4}\\ \\text{m}) = 2.0 \\cdot 10^6$ V/m $= 2000$ kV/m. Thin dielectrics produce enormous field strengths even at modest voltages. Hilfsmittel: apply E = U/d (S.13); convert the gap to metres.",
"source": "https://50ohm.de/NEA_e_feld.html#EB103",
"confidence": 8
},
"EB104": {
"revision": 5,
- "explanation": "Dielectric (breakdown) strength is a maximum field strength, so rearrange $E = U/d$ to $U_{\\max} = E \\cdot d$. Keep units consistent: $0.15\\ \\text{mm} = 0.015$ cm, so $U_{\\max} = 400\\ \\text{kV/cm} \\cdot 0.015\\ \\text{cm} = 6$ kV. Above that, the PTFE film punches through. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Dielectric (breakdown) strength is a maximum field strength, so rearrange $E = U/d$ to $U_{\\max} = E \\cdot d$. Keep units consistent: $0.15\\ \\text{mm} = 0.015$ cm, so $U_{\\max} = 400\\ \\text{kV/cm} \\cdot 0.015\\ \\text{cm} = 6$ kV. Above that, the PTFE film punches through. Hilfsmittel: rearrange E = U/d → U_max = E·d (S.13); keep units consistent.",
"source": "https://50ohm.de/NEA_e_feld.html#EB104",
"confidence": 8
},
@@ -5515,7 +5515,7 @@
},
"EB203": {
"revision": 4,
- "explanation": "For a toroid the magnetic path length is the mean circumference, $l_m = \\pi d$, and Ampere's law gives $H = N I / l_m = N I / (\\pi d)$. Here $H = (6 \\cdot 2.5) / (\\pi \\cdot 0.026\\ \\text{m}) = 15 / 0.0817 \\approx 183.6$ A/m. Watch the units — using the diameter in cm would shift the answer by $100\\times$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "For a toroid the magnetic path length is the mean circumference, $l_m = \\pi d$, and Ampere's law gives $H = N I / l_m = N I / (\\pi d)$. Here $H = (6 \\cdot 2.5) / (\\pi \\cdot 0.026\\ \\text{m}) = 15 / 0.0817 \\approx 183.6$ A/m. Watch the units — using the diameter in cm would shift the answer by $100\\times$. Hilfsmittel: apply H = N·I/l_m with l_m = π·d (Magnetische Feldstärke, S.13).",
"source": "https://50ohm.de/NEA_h_feld.html#EB203",
"confidence": 8
},
@@ -5563,7 +5563,7 @@
},
"EB305": {
"revision": 4,
- "explanation": "By convention the polarisation of a wave is the direction of its electric-field vector ($E$). A vertical $E$-field is vertical polarisation, horizontal is horizontal. It is defined by the E-field, not the magnetic field, the travel direction, or the near field. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "By convention the polarisation of a wave is the direction of its electric-field vector ($E$). A vertical $E$-field is vertical polarisation, horizontal is horizontal. It is defined by the E-field, not the magnetic field, the travel direction, or the near field.",
"source": "https://50ohm.de/NEA_polarisation_2.html#EB305",
"confidence": 8
},
@@ -5599,103 +5599,103 @@
},
"EB311": {
"revision": 3,
- "explanation": "Wavelength and frequency are linked by $\\lambda = c/f$. Using the handy form $\\lambda_{\\text{m}} \\approx 300 / f_{\\text{MHz}}$: $300 / 1.84 \\approx 163$ m. (At $1.84$ MHz this is the $160$ m band, and $163$ m fits.) Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Wavelength and frequency are linked by $\\lambda = c/f$. Using the handy form $\\lambda_{\\text{m}} \\approx 300 / f_{\\text{MHz}}$: $300 / 1.84 \\approx 163$ m. (At $1.84$ MHz this is the $160$ m band, and $163$ m fits.) Hilfsmittel: apply c = f·λ (λ[m] ≈ 300/f[MHz]), S.17.",
"source": "https://50ohm.de/NEA_wellenlaenge_2.html#EB311",
"confidence": 8
},
"EB312": {
"revision": 3,
- "explanation": "Use $\\lambda = c/f$. For radio exam mental math, with $c \\approx 300$ million m/s and frequency in MHz, $\\lambda_m \\approx 300/f_{MHz}$. At 21 MHz, $300/21 \\approx 14.29$ m. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Use $\\lambda = c/f$. For radio exam mental math, with $c \\approx 300$ million m/s and frequency in MHz, $\\lambda_m \\approx 300/f_{MHz}$. At 21 MHz, $300/21 \\approx 14.29$ m. Hilfsmittel: apply c = f·λ (λ[m] ≈ 300/f[MHz]), S.17.",
"source": "https://50ohm.de/NEA_wellenlaenge_2.html#EB312",
"confidence": 8
},
"EB313": {
"revision": 3,
- "explanation": "Wavelength and frequency are inversely related: $\\lambda = c/f$. With frequency in MHz, use $\\lambda_m \\approx 300/f_{MHz}$; for 28.5 MHz this gives $300/28.5 \\approx 10.5$ m. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Wavelength and frequency are inversely related: $\\lambda = c/f$. With frequency in MHz, use $\\lambda_m \\approx 300/f_{MHz}$; for 28.5 MHz this gives $300/28.5 \\approx 10.5$ m. Hilfsmittel: apply c = f·λ (λ[m] ≈ 300/f[MHz]), S.17.",
"source": "https://50ohm.de/NEA_wellenlaenge_2.html#EB313",
"confidence": 8
},
"EB314": {
"revision": 3,
- "explanation": "Rearrange $\\lambda = c/f$ to $f \\approx 300 / \\lambda_{\\text{m}}$ (MHz with metres): $f = 300 / 80.0 = 3.75$ MHz. That places $80$ m wavelength in the $80$ m band, as expected. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Rearrange $\\lambda = c/f$ to $f \\approx 300 / \\lambda_{\\text{m}}$ (MHz with metres): $f = 300 / 80.0 = 3.75$ MHz. That places $80$ m wavelength in the $80$ m band, as expected. Hilfsmittel: rearrange c = f·λ (λ[m] ≈ 300/f[MHz]), S.17: f[MHz] ≈ 300/λ[m].",
"source": "https://50ohm.de/NEA_wellenlaenge_2.html#EB314",
"confidence": 8
},
"EB315": {
"revision": 3,
- "explanation": "Convert first: $30\\ \\text{mm} = 0.03$ m. Then $f = c/\\lambda = 3 \\cdot 10^8 / 0.03 = 1 \\cdot 10^{10}$ Hz $= 10$ GHz. Centimetre wavelengths mean microwave (GHz) frequencies. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Convert first: $30\\ \\text{mm} = 0.03$ m. Then $f = c/\\lambda = 3 \\cdot 10^8 / 0.03 = 1 \\cdot 10^{10}$ Hz $= 10$ GHz. Centimetre wavelengths mean microwave (GHz) frequencies. Hilfsmittel: apply f = c/λ (c = f·λ (λ[m] ≈ 300/f[MHz]), S.17).",
"source": "https://50ohm.de/NEA_wellenlaenge_2.html#EB315",
"confidence": 8
},
"EB316": {
"revision": 3,
- "explanation": "With $\\lambda = 10\\ \\text{cm} = 0.1$ m, $f = c/\\lambda = 3 \\cdot 10^8 / 0.1 = 3 \\cdot 10^9$ Hz $= 3$ GHz. ($10$ cm is the $13$ cm-ish microwave region, so GHz is right.) Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "With $\\lambda = 10\\ \\text{cm} = 0.1$ m, $f = c/\\lambda = 3 \\cdot 10^8 / 0.1 = 3 \\cdot 10^9$ Hz $= 3$ GHz. ($10$ cm is the $13$ cm-ish microwave region, so GHz is right.) Hilfsmittel: apply f = c/λ (c = f·λ (λ[m] ≈ 300/f[MHz]), S.17).",
"source": "https://50ohm.de/NEA_wellenlaenge_2.html#EB316",
"confidence": 8
},
"EB401": {
"revision": 5,
- "explanation": "For a sine wave, RMS is the heating-equivalent DC value and the peak is larger by $\\sqrt{2}$. Mains 230 V is RMS, so $U_{peak} = 230 V \\cdot \\sqrt{2} \\approx 325 V$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "For a sine wave, RMS is the heating-equivalent DC value and the peak is larger by $\\sqrt{2}$. Mains 230 V is RMS, so $U_{peak} = 230 V \\cdot \\sqrt{2} \\approx 325 V$. Hilfsmittel: apply Û = U_eff·√2 (Wechselspannung, S.12).",
"source": "https://50ohm.de/NEA_spitze_effektiv_wert.html#EB401",
"confidence": 8
},
"EB402": {
"revision": 5,
- "explanation": "Peak-to-peak voltage is the full swing from negative peak to positive peak: $U_{pp} = 2 U_{peak}$. From 230 V RMS, $U_{peak} = 230\\sqrt{2} \\approx 325 V$, so $U_{pp} \\approx 650 V$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Peak-to-peak voltage is the full swing from negative peak to positive peak: $U_{pp} = 2 U_{peak}$. From 230 V RMS, $U_{peak} = 230\\sqrt{2} \\approx 325 V$, so $U_{pp} \\approx 650 V$. Hilfsmittel: Û = U_eff·√2 then U_SS = 2·Û (Wechselspannung, S.12).",
"source": "https://50ohm.de/NEA_spitze_effektiv_wert.html#EB402",
"confidence": 8
},
"EB403": {
"revision": 5,
- "explanation": "For sine waves, $U_{peak} = U_{RMS}\\sqrt{2}$ and $U_{pp} = 2U_{peak}$. With 12 V RMS, the peak is about 17 V and peak-to-peak is about 34 V. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "For sine waves, $U_{peak} = U_{RMS}\\sqrt{2}$ and $U_{pp} = 2U_{peak}$. With 12 V RMS, the peak is about 17 V and peak-to-peak is about 34 V. Hilfsmittel: Û = U_eff·√2 then U_SS = 2·Û (Wechselspannung, S.12).",
"source": "https://50ohm.de/NEA_spitze_effektiv_wert.html#EB403",
"confidence": 8
},
"EB404": {
"revision": 5,
- "explanation": "For sine waves, RMS is peak divided by $\\sqrt{2}$ because RMS represents equal heating power. $12 V / 1.414 \\approx 8.5 V$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "For sine waves, RMS is peak divided by $\\sqrt{2}$ because RMS represents equal heating power. $12 V / 1.414 \\approx 8.5 V$. Hilfsmittel: rearrange Û = U_eff·√2 → U_eff = Û/√2 (Wechselspannung, S.12).",
"source": "https://50ohm.de/NEA_spitze_effektiv_wert.html#EB404",
"confidence": 8
},
"EB405": {
"revision": 5,
- "explanation": "The DC voltage that heats a resistor the same as a sine is the sine's RMS (effective) value, and either polarity heats equally. For a $1$ V peak sine, $U_{\\text{RMS}} = 1/\\sqrt{2} \\approx 0.7$ V, so $+0.7$ V and $-0.7$ V both dissipate the same power as the AC waveform. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "The DC voltage that heats a resistor the same as a sine is the sine's RMS (effective) value, and either polarity heats equally. For a $1$ V peak sine, $U_{\\text{RMS}} = 1/\\sqrt{2} \\approx 0.7$ V, so $+0.7$ V and $-0.7$ V both dissipate the same power as the AC waveform. Hilfsmittel: U_eff = Û/√2 (Û = U_eff·√2, Wechselspannung, S.12).",
"source": "https://50ohm.de/NEA_spitze_effektiv_wert.html#EB405",
"confidence": 8
},
"EB406": {
"revision": 5,
- "explanation": "On an oscilloscope, peak-to-peak voltage is the vertical distance from trough to crest. Count vertical divisions and multiply by volts/div; the shown trace reads 12 V peak-to-peak. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "On an oscilloscope, peak-to-peak voltage is the vertical distance from trough to crest. Count vertical divisions and multiply by volts/div; the shown trace reads 12 V peak-to-peak. Hilfsmittel: read U_SS off the grid; relate via U_SS = 2·Û, Û = U_eff·√2 (Wechselspannung, S.12).",
"source": "https://50ohm.de/NEA_spitze_effektiv_wert.html#EB406",
"confidence": 8
},
"EB407": {
"revision": 4,
- "explanation": "Peak-to-peak is the complete positive-to-negative swing. If the diagram gives a 20 V peak from zero to one crest, the full swing is $2 \\cdot 20 V = 40 V$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Peak-to-peak is the complete positive-to-negative swing. If the diagram gives a 20 V peak from zero to one crest, the full swing is $2 \\cdot 20 V = 40 V$. Hilfsmittel: U_SS = 2·Û (Wechselspannung, S.12).",
"source": "https://50ohm.de/NEA_spitze_effektiv_wert.html#EB407",
"confidence": 8
},
"EB408": {
"revision": 5,
- "explanation": "Frequency is cycles per second, so it is the reciprocal of period: $f = 1/T$. With $T = 50 µs = 50 \\cdot 10^{-6} s$, $f = 1/T = 20,000 Hz = 20 kHz$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Frequency is cycles per second, so it is the reciprocal of period: $f = 1/T$. With $T = 50 µs = 50 \\cdot 10^{-6} s$, $f = 1/T = 20,000 Hz = 20 kHz$. Hilfsmittel: apply f = 1/T (Wechselspannung, S.12).",
"source": "https://50ohm.de/NEA_oszilloskop_1.html#EB408",
"confidence": 8
},
"EB409": {
"revision": 5,
- "explanation": "First read one full cycle on the oscilloscope grid to get the period $T$, then use $f = 1/T$. The trace period is about $12 µs$, so $f \\approx 1/(12 \\cdot 10^{-6}) = 83.3 kHz$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "First read one full cycle on the oscilloscope grid to get the period $T$, then use $f = 1/T$. The trace period is about $12 µs$, so $f \\approx 1/(12 \\cdot 10^{-6}) = 83.3 kHz$. Hilfsmittel: read T off the grid, then f = 1/T (Wechselspannung, S.12).",
"source": "https://50ohm.de/NEA_oszilloskop_1.html#EB409",
"confidence": 8
},
"EB410": {
"revision": 5,
- "explanation": "Oscilloscope timebase reading is divisions times time/div. Four divisions at 5 ms/div gives $T = 20 ms = 0.020 s$, so $f = 1/T = 50 Hz$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Oscilloscope timebase reading is divisions times time/div. Four divisions at 5 ms/div gives $T = 20 ms = 0.020 s$, so $f = 1/T = 50 Hz$. Hilfsmittel: T = divisions × time/div, then f = 1/T (Wechselspannung, S.12).",
"source": "https://50ohm.de/NEA_oszilloskop_1.html#EB410",
"confidence": 8
},
"EB411": {
"revision": 5,
- "explanation": "Read the period from the grid: $4 \\cdot 0.03 µs = 0.12 µs$. Then $f = 1/T = 1/(0.12 \\cdot 10^{-6} s) \\approx 8.33 MHz$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Read the period from the grid: $4 \\cdot 0.03 µs = 0.12 µs$. Then $f = 1/T = 1/(0.12 \\cdot 10^{-6} s) \\approx 8.33 MHz$. Hilfsmittel: read T off the grid, then f = 1/T (Wechselspannung, S.12).",
"source": "https://50ohm.de/NEA_oszilloskop_1.html#EB411",
"confidence": 8
},
@@ -5719,67 +5719,67 @@
},
"EB504": {
"revision": 3,
- "explanation": "Start from $P = U \\cdot I$ and substitute Ohm's law $I = U/R$ to eliminate the unknown current: $P = U^2/R$. Solving for the voltage gives $U = \\sqrt{P \\cdot R}$. (Option C, $\\sqrt{P/R}$, actually gives the current, not the voltage.) Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Start from $P = U \\cdot I$ and substitute Ohm's law $I = U/R$ to eliminate the unknown current: $P = U^2/R$. Solving for the voltage gives $U = \\sqrt{P \\cdot R}$. (Option C, $\\sqrt{P/R}$, actually gives the current, not the voltage.) Hilfsmittel: combine P = U·I with I = U/R to get P = U²/R, hence U = √(P·R) (Leistung, S.12).",
"source": "https://50ohm.de/NEA_leistung_2.html#EB504",
"confidence": 8
},
"EB505": {
"revision": 3,
- "explanation": "For an ohmic load, use RMS values in the power formulas. From $P = I^2R$ you get $I = \\sqrt{P/R}$, and from $P = U^2/R$ you get $U = \\sqrt{PR}$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "For an ohmic load, use RMS values in the power formulas. From $P = I^2R$ you get $I = \\sqrt{P/R}$, and from $P = U^2/R$ you get $U = \\sqrt{PR}$. Hilfsmittel: from P = I²·R and P = U²/R: I = √(P/R), U = √(P·R) (Leistung, S.12).",
"source": "https://50ohm.de/NEA_leistung_2.html#EB505",
"confidence": 8
},
"EB506": {
"revision": 3,
- "explanation": "These are just Ohm's law plus power rearranged. From $P = U^2/R$, $R = U^2/P$; from $P = I^2R$, $R = P/I^2$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "These are just Ohm's law plus power rearranged. From $P = U^2/R$, $R = U^2/P$; from $P = I^2R$, $R = P/I^2$. Hilfsmittel: rearrange P = U²/R and P = I²·R for R (Leistung, S.12).",
"source": "https://50ohm.de/NEA_leistung_2.html#EB506",
"confidence": 8
},
"EB507": {
"revision": 4,
- "explanation": "For RF power in a resistive 50 Ohm load, use RMS voltage: $P = U^2/R$. With $U = 100 V$ and $R = 50 Ohm$, $P = 100^2/50 = 200 W$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "For RF power in a resistive 50 Ohm load, use RMS voltage: $P = U^2/R$. With $U = 100 V$ and $R = 50 Ohm$, $P = 100^2/50 = 200 W$. Hilfsmittel: apply P = U²/R (Leistung, S.12) with RMS voltage.",
"source": "https://50ohm.de/NEA_leistung_2.html#EB507",
"confidence": 8
},
"EB508": {
"revision": 4,
- "explanation": "Use RMS current in the resistive power formula $P = I^2R$. With $I = 2 A$ and $R = 50 Ohm$, $P = 2^2 \\cdot 50 = 200 W$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Use RMS current in the resistive power formula $P = I^2R$. With $I = 2 A$ and $R = 50 Ohm$, $P = 2^2 \\cdot 50 = 200 W$. Hilfsmittel: apply P = I²·R (Leistung, S.12) with RMS current.",
"source": "https://50ohm.de/NEA_leistung_2.html#EB508",
"confidence": 8
},
"EB509": {
"revision": 4,
- "explanation": "With the voltage and resistance known, use $P = U^2/R = (10\\ \\text{V})^2 / 100\\ \\Omega = 100/100 = 1.00$ W. The resistor must be rated at least this, so a $1$ W part is the minimum. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "With the voltage and resistance known, use $P = U^2/R = (10\\ \\text{V})^2 / 100\\ \\Omega = 100/100 = 1.00$ W. The resistor must be rated at least this, so a $1$ W part is the minimum. Hilfsmittel: apply P = U²/R (Leistung, S.12).",
"source": "https://50ohm.de/NEA_leistung_2.html#EB509",
"confidence": 8
},
"EB510": {
"revision": 4,
- "explanation": "A resistor has two ceilings — voltage and power — and the lower allowed voltage wins. The power limit gives $U = \\sqrt{P \\cdot R} = \\sqrt{1\\ \\text{W} \\cdot 10000\\ \\Omega} = 100$ V. Since $100$ V is well below the $700$ V breakdown rating, power is the binding limit: max $100$ V. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "A resistor has two ceilings — voltage and power — and the lower allowed voltage wins. The power limit gives $U = \\sqrt{P \\cdot R} = \\sqrt{1\\ \\text{W} \\cdot 10000\\ \\Omega} = 100$ V. Since $100$ V is well below the $700$ V breakdown rating, power is the binding limit: max $100$ V. Hilfsmittel: the power limit gives U = √(P·R) (Leistung, S.12); compare with the voltage rating, lower wins.",
"source": "https://50ohm.de/NEA_leistung_2.html#EB510",
"confidence": 8
},
"EB511": {
"revision": 4,
- "explanation": "Check both ceilings and take the stricter. Power limit: $U = \\sqrt{P \\cdot R} = \\sqrt{6\\ \\text{W} \\cdot 100000\\ \\Omega} \\approx 775$ V. That is below the $1000$ V voltage rating, so the power rating binds first — the maximum is $\\approx 775$ V. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Check both ceilings and take the stricter. Power limit: $U = \\sqrt{P \\cdot R} = \\sqrt{6\\ \\text{W} \\cdot 100000\\ \\Omega} \\approx 775$ V. That is below the $1000$ V voltage rating, so the power rating binds first — the maximum is $\\approx 775$ V. Hilfsmittel: the power limit gives U = √(P·R) (Leistung, S.12); compare with the voltage rating, lower wins.",
"source": "https://50ohm.de/NEA_leistung_2.html#EB511",
"confidence": 8
},
"EB512": {
"revision": 4,
- "explanation": "Solve $P = I^2R$ for current: $I = \\sqrt{P/R}$. With $P = 23.0 W$ and $R = 120 Ohm$, $I = \\sqrt{23/120} \\approx 0.438 A = 438 mA$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Solve $P = I^2R$ for current: $I = \\sqrt{P/R}$. With $P = 23.0 W$ and $R = 120 Ohm$, $I = \\sqrt{23/120} \\approx 0.438 A = 438 mA$. Hilfsmittel: rearrange P = I²·R → I = √(P/R) (Leistung, S.12).",
"source": "https://50ohm.de/NEA_leistung_2.html#EB512",
"confidence": 8
},
"EB513": {
"revision": 4,
- "explanation": "Convert the scope reading to RMS in steps: peak is half the peak-to-peak, $U_{\\text{pk}} = 25/2 = 12.5$ V; RMS of a sine is $U_{\\text{pk}}/\\sqrt{2} = 12.5/1.414 \\approx 8.84$ V. Then $I_{\\text{RMS}} = U/R = 8.84\\ \\text{V} / 1000\\ \\Omega \\approx 8.8$ mA. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Convert the scope reading to RMS in steps: peak is half the peak-to-peak, $U_{\\text{pk}} = 25/2 = 12.5$ V; RMS of a sine is $U_{\\text{pk}}/\\sqrt{2} = 12.5/1.414 \\approx 8.84$ V. Then $I_{\\text{RMS}} = U/R = 8.84\\ \\text{V} / 1000\\ \\Omega \\approx 8.8$ mA. Hilfsmittel: first Û = U_SS/2 and U_eff = Û/√2 (Wechselspannung, S.12), then I = U/R (Ohmsches Gesetz, S.11).",
"source": "https://50ohm.de/NEA_leistung_2.html#EB513",
"confidence": 8
},
"EB514": {
"revision": 4,
- "explanation": "Parallel resistors each dissipate independently, so power ratings add: $11 \\cdot 5\\ \\text{W} = 55$ W. (As a check, eleven $560\\ \\Omega$ in parallel give $560/11 \\approx 51\\ \\Omega \\approx 50\\ \\Omega$, the wanted dummy-load value.) The total handles much more than a single $5$ W resistor. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Parallel resistors each dissipate independently, so power ratings add: $11 \\cdot 5\\ \\text{W} = 55$ W. (As a check, eleven $560\\ \\Omega$ in parallel give $560/11 \\approx 51\\ \\Omega \\approx 50\\ \\Omega$, the wanted dummy-load value.) The total handles much more than a single $5$ W resistor.",
"source": "https://50ohm.de/NEA_leistung_2.html#EB514",
"confidence": 8
},
@@ -5959,7 +5959,7 @@
},
"EC306": {
"revision": 4,
- "explanation": "With turns $N$ and cross-section $A$ held fixed, $L = \\mu_0 \\mu_r N^2 A / l$ is inversely proportional to length $l$. Doubling the length halves the inductance: $12\\ \\mu\\text{H} / 2 = 6\\ \\mu\\text{H}$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "With turns $N$ and cross-section $A$ held fixed, $L = \\mu_0 \\mu_r N^2 A / l$ is inversely proportional to length $l$. Doubling the length halves the inductance: $12\\ \\mu\\text{H} / 2 = 6\\ \\mu\\text{H}$. Hilfsmittel: L = μ0·μr·N²·A/l is ∝ 1/l (Zylinderspule, S.13); doubling length halves L.",
"source": "https://50ohm.de/NEA_spule_1.html#EC306",
"confidence": 8
},
@@ -5971,13 +5971,13 @@
},
"EC401": {
"revision": 4,
- "explanation": "An ideal transformer scales voltage by turns ratio: $U_1/U_2 = N_1/N_2$. A 15:1 primary-to-secondary ratio steps 230 V down to $230/15 \\approx 15.3 V$, so about 15 V. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "An ideal transformer scales voltage by turns ratio: $U_1/U_2 = N_1/N_2$. A 15:1 primary-to-secondary ratio steps 230 V down to $230/15 \\approx 15.3 V$, so about 15 V. Hilfsmittel: apply U_2 = U_1·N_2/N_1 (Übersetzungsverhältnis ü = N_P/N_S = U_P/U_S = I_S/I_P = √(Z_P/Z_S), S.13).",
"source": "https://50ohm.de/NEA_uebertrager_1.html#EC401",
"confidence": 8
},
"EC402": {
"revision": 4,
- "explanation": "An ideal transformer scales voltage by the turns ratio: $U_2/U_1 = N_2/N_1$. With the primary having five times the secondary's turns, the secondary voltage is one fifth: $230\\ \\text{V} / 5 = 46$ V. More turns means more volts on that side. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "An ideal transformer scales voltage by the turns ratio: $U_2/U_1 = N_2/N_1$. With the primary having five times the secondary's turns, the secondary voltage is one fifth: $230\\ \\text{V} / 5 = 46$ V. More turns means more volts on that side. Hilfsmittel: apply U_2 = U_1·N_2/N_1 (Übersetzungsverhältnis ü = N_P/N_S = U_P/U_S = I_S/I_P = √(Z_P/Z_S), S.13).",
"source": "https://50ohm.de/NEA_uebertrager_1.html#EC402",
"confidence": 8
},
@@ -6085,7 +6085,7 @@
},
"EC516": {
"revision": 4,
- "explanation": "The resistor takes the difference between supply and LED voltage: $5.5\\ \\text{V} - 1.75\\ \\text{V} = 3.75$ V. So $R = 3.75\\ \\text{V} / 0.025\\ \\text{A} = 150\\ \\Omega$. Its power dissipation is $P = U \\cdot I = 3.75\\ \\text{V} \\cdot 0.025\\ \\text{A} \\approx 0.094$ W, so a $0.1$ W resistor is the smallest standard rating that survives. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "The resistor takes the difference between supply and LED voltage: $5.5\\ \\text{V} - 1.75\\ \\text{V} = 3.75$ V. So $R = 3.75\\ \\text{V} / 0.025\\ \\text{A} = 150\\ \\Omega$. Its power dissipation is $P = U \\cdot I = 3.75\\ \\text{V} \\cdot 0.025\\ \\text{A} \\approx 0.094$ W, so a $0.1$ W resistor is the smallest standard rating that survives. Hilfsmittel: first R = U/I with the LED drop subtracted (Ohmsches Gesetz, S.11), then P = U·I (Leistung, S.12).",
"source": "https://50ohm.de/NEA_diode_1.html#EC516",
"confidence": 8
},
@@ -6115,7 +6115,7 @@
},
"EC521": {
"revision": 3,
- "explanation": "With no load, the series resistor carries only the Zener current and must drop the supply down to the Zener voltage: $13.8\\ \\text{V} - 5\\ \\text{V} = 8.8$ V across it at $30$ mA. So $R = 8.8\\ \\text{V} / 0.030\\ \\text{A} \\approx 293\\ \\Omega$. (Option B's milliohm value would short the supply — a sanity-check fail.) Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "With no load, the series resistor carries only the Zener current and must drop the supply down to the Zener voltage: $13.8\\ \\text{V} - 5\\ \\text{V} = 8.8$ V across it at $30$ mA. So $R = 8.8\\ \\text{V} / 0.030\\ \\text{A} \\approx 293\\ \\Omega$. (Option B's milliohm value would short the supply — a sanity-check fail.) Hilfsmittel: apply R = U/I (Ohmsches Gesetz, S.11) to the resistor's drop and Zener current.",
"source": "https://50ohm.de/NEA_diode_1.html#EC521",
"confidence": 8
},
@@ -6181,7 +6181,7 @@
},
"EC610": {
"revision": 5,
- "explanation": "A silicon BJT conducts once its base-emitter junction is forward-biased past the silicon threshold, about $0.6$-$0.7$ V, so $U_{BE} \\approx 0.6$ V (positive for an NPN). A negative or zero $U_{BE}$ leaves the junction off, so those options do not turn it on. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "A silicon BJT conducts once its base-emitter junction is forward-biased past the silicon threshold, about $0.6$-$0.7$ V, so $U_{BE} \\approx 0.6$ V (positive for an NPN). A negative or zero $U_{BE}$ leaves the junction off, so those options do not turn it on.",
"source": "https://50ohm.de/NEA_transistor_1.html#EC610",
"confidence": 8
},
@@ -6217,31 +6217,31 @@
},
"ED101": {
"revision": 3,
- "explanation": "Series resistors carry the same current, so by $U = R \\cdot I$ the voltages split in the resistance ratio: $U_1/U_2 = R_1/R_2$. With $R_1 = 5 R_2$, $U_1 = 5\\,U_2$ — the larger resistor drops the larger voltage. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Series resistors carry the same current, so by $U = R \\cdot I$ the voltages split in the resistance ratio: $U_1/U_2 = R_1/R_2$. With $R_1 = 5 R_2$, $U_1 = 5\\,U_2$ — the larger resistor drops the larger voltage. Hilfsmittel: series voltages split as U1/U2 = R1/R2 (Spannungsteiler, S.12).",
"source": "https://50ohm.de/NEA_spannungsteiler_1.html#ED101",
"confidence": 8
},
"ED102": {
"revision": 3,
- "explanation": "In a series divider the voltage ratio equals the resistance ratio, $U_1/U_2 = R_1/R_2$. Here $R_1 = R_2/6$, so $U_1 = U_2/6$ — the smaller resistor drops proportionally less voltage. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "In a series divider the voltage ratio equals the resistance ratio, $U_1/U_2 = R_1/R_2$. Here $R_1 = R_2/6$, so $U_1 = U_2/6$ — the smaller resistor drops proportionally less voltage. Hilfsmittel: series voltages split as U1/U2 = R1/R2 (Spannungsteiler, S.12).",
"source": "https://50ohm.de/NEA_spannungsteiler_1.html#ED102",
"confidence": 8
},
"ED103": {
"revision": 3,
- "explanation": "Use the divider rule $U_2 = U \\cdot R_2/(R_1 + R_2)$. With $U = 9$ V, $R_1 = 10\\ \\text{k}\\Omega$, $R_2 = 20\\ \\text{k}\\Omega$: $U_2 = 9 \\cdot 20/(10+20) = 9 \\cdot 20/30 = 6.0$ V. The bigger resistor takes the bigger share. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Use the divider rule $U_2 = U \\cdot R_2/(R_1 + R_2)$. With $U = 9$ V, $R_1 = 10\\ \\text{k}\\Omega$, $R_2 = 20\\ \\text{k}\\Omega$: $U_2 = 9 \\cdot 20/(10+20) = 9 \\cdot 20/30 = 6.0$ V. The bigger resistor takes the bigger share. Hilfsmittel: apply the divider U2 = U·R2/(R1+R2) (Spannungsteiler, S.12).",
"source": "https://50ohm.de/NEA_spannungsteiler_1.html#ED103",
"confidence": 8
},
"ED104": {
"revision": 3,
- "explanation": "For two parallel resistors use $R_g = R_1R_2/(R_1+R_2)$, because conductances add in parallel. With 100 Ohm and 400 Ohm: $100\\cdot400/(100+400)=40000/500=80 Ohm$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "For two parallel resistors use $R_g = R_1R_2/(R_1+R_2)$, because conductances add in parallel. With 100 Ohm and 400 Ohm: $100\\cdot400/(100+400)=40000/500=80 Ohm$. Hilfsmittel: apply R_g = R1·R2/(R1+R2) (Reihen-/Parallelschaltung, S.12).",
"source": "https://50ohm.de/NEA_reihe_parallel_widerstand.html#ED104",
"confidence": 8
},
"ED105": {
"revision": 3,
- "explanation": "Parallel resistance is always lower than the smallest branch. For two branches use $R_g = R_1R_2/(R_1+R_2)$: $50\\cdot200/(50+200)=10000/250=40 Ohm$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Parallel resistance is always lower than the smallest branch. For two branches use $R_g = R_1R_2/(R_1+R_2)$: $50\\cdot200/(50+200)=10000/250=40 Ohm$. Hilfsmittel: apply R_g = R1·R2/(R1+R2) (Reihen-/Parallelschaltung, S.12).",
"source": "https://50ohm.de/NEA_reihe_parallel_widerstand.html#ED105",
"confidence": 8
},
@@ -6259,37 +6259,37 @@
},
"ED108": {
"revision": 3,
- "explanation": "Reduce mixed resistor networks one obvious block at a time. Series resistors add directly: $500 + 500 = 1000 Ohm$. That 1000 Ohm branch is parallel with another 1000 Ohm resistor, and two equal parallel resistors halve: $1000/2 = 500 Ohm$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Reduce mixed resistor networks one obvious block at a time. Series resistors add directly: $500 + 500 = 1000 Ohm$. That 1000 Ohm branch is parallel with another 1000 Ohm resistor, and two equal parallel resistors halve: $1000/2 = 500 Ohm$. Hilfsmittel: series adds, two equal parallel resistors halve (Reihen-/Parallelschaltung, S.12).",
"source": "https://50ohm.de/NEA_reihe_parallel_widerstand.html#ED108",
"confidence": 8
},
"ED109": {
"revision": 3,
- "explanation": "First combine the series path: $500 Ohm + 1.5 kOhm = 2 kOhm$. That branch is parallel to another 2 kOhm branch, and equal parallel resistors give half the value, so $R_g = 1 kOhm$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "First combine the series path: $500 Ohm + 1.5 kOhm = 2 kOhm$. That branch is parallel to another 2 kOhm branch, and equal parallel resistors give half the value, so $R_g = 1 kOhm$. Hilfsmittel: series adds, two equal parallel resistors halve (Reihen-/Parallelschaltung, S.12).",
"source": "https://50ohm.de/NEA_reihe_parallel_widerstand.html#ED109",
"confidence": 8
},
"ED110": {
"revision": 3,
- "explanation": "Parallel branches have the same voltage; for two equal 1 kOhm resistors the equivalent is $1 kOhm / 2 = 500 Ohm$. Add the remaining 500 Ohm series resistor and the total becomes $500 + 500 = 1000 Ohm$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Parallel branches have the same voltage; for two equal 1 kOhm resistors the equivalent is $1 kOhm / 2 = 500 Ohm$. Add the remaining 500 Ohm series resistor and the total becomes $500 + 500 = 1000 Ohm$. Hilfsmittel: two equal parallel resistors halve, then add the series part (Reihen-/Parallelschaltung, S.12).",
"source": "https://50ohm.de/NEA_reihe_parallel_widerstand.html#ED110",
"confidence": 8
},
"ED111": {
"revision": 3,
- "explanation": "Start with the parallel part: two equal 2 kOhm resistors in parallel give $2 kOhm / 2 = 1 kOhm$. That equivalent is then in series with R1, so $1 kOhm + 1 kOhm = 2 kOhm$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Start with the parallel part: two equal 2 kOhm resistors in parallel give $2 kOhm / 2 = 1 kOhm$. That equivalent is then in series with R1, so $1 kOhm + 1 kOhm = 2 kOhm$. Hilfsmittel: two equal parallel resistors halve, then add the series part (Reihen-/Parallelschaltung, S.12).",
"source": "https://50ohm.de/NEA_reihe_parallel_widerstand.html#ED111",
"confidence": 8
},
"ED112": {
"revision": 3,
- "explanation": "For two unequal parallel resistors use $R_g = R_1R_2/(R_1+R_2)$. Thus $3 kOhm || 1.5 kOhm = 4.5/4.5 kOhm = 1 kOhm$, and the series R1 adds another 1 kOhm for a total of 2 kOhm. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "For two unequal parallel resistors use $R_g = R_1R_2/(R_1+R_2)$. Thus $3 kOhm || 1.5 kOhm = 4.5/4.5 kOhm = 1 kOhm$, and the series R1 adds another 1 kOhm for a total of 2 kOhm. Hilfsmittel: parallel R_g = R1·R2/(R1+R2), then add the series part (Reihen-/Parallelschaltung, S.12).",
"source": "https://50ohm.de/NEA_reihe_parallel_widerstand.html#ED112",
"confidence": 8
},
"ED113": {
"revision": 3,
- "explanation": "Use reciprocal addition for three parallel resistors: $1/R = 1/10 kOhm + 1/2.5 kOhm + 1/500 Ohm$, giving 400 Ohm. The remaining 600 Ohm is in series, so the total is $400 + 600 = 1000 Ohm$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Use reciprocal addition for three parallel resistors: $1/R = 1/10 kOhm + 1/2.5 kOhm + 1/500 Ohm$, giving 400 Ohm. The remaining 600 Ohm is in series, so the total is $400 + 600 = 1000 Ohm$. Hilfsmittel: three in parallel via 1/R_g = Σ1/Ri, then add the series part (Reihen-/Parallelschaltung, S.12).",
"source": "https://50ohm.de/NEA_reihe_parallel_widerstand.html#ED113",
"confidence": 8
},
@@ -6313,49 +6313,49 @@
},
"ED117": {
"revision": 3,
- "explanation": "Capacitances in parallel add directly. Put everything in nF: $0.1\\ \\mu\\text{F} = 100$ nF, $C_2 = 150$ nF, $50000\\ \\text{pF} = 50$ nF. Sum $= 100 + 150 + 50 = 300\\ \\text{nF} = 0.3\\ \\mu\\text{F}$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Capacitances in parallel add directly. Put everything in nF: $0.1\\ \\mu\\text{F} = 100$ nF, $C_2 = 150$ nF, $50000\\ \\text{pF} = 50$ nF. Sum $= 100 + 150 + 50 = 300\\ \\text{nF} = 0.3\\ \\mu\\text{F}$. Hilfsmittel: parallel caps add: C_G = ΣCi (Kapazität, S.13).",
"source": "https://50ohm.de/NEA_reihe_parallel_kondensator.html#ED117",
"confidence": 8
},
"ED118": {
"revision": 4,
- "explanation": "Parallel capacitors add after unit conversion: $22\\ \\text{nF} + 0.033\\ \\mu\\text{F}\\,(33\\ \\text{nF}) + 15000\\ \\text{pF}\\,(15\\ \\text{nF}) = 70\\ \\text{nF} = 0.070\\ \\mu\\text{F}$. Parallel always increases total capacitance. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Parallel capacitors add after unit conversion: $22\\ \\text{nF} + 0.033\\ \\mu\\text{F}\\,(33\\ \\text{nF}) + 15000\\ \\text{pF}\\,(15\\ \\text{nF}) = 70\\ \\text{nF} = 0.070\\ \\mu\\text{F}$. Parallel always increases total capacitance. Hilfsmittel: parallel caps add: C_G = ΣCi (Kapazität, S.13).",
"source": "https://50ohm.de/NEA_reihe_parallel_kondensator.html#ED118",
"confidence": 8
},
"ED119": {
"revision": 4,
- "explanation": "Capacitors in series combine like parallel resistors; for $n$ equal ones, $C_g = C/n$. Three $0.33\\ \\mu\\text{F}$ in series give $0.33/3 = 0.110\\ \\mu\\text{F}$. Series capacitance is always smaller than the smallest capacitor. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Capacitors in series combine like parallel resistors; for $n$ equal ones, $C_g = C/n$. Three $0.33\\ \\mu\\text{F}$ in series give $0.33/3 = 0.110\\ \\mu\\text{F}$. Series capacitance is always smaller than the smallest capacitor. Hilfsmittel: series caps: 1/C_G = Σ1/Ci, equal ones → C/n (Kapazität, S.13).",
"source": "https://50ohm.de/NEA_reihe_parallel_kondensator.html#ED119",
"confidence": 8
},
"ED120": {
"revision": 4,
- "explanation": "For series capacitors, $1/C_g = \\sum 1/C_i$. Convert $200000\\ \\text{nF} = 200\\ \\mu\\text{F}$, then $1/C_g = 1/100 + 1/200 + 1/200 = 2/200 + 1/200 + 1/200 = 4/200$, so $C_g = 50\\ \\mu\\text{F}$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "For series capacitors, $1/C_g = \\sum 1/C_i$. Convert $200000\\ \\text{nF} = 200\\ \\mu\\text{F}$, then $1/C_g = 1/100 + 1/200 + 1/200 = 2/200 + 1/200 + 1/200 = 4/200$, so $C_g = 50\\ \\mu\\text{F}$. Hilfsmittel: series caps via 1/C_G = Σ1/Ci (Kapazität, S.13).",
"source": "https://50ohm.de/NEA_reihe_parallel_kondensator.html#ED120",
"confidence": 8
},
"ED121": {
"revision": 3,
- "explanation": "Capacitors behave opposite to resistors for series/parallel math: parallel capacitances add, while series capacitance uses reciprocals. Two equal 10 nF capacitors in series give $10/2 = 5 nF$; in parallel with C3 = 5 nF, the total is 10 nF. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Capacitors behave opposite to resistors for series/parallel math: parallel capacitances add, while series capacitance uses reciprocals. Two equal 10 nF capacitors in series give $10/2 = 5 nF$; in parallel with C3 = 5 nF, the total is 10 nF. Hilfsmittel: series caps halve (1/C_G = Σ1/Ci), parallel caps add (Kapazität, S.13).",
"source": "https://50ohm.de/NEA_reihe_parallel_kondensator.html#ED121",
"confidence": 8
},
"ED122": {
"revision": 3,
- "explanation": "First add the parallel capacitors: $C_2+C_3=1 uF+1 uF=2 uF$. That 2 uF equivalent is in series with C1 = 2 uF; two equal capacitors in series halve, so $C_g=1.0 uF$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "First add the parallel capacitors: $C_2+C_3=1 uF+1 uF=2 uF$. That 2 uF equivalent is in series with C1 = 2 uF; two equal capacitors in series halve, so $C_g=1.0 uF$. Hilfsmittel: parallel caps add, then series caps via 1/C_G = Σ1/Ci (Kapazität, S.13).",
"source": "https://50ohm.de/NEA_reihe_parallel_kondensator.html#ED122",
"confidence": 8
},
"ED123": {
"revision": 3,
- "explanation": "Parallel capacitors add directly, so C2 and C3 become $4 nF + 4 nF = 8 nF$. That 8 nF equivalent is in series with C1 = 8 nF, and equal series capacitors halve, giving $C_g=4 nF$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Parallel capacitors add directly, so C2 and C3 become $4 nF + 4 nF = 8 nF$. That 8 nF equivalent is in series with C1 = 8 nF, and equal series capacitors halve, giving $C_g=4 nF$. Hilfsmittel: parallel caps add, then equal series caps halve (Kapazität, S.13).",
"source": "https://50ohm.de/NEA_reihe_parallel_kondensator.html#ED123",
"confidence": 8
},
"ED124": {
"revision": 3,
- "explanation": "Convert first so the units match: $100000 pF = 100 nF$. C2 and C3 are parallel, so they add to 200 nF; that is in series with C1 = 200 nF, and two equal series capacitances give 100 nF. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Convert first so the units match: $100000 pF = 100 nF$. C2 and C3 are parallel, so they add to 200 nF; that is in series with C1 = 200 nF, and two equal series capacitances give 100 nF. Hilfsmittel: parallel caps add, then equal series caps halve (Kapazität, S.13).",
"source": "https://50ohm.de/NEA_reihe_parallel_kondensator.html#ED124",
"confidence": 8
},
@@ -6475,7 +6475,7 @@
},
"ED304": {
"revision": 3,
- "explanation": "This is a single-diode half-wave rectifier. The diode conducts only on the half-cycle where it is forward biased, so current flows through the load in one direction; the opposite half-cycle is blocked, leaving pulsating DC. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "This is a single-diode half-wave rectifier. The diode conducts only on the half-cycle where it is forward biased, so current flows through the load in one direction; the opposite half-cycle is blocked, leaving pulsating DC.",
"source": "https://50ohm.de/NEA_gleichrichter_1.html#ED304",
"confidence": 8
},
@@ -6559,13 +6559,13 @@
},
"EE203": {
"revision": 4,
- "explanation": "In upper sideband (USB), each audio component appears above the suppressed carrier by its audio frequency. A 1 kHz tone is $0.001 MHz$, so $21.250 MHz + 0.001 MHz = 21.251 MHz$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "In upper sideband (USB), each audio component appears above the suppressed carrier by its audio frequency. A 1 kHz tone is $0.001 MHz$, so $21.250 MHz + 0.001 MHz = 21.251 MHz$.",
"source": "https://50ohm.de/NEA_ssb_2.html#EE203",
"confidence": 8
},
"EE204": {
"revision": 4,
- "explanation": "Ideal SSB suppresses the carrier and keeps one sideband. LSB (lower sideband) places the audio below the carrier: $3.650\\ \\text{MHz} - 0.002\\ \\text{MHz} = 3.648$ MHz. Only that single component is transmitted, so options listing the carrier or upper sideband are wrong. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Ideal SSB suppresses the carrier and keeps one sideband. LSB (lower sideband) places the audio below the carrier: $3.650\\ \\text{MHz} - 0.002\\ \\text{MHz} = 3.648$ MHz. Only that single component is transmitted, so options listing the carrier or upper sideband are wrong.",
"source": "https://50ohm.de/NEA_ssb_2.html#EE204",
"confidence": 8
},
@@ -6637,7 +6637,7 @@
},
"EE403": {
"revision": 3,
- "explanation": "In SSB the RF bandwidth equals the audio (NF) bandwidth, because SSB is a pure frequency translation of the audio with no extra sidebands. So a $50$ Hz audio signal occupies about $50$ Hz of RF — narrow in, narrow out. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "In SSB the RF bandwidth equals the audio (NF) bandwidth, because SSB is a pure frequency translation of the audio with no extra sidebands. So a $50$ Hz audio signal occupies about $50$ Hz of RF — narrow in, narrow out.",
"source": "https://50ohm.de/NEA_digimode_ssb.html#EE403",
"confidence": 8
},
@@ -6649,7 +6649,7 @@
},
"EE405": {
"revision": 4,
- "explanation": "Use a reporting network: transmit with a mode designed for it (e.g. WSPR, or CW into the Reverse Beacon Network), then look up your callsign on the matching internet platform to see who copied you and where. The spots are gathered automatically by listening stations — no email handshake or made-up 'AUTO RSVP' procedure exists. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Use a reporting network: transmit with a mode designed for it (e.g. WSPR, or CW into the Reverse Beacon Network), then look up your callsign on the matching internet platform to see who copied you and where. The spots are gathered automatically by listening stations — no email handshake or made-up 'AUTO RSVP' procedure exists.",
"source": "https://50ohm.de/NEA_automatische_empfangsberichte.html#EE405",
"confidence": 8
},
@@ -6727,31 +6727,31 @@
},
"EF201": {
"revision": 4,
- "explanation": "A mixer's main outputs are the sum and the absolute difference of its two input frequencies. With $31.7$ MHz and $21$ MHz: sum $= 31.7 + 21 = 52.7$ MHz and difference $= |31.7 - 21| = 10.7$ MHz. (The $10.7$ MHz difference is the classic FM IF.) Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "A mixer's main outputs are the sum and the absolute difference of its two input frequencies. With $31.7$ MHz and $21$ MHz: sum $= 31.7 + 21 = 52.7$ MHz and difference $= |31.7 - 21| = 10.7$ MHz. (The $10.7$ MHz difference is the classic FM IF.) Hilfsmittel: the difference product matches f_ZF = |f1 − f2| (S.14); the sum product f1 + f2 is mixer knowledge, not in the sheet.",
"source": "https://50ohm.de/NEA_mischer.html#EF201",
"confidence": 8
},
"EF202": {
"revision": 4,
- "explanation": "Mixing produces sum and difference: $38.7 + 28 = 66.7$ MHz and $|38.7 - 28| = 10.7$ MHz. The wanted IF is usually the difference, $10.7$ MHz, with the sum filtered out afterwards. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Mixing produces sum and difference: $38.7 + 28 = 66.7$ MHz and $|38.7 - 28| = 10.7$ MHz. The wanted IF is usually the difference, $10.7$ MHz, with the sum filtered out afterwards. Hilfsmittel: the difference product matches f_ZF = |f1 − f2| (S.14); the sum product f1 + f2 is mixer knowledge, not in the sheet.",
"source": "https://50ohm.de/NEA_mischer.html#EF202",
"confidence": 8
},
"EF203": {
"revision": 4,
- "explanation": "The desired mixer products are the sum and difference frequencies: $30 + 39 = 69$ MHz and $|39 - 30| = 9$ MHz. The two input frequencies themselves are not the wanted output — only their sum and difference are. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "The desired mixer products are the sum and difference frequencies: $30 + 39 = 69$ MHz and $|39 - 30| = 9$ MHz. The two input frequencies themselves are not the wanted output — only their sum and difference are. Hilfsmittel: the difference product matches f_ZF = |f1 − f2| (S.14); the sum product f1 + f2 is mixer knowledge, not in the sheet.",
"source": "https://50ohm.de/NEA_mischer.html#EF203",
"confidence": 8
},
"EF204": {
"revision": 4,
- "explanation": "Sum and difference again: $145 + 136 = 281$ MHz and $|145 - 136| = 9$ MHz. So a $9$ MHz IF can be produced from these two VHF inputs, with the $281$ MHz sum rejected by the IF filter. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Sum and difference again: $145 + 136 = 281$ MHz and $|145 - 136| = 9$ MHz. So a $9$ MHz IF can be produced from these two VHF inputs, with the $281$ MHz sum rejected by the IF filter. Hilfsmittel: the difference product matches f_ZF = |f1 − f2| (S.14); the sum product f1 + f2 is mixer knowledge, not in the sheet.",
"source": "https://50ohm.de/NEA_mischer.html#EF204",
"confidence": 8
},
"EF205": {
"revision": 4,
- "explanation": "A mixer multiplies signals, producing first-order products at the sum and absolute difference: $f_{sum}=f_1+f_2$ and $f_{diff}=|f_1-f_2|$. For 145 MHz and 136 MHz these are 281 MHz and 9 MHz. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "A mixer multiplies signals, producing first-order products at the sum and absolute difference: $f_{sum}=f_1+f_2$ and $f_{diff}=|f_1-f_2|$. For 145 MHz and 136 MHz these are 281 MHz and 9 MHz. Hilfsmittel: the difference product matches f_ZF = |f1 − f2| (S.14); the sum product f1 + f2 is mixer knowledge, not in the sheet.",
"source": "https://50ohm.de/NEA_mischer.html#EF205",
"confidence": 8
},
@@ -6841,19 +6841,19 @@
},
"EF301": {
"revision": 4,
- "explanation": "A frequency multiplier chain multiplies stage by stage, so to find the oscillator frequency you work backward by dividing by each multiplier. Here $145.2 MHz / 2 / 3 / 2 = 12.1 MHz$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "A frequency multiplier chain multiplies stage by stage, so to find the oscillator frequency you work backward by dividing by each multiplier. Here $145.2 MHz / 2 / 3 / 2 = 12.1 MHz$.",
"source": "https://50ohm.de/NEA_frequenzvervielfacher_1.html#EF301",
"confidence": 8
},
"EF302": {
"revision": 4,
- "explanation": "Frequency multipliers scale the input frequency by their factor. To recover the starting oscillator frequency, reverse the chain by division: $21.360 MHz / 3 / 2 = 3.560 MHz$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Frequency multipliers scale the input frequency by their factor. To recover the starting oscillator frequency, reverse the chain by division: $21.360 MHz / 3 / 2 = 3.560 MHz$.",
"source": "https://50ohm.de/NEA_frequenzvervielfacher_1.html#EF302",
"confidence": 8
},
"EF303": {
"revision": 4,
- "explanation": "Trace the multiplier chain: the VFO at $3.51$ MHz passes two doublers, $\\times 2 \\times 2 = \\times 4$, giving $3.51 \\cdot 4 = 14.04$ MHz at output a. Frequency multipliers move a low, stable VFO up to the wanted band. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Trace the multiplier chain: the VFO at $3.51$ MHz passes two doublers, $\\times 2 \\times 2 = \\times 4$, giving $3.51 \\cdot 4 = 14.04$ MHz at output a. Frequency multipliers move a low, stable VFO up to the wanted band.",
"source": "https://50ohm.de/NEA_frequenzvervielfacher_1.html#EF303",
"confidence": 8
},
@@ -6883,7 +6883,7 @@
},
"EF308": {
"revision": 4,
- "explanation": "Speech intelligibility needs roughly the $300$ Hz-$2.7$ kHz band, so an audio amplifier for SSB voice needs at least about $2.5$ kHz of bandwidth. More than that wastes bandwidth; much less (1 kHz) would make speech hard to understand. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Speech intelligibility needs roughly the $300$ Hz-$2.7$ kHz band, so an audio amplifier for SSB voice needs at least about $2.5$ kHz of bandwidth. More than that wastes bandwidth; much less (1 kHz) would make speech hard to understand.",
"source": "https://50ohm.de/NEA_verstaerker.html#EF308",
"confidence": 8
},
@@ -6895,7 +6895,7 @@
},
"EF310": {
"revision": 4,
- "explanation": "An SSB voice signal only needs the width of one speech sideband, so the sideband-selecting filter is about $2.4$ kHz wide. The other values are unrelated ($455$ kHz and $10.7$ MHz are IF centre frequencies, not bandwidths; $800$ Hz is too narrow for voice). Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "An SSB voice signal only needs the width of one speech sideband, so the sideband-selecting filter is about $2.4$ kHz wide. The other values are unrelated ($455$ kHz and $10.7$ MHz are IF centre frequencies, not bandwidths; $800$ Hz is too narrow for voice).",
"source": "https://50ohm.de/NEA_ssb_2.html#EF310",
"confidence": 8
},
@@ -6985,7 +6985,7 @@
},
"EG102": {
"revision": 3,
- "explanation": "A wire HF antenna can in principle be any length: with a suitable matching network (Antennentuner) you can present $50\\ \\Omega$ to the transmitter even when the wire is not resonant. The wire's length sets its resonance and feed-point impedance, not whether it can be used at all — so the rigid $\\lambda/2$ or $\\lambda/4$ options are wrong. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "A wire HF antenna can in principle be any length: with a suitable matching network (Antennentuner) you can present $50\\ \\Omega$ to the transmitter even when the wire is not resonant. The wire's length sets its resonance and feed-point impedance, not whether it can be used at all — so the rigid $\\lambda/2$ or $\\lambda/4$ options are wrong.",
"source": "https://50ohm.de/NEA_antenne_laenge_resonanz.html#EG102",
"confidence": 8
},
@@ -7003,7 +7003,7 @@
},
"EG105": {
"revision": 3,
- "explanation": "A magnetic loop is electrically small, often around $\\lambda/10$ circumference or less. Its current is high and voltage is comparatively low, so the near field is dominated by the magnetic H-field; this is why it behaves differently from a full-size resonant wire antenna. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "A magnetic loop is electrically small, often around $\\lambda/10$ circumference or less. Its current is high and voltage is comparatively low, so the near field is dominated by the magnetic H-field; this is why it behaves differently from a full-size resonant wire antenna.",
"source": "https://50ohm.de/NEA_antennenformen_2.html#EG105",
"confidence": 8
},
@@ -7021,13 +7021,13 @@
},
"EG108": {
"revision": 3,
- "explanation": "A $5/8\\,\\lambda$ vertical concentrates more of its radiation toward the horizon than a $\\lambda/4$ whip, giving roughly 1-3 dB more gain in useful low-angle directions — ideal for mobile VHF/UHF where you want range along the ground. The other options (power handling, mounting, noise) are not the reason it is preferred. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "A $5/8\\,\\lambda$ vertical concentrates more of its radiation toward the horizon than a $\\lambda/4$ whip, giving roughly 1-3 dB more gain in useful low-angle directions — ideal for mobile VHF/UHF where you want range along the ground. The other options (power handling, mounting, noise) are not the reason it is preferred.",
"source": "https://50ohm.de/NEA_antennenformen_2.html#EG108",
"confidence": 8
},
"EG109": {
- "revision": 3,
- "explanation": "First the wavelength: $\\lambda = 300 / f_{\\text{MHz}} = 300/28.5 \\approx 10.53$ m. The electrical length of a $5/8\\,\\lambda$ radiator is $0.625 \\cdot 10.53 \\approx 6.58$ m. (This is the electrical length; a real whip is trimmed a few percent shorter for the velocity factor.) Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "revision": 4,
+ "explanation": "First the wavelength: $\\lambda = 300/f_{\\text{MHz}} = 300/28.5 \\approx 10.53\\,\\text{m}$. The electrical length of a $5/8\\,\\lambda$ radiator is $0.625 \\cdot 10.53 \\approx 6.58\\,\\text{m}$. (This is the electrical length; a real whip is built a few percent shorter to allow for conductor end effects — an empirical antenna shortening, distinct from the HF-line velocity factor.) Hilfsmittel: first λ = c/f (c = f·λ (λ[m] ≈ 300/f[MHz]), S.17), then take 5/8 of it (the 5/8 factor is antenna knowledge).",
"source": "https://50ohm.de/NEA_antenne_laenge_resonanz.html#EG109",
"confidence": 8
},
@@ -7135,7 +7135,7 @@
},
"EG213": {
"revision": 3,
- "explanation": "A ground-plane is unbalanced (asymmetric): one side is the radiator, the other is the radial/counterpoise system near earth potential, so the two halves are not mirror images. A folded dipole, a long Yagi, and a center-fed $\\lambda/2$ dipole are all symmetric about their feed point. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "A ground-plane is unbalanced (asymmetric): one side is the radiator, the other is the radial/counterpoise system near earth potential, so the two halves are not mirror images. A folded dipole, a long Yagi, and a center-fed $\\lambda/2$ dipole are all symmetric about their feed point.",
"source": "https://50ohm.de/NEA_antennenformen_2.html#EG213",
"confidence": 8
},
@@ -7171,7 +7171,7 @@
},
"EG219": {
"revision": 3,
- "explanation": "A vertical half-wave antenna radiates mainly perpendicular to the vertical conductor. Since the conductor is vertical, the strongest radiation leaves at low elevation angles, which is useful for DX because low-angle rays make longer ionospheric hops. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "A vertical half-wave antenna radiates mainly perpendicular to the vertical conductor. Since the conductor is vertical, the strongest radiation leaves at low elevation angles, which is useful for DX because low-angle rays make longer ionospheric hops.",
"source": "https://50ohm.de/NEA_antennenformen_2.html#EG219",
"confidence": 8
},
@@ -7183,7 +7183,7 @@
},
"EG221": {
"revision": 4,
- "explanation": "dBd is referenced to a half-wave dipole, and a dipole already has $2.15$ dBi of gain over isotropic. So convert by adding that offset: $5\\ \\text{dBd} + 2.15\\ \\text{dB} = 7.15$ dBi. Whenever you see dBi vs dBd, the gap is always $2.15$ dB. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "dBd is referenced to a half-wave dipole, and a dipole already has $2.15$ dBi of gain over isotropic. So convert by adding that offset: $5\\ \\text{dBd} + 2.15\\ \\text{dB} = 7.15$ dBi. Whenever you see dBi vs dBd, the gap is always $2.15$ dB. Hilfsmittel: add the 2,15 dB dipole offset: g_i = g_d + 2,15 dB (Antennen, Pegel, S.15).",
"source": "https://50ohm.de/NEA_antennengewinn.html#EG221",
"confidence": 8
},
@@ -7219,7 +7219,7 @@
},
"EG304": {
"revision": 3,
- "explanation": "A feeder is unbalanced (unsymmetrisch) when its two conductors are not geometrically equivalent — as in coax, where a central inner conductor sits inside an outer shield at a different potential. A balanced line uses two identical, symmetric conductors instead. Reflection, resonance, and length do not define balance. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "A feeder is unbalanced (unsymmetrisch) when its two conductors are not geometrically equivalent — as in coax, where a central inner conductor sits inside an outer shield at a different potential. A balanced line uses two identical, symmetric conductors instead. Reflection, resonance, and length do not define balance.",
"source": "https://50ohm.de/NEA_uebertragungsleitungen_2.html#EG304",
"confidence": 8
},
@@ -7237,67 +7237,67 @@
},
"EG307": {
"revision": 4,
- "explanation": "Attenuations in dB add along the signal path. Treat each cable or inline loss as a positive dB loss and sum them; the losses in the shown station layout total 5 dB before antenna gain is considered. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Attenuations in dB add along the signal path. Treat each cable or inline loss as a positive dB loss and sum them; the losses in the shown station layout total 5 dB before antenna gain is considered. Hilfsmittel: losses in dB add along the path; each is a = 10·log10(P_in/P_out) (Pegel, S.15).",
"source": "https://50ohm.de/NEA_kabeldaempfung_1.html#EG307",
"confidence": 8
},
"EG308": {
"revision": 4,
- "explanation": "With SWR $= 1$ nothing is reflected, so the missing power is pure cable loss. Going from $100$ W to $50$ W is a factor of $2$, and $10\\log_{10}(2) \\approx 3$ dB of attenuation. (Attenuation is a positive loss figure, and dBm would be an absolute power, not a loss — so those options are wrong.) Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "With SWR $= 1$ nothing is reflected, so the missing power is pure cable loss. Going from $100$ W to $50$ W is a factor of $2$, and $10\\log_{10}(2) \\approx 3$ dB of attenuation. (Attenuation is a positive loss figure, and dBm would be an absolute power, not a loss — so those options are wrong.) Hilfsmittel: apply a = 10·log10(P_in/P_out) (Pegel, S.15); ×2 power = 3 dB (table, S.15).",
"source": "https://50ohm.de/NEA_kabeldaempfung_1.html#EG308",
"confidence": 8
},
"EG309": {
"revision": 4,
- "explanation": "Attenuation in dB compares input to output power: $a = 10\\log_{10}(P_{\\text{in}}/P_{\\text{out}})$. Only a quarter remains, so the ratio is $4$, and $10\\log_{10}(4) \\approx 6$ dB. (Each halving is $3$ dB, and a quarter is two halvings, $3+3 = 6$ dB.) Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Attenuation in dB compares input to output power: $a = 10\\log_{10}(P_{\\text{in}}/P_{\\text{out}})$. Only a quarter remains, so the ratio is $4$, and $10\\log_{10}(4) \\approx 6$ dB. (Each halving is $3$ dB, and a quarter is two halvings, $3+3 = 6$ dB.) Hilfsmittel: apply a = 10·log10(P_in/P_out) (Pegel, S.15); ×4 power = 6 dB (table, S.15).",
"source": "https://50ohm.de/NEA_kabeldaempfung_1.html#EG309",
"confidence": 8
},
"EG310": {
"revision": 4,
- "explanation": "Using $a = 10\\log_{10}(P_{\\text{in}}/P_{\\text{out}})$ with only one tenth of the power left, the ratio is $10$ and $10\\log_{10}(10) = 10$ dB. The $\\times 10$-power $= +10$-dB anchor makes this one immediate. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Using $a = 10\\log_{10}(P_{\\text{in}}/P_{\\text{out}})$ with only one tenth of the power left, the ratio is $10$ and $10\\log_{10}(10) = 10$ dB. The $\\times 10$-power $= +10$-dB anchor makes this one immediate. Hilfsmittel: apply a = 10·log10(P_in/P_out) (Pegel, S.15); ×10 power = 10 dB (table, S.15).",
"source": "https://50ohm.de/NEA_kabeldaempfung_1.html#EG310",
"confidence": 8
},
"EG311": {
"revision": 4,
- "explanation": "For one cable at one frequency, attenuation in decibels is proportional to length. Scale directly: $20\\ \\text{dB} \\cdot (20\\ \\text{m} / 100\\ \\text{m}) = 4$ dB. Because dB already represents the loss logarithmically, you scale the dB value linearly with length — no powers needed. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "For one cable at one frequency, attenuation in decibels is proportional to length. Scale directly: $20\\ \\text{dB} \\cdot (20\\ \\text{m} / 100\\ \\text{m}) = 4$ dB. Because dB already represents the loss logarithmically, you scale the dB value linearly with length — no powers needed. Hilfsmittel: cable attenuation scales linearly with length; read the per-100 m loss from the Kabeldämpfungsdiagramm Koaxialkabel (S.22) and scale by the cable length.",
"source": "https://50ohm.de/NEA_kabeldaempfung_1.html#EG311",
"confidence": 8
},
"EG312": {
"revision": 4,
- "explanation": "Read the cable-loss chart for RG58 (full-PE, $4.95$ mm) at $145$ MHz: about $20$ dB per $100$ m. The cable here is exactly $100$ m long, so the attenuation is that chart value, $20$ dB. (Thin RG58 is quite lossy at VHF — a good reason to use thicker cable on $2$ m.) Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Read the cable-loss chart for RG58 (full-PE, $4.95$ mm) at $145$ MHz: about $20$ dB per $100$ m. The cable here is exactly $100$ m long, so the attenuation is that chart value, $20$ dB. (Thin RG58 is quite lossy at VHF — a good reason to use thicker cable on $2$ m.) Hilfsmittel: read the per-100 m loss for this cable/frequency from the Kabeldämpfungsdiagramm Koaxialkabel (S.22); here the length is 100 m, so it is the chart value directly.",
"source": "https://50ohm.de/NEA_kabeldaempfung_1.html#EG312",
"confidence": 8
},
"EG313": {
"revision": 4,
- "explanation": "From the chart, RG58 at $145$ MHz is about $20$ dB per $100$ m. Attenuation scales with length, so $15$ m gives $20 \\cdot 15/100 = 3$ dB. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "From the chart, RG58 at $145$ MHz is about $20$ dB per $100$ m. Attenuation scales with length, so $15$ m gives $20 \\cdot 15/100 = 3$ dB. Hilfsmittel: read the per-100 m loss from the Kabeldämpfungsdiagramm Koaxialkabel (S.22), then scale linearly by length.",
"source": "https://50ohm.de/NEA_kabeldaempfung_1.html#EG313",
"confidence": 8
},
"EG314": {
"revision": 4,
- "explanation": "From the chart, the thin RG174 ($2.8$ mm) at $145$ MHz is about $40$ dB per $100$ m — twice as lossy as RG58. Scaled to $50$ m: $40 \\cdot 50/100 = 20$ dB. Thinner cable means higher loss. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "From the chart, the thin RG174 ($2.8$ mm) at $145$ MHz is about $40$ dB per $100$ m — twice as lossy as RG58. Scaled to $50$ m: $40 \\cdot 50/100 = 20$ dB. Thinner cable means higher loss. Hilfsmittel: read the per-100 m loss from the Kabeldämpfungsdiagramm Koaxialkabel (S.22), then scale linearly by length.",
"source": "https://50ohm.de/NEA_kabeldaempfung_1.html#EG314",
"confidence": 8
},
"EG315": {
"revision": 4,
- "explanation": "From the chart, the thick $12.7$ mm PE-foam cable at $435$ MHz is about $7$ dB per $100$ m. For $40$ m: $7 \\cdot 40/100 = 2.8$ dB. Larger diameter and foam dielectric keep the loss low even at UHF. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "From the chart, the thick $12.7$ mm PE-foam cable at $435$ MHz is about $7$ dB per $100$ m. For $40$ m: $7 \\cdot 40/100 = 2.8$ dB. Larger diameter and foam dielectric keep the loss low even at UHF. Hilfsmittel: read the per-100 m loss from the Kabeldämpfungsdiagramm Koaxialkabel (S.22), then scale linearly by length.",
"source": "https://50ohm.de/NEA_kabeldaempfung_1.html#EG315",
"confidence": 8
},
"EG316": {
"revision": 4,
- "explanation": "From the chart, the $10.3$ mm PE-foam cable at $1296$ MHz is about $20.5$ dB per $100$ m — loss climbs steeply with frequency. For $40$ m: $20.5 \\cdot 40/100 \\approx 8.2$ dB. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "From the chart, the $10.3$ mm PE-foam cable at $1296$ MHz is about $20.5$ dB per $100$ m — loss climbs steeply with frequency. For $40$ m: $20.5 \\cdot 40/100 \\approx 8.2$ dB. Hilfsmittel: read the per-100 m loss from the Kabeldämpfungsdiagramm Koaxialkabel (S.22), then scale linearly by length.",
"source": "https://50ohm.de/NEA_kabeldaempfung_1.html#EG316",
"confidence": 8
},
"EG401": {
"revision": 5,
- "explanation": "At SWR $= 3$ the voltage reflection coefficient is $\\Gamma = (S-1)/(S+1) = (3-1)/(3+1) = 0.5$. Reflected power scales as $\\Gamma^2 = 0.25$, so $0.25 \\cdot 100$ W $= 25$ W travels back toward the transmitter. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "At SWR $= 3$ the voltage reflection coefficient is $\\Gamma = (S-1)/(S+1) = (3-1)/(3+1) = 0.5$. Reflected power scales as $\\Gamma^2 = 0.25$, so $0.25 \\cdot 100$ W $= 25$ W travels back toward the transmitter. Hilfsmittel: first |r| = (s−1)/(s+1) (Reflexionsfaktor, S.17), then reflected power P_r = P_v·|r|² (S.17).",
"source": "https://50ohm.de/NEA_swr_2.html#EG401",
"confidence": 8
},
@@ -7315,7 +7315,7 @@
},
"EG404": {
"revision": 3,
- "explanation": "In normal coax operation, RF current flows on the inner conductor and the inside of the shield. Current on the outside of the shield is common-mode current, called Mantelstrom or mantle current; it can make the feed line radiate as part of the antenna. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "In normal coax operation, RF current flows on the inner conductor and the inside of the shield. Current on the outside of the shield is common-mode current, called Mantelstrom or mantle current; it can make the feed line radiate as part of the antenna.",
"source": "https://50ohm.de/NEA_mantelwellen_1.html#EG404",
"confidence": 8
},
@@ -7351,61 +7351,61 @@
},
"EG502": {
"revision": 4,
- "explanation": "Build EIRP in two steps. First get the real power at the antenna by subtracting feed-line losses from the transmitter output, $P_{\\text{Sender}} - P_{\\text{Verluste}}$; then multiply by the antenna gain factor $G$. Adding gain (options C/D) is the classic error — power and a linear gain factor multiply, not add (you would only add if everything were in dB). Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Build EIRP in two steps. First get the real power at the antenna by subtracting feed-line losses from the transmitter output, $P_{\\text{Sender}} - P_{\\text{Verluste}}$; then multiply by the antenna gain factor $G$. Adding gain (options C/D) is the classic error — power and a linear gain factor multiply, not add (you would only add if everything were in dB). Hilfsmittel: subtract feed-line loss, then multiply by the gain factor: the sheet gives P_ERP = P_S·10^((g_d−a)/10dB) and P_EIRP = P_ERP·1,64 (S.15).",
"source": "https://50ohm.de/NEA_aequivalente_isotrope_strahlungsleistung_eirp_2.html#EG502",
"confidence": 8
},
"EG503": {
"revision": 5,
- "explanation": "Convert the gain from dB to a factor: $26\\ \\text{dBi} \\to 10^{26/10} = 10^{2.6} \\approx 398$. Then EIRP $= 0.25\\ \\text{W} \\cdot 398 \\approx 100$ W. Shortcut: $26$ dB $= 20 + 6$ dB $= \\times 100 \\cdot \\times 4 = \\times 400$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Convert the gain from dB to a factor: $26\\ \\text{dBi} \\to 10^{26/10} = 10^{2.6} \\approx 398$. Then EIRP $= 0.25\\ \\text{W} \\cdot 398 \\approx 100$ W. Shortcut: $26$ dB $= 20 + 6$ dB $= \\times 100 \\cdot \\times 4 = \\times 400$. Hilfsmittel: first dBi→factor with G = 10^(g/10dB) (Pegel, S.15), then P_EIRP = P·G.",
"source": "https://50ohm.de/NEA_aequivalente_isotrope_strahlungsleistung_eirp_2.html#EG503",
"confidence": 8
},
"EG504": {
"revision": 5,
- "explanation": "Gain factor first: $36\\ \\text{dBi} \\to 10^{3.6} \\approx 3981$. EIRP $= 5\\ \\text{W} \\cdot 3981 \\approx 20000$ W. Shortcut: $36$ dB $= 30 + 6$ dB $= \\times 1000 \\cdot \\times 4 = \\times 4000$, and $5 \\cdot 4000 = 20000$ W. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Gain factor first: $36\\ \\text{dBi} \\to 10^{3.6} \\approx 3981$. EIRP $= 5\\ \\text{W} \\cdot 3981 \\approx 20000$ W. Shortcut: $36$ dB $= 30 + 6$ dB $= \\times 1000 \\cdot \\times 4 = \\times 4000$, and $5 \\cdot 4000 = 20000$ W. Hilfsmittel: first dBi→factor with G = 10^(g/10dB) (Pegel, S.15), then P_EIRP = P·G.",
"source": "https://50ohm.de/NEA_aequivalente_isotrope_strahlungsleistung_eirp_2.html#EG504",
"confidence": 8
},
"EG505": {
"revision": 5,
- "explanation": "Work in decibels, then convert once. Net gain toward isotropic $= 11\\ \\text{dBi} - 1\\ \\text{dB (cable)} = 10$ dB, which is a factor of $10$. So EIRP $= 100\\ \\text{W} \\cdot 10 = 1000$ W. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Work in decibels, then convert once. Net gain toward isotropic $= 11\\ \\text{dBi} - 1\\ \\text{dB (cable)} = 10$ dB, which is a factor of $10$. So EIRP $= 100\\ \\text{W} \\cdot 10 = 1000$ W. Hilfsmittel: net dB = gain − cable loss, then factor via G = 10^(g/10dB) (Pegel, S.15); ×10 = 10 dB.",
"source": "https://50ohm.de/NEA_aequivalente_isotrope_strahlungsleistung_eirp_2.html#EG505",
"confidence": 8
},
"EG506": {
"revision": 4,
- "explanation": "A dipole has $2.15$ dBi gain (factor $1.64$), and here the cable loss is also $2.15$ dB (factor $1.64$). Gain and loss are equal and opposite, so they cancel exactly: EIRP $= 75$ W, the same as the transmitter output. A neat reminder that a lossy-fed dipole radiates about its raw power in EIRP terms. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "A dipole has $2.15$ dBi gain (factor $1.64$), and here the cable loss is also $2.15$ dB (factor $1.64$). Gain and loss are equal and opposite, so they cancel exactly: EIRP $= 75$ W, the same as the transmitter output. A neat reminder that a lossy-fed dipole radiates about its raw power in EIRP terms. Hilfsmittel: g_i = g_d + 2,15 dB and the equal cable loss cancel (Pegel, S.15); EIRP stays at the TX power.",
"source": "https://50ohm.de/NEA_aequivalente_isotrope_strahlungsleistung_eirp_2.html#EG506",
"confidence": 8
},
"EG507": {
"revision": 5,
- "explanation": "Two effects in turn: $10$ dB of cable loss cuts $100$ W down to $10$ W at the antenna, and the dipole then adds $2.15$ dBi (factor $1.64$) toward isotropic. EIRP $= 10\\ \\text{W} \\cdot 1.64 \\approx 16.4$ W. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Two effects in turn: $10$ dB of cable loss cuts $100$ W down to $10$ W at the antenna, and the dipole then adds $2.15$ dBi (factor $1.64$) toward isotropic. EIRP $= 10\\ \\text{W} \\cdot 1.64 \\approx 16.4$ W. Hilfsmittel: 10 dB loss → ÷10 power, then ×1,64 for the dipole (g_i = 2,15 dB; Antennen, Pegel, S.15).",
"source": "https://50ohm.de/NEA_aequivalente_isotrope_strahlungsleistung_eirp_2.html#EG507",
"confidence": 8
},
"EG508": {
"revision": 5,
- "explanation": "The gain is given in dBd, so first convert: $5\\ \\text{dBd} = 7.15$ dBi. Net of the $2$ dB cable loss, $7.15 - 2 = 5.15$ dB, a factor of $10^{0.515} \\approx 3.28$. EIRP $= 5\\ \\text{W} \\cdot 3.28 \\approx 16.4$ W. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "The gain is given in dBd, so first convert: $5\\ \\text{dBd} = 7.15$ dBi. Net of the $2$ dB cable loss, $7.15 - 2 = 5.15$ dB, a factor of $10^{0.515} \\approx 3.28$. EIRP $= 5\\ \\text{W} \\cdot 3.28 \\approx 16.4$ W. Hilfsmittel: first dBd→dBi (g_i = g_d + 2,15 dB), subtract cable loss, then factor via G = 10^(g/10dB) (Pegel, S.15).",
"source": "https://50ohm.de/NEA_aequivalente_isotrope_strahlungsleistung_eirp_2.html#EG508",
"confidence": 8
},
"EG509": {
"revision": 5,
- "explanation": "Convert the dipole-referenced gain: $11\\ \\text{dBd} = 13.15$ dBi. After $1$ dB cable loss, $13.15 - 1 = 12.15$ dB $\\to 10^{1.215} \\approx 16.4$. EIRP $= 0.6\\ \\text{W} \\cdot 16.4 \\approx 9.8$ W. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Convert the dipole-referenced gain: $11\\ \\text{dBd} = 13.15$ dBi. After $1$ dB cable loss, $13.15 - 1 = 12.15$ dB $\\to 10^{1.215} \\approx 16.4$. EIRP $= 0.6\\ \\text{W} \\cdot 16.4 \\approx 9.8$ W. Hilfsmittel: first dBd→dBi (g_i = g_d + 2,15 dB), subtract cable loss, then factor via G = 10^(g/10dB) (Pegel, S.15).",
"source": "https://50ohm.de/NEA_aequivalente_isotrope_strahlungsleistung_eirp_2.html#EG509",
"confidence": 8
},
"EG510": {
"revision": 5,
- "explanation": "With $0$ dBd the antenna equals a dipole, i.e. $2.15$ dBi. Subtract the $1.5$ dB cable loss: $2.15 - 1.5 = 0.65$ dB $\\to 10^{0.065} \\approx 1.16$. EIRP $= 8.5\\ \\text{W} \\cdot 1.16 \\approx 9.9$ W. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "With $0$ dBd the antenna equals a dipole, i.e. $2.15$ dBi. Subtract the $1.5$ dB cable loss: $2.15 - 1.5 = 0.65$ dB $\\to 10^{0.065} \\approx 1.16$. EIRP $= 8.5\\ \\text{W} \\cdot 1.16 \\approx 9.9$ W. Hilfsmittel: 0 dBd = 2,15 dBi (g_i = g_d + 2,15 dB), subtract cable loss, then factor via G = 10^(g/10dB) (Pegel, S.15).",
"source": "https://50ohm.de/NEA_aequivalente_isotrope_strahlungsleistung_eirp_2.html#EG510",
"confidence": 8
},
"EG511": {
"revision": 4,
- "explanation": "The BEMFV $\\S\\,9$ notification (Anzeige) for a fixed station is required once radiated power reaches $10$ W EIRP, so stay at or below it. Convert the gain: $5.15\\ \\text{dBi} \\to 10^{0.515} \\approx 3.28$. Working back, $P_{\\text{max}} = 10\\ \\text{W} / 3.28 \\approx 3$ W of transmitter power keeps you under the limit. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "The BEMFV $\\S\\,9$ notification (Anzeige) for a fixed station is required once radiated power reaches $10$ W EIRP, so stay below it. Convert the gain: $5.15\\ \\text{dBi} \\to 10^{0.515} \\approx 3.28$. Working back, $P_{\\text{max}} = 10\\ \\text{W} / 3.28 \\approx 3$ W of transmitter power keeps you under the limit. Hilfsmittel: convert the gain to a factor via G = 10^(g/10dB) (Pegel, S.15), then back out P = EIRP/G from the 10 W limit.",
"source": "https://50ohm.de/NEA_aequivalente_isotrope_strahlungsleistung_eirp_2.html#EG511",
"confidence": 9
},
@@ -7609,13 +7609,13 @@
},
"EI103": {
"revision": 4,
- "explanation": "For an analog meter, first read the pointer as a fraction of full scale, then multiply by the selected range. The pointer is at about 29 percent of full scale, so on the 10 V range the value is $0.29 \\cdot 10 V = 2.9 V$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "For an analog meter, first read the pointer as a fraction of full scale, then multiply by the selected range. The pointer is at about 29 percent of full scale, so on the 10 V range the value is $0.29 \\cdot 10 V = 2.9 V$.",
"source": "https://50ohm.de/NEA_zeigerinstrumente_ablesen.html#EI103",
"confidence": 8
},
"EI104": {
"revision": 4,
- "explanation": "The pointer fraction is independent of the selected range. With the pointer at about 29 percent of full scale and the range set to 300 V, the reading is $0.29 \\cdot 300 V \\approx 88 V$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "The pointer fraction is independent of the selected range. With the pointer at about 29 percent of full scale and the range set to 300 V, the reading is $0.29 \\cdot 300 V \\approx 88 V$.",
"source": "https://50ohm.de/NEA_zeigerinstrumente_ablesen.html#EI104",
"confidence": 8
},
@@ -7627,7 +7627,7 @@
},
"EI202": {
"revision": 3,
- "explanation": "Find a resonant frequency either by measuring $L$ and $C$ and computing $f_0 = 1/(2\\pi\\sqrt{LC})$, or directly by sweeping the circuit with a VNA and watching for the resonance dip/peak. A plain frequency counter or DMM cannot find a passive circuit's resonance on its own. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Find a resonant frequency either by measuring $L$ and $C$ and computing $f_0 = 1/(2\\pi\\sqrt{LC})$, or directly by sweeping the circuit with a VNA and watching for the resonance dip/peak. A plain frequency counter or DMM cannot find a passive circuit's resonance on its own. Hilfsmittel: measure L and C and compute f₀ = 1/(2π·√(L·C)), S.14, or sweep with a VNA.",
"source": "https://50ohm.de/NEA_vna_1.html#EI202",
"confidence": 8
},
@@ -7657,13 +7657,13 @@
},
"EI301": {
"revision": 5,
- "explanation": "On an oscilloscope, time is read horizontally. One full sine period spans 8 divisions, and the timebase is 0.5 ms/div, so $T = 8 \\cdot 0.5 ms = 4 ms$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "On an oscilloscope, time is read horizontally. One full sine period spans 8 divisions, and the timebase is 0.5 ms/div, so $T = 8 \\cdot 0.5 ms = 4 ms$.",
"source": "https://50ohm.de/NEA_oszilloskop_1.html#EI301",
"confidence": 8
},
"EI302": {
"revision": 5,
- "explanation": "Frequency is the reciprocal of period: $f=1/T$. From the oscilloscope reading $T=4 ms = 0.004 s$, so $f=1/0.004 s = 250 Hz$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Frequency is the reciprocal of period: $f=1/T$. From the oscilloscope reading $T=4 ms = 0.004 s$, so $f=1/0.004 s = 250 Hz$. Hilfsmittel: apply f = 1/T (Wechselspannung, S.12).",
"source": "https://50ohm.de/NEA_oszilloskop_1.html#EI302",
"confidence": 8
},
@@ -7729,7 +7729,7 @@
},
"EI504": {
"revision": 4,
- "explanation": "A 10:1 prescaler (Frequenzteiler) divides the input frequency by ten before the counter sees it, so the display reads one tenth of the true value. Multiply back: $10 \\cdot 14.5625\\ \\text{MHz} = 145.625$ MHz. Prescalers are used precisely so a slower counter can measure high (VHF) frequencies. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "A 10:1 prescaler (Frequenzteiler) divides the input frequency by ten before the counter sees it, so the display reads one tenth of the true value. Multiply back: $10 \\cdot 14.5625\\ \\text{MHz} = 145.625$ MHz. Prescalers are used precisely so a slower counter can measure high (VHF) frequencies.",
"source": "https://50ohm.de/NEA_frequenzmessung_1.html#EI504",
"confidence": 8
},
@@ -7981,7 +7981,7 @@
},
"EJ218": {
"revision": 3,
- "explanation": "For FT8, JS8, PSK31 and similar modes, set the audio (NF) drive low enough that the ALC does not engage at all. ALC action on these constant-envelope digital signals causes distortion and splatter, so the clean operating point is just below the ALC threshold — not at maximum. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "For FT8, JS8, PSK31 and similar modes, set the audio (NF) drive low enough that the ALC does not engage at all. ALC action on these constant-envelope digital signals causes distortion and splatter, so the clean operating point is just below the ALC threshold — not at maximum.",
"source": "https://50ohm.de/NEA_digimod_uebersteuerung.html#EJ218",
"confidence": 8
},
@@ -8017,7 +8017,7 @@
},
"EK105": {
"revision": 3,
- "explanation": "At 80 m, a 3.65 m distance is still in the reactive near field. In this region electric and magnetic fields are not yet locked into a stable far-field wave impedance, so the simple far-field power-density approximation is invalid; use measurement, simulation, or near-field calculation. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "At 80 m, a 3.65 m distance is still in the reactive near field. In this region electric and magnetic fields are not yet locked into a stable far-field wave impedance, so the simple far-field power-density approximation is invalid; use measurement, simulation, or near-field calculation. Hilfsmittel: the far-field relation is invalid this close (valid only for d > λ/(2π); E = √(30Ω·P_EIRP)/d (Feldstärke im Fernfeld, S.15)).",
"source": "https://50ohm.de/NEA_naeherungsformel_1.html#EK105",
"confidence": 8
},
@@ -8035,7 +8035,7 @@
},
"EK108": {
"revision": 3,
- "explanation": "Build the net gain: $7.5\\ \\text{dBd} = 9.65$ dBi, minus $1.5$ dB cable loss $= 8.15$ dBi $\\to$ factor $\\approx 6.53$, so EIRP $= 100\\ \\text{W} \\cdot 6.53 \\approx 653$ W. The far-field distance for a field-strength limit is $d = \\sqrt{30 \\cdot P_{\\text{EIRP}}} / E = \\sqrt{30 \\cdot 653} / 28 \\approx 140 / 28 \\approx 5.0$ m. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Build the net gain: $7.5\\ \\text{dBd} = 9.65$ dBi, minus $1.5$ dB cable loss $= 8.15$ dBi $\\to$ factor $\\approx 6.53$, so EIRP $= 100\\ \\text{W} \\cdot 6.53 \\approx 653$ W. The far-field distance for a field-strength limit is $d = \\sqrt{30 \\cdot P_{\\text{EIRP}}} / E = \\sqrt{30 \\cdot 653} / 28 \\approx 140 / 28 \\approx 5.0$ m. Hilfsmittel: convert dBd→dBi (+2,15 dB) and subtract cable loss for the EIRP factor (S.15), then d = √(30·P_EIRP)/E (E = √(30Ω·P_EIRP)/d (Feldstärke im Fernfeld, S.15)).",
"source": "https://50ohm.de/NEA_naeherungsformel_1.html#EK108",
"confidence": 8
},
@@ -8125,19 +8125,19 @@
},
"NA201": {
"revision": 2,
- "explanation": "Electric potential difference is measured in volts; amperes measure current, ohms resistance, and ampere-hours charge capacity. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Electric potential difference is measured in volts; amperes measure current, ohms resistance, and ampere-hours charge capacity.",
"source": "https://50ohm.de/NEA_spannung.html#NA201",
"confidence": 8
},
"NA202": {
"revision": 2,
- "explanation": "Electric current is the rate of flow of charge, and the SI unit for current is the ampere. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Electric current is the rate of flow of charge, and the SI unit for current is the ampere.",
"source": "https://50ohm.de/NEA_strom.html#NA202",
"confidence": 8
},
"NA203": {
"revision": 2,
- "explanation": "Electrical resistance is measured in ohms, the unit that relates voltage and current through Ohm's law. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Electrical resistance is measured in ohms, the unit that relates voltage and current through Ohm's law.",
"source": "https://50ohm.de/NEA_ohmsches_gesetz.html#NA203",
"confidence": 8
},
@@ -8161,37 +8161,37 @@
},
"NA207": {
"revision": 2,
- "explanation": "One hertz means one cycle per second, so dimensionally $Hz = 1/s$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "One hertz means one cycle per second, so dimensionally $Hz = 1/s$.",
"source": "https://50ohm.de/NEA_frequenz.html#NA207",
"confidence": 8
},
"NA208": {
"revision": 3,
- "explanation": "Milli means $10^{-3}$, so one volt is 1000 millivolts and 4.2 V is 4200 mV. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Milli means $10^{-3}$, so one volt is 1000 millivolts and 4.2 V is 4200 mV. Hilfsmittel: use the Zehnerpotenzen/SI-Präfix table (S.11): milli = 10⁻³.",
"source": "https://50ohm.de/NEA_spannung.html#NA208",
"confidence": 8
},
"NA209": {
"revision": 3,
- "explanation": "Milli means $10^{-3}$; therefore 42 mA is $42/1000$ A, or 0.042 A. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Milli means $10^{-3}$; therefore 42 mA is $42/1000$ A, or 0.042 A. Hilfsmittel: use the Zehnerpotenzen/SI-Präfix table (S.11): milli = 10⁻³.",
"source": "https://50ohm.de/NEA_strom.html#NA209",
"confidence": 8
},
"NA210": {
"revision": 3,
- "explanation": "Milli means one thousandth, so one watt contains 1000 milliwatts. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Milli means one thousandth, so one watt contains 1000 milliwatts. Hilfsmittel: use the Zehnerpotenzen/SI-Präfix table (S.11): milli = 10⁻³.",
"source": "https://50ohm.de/NEA_leistung.html#NA210",
"confidence": 8
},
"NA211": {
"revision": 4,
- "explanation": "$0.010\\,\\mathrm{W} \\cdot 1000\\,\\mathrm{mW/W} = 10\\,\\mathrm{mW}$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "$0.010\\,\\mathrm{W} \\cdot 1000\\,\\mathrm{mW/W} = 10\\,\\mathrm{mW}$. Hilfsmittel: use the Zehnerpotenzen/SI-Präfix table (S.11): milli = 10⁻³.",
"source": "https://50ohm.de/NEA_leistung.html#NA211",
"confidence": 8
},
"NA212": {
"revision": 3,
- "explanation": "Mega means $10^6$; $144000000$ Hz divided by $10^6$ is 144 MHz. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Mega means $10^6$; $144000000$ Hz divided by $10^6$ is 144 MHz. Hilfsmittel: use the Zehnerpotenzen/SI-Präfix table (S.11): mega = 10⁶.",
"source": "https://50ohm.de/NEA_frequenz.html#NA212",
"confidence": 8
},
@@ -8245,19 +8245,19 @@
},
"NB204": {
"revision": 4,
- "explanation": "Series-connected cells add their voltages; six 1.5 V cells give $6 \\cdot 1.5 V = 9 V$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Series-connected cells add their voltages; six 1.5 V cells give $6 \\cdot 1.5 V = 9 V$. Hilfsmittel: series voltages add: U_G = U1+U2+… (Reihenschaltung, S.12).",
"source": "https://50ohm.de/NEA_batterien_und_akkus.html#NB204",
"confidence": 8
},
"NB205": {
"revision": 4,
- "explanation": "The voltmeter is connected across two 1.5 V cells in series, so it reads their sum: 3 V. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "The voltmeter is connected across two 1.5 V cells in series, so it reads their sum: 3 V. Hilfsmittel: series voltages add: U_G = U1+U2 (Reihenschaltung, S.12).",
"source": "https://50ohm.de/NEA_spannungsmessung.html#NB205",
"confidence": 7
},
"NB206": {
"revision": 4,
- "explanation": "Both meter leads are on points with the same potential in the shown circuit, so the potential difference is 0 V. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Both meter leads are on points with the same potential in the shown circuit, so the potential difference is 0 V.",
"source": "https://50ohm.de/NEA_spannungsmessung.html#NB206",
"confidence": 7
},
@@ -8275,13 +8275,13 @@
},
"NB302": {
"revision": 3,
- "explanation": "Use $f = c/\\lambda$: $300000000 / 2.08$ is about 144 MHz. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Use $f = c/\\lambda$: $300000000 / 2.08$ is about 144 MHz. Hilfsmittel: apply f = c/λ (c = f·λ (λ[m] ≈ 300/f[MHz]), S.17).",
"source": "https://50ohm.de/NEA_wellenlaenge.html#NB302",
"confidence": 8
},
"NB303": {
"revision": 3,
- "explanation": "Use $\\lambda = c/f$: $300000000 / 433500000$ is about 0.69 m. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Use $\\lambda = c/f$: $300000000 / 433500000$ is about 0.69 m. Hilfsmittel: apply λ = c/f (c = f·λ (λ[m] ≈ 300/f[MHz]), S.17).",
"source": "https://50ohm.de/NEA_wellenlaenge.html#NB303",
"confidence": 8
},
@@ -8323,25 +8323,25 @@
},
"NB501": {
"revision": 2,
- "explanation": "Ohm's law relates voltage, current and resistance as $U = R \\cdot I$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Ohm's law relates voltage, current and resistance as $U = R \\cdot I$. Hilfsmittel: apply Ohm's law U = R·I (Ohmsches Gesetz, S.11).",
"source": "https://50ohm.de/NEA_ohmsches_gesetz.html#NB501",
"confidence": 8
},
"NB502": {
"revision": 2,
- "explanation": "Rearranging Ohm's law gives current as voltage divided by resistance: $I = U/R$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Rearranging Ohm's law gives current as voltage divided by resistance: $I = U/R$. Hilfsmittel: rearrange Ohm's law → I = U/R (Ohmsches Gesetz, S.11).",
"source": "https://50ohm.de/NEA_ohmsches_gesetz.html#NB502",
"confidence": 8
},
"NB503": {
"revision": 2,
- "explanation": "Rearranging $U = R \\cdot I$ for resistance gives $R = U/I$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Rearranging $U = R \\cdot I$ for resistance gives $R = U/I$. Hilfsmittel: rearrange Ohm's law → R = U/I (Ohmsches Gesetz, S.11).",
"source": "https://50ohm.de/NEA_ohmsches_gesetz.html#NB503",
"confidence": 8
},
"NB504": {
"revision": 3,
- "explanation": "Using Ohm's law with the shown resistance, $U = R \\cdot I$ gives 9.000 V for 90 mA. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Using Ohm's law with the shown resistance, $U = R \\cdot I$ gives 9.000 V for 90 mA. Hilfsmittel: apply Ohm's law U = R·I (Ohmsches Gesetz, S.11).",
"source": "https://50ohm.de/NEA_ohmsches_gesetz.html#NB504",
"confidence": 8
},
@@ -8353,37 +8353,37 @@
},
"NB601": {
"revision": 3,
- "explanation": "DC input power is $P = U \\cdot I$, so $13.8 V \\cdot 1.5 A = 20.7 W$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "DC input power is $P = U \\cdot I$, so $13.8 V \\cdot 1.5 A = 20.7 W$. Hilfsmittel: apply P = U·I (Leistung, S.12).",
"source": "https://50ohm.de/NEA_leistung.html#NB601",
"confidence": 8
},
"NB602": {
"revision": 3,
- "explanation": "Power converted to heat is $P = U \\cdot I$; 50 V times 0.050 A gives 2.5 W. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Power converted to heat is $P = U \\cdot I$; 50 V times 0.050 A gives 2.5 W. Hilfsmittel: apply P = U·I (Leistung, S.12).",
"source": "https://50ohm.de/NEA_leistung.html#NB602",
"confidence": 8
},
"NB603": {
"revision": 3,
- "explanation": "20 mA is 0.020 A, and $3.2 V \\cdot 0.020 A = 0.064 W = 64.0 mW$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "20 mA is 0.020 A, and $3.2 V \\cdot 0.020 A = 0.064 W = 64.0 mW$. Hilfsmittel: apply P = U·I (Leistung, S.12).",
"source": "https://50ohm.de/NEA_leistung.html#NB603",
"confidence": 8
},
"NB604": {
"revision": 3,
- "explanation": "From $P = U \\cdot I$, current is $I = P/U = 100 W / 12 V = 8.33 A$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "From $P = U \\cdot I$, current is $I = P/U = 100 W / 12 V = 8.33 A$. Hilfsmittel: rearrange P = U·I → I = P/U (Leistung, S.12).",
"source": "https://50ohm.de/NEA_leistung.html#NB604",
"confidence": 8
},
"NB605": {
"revision": 3,
- "explanation": "A 3 W load at 12 V draws $I = P/U = 3/12 = 0.25 A$, which is 250 mA. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "A 3 W load at 12 V draws $I = P/U = 3/12 = 0.25 A$, which is 250 mA. Hilfsmittel: rearrange P = U·I → I = P/U (Leistung, S.12).",
"source": "https://50ohm.de/NEA_leistung.html#NB605",
"confidence": 8
},
"NB606": {
"revision": 3,
- "explanation": "A 48 W load at 12 V draws $I = P/U = 48/12 = 4 A$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "A 48 W load at 12 V draws $I = P/U = 48/12 = 4 A$. Hilfsmittel: rearrange P = U·I → I = P/U (Leistung, S.12).",
"source": "https://50ohm.de/NEA_leistung.html#NB606",
"confidence": 8
},
@@ -8449,19 +8449,19 @@
},
"NC108": {
"revision": 2,
- "explanation": "In the resistor tolerance colour code, silver denotes a tolerance of plus or minus 10 percent. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "In the resistor tolerance colour code, silver denotes a tolerance of plus or minus 10 percent. Hilfsmittel: read the Widerstands-Farbcode table (S.11): silver = ±10 %.",
"source": "https://50ohm.de/NEA_widerstandsfarbcode.html#NC108",
"confidence": 8
},
"NC109": {
"revision": 2,
- "explanation": "In the resistor tolerance colour code, gold denotes a tolerance of plus or minus 5 percent. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "In the resistor tolerance colour code, gold denotes a tolerance of plus or minus 5 percent. Hilfsmittel: read the Widerstands-Farbcode table (S.11): gold = ±5 %.",
"source": "https://50ohm.de/NEA_widerstandsfarbcode.html#NC109",
"confidence": 8
},
"NC110": {
"revision": 2,
- "explanation": "In the resistor tolerance colour code, brown denotes a tolerance of plus or minus 1 percent. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "In the resistor tolerance colour code, brown denotes a tolerance of plus or minus 1 percent. Hilfsmittel: read the Widerstands-Farbcode table (S.11): brown = ±1 %.",
"source": "https://50ohm.de/NEA_widerstandsfarbcode.html#NC110",
"confidence": 8
},
@@ -8515,7 +8515,7 @@
},
"ND102": {
"revision": 5,
- "explanation": "Mobile amateur transceivers are normally designed for vehicle electrical systems, so an external supply is usually around 13.8 V DC. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Mobile amateur transceivers are normally designed for vehicle electrical systems, so an external supply is usually around 13.8 V DC.",
"source": "https://50ohm.de/NEA_netzgeraet_1.html#ND102",
"confidence": 7
},
@@ -8683,7 +8683,7 @@
},
"NE305": {
"revision": 4,
- "explanation": "A 15 kHz-wide emission extends about half its bandwidth on each side of the centre frequency, so it needs at least 7.5 kHz clearance. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "A 15 kHz-wide emission extends about half its bandwidth on each side of the centre frequency, so it needs at least 7.5 kHz clearance.",
"source": "https://50ohm.de/NEA_bandbreite.html#NE305",
"confidence": 7
},
@@ -8923,7 +8923,7 @@
},
"NG104": {
"revision": 3,
- "explanation": "A Marconi antenna is a quarter-wave vertical worked against earth or a counterpoise, so it is a $\\lambda/4$ vertical antenna. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "A Marconi antenna is a quarter-wave vertical worked against earth or a counterpoise, so it is a $\\lambda/4$ vertical antenna.",
"source": "https://50ohm.de/NEA_rundstrahler.html#NG104",
"confidence": 7
},
@@ -9019,7 +9019,7 @@
},
"NG301": {
"revision": 3,
- "explanation": "An SWR of 1:1 means no reflected power from mismatch, which is the best possible match. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "An SWR of 1:1 means no reflected power from mismatch, which is the best possible match. Hilfsmittel: SWR 1 gives reflection factor |r| = (s−1)/(s+1) = 0 (Reflexion, S.17), so no reflected power — the ideal match.",
"source": "https://50ohm.de/NEA_swr.html#NG301",
"confidence": 8
},
@@ -9091,7 +9091,7 @@
},
"NH303": {
"revision": 3,
- "explanation": "The best VHF path is the station with the clearest quasi-optical path in the terrain profile; in the shown figure that is $\\text{E}_3$. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "The best VHF path is the station with the clearest quasi-optical path in the terrain profile; in the shown figure that is $\\text{E}_3$.",
"source": "https://50ohm.de/NEA_funkhorizont.html#NH303",
"confidence": 7
},
@@ -9151,7 +9151,7 @@
},
"NI203": {
"revision": 3,
- "explanation": "An ideal match has no reflected wave, so the SWR meter should read 1:1. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "An ideal match has no reflected wave, so the SWR meter should read 1:1.",
"source": "https://50ohm.de/NEA_swr.html#NI203",
"confidence": 8
},
@@ -9211,7 +9211,7 @@
},
"NK301": {
"revision": 4,
- "explanation": "Common electrical-safety practice treats contact above 50 V AC or 120 V DC as hazardous under normal dry conditions. Hilfsmittel: the formula needed here is printed in the Formelsammlung of the official exam aids (Hilfsmittel), so in the exam you apply it rather than recalling it from memory.",
+ "explanation": "Common electrical-safety practice treats contact above 50 V AC or 120 V DC as hazardous under normal dry conditions.",
"source": "https://50ohm.de/NEA_gefahren.html#NK301",
"confidence": 8
},
@@ -9937,259 +9937,259 @@
},
"VD701": {
"revision": 3,
- "explanation": "The Radio Regulations are the international framework, but you operate under German implementation. Memorise: RR may allow a band internationally; AFuV Anlage 1 tells you whether you may use it in Germany. Hilfsmittel: the frequency-allocation table in the official exam aids (Hilfsmittel) lists these band limits and usage parameters directly, so in the exam this is a table lookup rather than a memory item.",
+ "explanation": "The Radio Regulations are the international framework, but you operate under German implementation. Memorise: RR may allow a band internationally; AFuV Anlage 1 tells you whether you may use it in Germany.",
"source": "https://50ohm.de/NEA_frequenzzuteilung.html#VD701",
"confidence": 10
},
"VD702": {
"revision": 2,
- "explanation": "AFuV Anlage 1 is the national table for German amateur frequency ranges and detailed usage conditions. Hilfsmittel: the frequency-allocation table in the official exam aids (Hilfsmittel) lists these band limits and usage parameters directly, so in the exam this is a table lookup rather than a memory item.",
+ "explanation": "AFuV Anlage 1 is the national table for German amateur frequency ranges and detailed usage conditions.",
"source": "https://50ohm.de/NEA_frequenzzuteilung.html#VD702",
"confidence": 10
},
"VD703": {
"revision": 2,
- "explanation": "CB radio is outside the amateur frequency allocations, so an amateur station is not authorised to transmit there. Hilfsmittel: the frequency-allocation table in the official exam aids (Hilfsmittel) lists these band limits and usage parameters directly, so in the exam this is a table lookup rather than a memory item.",
+ "explanation": "CB radio is outside the amateur frequency allocations, so an amateur station is not authorised to transmit there.",
"source": "https://50ohm.de/NEA_nur_mit_afu_stellen.html#VD703",
"confidence": 10
},
"VD704": {
"revision": 3,
- "explanation": "Primary means priority user. A primary service may demand protection from secondary users, so secondary stations must move or stop if they cause trouble. Hilfsmittel: the frequency-allocation table in the official exam aids (Hilfsmittel) lists these band limits and usage parameters directly, so in the exam this is a table lookup rather than a memory item.",
+ "explanation": "Primary means priority user. A primary service may demand protection from secondary users, so secondary stations must move or stop if they cause trouble. Hilfsmittel: a text lookup — the definition of a primärer Funkdienst is in Anlage 1, Nutzungsbedingungen ¶3 (S. 1).",
"source": "https://50ohm.de/NEA_primaerer_sekundaerer_funkdienst.html#VD704",
"confidence": 10
},
"VD705": {
"revision": 3,
- "explanation": "Secondary has the weaker position: do not interfere, and do not expect protection. Good memory hook: secondary means 'no trouble caused, no protection claimed'. Hilfsmittel: the frequency-allocation table in the official exam aids (Hilfsmittel) lists these band limits and usage parameters directly, so in the exam this is a table lookup rather than a memory item.",
+ "explanation": "Secondary has the weaker position: do not interfere, and do not expect protection. Good memory hook: secondary means 'no trouble caused, no protection claimed'. Hilfsmittel: a text lookup — the definition of a sekundärer Funkdienst is in Anlage 1, Nutzungsbedingungen ¶3 (S. 1).",
"source": "https://50ohm.de/NEA_primaerer_sekundaerer_funkdienst.html#VD705",
"confidence": 10
},
"VD706": {
"revision": 2,
- "explanation": "AFuV Anlage 1 lists 7000-7200 kHz with primary status for the amateur service. Hilfsmittel: the frequency-allocation table in the official exam aids (Hilfsmittel) lists these band limits and usage parameters directly, so in the exam this is a table lookup rather than a memory item.",
+ "explanation": "AFuV Anlage 1 lists 7000-7200 kHz with primary status for the amateur service. Hilfsmittel: a table lookup — the Status column of the allocation table (Anlage 1, Tabellarische Übersicht, S. 2) marks 7000–7200 kHz as primary.",
"source": "https://50ohm.de/NEA_primaerer_sekundaerer_funkdienst.html#VD706",
"confidence": 10
},
"VD707": {
"revision": 2,
- "explanation": "A coastal station on its fixed assigned frequency cannot simply move; even in a shared primary band the amateur station must stop using that frequency unless a real emergency exists. Hilfsmittel: the frequency-allocation table in the official exam aids (Hilfsmittel) lists these band limits and usage parameters directly, so in the exam this is a table lookup rather than a memory item.",
+ "explanation": "A coastal station on its fixed assigned frequency cannot simply move; even in a shared primary band the amateur station must stop using that frequency unless a real emergency exists.",
"source": "https://50ohm.de/NEA_primaerer_sekundaerer_funkdienst.html#VD707",
"confidence": 9
},
"VD708": {
"revision": 2,
- "explanation": "The 433.05-434.79 MHz ISM designation means non-communication industrial, scientific, medical, domestic or similar applications also use that range. Hilfsmittel: the frequency-allocation table in the official exam aids (Hilfsmittel) lists these band limits and usage parameters directly, so in the exam this is a table lookup rather than a memory item.",
+ "explanation": "The 433.05-434.79 MHz ISM designation means non-communication industrial, scientific, medical, domestic or similar applications also use that range.",
"source": "https://50ohm.de/NEA_primaerer_sekundaerer_funkdienst.html#VD708",
"confidence": 10
},
"VD709": {
"revision": 3,
- "explanation": "This is the German 160 m band. Memorise it as the MF/low-HF entry: 1810-2000 kHz, just below the broadcast-style 2 MHz area. Hilfsmittel: the frequency-allocation table in the official exam aids (Hilfsmittel) lists these band limits and usage parameters directly, so in the exam this is a table lookup rather than a memory item.",
+ "explanation": "This is the German 160 m band. Memorise it as the MF/low-HF entry: 1810-2000 kHz, just below the broadcast-style 2 MHz area. Hilfsmittel: a table lookup, not a memory item — the band edges are listed in the allocation table (Anlage 1, Tabellarische Übersicht, S. 2).",
"source": "https://50ohm.de/NEA_amateurfunkbaender.html#VD709",
"confidence": 10
},
"VD710": {
"revision": 3,
- "explanation": "This is the 80 m band. The exam table value is 3.5-3.8 MHz; remember it as the first classic HF band after 160 m. Hilfsmittel: the frequency-allocation table in the official exam aids (Hilfsmittel) lists these band limits and usage parameters directly, so in the exam this is a table lookup rather than a memory item.",
+ "explanation": "This is the 80 m band. The exam table value is 3.5-3.8 MHz; remember it as the first classic HF band after 160 m. Hilfsmittel: a table lookup, not a memory item — the band edges are listed in the allocation table (Anlage 1, Tabellarische Übersicht, S. 2).",
"source": "https://50ohm.de/NEA_amateurfunkbaender.html#VD710",
"confidence": 10
},
"VD711": {
"revision": 3,
- "explanation": "This is the 40 m band. The German amateur range is 7.0-7.2 MHz; the clean round 7 MHz start makes it easy to anchor. Hilfsmittel: the frequency-allocation table in the official exam aids (Hilfsmittel) lists these band limits and usage parameters directly, so in the exam this is a table lookup rather than a memory item.",
+ "explanation": "This is the 40 m band. The German amateur range is 7.0-7.2 MHz; the clean round 7 MHz start makes it easy to anchor. Hilfsmittel: a table lookup, not a memory item — the band edges are listed in the allocation table (Anlage 1, Tabellarische Übersicht, S. 2).",
"source": "https://50ohm.de/NEA_amateurfunkbaender.html#VD711",
"confidence": 10
},
"VD712": {
"revision": 3,
- "explanation": "This is the 30 m WARC band, and it is deliberately narrow: 10.100-10.150 MHz. Memory hook: 30 m is the tiny 50 kHz slice at 10.1 MHz. Hilfsmittel: the frequency-allocation table in the official exam aids (Hilfsmittel) lists these band limits and usage parameters directly, so in the exam this is a table lookup rather than a memory item.",
+ "explanation": "This is the 30 m WARC band, and it is deliberately narrow: 10.100-10.150 MHz. Memory hook: 30 m is the tiny 50 kHz slice at 10.1 MHz. Hilfsmittel: a table lookup, not a memory item — the band edges are listed in the allocation table (Anlage 1, Tabellarische Übersicht, S. 2).",
"source": "https://50ohm.de/NEA_amateurfunkbaender.html#VD712",
"confidence": 10
},
"VD713": {
"revision": 3,
- "explanation": "This is the 20 m band. The table value is 14.000-14.350 MHz; remember 14 MHz as the main long-distance HF anchor. Hilfsmittel: the frequency-allocation table in the official exam aids (Hilfsmittel) lists these band limits and usage parameters directly, so in the exam this is a table lookup rather than a memory item.",
+ "explanation": "This is the 20 m band. The table value is 14.000-14.350 MHz; remember 14 MHz as the main long-distance HF anchor. Hilfsmittel: a table lookup, not a memory item — the band edges are listed in the allocation table (Anlage 1, Tabellarische Übersicht, S. 2).",
"source": "https://50ohm.de/NEA_amateurfunkbaender.html#VD713",
"confidence": 10
},
"VD714": {
"revision": 3,
- "explanation": "This is the 17 m WARC band. Like the other WARC bands, it is narrow: 18.068-18.168 MHz, a 100 kHz slice. Hilfsmittel: the frequency-allocation table in the official exam aids (Hilfsmittel) lists these band limits and usage parameters directly, so in the exam this is a table lookup rather than a memory item.",
+ "explanation": "This is the 17 m WARC band. Like the other WARC bands, it is narrow: 18.068-18.168 MHz, a 100 kHz slice. Hilfsmittel: a table lookup, not a memory item — the band edges are listed in the allocation table (Anlage 1, Tabellarische Übersicht, S. 2).",
"source": "https://50ohm.de/NEA_amateurfunkbaender.html#VD714",
"confidence": 10
},
"VD715": {
"revision": 3,
- "explanation": "This is the 15 m band. The range starts cleanly at 21 MHz and runs to 21.45 MHz. Hilfsmittel: the frequency-allocation table in the official exam aids (Hilfsmittel) lists these band limits and usage parameters directly, so in the exam this is a table lookup rather than a memory item.",
+ "explanation": "This is the 15 m band. The range starts cleanly at 21 MHz and runs to 21.45 MHz. Hilfsmittel: a table lookup, not a memory item — the band edges are listed in the allocation table (Anlage 1, Tabellarische Übersicht, S. 2).",
"source": "https://50ohm.de/NEA_amateurfunkbaender.html#VD715",
"confidence": 10
},
"VD716": {
"revision": 3,
- "explanation": "This is the 12 m WARC band. Memory hook: another narrow 100 kHz WARC slice, 24.89-24.99 MHz, just below 25 MHz. Hilfsmittel: the frequency-allocation table in the official exam aids (Hilfsmittel) lists these band limits and usage parameters directly, so in the exam this is a table lookup rather than a memory item.",
+ "explanation": "This is the 12 m WARC band. Memory hook: another narrow 100 kHz WARC slice, 24.89-24.99 MHz, just below 25 MHz. Hilfsmittel: a table lookup, not a memory item — the band edges are listed in the allocation table (Anlage 1, Tabellarische Übersicht, S. 2).",
"source": "https://50ohm.de/NEA_amateurfunkbaender.html#VD716",
"confidence": 10
},
"VD717": {
"revision": 3,
- "explanation": "This is the 10 m band. It is the broad class-N HF band too: 28-29.7 MHz. Hilfsmittel: the frequency-allocation table in the official exam aids (Hilfsmittel) lists these band limits and usage parameters directly, so in the exam this is a table lookup rather than a memory item.",
+ "explanation": "This is the 10 m band. It is the broad class-N HF band too: 28-29.7 MHz. Hilfsmittel: a table lookup, not a memory item — the band edges are listed in the allocation table (Anlage 1, Tabellarische Übersicht, S. 2).",
"source": "https://50ohm.de/NEA_amateurfunkbaender.html#VD717",
"confidence": 10
},
"VD718": {
"revision": 3,
- "explanation": "This is the 6 m band in Germany: 50-52 MHz. Remember it as the bridge between HF-like propagation and VHF regulation. Hilfsmittel: the frequency-allocation table in the official exam aids (Hilfsmittel) lists these band limits and usage parameters directly, so in the exam this is a table lookup rather than a memory item.",
+ "explanation": "This is the 6 m band in Germany: 50-52 MHz. Remember it as the bridge between HF-like propagation and VHF regulation. Hilfsmittel: a table lookup, not a memory item — the band edges are listed in the allocation table (Anlage 1, Tabellarische Übersicht, S. 2).",
"source": "https://50ohm.de/NEA_amateurfunkbaender.html#VD718",
"confidence": 10
},
"VD719": {
"revision": 3,
- "explanation": "This is the 2 m VHF band. The exam value is compact and round: 144-146 MHz. Hilfsmittel: the frequency-allocation table in the official exam aids (Hilfsmittel) lists these band limits and usage parameters directly, so in the exam this is a table lookup rather than a memory item.",
+ "explanation": "This is the 2 m VHF band. The exam value is compact and round: 144-146 MHz. Hilfsmittel: a table lookup, not a memory item — the band edges are listed in the allocation table (Anlage 1, Tabellarische Übersicht, S. 2).",
"source": "https://50ohm.de/NEA_amateurfunkbaender.html#VD719",
"confidence": 10
},
"VD720": {
"revision": 3,
- "explanation": "This is the 70 cm UHF band. Memorise the German allocation as the round 10 MHz block from 430 to 440 MHz. Hilfsmittel: the frequency-allocation table in the official exam aids (Hilfsmittel) lists these band limits and usage parameters directly, so in the exam this is a table lookup rather than a memory item.",
+ "explanation": "This is the 70 cm UHF band. Memorise the German allocation as the round 10 MHz block from 430 to 440 MHz. Hilfsmittel: a table lookup, not a memory item — the band edges are listed in the allocation table (Anlage 1, Tabellarische Übersicht, S. 2).",
"source": "https://50ohm.de/NEA_amateurfunkbaender.html#VD720",
"confidence": 10
},
"VD721": {
"revision": 3,
- "explanation": "This is the 23 cm band. The German amateur range is 1240-1300 MHz; remember it as the first microwave-style exam range after 70 cm. Hilfsmittel: the frequency-allocation table in the official exam aids (Hilfsmittel) lists these band limits and usage parameters directly, so in the exam this is a table lookup rather than a memory item.",
+ "explanation": "This is the 23 cm band. The German amateur range is 1240-1300 MHz; remember it as the first microwave-style exam range after 70 cm. Hilfsmittel: a table lookup, not a memory item — the band edges are listed in the allocation table (Anlage 1, Tabellarische Übersicht, S. 2).",
"source": "https://50ohm.de/NEA_amateurfunkbaender.html#VD721",
"confidence": 10
},
"VD722": {
"revision": 3,
- "explanation": "This is the 13 cm band. The exam range is 2320-2450 MHz, ending at the familiar 2.45 GHz ISM neighbourhood. Hilfsmittel: the frequency-allocation table in the official exam aids (Hilfsmittel) lists these band limits and usage parameters directly, so in the exam this is a table lookup rather than a memory item.",
+ "explanation": "This is the 13 cm band. The exam range is 2320-2450 MHz, ending at the familiar 2.45 GHz ISM neighbourhood. Hilfsmittel: a table lookup, not a memory item — the band edges are listed in the allocation table (Anlage 1, Tabellarische Übersicht, S. 2).",
"source": "https://50ohm.de/NEA_amateurfunkbaender.html#VD722",
"confidence": 10
},
"VD723": {
"revision": 3,
- "explanation": "Class N is the compact starter set: 10 m, 2 m and 70 cm. Memorise the three islands: 28-29.7 MHz, 144-146 MHz and 430-440 MHz. Hilfsmittel: the frequency-allocation table in the official exam aids (Hilfsmittel) lists these band limits and usage parameters directly, so in the exam this is a table lookup rather than a memory item.",
+ "explanation": "Class N is the compact starter set: 10 m, 2 m and 70 cm. Memorise the three islands: 28-29.7 MHz, 144-146 MHz and 430-440 MHz. Hilfsmittel: a table lookup, not a memory item — the bands open to class N are listed in the allocation table (Anlage 1, Tabellarische Übersicht, S. 2).",
"source": "https://50ohm.de/NEA_frequenz.html#VD723",
"confidence": 10
},
"VD724": {
"revision": 2,
- "explanation": "For class N on 2 m and 70 cm, AFuV Anlage 1 uses an EIRP cap of 10 W rather than a transmitter-output cap. Hilfsmittel: the frequency-allocation table in the official exam aids (Hilfsmittel) lists these band limits and usage parameters directly, so in the exam this is a table lookup rather than a memory item.",
+ "explanation": "For class N on 2 m and 70 cm, AFuV Anlage 1 uses an EIRP cap of 10 W rather than a transmitter-output cap. Hilfsmittel: the table prints 6,1 W ERP for class N on 2 m/70 cm (Anlage 1, S. 2); convert to EIRP with P_EIRP = P_ERP·1,64 (S. 15) → ≈10 W EIRP — not a direct table value.",
"source": "https://50ohm.de/NEA_sendeleistung_klasse_n.html#VD724",
"confidence": 10
},
"VD725": {
"revision": 2,
- "explanation": "EIRP is transmitter power times antenna gain relative to isotropic: $5 W \\cdot 2.5 = 12.5 W$, which exceeds the 10 W class N limit. Hilfsmittel: the frequency-allocation table in the official exam aids (Hilfsmittel) lists these band limits and usage parameters directly, so in the exam this is a table lookup rather than a memory item.",
+ "explanation": "EIRP is transmitter power times antenna gain relative to isotropic: $5 W \\cdot 2.5 = 12.5 W$, which exceeds the 10 W class N limit. Hilfsmittel: EIRP = transmitter power × isotropic gain factor (5 W × 2,5 = 12,5 W); the 10 W EIRP class-N limit comes from 6,1 W ERP × 1,64 (Anlage 1 S. 2 + conversion S. 15).",
"source": "https://50ohm.de/NEA_sendeleistung_klasse_n.html#VD725",
"confidence": 10
},
"VD726": {
"revision": 2,
- "explanation": "EIRP is transmitter power times antenna gain relative to isotropic: $5 W \\cdot 1.8 = 9 W$, which stays below the 10 W class N limit. Hilfsmittel: the frequency-allocation table in the official exam aids (Hilfsmittel) lists these band limits and usage parameters directly, so in the exam this is a table lookup rather than a memory item.",
+ "explanation": "EIRP is transmitter power times antenna gain relative to isotropic: $5 W \\cdot 1.8 = 9 W$, which stays below the 10 W class N limit. Hilfsmittel: EIRP = transmitter power × isotropic gain factor (5 W × 1,8 = 9 W); the 10 W EIRP class-N limit comes from 6,1 W ERP × 1,64 (Anlage 1 S. 2 + conversion S. 15).",
"source": "https://50ohm.de/NEA_sendeleistung_klasse_n.html#VD726",
"confidence": 10
},
"VD727": {
"revision": 2,
- "explanation": "AFuV Anlage 1 permits class E operation from 1810 to 1850 kHz with a maximum of 100 W PEP. Hilfsmittel: the frequency-allocation table in the official exam aids (Hilfsmittel) lists these band limits and usage parameters directly, so in the exam this is a table lookup rather than a memory item.",
+ "explanation": "AFuV Anlage 1 permits class E operation from 1810 to 1850 kHz with a maximum of 100 W PEP. Hilfsmittel: a table lookup, not a memory item — the per-class power limits are listed in the allocation table (Anlage 1, Tabellarische Übersicht, S. 2).",
"source": "https://50ohm.de/NEA_ausgangsleistung.html#VD727",
"confidence": 10
},
"VD728": {
"revision": 2,
- "explanation": "AFuV Anlage 1 lists 750 W PEP for class A in the 3.5-3.8 MHz band. Hilfsmittel: the frequency-allocation table in the official exam aids (Hilfsmittel) lists these band limits and usage parameters directly, so in the exam this is a table lookup rather than a memory item.",
+ "explanation": "AFuV Anlage 1 lists 750 W PEP for class A in the 3.5-3.8 MHz band. Hilfsmittel: a table lookup, not a memory item — the per-class power limits are listed in the allocation table (Anlage 1, Tabellarische Übersicht, S. 2).",
"source": "https://50ohm.de/NEA_ausgangsleistung.html#VD728",
"confidence": 10
},
"VD729": {
"revision": 2,
- "explanation": "AFuV Anlage 1 gives 3.5-3.8 MHz limits of 750 W PEP for class A and 100 W PEP for class E. Hilfsmittel: the frequency-allocation table in the official exam aids (Hilfsmittel) lists these band limits and usage parameters directly, so in the exam this is a table lookup rather than a memory item.",
+ "explanation": "AFuV Anlage 1 gives 3.5-3.8 MHz limits of 750 W PEP for class A and 100 W PEP for class E. Hilfsmittel: a table lookup, not a memory item — the per-class power limits are listed in the allocation table (Anlage 1, Tabellarische Übersicht, S. 2).",
"source": "https://50ohm.de/NEA_ausgangsleistung.html#VD729",
"confidence": 10
},
"VD730": {
"revision": 2,
- "explanation": "AFuV Anlage 1 limits class A in 10.1-10.15 MHz to 150 W PEP. Hilfsmittel: the frequency-allocation table in the official exam aids (Hilfsmittel) lists these band limits and usage parameters directly, so in the exam this is a table lookup rather than a memory item.",
+ "explanation": "AFuV Anlage 1 limits class A in 10.1-10.15 MHz to 150 W PEP. Hilfsmittel: a table lookup, not a memory item — the per-class power limits are listed in the allocation table (Anlage 1, Tabellarische Übersicht, S. 2).",
"source": "https://50ohm.de/NEA_ausgangsleistung.html#VD730",
"confidence": 10
},
"VD731": {
"revision": 2,
- "explanation": "AFuV Anlage 1 lists 750 W PEP for class A on both 14.000-14.350 MHz and 18.068-18.168 MHz. Hilfsmittel: the frequency-allocation table in the official exam aids (Hilfsmittel) lists these band limits and usage parameters directly, so in the exam this is a table lookup rather than a memory item.",
+ "explanation": "AFuV Anlage 1 lists 750 W PEP for class A on both 14.000-14.350 MHz and 18.068-18.168 MHz. Hilfsmittel: a table lookup, not a memory item — the per-class power limits are listed in the allocation table (Anlage 1, Tabellarische Übersicht, S. 2).",
"source": "https://50ohm.de/NEA_ausgangsleistung.html#VD731",
"confidence": 10
},
"VD732": {
"revision": 2,
- "explanation": "AFuV Anlage 1 lists 750 W PEP for class A on both 21.000-21.450 MHz and 24.890-24.990 MHz. Hilfsmittel: the frequency-allocation table in the official exam aids (Hilfsmittel) lists these band limits and usage parameters directly, so in the exam this is a table lookup rather than a memory item.",
+ "explanation": "AFuV Anlage 1 lists 750 W PEP for class A on both 21.000-21.450 MHz and 24.890-24.990 MHz. Hilfsmittel: a table lookup, not a memory item — the per-class power limits are listed in the allocation table (Anlage 1, Tabellarische Übersicht, S. 2).",
"source": "https://50ohm.de/NEA_ausgangsleistung.html#VD732",
"confidence": 10
},
"VD733": {
"revision": 2,
- "explanation": "AFuV Anlage 1 gives 21 MHz and 28 MHz limits of 750 W PEP for class A and 100 W PEP for class E. Hilfsmittel: the frequency-allocation table in the official exam aids (Hilfsmittel) lists these band limits and usage parameters directly, so in the exam this is a table lookup rather than a memory item.",
+ "explanation": "AFuV Anlage 1 gives 21 MHz and 28 MHz limits of 750 W PEP for class A and 100 W PEP for class E. Hilfsmittel: a table lookup, not a memory item — the per-class power limits are listed in the allocation table (Anlage 1, Tabellarische Übersicht, S. 2).",
"source": "https://50ohm.de/NEA_ausgangsleistung.html#VD733",
"confidence": 10
},
"VD734": {
"revision": 2,
- "explanation": "AFuV Anlage 1 gives 144-146 MHz and 430-440 MHz limits of 750 W PEP for class A and 75 W PEP for class E. Hilfsmittel: the frequency-allocation table in the official exam aids (Hilfsmittel) lists these band limits and usage parameters directly, so in the exam this is a table lookup rather than a memory item.",
+ "explanation": "AFuV Anlage 1 gives 144-146 MHz and 430-440 MHz limits of 750 W PEP for class A and 75 W PEP for class E. Hilfsmittel: a table lookup, not a memory item — the per-class power limits are listed in the allocation table (Anlage 1, Tabellarische Übersicht, S. 2).",
"source": "https://50ohm.de/NEA_ausgangsleistung.html#VD734",
"confidence": 10
},
"VD735": {
"revision": 2,
- "explanation": "AFuV Anlage 1 allows class A up to 750 W PEP at 1240-1300 MHz but adds a special 5 W EIRP cap in 1247-1263 MHz. Hilfsmittel: the frequency-allocation table in the official exam aids (Hilfsmittel) lists these band limits and usage parameters directly, so in the exam this is a table lookup rather than a memory item.",
+ "explanation": "AFuV Anlage 1 allows class A up to 750 W PEP at 1240-1300 MHz but adds a special 5 W EIRP cap in 1247-1263 MHz. Hilfsmittel: 750 W PEP for class A at 1240–1300 MHz is in the table (Anlage 1, S. 2); the special 1247–1263 MHz limit is 3,05 W ERP in the Zusätzliche Nutzungsbestimmungen (S. 3) — i.e. ≈5 W EIRP via ×1,64 (S. 15).",
"source": "https://50ohm.de/NEA_ausgangsleistung.html#VD735",
"confidence": 10
},
"VD736": {
"revision": 2,
- "explanation": "For class A between 1300 MHz and 250 GHz, AFuV Anlage 1 lists a maximum transmitter output of 75 W PEP. Hilfsmittel: the frequency-allocation table in the official exam aids (Hilfsmittel) lists these band limits and usage parameters directly, so in the exam this is a table lookup rather than a memory item.",
+ "explanation": "For class A between 1300 MHz and 250 GHz, AFuV Anlage 1 lists a maximum transmitter output of 75 W PEP. Hilfsmittel: a table lookup, not a memory item — the per-class power limits are listed in the allocation table (Anlage 1, Tabellarische Übersicht, S. 2).",
"source": "https://50ohm.de/NEA_ausgangsleistung.html#VD736",
"confidence": 10
},
"VD737": {
"revision": 2,
- "explanation": "For class E between 1300 MHz and 250 GHz, AFuV Anlage 1 lists a maximum transmitter output of 5 W PEP. Hilfsmittel: the frequency-allocation table in the official exam aids (Hilfsmittel) lists these band limits and usage parameters directly, so in the exam this is a table lookup rather than a memory item.",
+ "explanation": "For class E between 1300 MHz and 250 GHz, AFuV Anlage 1 lists a maximum transmitter output of 5 W PEP. Hilfsmittel: a table lookup, not a memory item — the per-class power limits are listed in the allocation table (Anlage 1, Tabellarische Übersicht, S. 2).",
"source": "https://50ohm.de/NEA_ausgangsleistung.html#VD737",
"confidence": 10
},
"VD738": {
"revision": 2,
- "explanation": "AFuV Anlage 1 sets the narrow 800 Hz occupied-bandwidth limit for 135.7-137.8 kHz, 472-479 kHz and 10.100-10.150 MHz. Hilfsmittel: the frequency-allocation table in the official exam aids (Hilfsmittel) lists these band limits and usage parameters directly, so in the exam this is a table lookup rather than a memory item.",
+ "explanation": "AFuV Anlage 1 sets the narrow 800 Hz occupied-bandwidth limit for 135.7-137.8 kHz, 472-479 kHz and 10.100-10.150 MHz. Hilfsmittel: a lookup — the maximum occupied bandwidth per band is in the Zusätzliche Nutzungsbestimmungen (Anlage 1, S. 3), not in the page-2 table.",
"source": "https://50ohm.de/NEA_bandbreite.html#VD738",
"confidence": 10
},
"VD739": {
"revision": 2,
- "explanation": "AFuV Anlage 1 gives 3.5-3.8 MHz a maximum occupied bandwidth of 2.7 kHz. Hilfsmittel: the frequency-allocation table in the official exam aids (Hilfsmittel) lists these band limits and usage parameters directly, so in the exam this is a table lookup rather than a memory item.",
+ "explanation": "AFuV Anlage 1 gives 3.5-3.8 MHz a maximum occupied bandwidth of 2.7 kHz. Hilfsmittel: a lookup — the maximum occupied bandwidth per band is in the Zusätzliche Nutzungsbestimmungen (Anlage 1, S. 3), not in the page-2 table.",
"source": "https://50ohm.de/NEA_bandbreite.html#VD739",
"confidence": 10
},
"VD740": {
"revision": 2,
- "explanation": "AFuV Anlage 1 gives 28.000-29.000 MHz a maximum occupied bandwidth of 7 kHz. Hilfsmittel: the frequency-allocation table in the official exam aids (Hilfsmittel) lists these band limits and usage parameters directly, so in the exam this is a table lookup rather than a memory item.",
+ "explanation": "AFuV Anlage 1 gives 28.000-29.000 MHz a maximum occupied bandwidth of 7 kHz. Hilfsmittel: a lookup — the maximum occupied bandwidth per band is in the Zusätzliche Nutzungsbestimmungen (Anlage 1, S. 3), not in the page-2 table.",
"source": "https://50ohm.de/NEA_bandbreite.html#VD740",
"confidence": 10
},
"VD741": {
"revision": 2,
- "explanation": "AFuV Anlage 1 gives 144-146 MHz a maximum occupied bandwidth of 40 kHz. Hilfsmittel: the frequency-allocation table in the official exam aids (Hilfsmittel) lists these band limits and usage parameters directly, so in the exam this is a table lookup rather than a memory item.",
+ "explanation": "AFuV Anlage 1 gives 144-146 MHz a maximum occupied bandwidth of 40 kHz. Hilfsmittel: a lookup — the maximum occupied bandwidth per band is in the Zusätzliche Nutzungsbestimmungen (Anlage 1, S. 3), not in the page-2 table.",
"source": "https://50ohm.de/NEA_bandbreite.html#VD741",
"confidence": 10
},
"VD742": {
"revision": 2,
- "explanation": "AFuV Anlage 1 gives 430-440 MHz a 2 MHz occupied-bandwidth limit, with 7 MHz allowed for AM television transmissions. Hilfsmittel: the frequency-allocation table in the official exam aids (Hilfsmittel) lists these band limits and usage parameters directly, so in the exam this is a table lookup rather than a memory item.",
+ "explanation": "AFuV Anlage 1 gives 430-440 MHz a 2 MHz occupied-bandwidth limit, with 7 MHz allowed for AM television transmissions. Hilfsmittel: a lookup — the maximum occupied bandwidth per band is in the Zusätzliche Nutzungsbestimmungen (Anlage 1, S. 3), not in the page-2 table.",
"source": "https://50ohm.de/NEA_bandbreite.html#VD742",
"confidence": 10
},
"VD743": {
"revision": 2,
- "explanation": "AFuV Anlage 1 caps class N in the 10 m band at 10 W ERP. Hilfsmittel: the frequency-allocation table in the official exam aids (Hilfsmittel) lists these band limits and usage parameters directly, so in the exam this is a table lookup rather than a memory item.",
+ "explanation": "AFuV Anlage 1 caps class N in the 10 m band at 10 W ERP. Hilfsmittel: a table lookup — the 10 m class-N limit (10 W ERP) is in the allocation table (Anlage 1, Tabellarische Übersicht, S. 2).",
"source": "https://50ohm.de/NEA_sendeleistung_klasse_n.html#VD743",
"confidence": 10
},
@@ -10499,4 +10499,4 @@
"source": "https://50ohm.de/NEA_antennen_baurecht_haftung.html#VE707",
"confidence": 8
}
-}
+}
\ No newline at end of file
diff --git a/references/Hilfsmittel.md b/references/Hilfsmittel.md
new file mode 100644
index 0000000..d4c4793
--- /dev/null
+++ b/references/Hilfsmittel.md
@@ -0,0 +1,212 @@
+# Hilfsmittel — official exam aid (Formelsammlung + tables)
+
+Catalog of everything in the BNetzA exam aid sheet that candidates are
+**allowed to use during the exam**, with the **printed page numbers** as
+they appear on each page. This file is the source of truth for the
+`Hilfsmittel:` notes appended to explanations (see
+`EXPLANATIONS.md`): a note may only cite a formula/table that is listed
+here, and should name it and its page.
+
+- Source PDF: `Hilfsmittel_12062024.pdf`
+ (https://www.bundesnetzagentur.de/SharedDocs/Downloads/DE/Sachgebiete/Telekommunikation/Unternehmen_Institutionen/Frequenzen/Amateurfunk/AntraegeFormulare/Hilfsmittel_12062024.pdf?__blob=publicationFile&v=3)
+- **Page numbering:** the printed page number = PDF page − 2 (the cover
+ and the "Hinweis" page are unnumbered). All "S. N" below are the
+ printed numbers.
+- Errata (per the PDF "Hinweis" page): on S. 15 the parabolic-dish gain
+ formula, on S. 17 the SWR formula, and on S. 21 the specific-resistance
+ table (Zink → Zinn) were corrected against the 3rd-edition catalog.
+
+If the fact a question needs is **not** in this list (e.g. diode forward
+voltages ~0.6 V / ~0.3 V, tan δ = 1/Q, semiconductor behaviour,
+definitions), it is a memory item — **do not** add a Hilfsmittel note.
+
+---
+
+## S. 1–3 — Frequenzbereichszuweisung (Anlage 1, Amateurfunkverordnung)
+
+- **S. 1 — Nutzungsbedingungen (¶1–4):** general usage rules (incl. the
+ **50 W ERP** cap for remote/automatic terrestrial stations above 30 MHz,
+ with a possible **1000 W ERP** exception for links) and, in ¶3, the
+ **definitions** of a *primärer* and *sekundärer Funkdienst* and their
+ protection/priority rights. This is a *text* page, not a table. (No
+ 750 W figure appears here — 750 W PEP is a class-A limit in the S. 2
+ table.)
+- **S. 2 — Tabellarische Übersicht:** per frequency band: status
+ (primary/secondary), the allowed frequency ranges, max. **power per
+ licence class (A/E/N) — in PEP or ERP exactly as printed** (not EIRP),
+ and additional-usage references.
+ Note the units: class N on **10 m is 10 W ERP**, but on **2 m/70 cm it
+ is 6,1 W ERP** (= ≈10 W EIRP after ×1,64, see S. 15) — the table does
+ *not* print "10 W EIRP". Power limits are in PEP/ERP, never EIRP.
+- **S. 3 — Zusätzliche Nutzungsbestimmungen (numbered conditions):**
+ max. occupied **bandwidth** per band (800 Hz, 2,7 kHz, 7 kHz, 40 kHz,
+ 2 MHz …), and special power limits such as the 1247–1263 MHz
+ **3,05 W ERP** sub-band cap (= ≈5 W EIRP). The bandwidth values live in
+ these numbered conditions, not in the S. 2 table.
+
+Not in Anlage 1: the general HF/VHF/UHF (KW/UKW) range *terminology*,
+the ISM-band *definition*, and foreign-country prefixes (Landeskenner) —
+those are memory items.
+
+## S. 4–7 — Rufzeichenplan für den Amateurfunkdienst in Deutschland
+
+- **S. 4** — intro: structure of German call signs (prefix DA–DR excl.
+ DE/DI, digit, 2–3-char suffix).
+- **S. 5 — Tabelle "Rufzeichen mit 2- oder 3-buchstabigen Suffixen":**
+ the main **call-sign-series → class/use** table. DA0 = Klubstation;
+ DL1–DL9 = person-bound class A; DO1–DO9 = class E; DN9 = class N;
+ DP0–DP1 = exterritorial class A; etc.
+- **S. 6** — club-station suffixes (1-char) table.
+- **S. 7** — club-station suffixes (4–7-digit); **§ 5** Kurzzeitzulassungen
+ ausländischer Funkamateure (visitors in Germany sign **DL/** class A,
+ **DO/** class E); **§ 6** Kennungen zum Betrieb leistungsschwacher Sender
+ (beacon/peil identifiers **MO, MOE, MOI, MOS, MOH, MO5**); §§ 7–8 rules.
+
+## S. 8 — International gebräuchliche Rufzeichenzusätze / Ausbildung / Remote
+
+- **§ 9** operating suffixes appended with `/`: `/m` (mobile, incl.
+ inland-waterway), `/mm` (maritime mobile), `/am` (aeronautical mobile),
+ `/p` (portable/temporary fixed; optional in Germany), etc.
+- **§ 10** Ausbildungsfunkbetrieb (`/Trainee` speech, `/T` CW/digital).
+- **§ 11** Remotebetrieb (`Remote` speech, `/R` CW/digital).
+
+Foreign-country prefixes (e.g. Swiss HB3/HB9 for a German operating
+*abroad*) are **not** in this plan — memory item.
+
+## S. 9–10 — IARU-Bandplan
+
+- **S. 9 — 2 m** and **S. 10 — 70 cm**: per segment, max bandwidth,
+ preferred mode and usage (CW, SSB, FM, digital, beacons, satellite…).
+
+## S. 11 — Formelsammlung: Grundlagen / Widerstände
+
+- Zehnerpotenzen & SI-Präfixe (p…T) — table.
+- Zweierpotenzen / Bit (2^0…2^12) — table.
+- **Ohmsches Gesetz:** `U = R·I`, `R = U/I`, `I = U/R`.
+- **Innenwiderstand:** `R_i = ΔU/ΔI`.
+- **Widerstand von Drähten:** `R = ρ·l/A_Dr`, `A_Dr = d²·π/4 = r²·π`.
+- **Widerstands-Farbcode** — table (Farbe / Wert / Multiplikator / Toleranz).
+
+## S. 12 — Widerstandsnetzwerke / Leistung / Wechselspannung
+
+- **Reihenschaltung:** `R_G = R1+R2+…+R_N`; 2 R: `R_G = R1+R2`.
+- **Parallelschaltung:** `1/R_G = 1/R1+1/R2+…`; 2 R: `R_G = R1·R2/(R1+R2)`.
+- **Spannungsteiler (unbelastet):** `U_G = U1+U2`, `U1/U2 = R1/R2`,
+ `U2/U_G = R2/(R1+R2)`.
+- **Stromteiler:** `I_G = I1+I2`, `I2/I1 = R1/R2`.
+- **Vorzugsreihen** E6/E12/E24 — table.
+- **Leistung:** `P = U·I = U²/R = I²·R`; `U = P/I = √(P·R)`;
+ `I = P/U = √(P/R)`.
+- **Arbeit/Energie:** `W = P·t`.
+- **Wirkungsgrad:** `η = P_ab/P_zu (·100 %)`; `P_ab = P_zu − P_V`.
+- **Wechselspannung:** `Û = U_eff·√2`, `U_SS = 2·Û`, `ω = 2·π·f`,
+ `Z = √(R²+X²)`, `T = 1/f`, `f = 1/T`.
+
+## S. 13 — Induktivität / Transformator / Kapazität
+
+- **Induktiver Blindwiderstand:** `X_L = ω·L`.
+- **L Reihe:** `L_G = L1+…+L_N`; **L Parallel:** `1/L_G = 1/L1+…`.
+- **Ringspule / lange Zylinderspule:** `L = μ0·μr·N²·A_S / l(_m)`.
+- **Ringkernspulen (auch mehrlagig):** `L = N²·A_L`.
+- **Magnetische Feldstärke (Ringspule):** `H = I·N/l_m`.
+- **Magnetische Flussdichte:** `B_m = μr·μ0·H`.
+- **Transformator-Übersetzungsverhältnis:**
+ `ü = N_P/N_S = U_P/U_S = I_S/I_P = √(Z_P/Z_S)`.
+- **Belastbarkeit von Wicklungen:** `I = S·A_Dr` mit `S ≈ 2,5 A/mm²`.
+- **Kapazitiver Blindwiderstand:** `X_C = 1/(ω·C)`.
+- **C Reihe:** `1/C_G = 1/C1+…`; **C Parallel:** `C_G = C1+…`.
+- **E-Feld im homogenen Feld:** `E = U/d`.
+- **Kapazität Kondensator:** `C = ε0·εr·A/d`.
+
+## S. 14 — Filter / Schwingkreis / Transistor / ZF & Spiegelfrequenz
+
+- **RC TP/HP:** `f_g = 1/(2π·R·C)`; **RL TP/HP:** `f_g = R/(2π·L)`
+ (`f_g` = −3 dB-Grenzfrequenz).
+- **Schwingkreis:** `f0 = 1/(2π·√(L·C))`, Resonanz `X_C = X_L`.
+- **Reihenschwingkreis:** `B = R_s/(2π·L)`, `Q = f0/B = X_L/R_s`.
+- **Parallelschwingkreis:** `B = 1/(2π·R_p·C)`, `Q = f0/B = R_p/X_L`.
+- **Transistor (DC):** `B = I_C/I_B`, `I_E = I_C+I_B`.
+- **Transistor (AC):** `v_I = β = ΔI_C/ΔI_B`, `v_U = β = ΔU_CE/ΔU_BE`,
+ `v_P = β² = v_U·v_I`. *(Transcribed verbatim. The printed
+ `v_U = β` is physically dubious — voltage gain is not generally the
+ current gain — but the aid prints it this way; do not silently
+ "correct" it.)*
+- **Zwischenfrequenz:** `f_ZF = |f_E − f_OSZ|` (the sheet notes that the
+ `f_ZF = f_E + f_OSZ` case is not considered).
+- **Spiegelfrequenz:** `f_S = 2·f_OSZ − f_E`.
+
+## S. 15 — Pegel / Antennen
+
+- **Pegel:** `p = 10·log10(P/1mW)` dBm; `p = 10·log10(P/1W)` dBW;
+ `u = 20·log10(U/0,775V)` dBu.
+- **Verstärkung/Gewinn:** `g = 10·log10(P2/P1)` dB;
+ `g = 20·log10(U2/U1)` dB.
+- **Dämpfung/Verluste:** `a = 10·log10(P1/P2)`; `a = 20·log10(U1/U2)`.
+- **dB ↔ Verhältnis-Tabelle** — exactly **11 rows** (Leistungs- /
+ Spannungsverhältnis): −20 (0,01/0,1), −10 (0,1/0,32), −6 (0,25/0,5),
+ −3 (0,5/0,71), −1 (0,79/0,89), 0 (1/1), +1 (1,26/1,12), +3 (2/1,41),
+ +6 (4/2), +10 (10/3,16), +20 (100/10). **There is no 16 dB row** (or
+ any value off this list): combine rows, e.g. 16 dB = +10 dB (×10) +
+ +6 dB (×4) → ×40.
+- **ERP:** `p_ERP = p_S − a + g_d`; `P_ERP = P_S·10^((g_d−a)/10dB)`.
+- **EIRP:** `p_EIRP = p_ERP + 2,15 dB`; `P_EIRP = P_ERP·1,64`.
+- **Feldstärke Fernfeld:** `E = √(30Ω·P_A·G_i)/d = √(30Ω·P_EIRP)/d`
+ (ab `d > λ/(2π)`).
+- **Gewinn:** `G_i = G_d·1,64`, `g_i = g_d + 2,15 dB`, `G = 10^(g/10dB)`.
+- Halbwellendipol `G_i=1,64 / g_i=2,15 dB`; λ/4-Vertikal mit
+ Bodenreflexion `G_i=3,28 / g_i=5,15 dB`; Parabolspiegel
+ `g_i = 10·log10[(π·d/λ)²·η] dB`.
+
+## S. 16 — Rauschen / Amplitudenmodulation / Frequenzmodulation
+
+- **Thermisches Rauschen:** `P_R = k·T_K·B`; `Δp_R = 10·log10(B1/B2)`;
+ `U_R = 2·√(P_R·R)`.
+- **Rauschzahl:** `F = (P_S/P_N)_Eingang/(P_S/P_N)_Ausgang`;
+ `a_F = 10·log10(F)`; `a_F = SNR_Eingang − SNR_Ausgang`.
+- **SNR:** `SNR = 10·log10(P_S/P_N) = 20·log10(U_S/U_N)`.
+- **Shannon-Hartley:** `C = B/1Hz · log2(1 + P_S/P_N)` bit/s.
+- **log2:** `log2(x) = log10(x)/log10(2)`.
+- **AM Modulationsgrad:** `m = Û_mod/Û_T`; **AM Bandbreite:**
+ `B = 2·f_mod,max`.
+- **FM Modulationsindex:** `m = Δf_T/f_mod`; **Carson-Bandbreite:**
+ `B ≈ 2·(Δf_T + f_mod,max)`.
+
+## S. 17 — Wellenlänge & Frequenz / Reflexion / Wellenwiderstand
+
+- **c = f·λ**, `f = c/λ`, `λ = c/f`; `c0 ≈ 3·10⁸ m/s`;
+ `f[MHz] ≈ 300/λ[m]`, `λ[m] ≈ 300/f[MHz]`.
+- **Verkürzungsfaktor:** `k_v = l_G/l_E = 1/√εr = c/c0`.
+- **Stehwellenverhältnis (SWR):**
+ `s = U_max/U_min = (U_v+U_r)/(U_v−U_r) = (√P_v+√P_r)/(√P_v−√P_r) = (1+|r|)/(1−|r|)`;
+ `s = R2/Z` (R2>Z) bzw. `s = Z/R2` (R22,5)
+ `Z = 120Ω/√εr·ln(2a/d)`; Viertelwellentransformator `Z = √(Z_E·Z_A)`.
+
+## S. 18 — Weitere Formeln
+
+- **MUF:** `MUF ≈ f_c/sin(α)`; `f_opt = MUF·0,85`.
+- **Empfindlichkeit von Messsystemen:** `E_MESS = R_i/U_i = 1/I_i`.
+- **Relativer maximaler Fehler:** `F_W = ±(G/100)·(W_E/W_M)`.
+- **Abtasttheorem:** `f_abtast > 2·f_max`; für Nicht-Basisband-Signale
+ `f_abtast > 2·(f_max − f_min)` **wenn `f_abtast < f_min` oder
+ `f_abtast > f_max`**.
+- **Datenübertragungs-/Symbolrate:** `C = R_S·n`.
+
+## S. 19–22 — Konstanten und Tabellen
+
+- **S. 19–20** — symbol/quantity glossary and physical constants
+ (k, c0, μ0, ε0, e, …).
+- **S. 21** — material tables: **relative permittivity ε_r** and
+ **specific resistance ρ** (per the S. 21 erratum, Zink → Zinn).
+- **S. 22 — Kabeldämpfungsdiagramm Koaxialkabel:** a chart of cable
+ attenuation in **dB per 100 m vs. frequency** for common coax types
+ (RG58 ≈ 4,95 mm, RG174 ≈ 2,8 mm, PE-foam 10,3/12,7 mm, …). Read the
+ per-100 m loss at the operating frequency, then **scale linearly with
+ cable length**. This chart *is* part of the aid — cable-loss questions
+ are Hilfsmittel lookups.
+
+