diff --git a/explanations.json b/explanations.json
index a2763aa..79e9f10 100644
--- a/explanations.json
+++ b/explanations.json
@@ -24,56 +24,56 @@
"confidence": 8
},
"AA105": {
- "revision": 3,
- "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.",
+ "revision": 5,
+ "explanation": "For power ratios, gain in dB is $10\\log_{10}(P_2/P_1)$; $10\\log_{10}(40)$ is about 16 dB. Hilfsmittel: apply $g = 10\\cdot\\log_{10}(P_2/P_1)$ (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: 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.",
+ "revision": 4,
+ "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 $g = 10\\cdot\\log_{10}(P_2/P_1)$ (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: apply p = 10·log10(P/1W) dBW (Pegel, S.15); +10 dB simply adds 10 to the dBW level.",
+ "revision": 4,
+ "explanation": "1 W is 0 dBW, and a 10 dB amplifier raises the power level by 10 dB, giving 10 dBW. Hilfsmittel: apply $p = 10\\cdot\\log_{10}(P/1\\,\\text{W})$ 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: invert p = 10·log10(P/1W) dBW (Pegel, S.15): P = 1W·10^(p/10).",
+ "revision": 4,
+ "explanation": "dBW is referenced to 1 W, so 20 dBW means $10^{20/10}\\,\\text{W} = 100\\,\\text{W} = 10^2\\,\\text{W}$. Hilfsmittel: invert $p = 10\\cdot\\log_{10}(P/1\\,\\text{W})$ dBW (Pegel, S.15): $P = 1\\,\\text{W}\\cdot 10^{p/10}$.",
"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: apply p = 10·log10(P/1mW) dBm (Pegel, S.15) after converting 10 W to 10000 mW.",
+ "revision": 5,
+ "explanation": "The amplifier output is 10 W; in dBm that is $10\\,\\text{W} = 10000\\,\\text{mW}$, and $10\\log_{10}(10000) = 40\\,\\text{dBm}$. Hilfsmittel: apply $p = 10\\cdot\\log_{10}(P/1\\,\\text{mW})$ 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: use p = 10·log10(P/1mW) dBm (Pegel, S.15); the dB↔ratio table (S.15) gives the +3/+20 dB factors.",
+ "revision": 4,
+ "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 $p = 10\\cdot\\log_{10}(P/1\\,\\text{mW})$ 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: apply the voltage form g = 20·log10(U2/U1) (Pegel, S.15).",
+ "revision": 5,
+ "explanation": "For voltage ratios use $20\\log_{10}(U_2/U_1)$; $20\\log_{10}(15)$ is about 23.5 dB. Hilfsmittel: apply the voltage form $g = 20\\cdot\\log_{10}(U_2/U_1)$ (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: invert g = 20·log10(U2/U1) (Pegel, S.15): ratio = 10^(120/20) = 10⁶.",
+ "revision": 4,
+ "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 $g = 20\\cdot\\log_{10}(U_2/U_1)$ (Pegel, S.15): $\\text{ratio} = 10^{120/20} = 10^6$.",
"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.",
+ "revision": 4,
+ "explanation": "Each S-step is 6 dB; S4 to S7 is three steps, so $3 \\times 6\\,\\text{dB} = 18\\,\\text{dB}$.",
"source": "https://50ohm.de/NEA_s_meter.html#AA113",
"confidence": 8
},
@@ -84,26 +84,26 @@
"confidence": 8
},
"AA115": {
- "revision": 3,
- "explanation": "1 ppm is one part in one million; 435 MHz divided by 10^6 is 435 Hz.",
+ "revision": 4,
+ "explanation": "1 ppm is one part in one million; $435\\,\\text{MHz}$ divided by $10^6$ is $435\\,\\text{Hz}$.",
"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.",
+ "revision": 4,
+ "explanation": "10 ppm at 14.200000 MHz is $14.2\\,\\text{MHz} \\times 10/10^6 = 142\\,\\text{Hz}$, so the possible frequency is 14.200000 MHz plus or minus 0.000142 MHz.",
"source": "https://50ohm.de/NEA_frequenzgenauigkeit.html#AA116",
"confidence": 8
},
"AB101": {
- "revision": 2,
- "explanation": "Use $R = rho l/A$ with copper $rho = 0.018 ohm mm2/m$ and $A = pi(0.1 mm)^2 = 0.0314 mm2$; $0.018 x 1.8 / 0.0314$ is about 1.02 ohm.",
+ "revision": 5,
+ "explanation": "Use $R = \\rho l/A$ with copper $\\rho = 0.018\\,\\Omega\\,\\text{mm}^2/\\text{m}$ and $A = \\pi(0.1\\,\\text{mm})^2 = 0.0314\\,\\text{mm}^2$; $0.018 \\cdot 1.8 / 0.0314$ is about 1.02 ohm.",
"source": "https://50ohm.de/NEA_leiterwiderstand.html#AB101",
"confidence": 8
},
"AB102": {
- "revision": 2,
- "explanation": "Rearrange $R = rho l/A$ to $l = RA/rho$; $1.5 ohm x 0.5 mm2 / 0.018 ohm mm2/m$ is about 41.7 m.",
+ "revision": 5,
+ "explanation": "Rearrange $R = \\rho l/A$ to $l = RA/\\rho$; $1.5\\,\\Omega \\cdot 0.5\\,\\text{mm}^2 / 0.018\\,\\Omega\\,\\text{mm}^2/\\text{m}$ is about 41.7 m.",
"source": "https://50ohm.de/NEA_leiterwiderstand.html#AB102",
"confidence": 8
},
@@ -174,32 +174,32 @@
"confidence": 8
},
"AB205": {
- "revision": 2,
- "explanation": "The load current is $4.8 V / 1.2 ohm = 4 A$ and the source drops 0.2 V internally, so $R_i = 0.2 V / 4 A = 0.05 ohm$.",
+ "revision": 3,
+ "explanation": "The load current is $4.8\\,\\text{V} / 1.2\\,\\Omega = 4\\,\\text{A}$ and the source drops 0.2 V internally, so $R_i = 0.2\\,\\text{V} / 4\\,\\text{A} = 0.05\\,\\Omega$.",
"source": "https://50ohm.de/NEA_innenwiderstand.html#AB205",
"confidence": 8
},
"AB206": {
- "revision": 2,
- "explanation": "The internal voltage drop is $13.5 V - 12.4 V = 1.1 V$; dividing by 0.9 A gives about 1.22 ohm.",
+ "revision": 3,
+ "explanation": "The internal voltage drop is $13.5\\,\\text{V} - 12.4\\,\\text{V} = 1.1\\,\\text{V}$; dividing by 0.9 A gives about 1.22 ohm.",
"source": "https://50ohm.de/NEA_innenwiderstand.html#AB206",
"confidence": 8
},
"AB207": {
- "revision": 2,
- "explanation": "The terminal voltage falls by 0.5 V at 2 A, so $R_i = 0.5 V / 2 A = 0.25 ohm$.",
+ "revision": 3,
+ "explanation": "The terminal voltage falls by 0.5 V at 2 A, so $R_i = 0.5\\,\\text{V} / 2\\,\\text{A} = 0.25\\,\\Omega$.",
"source": "https://50ohm.de/NEA_innenwiderstand.html#AB207",
"confidence": 8
},
"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: apply R_i = ΔU/ΔI (Innenwiderstand, S.11).",
+ "revision": 4,
+ "explanation": "The voltage drop is 0.2 V at 20 A, so $R_i = 0.2\\,\\text{V} / 20\\,\\text{A} = 0.01\\,\\Omega = 10\\,\\text{m}\\Omega$. Hilfsmittel: apply $R_i = \\Delta U/\\Delta I$ (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: series voltages add: U_G = U1+U2 (Spannungsteiler, S.12); the Ah capacity stays that of one cell (outside the sheet).",
+ "revision": 4,
+ "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: $U_G = U_1 + U_2$ (Spannungsteiler, S.12); the Ah capacity stays that of one cell (outside the sheet).",
"source": "https://50ohm.de/NEA_akku.html#AB209",
"confidence": 7
},
@@ -210,8 +210,8 @@
"confidence": 8
},
"AB211": {
- "revision": 2,
- "explanation": "Discharging only down to 10 percent leaves 90 percent usable: $0.9 x 60 Ah = 54 Ah$; $54 Ah / 0.8 A = 67.5 h$.",
+ "revision": 4,
+ "explanation": "Discharging only down to 10 percent leaves 90 percent usable: $0.9 \\cdot 60\\,\\text{Ah} = 54\\,\\text{Ah}$; $54\\,\\text{Ah} / 0.8\\,\\text{A} = 67.5\\,\\text{h}$.",
"source": "https://50ohm.de/NEA_akku.html#AB211",
"confidence": 8
},
@@ -222,32 +222,32 @@
"confidence": 8
},
"AB213": {
- "revision": 2,
- "explanation": "Input power is $12 V x 2 A = 24 W$ and output power is $5 V x 3 A = 15 W$; efficiency is $15/24 = 62.5%$.",
+ "revision": 4,
+ "explanation": "Input power is $12\\,\\text{V} \\cdot 2\\,\\text{A} = 24\\,\\text{W}$ and output power is $5\\,\\text{V} \\cdot 3\\,\\text{A} = 15\\,\\text{W}$; efficiency is $15/24 = 62.5\\%$.",
"source": "https://50ohm.de/NEA_spannungswandler.html#AB213",
"confidence": 8
},
"AB214": {
- "revision": 2,
- "explanation": "Input power is $5 V x 3 A = 15 W$ and output power is $12 V x 1 A = 12 W$; efficiency is $12/15 = 80%$.",
+ "revision": 4,
+ "explanation": "Input power is $5\\,\\text{V} \\cdot 3\\,\\text{A} = 15\\,\\text{W}$ and output power is $12\\,\\text{V} \\cdot 1\\,\\text{A} = 12\\,\\text{W}$; efficiency is $12/15 = 80\\%$.",
"source": "https://50ohm.de/NEA_spannungswandler.html#AB214",
"confidence": 8
},
"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: first I_eff = Î/√2 (from Û = U_eff·√2, Wechselspannung, S.12), then P = I_eff²·R (Leistung, S.12).",
+ "revision": 5,
+ "explanation": "For a sine current, $I_\\mathrm{eff} = I_\\mathrm{max} / \\sqrt{2}$; power is $I_\\mathrm{eff}^2 R = (0.5/\\sqrt{2})^2 \\cdot 20 = 2.5\\,\\text{W}$. Hilfsmittel: first $I_\\mathrm{eff} = \\hat{I}/\\sqrt{2}$ (from $\\hat{U} = U_\\mathrm{eff}\\cdot\\sqrt{2}$, Wechselspannung, S.12), then $P = I_\\mathrm{eff}^2\\cdot R$ (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.",
+ "revision": 3,
+ "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
},
"AB303": {
- "revision": 2,
- "explanation": "The two sine waves are shifted by one eighth of a full cycle; $360 degrees / 8 = 45 degrees$.",
+ "revision": 3,
+ "explanation": "The two sine waves are shifted by one eighth of a full cycle; $360^\\circ / 8 = 45^\\circ$.",
"source": "https://50ohm.de/NEA_phase.html#AB303",
"confidence": 7
},
@@ -300,26 +300,26 @@
"confidence": 8
},
"AB409": {
- "revision": 2,
- "explanation": "Noise power scales with bandwidth; changing from 2.5 kHz to 0.5 kHz is a factor of 5 reduction, and 10 log10(5) is about 7 dB.",
+ "revision": 3,
+ "explanation": "Noise power scales with bandwidth; changing from 2.5 kHz to 0.5 kHz is a factor of 5 reduction, and $10\\log_{10}(5)$ is about 7 dB.",
"source": "https://50ohm.de/NEA_rauschen.html#AB409",
"confidence": 8
},
"AB501": {
- "revision": 3,
- "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.",
+ "revision": 6,
+ "explanation": "Stored energy in watt-hours is voltage times ampere-hour capacity: $12\\,\\text{V} \\cdot 5\\,\\text{Ah} = 60\\,\\text{Wh}$. Hilfsmittel: $W = U\\cdot I\\cdot t$, i.e. $W = P\\cdot t$ with $P = U\\cdot I$ (Arbeit/Leistung, S.12); $U\\cdot\\text{Ah}$ 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: first P = U·I, then W = P·t (Leistung/Arbeit, S.12).",
+ "revision": 6,
+ "explanation": "Power is $230\\,\\text{V} \\cdot 0.63\\,\\text{A} = 144.9\\,\\text{W}$; over 7 h this is $144.9\\,\\text{W} \\cdot 7\\,\\text{h} = 1014\\,\\text{Wh}$, about 1.01 kWh. Hilfsmittel: first $P = U\\cdot I$, then $W = P\\cdot t$ (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: first P = U²/R (Leistung, S.12), then W = P·t (S.12); 1 Wh = 3600 J.",
+ "revision": 5,
+ "explanation": "The resistor has $P = U^2/R = 10^2/100 = 1\\,\\text{W}$; over one hour that is 1 Wh, equal to 3600 J. Hilfsmittel: first $P = U^2/R$ (Leistung, S.12), then $W = P\\cdot t$ (S.12); 1 Wh = 3600 J.",
"source": "https://50ohm.de/NEA_ladung_energie.html#AB503",
"confidence": 7
},
@@ -336,8 +336,8 @@
"confidence": 8
},
"AC102": {
- "revision": 2,
- "explanation": "Capacitive reactance is negative in AC sign convention, and its magnitude is $1/(2 pi f C)$, so it depends on frequency and capacitance.",
+ "revision": 3,
+ "explanation": "Capacitive reactance is negative in AC sign convention, and its magnitude is $1/(2 \\pi f C)$, so it depends on frequency and capacitance.",
"source": "https://50ohm.de/NEA_kondensator_2.html#AC102",
"confidence": 8
},
@@ -348,32 +348,32 @@
"confidence": 8
},
"AC104": {
- "revision": 2,
- "explanation": "Use $X_C = 1/(2 pi f C)$; with 100 MHz and 10 pF this gives about 159 ohm.",
+ "revision": 3,
+ "explanation": "Use $X_C = 1/(2 \\pi f C)$; with 100 MHz and 10 pF this gives about 159 ohm.",
"source": "https://50ohm.de/NEA_kondensator_2.html#AC104",
"confidence": 8
},
"AC105": {
- "revision": 2,
- "explanation": "Use $X_C = 1/(2 pi f C)$; with 145 MHz and 50 pF this is about 22 ohm.",
+ "revision": 3,
+ "explanation": "Use $X_C = 1/(2 \\pi f C)$; with 145 MHz and 50 pF this is about 22 ohm.",
"source": "https://50ohm.de/NEA_kondensator_2.html#AC105",
"confidence": 8
},
"AC106": {
- "revision": 2,
- "explanation": "Use $X_C = 1/(2 pi f C)$; with 100 MHz and 100 pF this gives about 15.9 ohm.",
+ "revision": 3,
+ "explanation": "Use $X_C = 1/(2 \\pi f C)$; with 100 MHz and 100 pF this gives about 15.9 ohm.",
"source": "https://50ohm.de/NEA_kondensator_2.html#AC106",
"confidence": 8
},
"AC107": {
- "revision": 2,
- "explanation": "Use $X_C = 1/(2 pi f C)$; with 435 MHz and 100 pF this gives about 3.7 ohm.",
+ "revision": 3,
+ "explanation": "Use $X_C = 1/(2 \\pi f C)$; with 435 MHz and 100 pF this gives about 3.7 ohm.",
"source": "https://50ohm.de/NEA_kondensator_2.html#AC107",
"confidence": 8
},
"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: 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).",
+ "revision": 6,
+ "explanation": "First find reactance from $X_C = U/I = 16\\,\\text{V} / 0.032\\,\\text{A} = 500\\,\\Omega$; 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 $C = 1/(2\\pi\\cdot f\\cdot X_C)$ ($X_C = 1/(\\omega C)$, $\\omega = 2\\pi f$, S.13/12).",
"source": "https://50ohm.de/NEA_kondensator_2.html#AC108",
"confidence": 8
},
@@ -402,8 +402,8 @@
"confidence": 8
},
"AC202": {
- "revision": 2,
- "explanation": "Inductive reactance is positive in AC sign convention and has magnitude $X_L = 2 pi f L$, so it depends on frequency and inductance.",
+ "revision": 3,
+ "explanation": "Inductive reactance is positive in AC sign convention and has magnitude $X_L = 2 \\pi f L$, so it depends on frequency and inductance.",
"source": "https://50ohm.de/NEA_spule_2.html#AC202",
"confidence": 8
},
@@ -414,32 +414,32 @@
"confidence": 8
},
"AC204": {
- "revision": 2,
- "explanation": "Use $X_L = 2 pi f L$; $2 pi x 100 MHz x 3 microhenry$ is about 1885 ohm.",
+ "revision": 4,
+ "explanation": "Use $X_L = 2 \\pi f L$; $2 \\pi \\cdot 100\\,\\text{MHz} \\cdot 3\\,\\mu\\text{H}$ is about 1885 ohm.",
"source": "https://50ohm.de/NEA_spule_2.html#AC204",
"confidence": 8
},
"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: apply L = N²·A_L (Ringkernspulen L = N²·A_L, S.13).",
+ "revision": 5,
+ "explanation": "For a core with AL value, $L = N^2 \\cdot A_L$; $14^2 \\cdot 1.5\\,\\text{nH} = 294\\,\\text{nH} = 0.294\\,\\mu\\text{H}$. Hilfsmittel: apply $L = N^2\\cdot A_L$ (Ringkernspulen $L = N^2\\cdot A_L$, 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: apply L = N²·A_L (Ringkernspulen L = N²·A_L, S.13).",
+ "revision": 5,
+ "explanation": "Use $L = N^2 \\cdot A_L$; $300^2 \\cdot 1250\\,\\text{nH} = 112500000\\,\\text{nH} = 112.5\\,\\text{mH}$. Hilfsmittel: apply $L = N^2\\cdot A_L$ (Ringkernspulen $L = N^2\\cdot A_L$, 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: rearrange L = N²·A_L → N = √(L/A_L) (Ringkernspulen L = N²·A_L, S.13).",
+ "revision": 5,
+ "explanation": "Rearrange to $N = \\sqrt{L/A_L}$; $\\sqrt{2\\,\\text{mH} / 250\\,\\text{nH}} = \\sqrt{8000}$, about 89 turns. Hilfsmittel: rearrange $L = N^2\\cdot A_L \\to N = \\sqrt{L/A_L}$ (Ringkernspulen $L = N^2\\cdot A_L$, 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: rearrange L = N²·A_L → N = √(L/A_L) (Ringkernspulen L = N²·A_L, S.13).",
+ "revision": 6,
+ "explanation": "Rearrange to $N = \\sqrt{L/A_L}$; $\\sqrt{12\\,\\mu\\text{H} / 30\\,\\text{nH}} = \\sqrt{400} = 20\\,\\text{turns}$. Hilfsmittel: rearrange $L = N^2\\cdot A_L \\to N = \\sqrt{L/A_L}$ (Ringkernspulen $L = N^2\\cdot A_L$, S.13).",
"source": "https://50ohm.de/NEA_spule_2.html#AC208",
"confidence": 8
},
@@ -468,26 +468,26 @@
"confidence": 8
},
"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 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).",
+ "revision": 6,
+ "explanation": "With losses neglected, primary and secondary power are equal: $6\\,\\text{V} \\cdot 1.15\\,\\text{A} = 6.9\\,\\text{W}$, and $6.9\\,\\text{W} / 230\\,\\text{V} = 0.030\\,\\text{A}$. Hilfsmittel: the transformer relation $U_P\\cdot I_P = U_S\\cdot I_S$ (from $\\ddot{u} = U_P/U_S = I_S/I_P$, S.13) gives the primary current; equivalently equal powers $P = U\\cdot I$ (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: 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).",
+ "revision": 4,
+ "explanation": "Impedance transforms with the square of the turns ratio; with 1:4, the input sees $16\\,\\text{k}\\Omega / 4^2 = 1\\,\\text{k}\\Omega$. Hilfsmittel: impedance transforms with the turns ratio squared (Übersetzungsverhältnis $\\ddot{u} = N_P/N_S = U_P/U_S = I_S/I_P = \\sqrt{Z_P/Z_S}$, 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: 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).",
+ "revision": 4,
+ "explanation": "The same 1:4 transformer reflects the secondary load by a factor of 16, so $6.4\\,\\text{k}\\Omega / 16 = 0.4\\,\\text{k}\\Omega$ at a-b. Hilfsmittel: impedance transforms with the turns ratio squared (Übersetzungsverhältnis $\\ddot{u} = N_P/N_S = U_P/U_S = I_S/I_P = \\sqrt{Z_P/Z_S}$, S.13).",
"source": "https://50ohm.de/NEA_uebertrager_2.html#AC304",
"confidence": 7
},
"AC305": {
- "revision": 2,
- "explanation": "The impedance ratio is $450/50 = 9$; turns ratio is the square root of impedance ratio, so $sqrt(9) = 3$.",
+ "revision": 3,
+ "explanation": "The impedance ratio is $450/50 = 9$; turns ratio is the square root of impedance ratio, so $\\sqrt{9} = 3$.",
"source": "https://50ohm.de/NEA_uebertrager_2.html#AC305",
"confidence": 8
},
@@ -498,8 +498,8 @@
"confidence": 8
},
"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: first A = d²·π/4 (S.11), then I = S·A with S ≈ 2,5 A/mm² (Belastbarkeit von Wicklungen, S.13).",
+ "revision": 7,
+ "explanation": "The wire area is $\\pi d^2/4 = \\pi \\cdot 0.5^2/4 = 0.196\\,\\text{mm}^2$; at $2.5\\,\\text{A/mm}^2$ the current is about $0.49\\,\\text{A}$. Hilfsmittel: first $A = d^2\\pi/4$ (S.11), then $I = S\\cdot A$ with $S \\approx 2{,}5\\,\\text{A/mm}^2$ (Belastbarkeit von Wicklungen, S.13).",
"source": "https://50ohm.de/NEA_uebertrager_2.html#AC307",
"confidence": 8
},
@@ -636,8 +636,8 @@
"confidence": 8
},
"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: 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.",
+ "revision": 4,
+ "explanation": "Base current is $5\\,\\text{mA} / 298 = 16.8\\,\\mu\\text{A}$; with about 0.6 V base-emitter drop, $R_1 = (12 - 0.6)\\,\\text{V} / 16.8\\,\\mu\\text{A}$, 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
},
@@ -648,14 +648,14 @@
"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: 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.",
+ "revision": 4,
+ "explanation": "Base current is $2\\,\\text{mA}/200 = 10\\,\\mu\\text{A}$; R2 carries ten times that, so R1 carries 110 microampere. With 1 V at the emitter, the base is about 1.6 V, giving $R_1 = 8.4\\,\\text{V}/110\\,\\mu\\text{A} = 76.4\\,\\text{k}\\Omega$. Hilfsmittel: first base current $I_B = I_C/B$ ($B = I_C/I_B$, S. 14); $R_2$ carries ten times that and $R_1$ 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: 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.",
+ "revision": 4,
+ "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, $R_1 = 9.4\\,\\text{V}/110\\,\\mu\\text{A} = 85.5\\,\\text{k}\\Omega$. 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
},
@@ -672,20 +672,20 @@
"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: apply the divider U_G = U·R/(R1+R2) (Spannungsteiler, S.12).",
+ "revision": 5,
+ "explanation": "The gate draws negligible current, so the divider gives $U_G = 44\\,\\text{V} \\cdot 1\\,\\text{k}\\Omega/(10\\,\\text{k}\\Omega + 1\\,\\text{k}\\Omega) = 4\\,\\text{V}$; with the source at reference, that is $U_\\mathrm{GS}$. Hilfsmittel: apply the divider $U_G = U\\cdot R/(R_1+R_2)$ (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: rearrange the divider for R2 (Spannungsteiler, S.12).",
+ "revision": 5,
+ "explanation": "For a divider, $R_2 = R_1 U_G/(U_B - U_G)$; $10\\,\\text{k}\\Omega \\cdot 2.8/(44 - 2.8)$ gives about 680 ohm. Hilfsmittel: rearrange the divider for $R_2$ (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: apply P = I²·R (Leistung, S.12).",
+ "revision": 5,
+ "explanation": "Conduction loss is $P = I^2 R$; $25^2 \\cdot 0.004\\,\\Omega = 2.5\\,\\text{W}$. Hilfsmittel: apply $P = I^2\\cdot R$ (Leistung, S.12).",
"source": "https://50ohm.de/NEA_transistor_2.html#AC523",
"confidence": 8
},
@@ -720,68 +720,68 @@
"confidence": 8
},
"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: series caps via 1/C_G = Σ1/Ci (Kapazität, S.13).",
+ "revision": 4,
+ "explanation": "Series capacitors add by reciprocals: $1/C = 1/100\\,\\text{pF} + 1/47\\,\\text{pF} + 1/22\\,\\text{pF}$, giving about 13.0 pF. Hilfsmittel: series caps via $1/C_G = \\sum 1/C_i$ (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: inductances in series add: L_G = ΣLi (Induktivität, S.13).",
+ "revision": 5,
+ "explanation": "Series inductances add directly: $2200\\,\\text{nH}$ is $2.2\\,\\mu\\text{H}$, $0.033\\,\\text{mH}$ is $33\\,\\mu\\text{H}$, so the sum is $2.2 + 33 + 150 = 185.2\\,\\mu\\text{H}$. Hilfsmittel: inductances in series add: $L_G = \\sum L_i$ (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: parallel caps add: C_G = ΣCi (Kapazität, S.13).",
+ "revision": 4,
+ "explanation": "The shown capacitances are effectively parallel, so they add: $100\\,\\text{pF} + 1500\\,\\text{pF} + 220\\,\\text{pF} + 1\\,\\text{pF} = 1821\\,\\text{pF}$. Hilfsmittel: parallel caps add: $C_G = \\sum C_i$ (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: 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).",
+ "revision": 4,
+ "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 $X_C = 1/(\\omega C)$ ($X_C = 1/(\\omega C)$ bzw. $X_L = \\omega L$, S.13), then $|Z| = \\sqrt{R^2+X^2}$ ($Z = \\sqrt{R^2+X^2}$, 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: 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).",
+ "revision": 4,
+ "explanation": "At 1 MHz and 100 microhenry, $X_L$ is about 628 ohm; $|Z| = \\sqrt{100^2 + 628^2}$, about 636 ohm. Hilfsmittel: first $X_L = \\omega L$ ($X_C = 1/(\\omega C)$ bzw. $X_L = \\omega L$, S.13), then $|Z| = \\sqrt{R^2+X^2}$ ($Z = \\sqrt{R^2+X^2}$, 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: resistors add in series / combine in parallel (Reihen-/Parallelschaltung, S.12); voltages via U = R·I (Ohmsches Gesetz, S.11).",
+ "revision": 3,
+ "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 $U = R\\cdot I$ (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: parallel then series (Reihen-/Parallelschaltung, S.12); current via I = U/R (Ohmsches Gesetz, S.11).",
+ "revision": 4,
+ "explanation": "R2 and R3 in parallel give $5\\,\\text{k}\\Omega$, in series with R1 gives $15\\,\\text{k}\\Omega$; $15\\,\\text{V} / 15\\,\\text{k}\\Omega = 1\\,\\text{mA}$ total, split equally so R3 has $0.5\\,\\text{mA}$. 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: reduce the network (Reihen-/Parallelschaltung, S.12), then P = U²/R (Leistung, S.12).",
+ "revision": 4,
+ "explanation": "The total current is 1 mA, so the parallel section has 5 V across it; R2 power is $5^2/10000 = 0.0025\\,\\text{W} = 2.5\\,\\text{mW}$. Hilfsmittel: reduce the network (Reihen-/Parallelschaltung, S.12), then $P = U^2/R$ (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: series adds, parallel = R1·R2/(R1+R2) (Reihen-/Parallelschaltung, S.12).",
+ "revision": 4,
+ "explanation": "The input is $200\\,\\Omega$ plus $100\\,\\Omega$ in parallel with $200\\,\\Omega + R$; at $R = 0$ this is about $267\\,\\Omega$, and at $R = 1\\,\\text{k}\\Omega$ it is about $292\\,\\Omega$. Hilfsmittel: series adds, parallel $= R_1 R_2/(R_1+R_2)$ (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: series adds, then two equal branches in parallel halve (Reihen-/Parallelschaltung, S.12).",
+ "revision": 3,
+ "explanation": "Each side branch is $2.2\\,\\text{k}\\Omega + 220\\,\\Omega = 2420\\,\\Omega$, and two equal branches in parallel give half that value, $1210\\,\\Omega$. 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 bridge balances when the two divider ratios match: U1/U2 = R1/R2 (Spannungsteiler, S.12).",
+ "revision": 4,
+ "explanation": "A bridge has zero branch voltage when the two divider ratios are equal, which gives $R_1/R_2 = R_3/R_4$. Hilfsmittel: the bridge balances when the two divider ratios match: $U_1/U_2 = R_1/R_2$ (Spannungsteiler, S.12).",
"source": "https://50ohm.de/NEA_brueckenschaltung.html#AD111",
"confidence": 7
},
@@ -798,8 +798,8 @@
"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: first the load in parallel with R2 (Reihen-/Parallelschaltung, S.12), then the divider (Spannungsteiler, S.12).",
+ "revision": 6,
+ "explanation": "The load is parallel to R2: $2.2\\,\\text{k}\\Omega \\parallel 8.2\\,\\text{k}\\Omega$ is about 1.73 kOhm. The divider output is $12\\,\\text{V} \\cdot 1.73/(10 + 1.73)$, about 1.8 V. Hilfsmittel: first the load in parallel with $R_2$ (Reihen-/Parallelschaltung, S.12), then the divider (Spannungsteiler, S.12).",
"source": "https://50ohm.de/NEA_spannungsteiler_2.html#AD114",
"confidence": 7
},
@@ -810,20 +810,20 @@
"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: apply f_g = 1/(2π·R·C) (RC-Hochpass, S.14).",
+ "revision": 5,
+ "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 $f_g = 1/(2\\pi\\cdot R\\cdot C)$ (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: apply f_g = 1/(2π·R·C) (RC-Tiefpass, S.14).",
+ "revision": 5,
+ "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 $f_g = 1/(2\\pi\\cdot R\\cdot C)$ (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: apply f_g = 1/(2π·R·C) (RC-Tiefpass, S.14).",
+ "revision": 5,
+ "explanation": "The relevant low-pass is R1 with C1; C2 is supply decoupling and the amplifier input is very high impedance. $1/(2 \\pi \\cdot 4.7\\,\\text{k}\\Omega \\cdot 6.8\\,\\text{nF})$ is about 5 kHz. Hilfsmittel: apply $f_g = 1/(2\\pi\\cdot R\\cdot C)$ (RC-Tiefpass, S.14).",
"source": "https://50ohm.de/NEA_schwingkreis_2.html#AD203",
"confidence": 7
},
@@ -846,44 +846,44 @@
"confidence": 8
},
"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: at resonance X_C = X_L cancel, leaving R (Schwingkreis, S.14).",
+ "revision": 3,
+ "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: apply f₀ = 1/(2π·√(L·C)), S.14.",
+ "revision": 4,
+ "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 $f_0 = 1/(2\\pi\\sqrt{L\\cdot C})$, 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: apply f₀ = 1/(2π·√(L·C)), S.14 (R does not set f₀).",
+ "revision": 5,
+ "explanation": "The resistor does not set the ideal resonant frequency; $1/(2 \\pi \\sqrt{10\\,\\mu\\text{H} \\cdot 1\\,\\text{nF}})$ is about 1.592 MHz. Hilfsmittel: apply $f_0 = 1/(2\\pi\\sqrt{L\\cdot C})$, S.14 (R does not set $f_0$).",
"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: apply f₀ = 1/(2π·√(L·C)), S.14.",
+ "revision": 4,
+ "explanation": "Using $f = 1/(2 \\pi \\sqrt{L C})$ with 100 microhenry and 0.01 microfarad gives about 159 kHz. Hilfsmittel: apply $f_0 = 1/(2\\pi\\sqrt{L\\cdot C})$, 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: apply f₀ = 1/(2π·√(L·C)), S.14.",
+ "revision": 5,
+ "explanation": "For the parallel resonant circuit, $f = 1/(2 \\pi \\sqrt{2.2\\,\\mu\\text{H} \\cdot 56\\,\\text{pF}})$, giving about 14.34 MHz. Hilfsmittel: apply $f_0 = 1/(2\\pi\\sqrt{L\\cdot C})$, 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: first add the parallel caps (S.13), then f₀ = 1/(2π·√(L·C)), S.14.",
+ "revision": 4,
+ "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 $f_0 = 1/(2\\pi\\sqrt{L\\cdot C})$, S.14.",
"source": "https://50ohm.de/NEA_schwingkreis_2.html#AD212",
"confidence": 7
},
"AD213": {
- "revision": 2,
- "explanation": "Resonant frequency is inversely proportional to $sqrt(L C)$, so using a smaller inductance raises the frequency.",
+ "revision": 3,
+ "explanation": "Resonant frequency is inversely proportional to $\\sqrt{L C}$, so using a smaller inductance raises the frequency.",
"source": "https://50ohm.de/NEA_schwingkreis_2.html#AD213",
"confidence": 8
},
@@ -942,26 +942,26 @@
"confidence": 8
},
"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: apply B = R/(2π·L) (Reihenschwingkreis, S.14).",
+ "revision": 5,
+ "explanation": "For a series resonant circuit, bandwidth is $B = R/(2 \\pi L)$; $10/(2 \\pi \\cdot 100\\,\\mu\\text{H})$ is about 15.9 kHz. Hilfsmittel: apply $B = R/(2\\pi L)$ (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: apply B = 1/(2π·R·C) (Parallelschwingkreis, S.14).",
+ "revision": 4,
+ "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 $B = 1/(2\\pi R C)$ (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: apply Q = f₀/B (Schwingkreis, S.14).",
+ "revision": 4,
+ "explanation": "For the series circuit, Q is resonant frequency divided by bandwidth; about $159\\,\\text{kHz} / 15.9\\,\\text{kHz} = 10$. Hilfsmittel: apply $Q = f_0/B$ (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: apply Q = f₀/B (Schwingkreis, S.14).",
+ "revision": 4,
+ "explanation": "For the parallel circuit, Q is resonant frequency divided by bandwidth; about $14.34\\,\\text{MHz} / 2.84\\,\\text{MHz} \\approx 5$. Hilfsmittel: apply $Q = f_0/B$ (Schwingkreis, S.14).",
"source": "https://50ohm.de/NEA_schwingkreis_2.html#AD226",
"confidence": 8
},
@@ -984,20 +984,20 @@
"confidence": 8
},
"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: 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.",
+ "revision": 6,
+ "explanation": "In each series string the cell voltages add, giving $30 \\cdot 0.6\\,\\text{V} = 18\\,\\text{V}$; four identical strings in parallel add their short-circuit currents to 4 A. Hilfsmittel: series voltages add ($U_G = U_1+U_2$) and parallel currents add ($I_G = I_1+I_2$), 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 cap charges to the peak: Û = U_eff·√2 (Wechselspannung, S.12).",
+ "revision": 5,
+ "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: $\\hat{U} = U_\\mathrm{eff}\\cdot\\sqrt{2}$ (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: 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).",
+ "revision": 5,
+ "explanation": "A 20:1 transformer gives $230\\,\\text{V} / 20 = 11.5\\,\\text{V}$ RMS; the peak is about $16.3\\,\\text{V}$, and adding 50 percent safety gives about $24.4\\,\\text{V}$, so choose at least $25\\,\\text{V}$. Hilfsmittel: first $U_S = U_P\\cdot N_S/N_P$ (Übersetzungsverhältnis $\\ddot{u} = N_P/N_S = U_P/U_S = I_S/I_P = \\sqrt{Z_P/Z_S}$, S.13), then $\\hat{U} = U_\\mathrm{eff}\\cdot\\sqrt{2}$ (Wechselspannung, S.12).",
"source": "https://50ohm.de/NEA_gleichrichter_2.html#AD303",
"confidence": 7
},
@@ -1014,8 +1014,8 @@
"confidence": 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: 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).",
+ "revision": 5,
+ "explanation": "The secondary peak is the mains peak divided by 8: $230\\,\\text{V} \\cdot 1.414 / 8$ is about 40.6 V, which is the unloaded capacitor voltage. Hilfsmittel: first scale by the turns ratio (Übersetzungsverhältnis $\\ddot{u} = N_P/N_S = U_P/U_S = I_S/I_P = \\sqrt{Z_P/Z_S}$, S.13), then $\\hat{U} = U_\\mathrm{eff}\\cdot\\sqrt{2}$ (Wechselspannung, S.12).",
"source": "https://50ohm.de/NEA_brueckengleichrichter.html#AD306",
"confidence": 7
},
@@ -1086,26 +1086,26 @@
"confidence": 7
},
"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: first I = U/R (Ohmsches Gesetz, S.11), then the drop·I via P = U·I (Leistung, S.12).",
+ "revision": 6,
+ "explanation": "The load current is $5\\,\\text{V} / 10\\,\\Omega = 0.5\\,\\text{A}$ and the regulator drops $13.8\\,\\text{V} - 5\\,\\text{V} = 8.8\\,\\text{V}$; loss is $8.8\\,\\text{V} \\cdot 0.5\\,\\text{A} = 4.4\\,\\text{W}$. Hilfsmittel: first $I = U/R$ (Ohmsches Gesetz, S.11), then (voltage drop) $\\times\\,I$ via $P = U\\cdot I$ (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: apply P = (U_in − U_out)·I, i.e. P = U·I (Leistung, S.12).",
+ "revision": 6,
+ "explanation": "A linear regulator dissipates the voltage drop times current: $(13.8\\,\\text{V} - 9\\,\\text{V}) \\cdot 0.9\\,\\text{A} = 4.32\\,\\text{W}$. Hilfsmittel: apply $P = (U_\\mathrm{in} - U_\\mathrm{out})\\cdot I$, i.e. $P = U\\cdot I$ (Leistung, S.12).",
"source": "https://50ohm.de/NEA_spannungsstabilisierung.html#AD319",
"confidence": 8
},
"AD320": {
- "revision": 2,
- "explanation": "Efficiency is output power over input power: $5 V x 0.450 A$ divided by $13.8 V x 0.455 A$ is about 0.36.",
+ "revision": 4,
+ "explanation": "Efficiency is output power over input power: $5\\,\\text{V} \\cdot 0.450\\,\\text{A}$ divided by $13.8\\,\\text{V} \\cdot 0.455\\,\\text{A}$ is about 0.36.",
"source": "https://50ohm.de/NEA_spannungsstabilisierung.html#AD320",
"confidence": 8
},
"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: apply η = P_ab/P_zu (Wirkungsgrad, S.12) with P = U·I for each side.",
+ "revision": 5,
+ "explanation": "The load power is $4.7\\,\\text{V} \\cdot 10\\,\\text{mA} = 47\\,\\text{mW}$; input power is $13.8\\,\\text{V} \\cdot (10 + 15) \\text{mA} = 345\\,\\text{mW}$, so efficiency is about 0.14. Hilfsmittel: apply $\\eta = P_\\mathrm{ab}/P_\\mathrm{zu}$ (Wirkungsgrad, S.12) with $P = U\\cdot I$ for each side.",
"source": "https://50ohm.de/NEA_spannungsstabilisierung.html#AD321",
"confidence": 7
},
@@ -1272,44 +1272,44 @@
"confidence": 8
},
"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: P_zu = U·I (Leistung, S.12), then P_ab = η·P_zu (Wirkungsgrad, S.12); the ~40 % class-A efficiency is outside knowledge.",
+ "revision": 5,
+ "explanation": "The DC input power is $50\\,\\text{V} \\cdot 2\\,\\text{A} = 100\\,\\text{W}$; class A efficiency is about 40%, so expected RF output is about $0.4 \\cdot 100\\,\\text{W} = 40\\,\\text{W}$. Hilfsmittel: $P_\\mathrm{zu} = U\\cdot I$ (Leistung, S.12), then $P_\\mathrm{ab} = \\eta\\cdot P_\\mathrm{zu}$ (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: P_zu = U·I (Leistung, S.12), then P_ab = η·P_zu (Wirkungsgrad, S.12); the ~85 % class-C efficiency is outside knowledge.",
+ "revision": 5,
+ "explanation": "The DC input power is $50\\,\\text{V} \\cdot 2\\,\\text{A} = 100\\,\\text{W}$; using about 85% efficiency for class C gives about $0.85 \\cdot 100\\,\\text{W} = 85\\,\\text{W}$ RF output. Hilfsmittel: $P_\\mathrm{zu} = U\\cdot I$ (Leistung, S.12), then $P_\\mathrm{ab} = \\eta\\cdot P_\\mathrm{zu}$ (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: 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).",
+ "revision": 6,
+ "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 $G = 10^{g/(10\\,\\text{dB})}$, the inverse of $g = 10\\cdot\\log_{10}(P_2/P_1)$ (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: apply the voltage form g = 20·log10(U2/U1) (Pegel, S.15).",
+ "revision": 6,
+ "explanation": "For equal impedances, voltage gain in dB is $20 \\log_{10}(U_2/U_1)$; $20 \\log_{10}(4\\,\\text{mV} / 1\\,\\text{mV}) = 20 \\log_{10}(4) = 12\\,\\text{dB}$. Hilfsmittel: apply the voltage form $g = 20\\cdot\\log_{10}(U_2/U_1)$ (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: apply g = 10·log10(P2/P1) (Pegel, S.15).",
+ "revision": 6,
+ "explanation": "Power gain in dB is $10 \\log_{10}(P_2/P_1)$; $10 \\log_{10}(38\\,\\text{W} / 2.5\\,\\text{W}) = 10 \\log_{10}(15.2) = 11.8\\,\\text{dB}$. Hilfsmittel: apply $g = 10\\cdot\\log_{10}(P_2/P_1)$ (Pegel, S.15).",
"source": "https://50ohm.de/NEA_verstaerkungsleistung.html#AD428",
"confidence": 8
},
"AD429": {
- "revision": 2,
- "explanation": "Efficiency is useful RF output divided by DC input: $10 W / 25 W = 0.40$, or 40%.",
+ "revision": 3,
+ "explanation": "Efficiency is useful RF output divided by DC input: $10\\,\\text{W} / 25\\,\\text{W} = 0.40$, or 40%.",
"source": "https://50ohm.de/NEA_verstaerker_wirkungsgrad.html#AD429",
"confidence": 8
},
"AD430": {
- "revision": 2,
- "explanation": "The DC input power is $12.5 V x 16 A = 200 W$; efficiency is $90 W / 200 W = 0.45$, or 45%.",
+ "revision": 4,
+ "explanation": "The DC input power is $12.5\\,\\text{V} \\cdot 16\\,\\text{A} = 200\\,\\text{W}$; efficiency is $90\\,\\text{W} / 200\\,\\text{W} = 0.45$, or 45%.",
"source": "https://50ohm.de/NEA_verstaerker_wirkungsgrad.html#AD430",
"confidence": 8
},
@@ -1530,8 +1530,8 @@
"confidence": 7
},
"AD704": {
- "revision": 2,
- "explanation": "The divider ratio is output frequency divided by the 12.5 kHz reference: 12.000 MHz / 12.5 kHz = 960 and 14.000 MHz / 12.5 kHz = 1120.",
+ "revision": 3,
+ "explanation": "The divider ratio is output frequency divided by the 12.5 kHz reference: $12.000\\,\\text{MHz} / 12.5\\,\\text{kHz} = 960$ and $14.000\\,\\text{MHz} / 12.5\\,\\text{kHz} = 1120$.",
"source": "https://50ohm.de/NEA_oszillator_pll.html#AD704",
"confidence": 7
},
@@ -1554,14 +1554,14 @@
"confidence": 7
},
"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: convert dB→power ratio via the inverse of g = 10·log10(P2/P1) (Pegel, S.15); the table (S.15) gives 20 dB → ×100.",
+ "revision": 4,
+ "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 $g = 10\\cdot\\log_{10}(P_2/P_1)$ (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: convert dB→power ratio (inverse of g = 10·log10(P2/P1), Pegel, S.15); table (S.15): 6 dB → ×4.",
+ "revision": 4,
+ "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 $g = 10\\cdot\\log_{10}(P_2/P_1)$, Pegel, S.15); table (S.15): 6 dB → ×4.",
"source": "https://50ohm.de/NEA_daempfungsglieder.html#AD804",
"confidence": 7
},
@@ -1686,8 +1686,8 @@
"confidence": 8
},
"AE304": {
- "revision": 3,
- "explanation": "Carson's rule estimates FM bandwidth as $B \\approx 2(\\Delta f + f_{mod,max})$. Here $\\Delta f$ is the deviation, German Hub/Frequenzhub. Higher audio frequency or larger Hub makes the RF bandwidth too large.",
+ "revision": 4,
+ "explanation": "Carson's rule estimates FM bandwidth as $B \\approx 2(\\Delta f + f_\\mathrm{mod,max})$. Here $\\Delta f$ is the deviation, German Hub/Frequenzhub. Higher audio frequency or larger Hub makes the RF bandwidth too large.",
"source": "https://50ohm.de/NEA_fm_3.html#AE304",
"confidence": 8
},
@@ -1710,14 +1710,14 @@
"confidence": 8
},
"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: apply Carson B ≈ 2·(Δf + f_mod,max) (FM, S.16).",
+ "revision": 7,
+ "explanation": "Carson's rule gives $B \\approx 2(\\Delta f + f_\\mathrm{mod,max})$, where $\\Delta f$ is FM deviation, German Hub/Frequenzhub. With $\\Delta f=2.5\\,\\text{kHz}$ and $f_\\mathrm{mod,max}=2.7\\,\\text{kHz}$, $B=2(2.5+2.7)=10.4\\,\\text{kHz}$. Hilfsmittel: apply Carson $B \\approx 2\\cdot(\\Delta f + f_\\mathrm{mod,max})$ (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: apply Carson B ≈ 2·(Δf + f_mod,max) (FM, S.16).",
+ "revision": 6,
+ "explanation": "Using Carson's rule, $B = 2 \\cdot (1.8\\,\\text{kHz} + 2.0\\,\\text{kHz}) = 7.6\\,\\text{kHz}$; the 145 MHz carrier frequency does not enter this bandwidth estimate. Hilfsmittel: apply Carson $B \\approx 2\\cdot(\\Delta f + f_\\mathrm{mod,max})$ (FM, S.16).",
"source": "https://50ohm.de/NEA_fm_3.html#AE309",
"confidence": 8
},
@@ -1728,14 +1728,14 @@
"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: rearrange Carson B ≈ 2·(Δf + f_mod) for f_mod (FM, S.16).",
+ "revision": 7,
+ "explanation": "Rearrange Carson's rule $B \\approx 2(\\Delta f + f_\\mathrm{mod})$. Here $\\Delta f$ is deviation, German Hub/Frequenzhub, so $f_\\mathrm{mod}=B/2-\\Delta f=10\\,\\text{kHz}/2-2.5\\,\\text{kHz}=2.5\\,\\text{kHz}$. Hilfsmittel: rearrange Carson $B \\approx 2\\cdot(\\Delta f + f_\\mathrm{mod})$ for $f_\\mathrm{mod}$ (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: rearrange Carson B ≈ 2·(Δf + f_mod) for Δf (FM, S.16).",
+ "revision": 7,
+ "explanation": "Rearrange Carson's rule $B \\approx 2(\\Delta f + f_\\mathrm{mod})$. The deviation $\\Delta f$, German Hub/Frequenzhub, is $B/2-f_\\mathrm{mod}=10\\,\\text{kHz}/2-2.7\\,\\text{kHz}=2.3\\,\\text{kHz}$. Hilfsmittel: rearrange Carson $B \\approx 2\\cdot(\\Delta f + f_\\mathrm{mod})$ for $\\Delta f$ (FM, S.16).",
"source": "https://50ohm.de/NEA_fm_3.html#AE312",
"confidence": 8
},
@@ -1776,8 +1776,8 @@
"confidence": 8
},
"AE406": {
- "revision": 2,
- "explanation": "Four symbol frequencies encode two bits per symbol; data rate is symbol rate times bits per symbol, so $23.4 x 2 = 46.8 bit/s$.",
+ "revision": 4,
+ "explanation": "Four symbol frequencies encode two bits per symbol; data rate is symbol rate times bits per symbol, so $23.4 \\cdot 2 = 46.8\\,\\text{bit/s}$.",
"source": "https://50ohm.de/NEA_mehrwertige_verfahren.html#AE406",
"confidence": 8
},
@@ -1842,8 +1842,8 @@
"confidence": 8
},
"AE417": {
- "revision": 2,
- "explanation": "At 0 dB SNR the linear SNR is 1, so $C = B x log2(1 + 1) = B$; 2.7 kHz therefore gives about 2.7 kbit/s.",
+ "revision": 3,
+ "explanation": "At 0 dB SNR the linear SNR is 1, so $C = B \\cdot \\log_2(1 + 1) = B$; 2.7 kHz therefore gives about 2.7 kbit/s.",
"source": "https://50ohm.de/NEA_shannon_hartley_gesetzt.html#AE417",
"confidence": 8
},
@@ -1854,14 +1854,14 @@
"confidence": 8
},
"AE419": {
- "revision": 2,
- "explanation": "30 dB SNR is a linear ratio of 1000, so capacity is $10 MHz x log2(1001)$, about $10 MHz x 10 = 100 Mbit/s$.",
+ "revision": 4,
+ "explanation": "30 dB SNR is a linear ratio of 1000, so capacity is $10\\,\\text{MHz} \\cdot \\log_2(1001)$, about $10\\,\\text{MHz} \\cdot 10 = 100\\,\\text{Mbit/s}$.",
"source": "https://50ohm.de/NEA_shannon_hartley_gesetzt.html#AE419",
"confidence": 8
},
"AE420": {
- "revision": 2,
- "explanation": "-20 dB SNR is a linear ratio of 0.01; $2700 x log2(1.01)$ is about 39 bit/s, so transmission is possible but very slow.",
+ "revision": 3,
+ "explanation": "-20 dB SNR is a linear ratio of 0.01; $2700 \\cdot \\log_2(1.01)$ is about 39 bit/s, so transmission is possible but very slow.",
"source": "https://50ohm.de/NEA_shannon_hartley_gesetzt.html#AE420",
"confidence": 8
},
@@ -1914,14 +1914,14 @@
"confidence": 8
},
"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: 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).",
+ "revision": 5,
+ "explanation": "The IF is $24.94\\,\\text{MHz} - 14.24\\,\\text{MHz} = 10.70\\,\\text{MHz}$; the image on the other side of the oscillator is $24.94\\,\\text{MHz} + 10.70\\,\\text{MHz} = 35.64\\,\\text{MHz}$. Hilfsmittel: first the IF $f_\\mathrm{ZF} = |f_E - f_\\mathrm{OSZ}|$ (Zwischenfrequenz $f_\\mathrm{ZF} = |f_E - f_\\mathrm{OSZ}|$, S.14), then the image Spiegelfrequenz $f_S = 2\\cdot f_\\mathrm{OSZ} - f_E$ (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: 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).",
+ "revision": 5,
+ "explanation": "With high-side injection the oscillator is $28.5\\,\\text{MHz} + 10.7\\,\\text{MHz} = 39.2\\,\\text{MHz}$; the image is another IF above that, $39.2\\,\\text{MHz} + 10.7\\,\\text{MHz} = 49.9\\,\\text{MHz}$. Hilfsmittel: first $f_\\mathrm{OSZ}$ from $f_\\mathrm{ZF} = |f_E - f_\\mathrm{OSZ}|$ (Zwischenfrequenz $f_\\mathrm{ZF} = |f_E - f_\\mathrm{OSZ}|$, S.14), then the image Spiegelfrequenz $f_S = 2\\cdot f_\\mathrm{OSZ} - f_E$ (S.14).",
"source": "https://50ohm.de/NEA_spiegelfrequenzen.html#AF108",
"confidence": 7
},
@@ -1980,20 +1980,20 @@
"confidence": 7
},
"AF118": {
- "revision": 2,
- "explanation": "High-side first conversion needs $21.1 MHz + 9 MHz = 30.1 MHz$ for the VFO; low-side conversion from 9 MHz to 460 kHz needs $9 MHz - 0.460 MHz = 8.54 MHz$ for the CO.",
+ "revision": 3,
+ "explanation": "High-side first conversion needs $21.1\\,\\text{MHz} + 9\\,\\text{MHz} = 30.1\\,\\text{MHz}$ for the VFO; low-side conversion from 9 MHz to 460 kHz needs $9\\,\\text{MHz} - 0.460\\,\\text{MHz} = 8.54\\,\\text{MHz}$ for the CO.",
"source": "https://50ohm.de/NEA_doppelueberlagerungsempfaenger_doppelsuper.html#AF118",
"confidence": 7
},
"AF119": {
- "revision": 2,
- "explanation": "With both oscillators above their input signals, the VFO is $28.00 MHz + 10.70 MHz = 38.70 MHz$ and the second oscillator is $10.70 MHz + 0.460 MHz = 11.16 MHz$.",
+ "revision": 3,
+ "explanation": "With both oscillators above their input signals, the VFO is $28.00\\,\\text{MHz} + 10.70\\,\\text{MHz} = 38.70\\,\\text{MHz}$ and the second oscillator is $10.70\\,\\text{MHz} + 0.460\\,\\text{MHz} = 11.16\\,\\text{MHz}$.",
"source": "https://50ohm.de/NEA_doppelueberlagerungsempfaenger_doppelsuper.html#AF119",
"confidence": 7
},
"AF120": {
- "revision": 2,
- "explanation": "The chain can mix $3.65 MHz + 46.35 MHz$ to a 50 MHz first IF, then $50 MHz - 41 MHz$ to 9 MHz, then $9.455 MHz - 9 MHz$ to 455 kHz.",
+ "revision": 3,
+ "explanation": "The chain can mix $3.65\\,\\text{MHz} + 46.35\\,\\text{MHz}$ to a 50 MHz first IF, then $50\\,\\text{MHz} - 41\\,\\text{MHz}$ to 9 MHz, then $9.455\\,\\text{MHz} - 9\\,\\text{MHz}$ to 455 kHz.",
"source": "https://50ohm.de/NEA_doppelueberlagerungsempfaenger_doppelsuper.html#AF120",
"confidence": 7
},
@@ -2004,14 +2004,14 @@
"confidence": 7
},
"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: 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)).",
+ "revision": 5,
+ "explanation": "The IF is $145.6\\,\\text{MHz} - 134.9\\,\\text{MHz} = 10.7\\,\\text{MHz}$; the image is the other signal 10.7 MHz from the oscillator, $134.9\\,\\text{MHz} - 10.7\\,\\text{MHz} = 124.2\\,\\text{MHz}$. Hilfsmittel: first $f_\\mathrm{ZF} = |f_E - f_\\mathrm{OSZ}|$ (Zwischenfrequenz $f_\\mathrm{ZF} = |f_E - f_\\mathrm{OSZ}|$, S.14), then the image mirrored about $f_\\mathrm{OSZ}$ (Spiegelfrequenz $f_S = 2\\cdot f_\\mathrm{OSZ} - f_E$ (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 image is mirrored about f_OSZ: Spiegelfrequenz f_S = 2·f_OSZ − f_E (S.14).",
+ "revision": 6,
+ "explanation": "The image is mirrored around the oscillator frequency: $2 \\cdot 39\\,\\text{MHz} - 28.3\\,\\text{MHz} = 49.7\\,\\text{MHz}$. Hilfsmittel: the image is mirrored about $f_\\mathrm{OSZ}$: Spiegelfrequenz $f_S = 2\\cdot f_\\mathrm{OSZ} - f_E$ (S.14).",
"source": "https://50ohm.de/NEA_spiegelfrequenzen.html#AF203",
"confidence": 8
},
@@ -2052,8 +2052,8 @@
"confidence": 7
},
"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 VFO is f_OSZ = f_IF ± f_rx (from f_ZF = |f_E − f_OSZ|, Zwischenfrequenz f_ZF = |f_E − f_OSZ| (S.14)).",
+ "revision": 5,
+ "explanation": "For a 50 MHz first IF and a 3 to 30 MHz receive range, the VFO can use either difference mixing $50 - f_\\mathrm{rx}$ = 47 to 20 MHz or sum mixing $50 + f_\\mathrm{rx}$ = 53 to 80 MHz. Hilfsmittel: the VFO is $f_\\mathrm{OSZ} = f_\\mathrm{IF} \\pm f_\\mathrm{rx}$ (from $f_\\mathrm{ZF} = |f_E - f_\\mathrm{OSZ}|$, Zwischenfrequenz, S.14).",
"source": "https://50ohm.de/NEA_doppelueberlagerungsempfaenger_doppelsuper.html#AF210",
"confidence": 7
},
@@ -2166,8 +2166,8 @@
"confidence": 8
},
"AF229": {
- "revision": 2,
- "explanation": "A noise factor of 2 corresponds to $10 log10(2) = 3 dB$, meaning the amplifier degrades the signal-to-noise ratio by 3 dB.",
+ "revision": 3,
+ "explanation": "A noise factor of 2 corresponds to $10 \\log_{10}(2) = 3\\,\\text{dB}$, meaning the amplifier degrades the signal-to-noise ratio by 3 dB.",
"source": "https://50ohm.de/NEA_snr_rauschzahl.html#AF229",
"confidence": 8
},
@@ -2220,8 +2220,8 @@
"confidence": 7
},
"AF307": {
- "revision": 3,
- "explanation": "The USB carrier frequency is symmetric to the LSB carrier around the 9 MHz filter center: $9.0000 MHz - (9.0015 - 9.0000) MHz = 8.9985 MHz$.",
+ "revision": 4,
+ "explanation": "The USB carrier frequency is symmetric to the LSB carrier around the 9 MHz filter center: $9.0000\\,\\text{MHz} - (9.0015 - 9.0000) \\text{MHz} = 8.9985\\,\\text{MHz}$.",
"source": "https://50ohm.de/NEA_modulatoren.html#AF307",
"confidence": 7
},
@@ -2262,8 +2262,8 @@
"confidence": 8
},
"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.",
+ "revision": 4,
+ "explanation": "Only the sequence $12\\,\\text{MHz} \\cdot 2 \\cdot 2 \\cdot 3 \\cdot 3$ 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
},
@@ -2388,8 +2388,8 @@
"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 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).",
+ "revision": 3,
+ "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 $U_G = U_B\\cdot R/(\\sum R)$ (Spannungsteiler, S. 12).",
"source": "https://50ohm.de/NEA_leistungsvertaerker.html#AF421",
"confidence": 7
},
@@ -2412,38 +2412,38 @@
"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: apply R = U/I (Ohmsches Gesetz, S.11) to the resistor's drop and current.",
+ "revision": 4,
+ "explanation": "The resistor must drop $13.5\\,\\text{V} - 4\\,\\text{V} = 9.5\\,\\text{V}$ at 10 mA, so $R = 9.5\\,\\text{V} / 0.010\\,\\text{A} = 950\\,\\Omega$. 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: apply R = U/I (Ohmsches Gesetz, S.11); then pick the nearest E-series value (S.12).",
+ "revision": 4,
+ "explanation": "The bias resistor drops $13.8\\,\\text{V} - 4\\,\\text{V} = 9.8\\,\\text{V}$ at 15 mA; $9.8\\,\\text{V} / 0.015\\,\\text{A} = 653\\,\\Omega$, 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: first I = U/R (Ohmsches Gesetz, S.11), then heat P = U·I (Leistung, S.12).",
+ "revision": 5,
+ "explanation": "With a 4 V MMIC drop, the resistor current is $(9\\,\\text{V} - 4\\,\\text{V}) / 470\\,\\Omega = 10.6\\,\\text{mA}$; MMIC heat is $4\\,\\text{V} \\cdot 10.6\\,\\text{mA} = 42.6\\,\\text{mW}$, about 43 mW. Hilfsmittel: first $I = U/R$ (Ohmsches Gesetz, S.11), then heat $P = U\\cdot I$ (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: overall gain in dB is the level difference (out − in), i.e. g = 10·log10(P2/P1) (Pegel, S.15).",
+ "revision": 4,
+ "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. $g = 10\\cdot\\log_{10}(P_2/P_1)$ (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 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.",
+ "revision": 4,
+ "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 $f_Q = (f_\\mathrm{rx} - f_\\mathrm{IF}) / 9$ gives 45.333 MHz and 45.556 MHz. Hilfsmittel: the down-conversion uses $f_Q = (f_\\mathrm{rx} - f_\\mathrm{IF})/9$, i.e. a difference ($f_\\mathrm{ZF} = |f_E - f_\\mathrm{OSZ}|$, Zwischenfrequenz, 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 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.",
+ "revision": 4,
+ "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 $(430 - 28) / 9 = 44.667\\,\\text{MHz}$ and $(434 - 30) / 9 = 44.889\\,\\text{MHz}$. Hilfsmittel: the down-conversion uses $f_Q = (f_\\mathrm{rx} - f_\\mathrm{IF})/9$, i.e. a difference ($f_\\mathrm{ZF} = |f_E - f_\\mathrm{OSZ}|$, Zwischenfrequenz, 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": 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.",
+ "revision": 5,
+ "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 $2^8 = 256$; divide the 1 V range by that.",
"source": "https://50ohm.de/NEA_digital_analog_umsetzer.html#AF610",
"confidence": 8
},
"AF611": {
- "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.",
+ "revision": 5,
+ "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 $2^{10} = 1024$; divide the 1 V range by that.",
"source": "https://50ohm.de/NEA_digital_analog_umsetzer.html#AF611",
"confidence": 8
},
@@ -2550,8 +2550,8 @@
"confidence": 8
},
"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: apply the Abtasttheorem f_abtast > 2·f_max (S.18).",
+ "revision": 4,
+ "explanation": "Nyquist requires a sampling rate greater than twice the highest signal frequency, so the smallest safe rate is just above $2 f_\\mathrm{max}$. Hilfsmittel: apply the Abtasttheorem $f_\\mathrm{abtast} > 2\\cdot f_\\mathrm{max}$ (S.18).",
"source": "https://50ohm.de/NEA_abtasttheorem.html#AF618",
"confidence": 8
},
@@ -2580,8 +2580,8 @@
"confidence": 8
},
"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 passband must stay below f_abtast/2 (Abtasttheorem, S.18).",
+ "revision": 3,
+ "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 $f_\\mathrm{abtast}/2$ (Abtasttheorem, S.18).",
"source": "https://50ohm.de/NEA_anti_alias_filter.html#AF623",
"confidence": 8
},
@@ -2592,8 +2592,8 @@
"confidence": 8
},
"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: reject everything above f_abtast/2 (Abtasttheorem, S.18).",
+ "revision": 3,
+ "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 $f_\\mathrm{abtast}/2$ (Abtasttheorem, S.18).",
"source": "https://50ohm.de/NEA_rekonstruktionsfilter.html#AF625",
"confidence": 8
},
@@ -2646,20 +2646,20 @@
"confidence": 8
},
"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: 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.",
+ "revision": 4,
+ "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 $f_\\mathrm{max} < f_\\mathrm{abtast}/2$ (Abtasttheorem, S. 18); mapping that magnitude limit to the symmetric $-f_s/2 \\ldots +f_s/2$ 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: 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.",
+ "revision": 4,
+ "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 $f_\\mathrm{max} < f_\\mathrm{abtast}/2$ (Abtasttheorem, S. 18); mapping that magnitude limit to the symmetric $-f_s/2 \\ldots +f_s/2$ 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: 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.",
+ "revision": 4,
+ "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 $f_\\mathrm{max} < f_\\mathrm{abtast}/2$ (Abtasttheorem, S. 18); mapping that magnitude limit to the symmetric $-f_s/2 \\ldots +f_s/2$ complex-I/Q span is outside knowledge.",
"source": "https://50ohm.de/NEA_iq_verfahren.html#AF636",
"confidence": 8
},
@@ -2748,8 +2748,8 @@
"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.",
+ "revision": 3,
+ "explanation": "A quarter-wave groundplane uses quarter-wave radiator and radials; $0.95 c/(4 \\cdot 7.1\\,\\text{MHz})$ is about 10.04 m.",
"source": "https://50ohm.de/NEA_verkuerzungsfaktor_2.html#AG104",
"confidence": 8
},
@@ -2976,32 +2976,32 @@
"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: 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.",
+ "revision": 5,
+ "explanation": "The rear ERP is transmit power times forward gain divided by front-to-back ratio: $100\\,\\text{W} \\times 10 / 100 = 10\\,\\text{W}$. Hilfsmittel: convert the dB figures via the dB↔ratio table (S. 15): 10 dB → ×10, 20 dB → ×100; the rear-ERP $= P\\cdot\\text{gain}/\\text{front-to-back}$ 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: 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.",
+ "revision": 5,
+ "explanation": "15 dBd is a factor 31.6 and 25 dB front-to-back is a factor 316; $6\\,\\text{W} \\cdot 31.6 / 316$ is about 0.6 W rear ERP. Hilfsmittel: first convert the dBd/dB figures to factors with the inverse of $g = 10\\cdot\\log_{10}(P_2/P_1)$ (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: apply g = 10·log10(P2/P1) (Pegel, S.15) to the forward/back power ratio.",
+ "revision": 5,
+ "explanation": "Front-to-back ratio is a power ratio: $10 \\log_{10}(15/0.6)$ is about 14 dB. Hilfsmittel: apply $g = 10\\cdot\\log_{10}(P_2/P_1)$ (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: use the voltage form g = 20·log10(U2/U1) (Pegel, S.15) for both ratios.",
+ "revision": 5,
+ "explanation": "For field strengths, use $20\\log_{10}$ 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 $g = 20\\cdot\\log_{10}(U_2/U_1)$ (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: −3 dB is half power, a 0,71 voltage ratio (dB↔ratio table, S.15); the half-power-beamwidth definition itself is antenna knowledge.",
+ "revision": 3,
+ "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
},
@@ -3042,26 +3042,26 @@
"confidence": 8
},
"AG226": {
- "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).",
+ "revision": 4,
+ "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 $\\lambda = c/f$ (S.17), then $g_i = 10\\cdot\\log_{10}[(\\pi d/\\lambda)^2\\cdot\\eta]$ (Parabolspiegel, S.15).",
"source": "https://50ohm.de/NEA_parbolspiegel_2.html#AG226",
"confidence": 8
},
"AG227": {
- "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).",
+ "revision": 4,
+ "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 $\\lambda = c/f$ (S.17), then $g_i = 10\\cdot\\log_{10}[(\\pi d/\\lambda)^2\\cdot\\eta]$ (Parabolspiegel, S.15).",
"source": "https://50ohm.de/NEA_parbolspiegel_2.html#AG227",
"confidence": 8
},
"AG228": {
- "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).",
+ "revision": 4,
+ "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 $\\lambda = c/f$ (S.17), then $g_i = 10\\cdot\\log_{10}[(\\pi d/\\lambda)^2\\cdot\\eta]$ (Parabolspiegel, S.15).",
"source": "https://50ohm.de/NEA_parbolspiegel_2.html#AG228",
"confidence": 8
},
"AG229": {
- "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).",
+ "revision": 4,
+ "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 $\\lambda = c/f$ (S.17), then $g_i = 10\\cdot\\log_{10}[(\\pi d/\\lambda)^2\\cdot\\eta]$ (Parabolspiegel, S.15).",
"source": "https://50ohm.de/NEA_parbolspiegel_2.html#AG229",
"confidence": 8
},
@@ -3090,14 +3090,14 @@
"confidence": 8
},
"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: apply the two-wire line Z = 120Ω/√εr·ln(2a/d) (Wellenwiderstand, S.17).",
+ "revision": 3,
+ "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 $Z = (120\\,\\Omega/\\sqrt{\\varepsilon_r})\\cdot\\ln(2a/d)$ (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: apply the coax line Z = 60Ω/√εr·ln(D/d) (Wellenwiderstand, S.17).",
+ "revision": 4,
+ "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 $Z = (60\\,\\Omega/\\sqrt{\\varepsilon_r})\\cdot\\ln(D/d)$ (Wellenwiderstand, S.17).",
"source": "https://50ohm.de/NEA_wellenwiderstand.html#AG306",
"confidence": 8
},
@@ -3162,8 +3162,8 @@
"confidence": 8
},
"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: 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).",
+ "revision": 3,
+ "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 $\\lambda = c/f$ ($c = f\\lambda$, $\\lambda[\\text{m}] \\approx 300/f[\\text{MHz}]$, S.17), take a quarter, then apply the Verkürzungsfaktor $k_v = 1/\\sqrt{\\varepsilon_r}$ (S.17).",
"source": "https://50ohm.de/NEA_uebertragungsleitungen_3.html#AG317",
"confidence": 8
},
@@ -3282,14 +3282,14 @@
"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: apply Z = √(Z_E·Z_A) (Viertelwellentransformator, S.17).",
+ "revision": 4,
+ "explanation": "For a quarter-wave transformer, $Z_t = \\sqrt{Z_1 Z_2}$; $\\sqrt{60 \\cdot 240}$ is 120 ohms. Hilfsmittel: apply $Z = \\sqrt{Z_E\\cdot Z_A}$ (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: apply Z = √(Z_E·Z_A) (Viertelwellentransformator, S.17).",
+ "revision": 4,
+ "explanation": "The quarter-wave transformer impedance is $\\sqrt{240 \\cdot 600}$, which is about 380 ohms. Hilfsmittel: apply $Z = \\sqrt{Z_E\\cdot Z_A}$ (Viertelwellentransformator, S.17).",
"source": "https://50ohm.de/NEA_impedanztransformation.html#AG418",
"confidence": 8
},
@@ -3366,8 +3366,8 @@
"confidence": 8
},
"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: 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.",
+ "revision": 4,
+ "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 $P_\\mathrm{ERP} = P_S\\cdot 10^{(g_d-a)/(10\\,\\text{dB})}$ (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
},
@@ -3474,8 +3474,8 @@
"confidence": 8
},
"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: apply MUF ≈ f_c/sin(α) then f_opt = MUF·0,85 (S.18).",
+ "revision": 5,
+ "explanation": "Using $\\text{MUF} = f_k/\\sin(45^\\circ)$ gives about 4.2 MHz; the optimum working frequency is about 85% of MUF, or 3.6 MHz. Hilfsmittel: apply $\\text{MUF} \\approx f_c/\\sin\\alpha$ then $f_\\mathrm{opt} = \\text{MUF}\\cdot 0{,}85$ (S.18).",
"source": "https://50ohm.de/NEA_muf_luf_2.html#AH209",
"confidence": 8
},
@@ -3648,8 +3648,8 @@
"confidence": 8
},
"AI104": {
- "revision": 3,
- "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).",
+ "revision": 5,
+ "explanation": "The meter input current is $I = U/R = 0.5\\,\\text{V} / 10\\,\\text{M}\\Omega = 50\\,\\text{nA}$. Hilfsmittel: apply $I = U/R$ (Ohmsches Gesetz, S.11).",
"source": "https://50ohm.de/NEA_strom_spannung_messung_3.html#AI104",
"confidence": 8
},
@@ -3684,8 +3684,8 @@
"confidence": 8
},
"AI205": {
- "revision": 2,
- "explanation": "R = 50 ohm and jX = 0 means a purely resistive 50 ohm antenna impedance, matching a normal 50 ohm VHF transmitter output.",
+ "revision": 3,
+ "explanation": "$R = 50\\,\\Omega$ and $jX = 0$ means a purely resistive 50 ohm antenna impedance, matching a normal 50 ohm VHF transmitter output.",
"source": "https://50ohm.de/NEA_vna_2.html#AI205",
"confidence": 8
},
@@ -3732,14 +3732,14 @@
"confidence": 8
},
"AI305": {
- "revision": 3,
- "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.",
+ "revision": 6,
+ "explanation": "From the trace the peak voltage is 100 V; $P_\\mathrm{PEP} = (100/\\sqrt{2})^2 / 50$ gives 100 W. Hilfsmittel: first $U_\\mathrm{eff} = \\hat{U}/\\sqrt{2}$ (from $\\hat{U} = U_\\mathrm{eff}\\cdot\\sqrt{2}$, Wechselspannung, S.12), then $P = U_\\mathrm{eff}^2/R$ (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: correct the 10:1 probe (×10), then U_eff = Û/√2 (Wechselspannung, S.12) and P = U_eff²/R (Leistung, S.12).",
+ "revision": 6,
+ "explanation": "A 10:1 probe means the real peak voltage is ten times the displayed value; $P_\\mathrm{PEP} = (60/\\sqrt{2})^2 / 50$ gives 36 W. Hilfsmittel: correct the 10:1 probe (×10), then $U_\\mathrm{eff} = \\hat{U}/\\sqrt{2}$ (Wechselspannung, S.12) and $P = U_\\mathrm{eff}^2/R$ (Leistung, S.12).",
"source": "https://50ohm.de/NEA_ssb_3.html#AI306",
"confidence": 8
},
@@ -3792,14 +3792,14 @@
"confidence": 8
},
"AI506": {
- "revision": 3,
- "explanation": "0.01% is $10^{-4}$; $29 MHz \\cdot 10^{-4} = 2900 Hz$.",
+ "revision": 4,
+ "explanation": "0.01% is $10^{-4}$; $29\\,\\text{MHz} \\cdot 10^{-4} = 2900\\,\\text{Hz}$.",
"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.",
+ "revision": 3,
+ "explanation": "0.00001% is $10^{-7}$; $14100\\,\\text{kHz} \\cdot 10^{-7}$ gives a maximum error of 1.410 Hz.",
"source": "https://50ohm.de/NEA_frequenzgenauigkeit.html#AI507",
"confidence": 8
},
@@ -3858,14 +3858,14 @@
"confidence": 8
},
"AI606": {
- "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.",
+ "revision": 5,
+ "explanation": "The two Schottky diodes form a voltage doubler, whose DC output is $U_\\mathrm{DC} \\approx 2\\hat{U} - 2U_F$. So the RF peak at the 5 Ω tap is $\\hat{U}_\\mathrm{tap} = U_\\mathrm{DC}/2 + U_F = 15.3/2 + 0.23 = 7.88\\,\\text{V}$. 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 $U_\\mathrm{eff} = \\hat{U}/\\sqrt{2} = 55.7\\,\\text{V}$ and $P = U_\\mathrm{eff}^2/R = 55.7^2/50 \\approx 62\\,\\text{W} \\approx 60\\,\\text{W}$. (Treating the circuit as a plain peak detector — forgetting the doubler's factor of 2 — would wrongly give about 240 W.) Hilfsmittel: $\\hat{U} = U_\\mathrm{eff}\\cdot\\sqrt{2}$ and $P = U_\\mathrm{eff}^2/R$ (both Leistung/Wechselspannung, S.12); the voltage-doubler behaviour, the Schottky drop and the $5\\,\\Omega$-tap ×10 factor are added knowledge.",
"source": "https://50ohm.de/NEA_sender_messungen.html#AI606",
"confidence": 8
},
"AI607": {
- "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.",
+ "revision": 5,
+ "explanation": "The two Schottky diodes form a voltage doubler, whose DC output is $U_\\mathrm{DC} \\approx 2\\hat{U} - 2U_F$, so the RF peak at the input is $\\hat{U} = U_\\mathrm{DC}/2 + U_F = 15.3/2 + 0.23 = 7.88\\,\\text{V}$, i.e. $U_\\mathrm{eff} = \\hat{U}/\\sqrt{2} = 5.57\\,\\text{V}$. The circuit input impedance is $56\\,\\Omega \\parallel 470\\,\\Omega \\approx 50\\,\\Omega$, so $P = U_\\mathrm{eff}^2/R = 5.57^2/50 \\approx 0.62\\,\\text{W} \\approx 600\\,\\text{mW}$. (Forgetting the doubler's factor of 2 would wrongly give about 2.4 W.) Hilfsmittel: $\\hat{U} = U_\\mathrm{eff}\\cdot\\sqrt{2}$ and $P = U_\\mathrm{eff}^2/R$ (both Leistung/Wechselspannung, S.12); the voltage-doubler behaviour, the Schottky drop and the $56\\,\\Omega \\parallel 470\\,\\Omega \\approx 50\\,\\Omega$ input are added knowledge.",
"source": "https://50ohm.de/NEA_sender_messungen.html#AI607",
"confidence": 8
},
@@ -3882,14 +3882,14 @@
"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: 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.",
+ "revision": 5,
+ "explanation": "For 1 W into 50 ohms, $U_\\mathrm{rms}=\\sqrt{50}$ 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, $U_\\mathrm{eff} = \\sqrt{P\\cdot R}$ (Leistung, S.12) and $\\hat{U} = U_\\mathrm{eff}\\cdot\\sqrt{2}$ (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: add the diode drop (knowledge), convert peak→RMS (Û = U_eff·√2, Wechselspannung, S.12), then P = U_eff²/R (Leistung, S.12).",
+ "revision": 3,
+ "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 ($\\hat{U} = U_\\mathrm{eff}\\cdot\\sqrt{2}$, Wechselspannung, S.12), then $P = U_\\mathrm{eff}^2/R$ (Leistung, S.12).",
"source": "https://50ohm.de/NEA_sender_messungen.html#AI611",
"confidence": 8
},
@@ -4032,32 +4032,32 @@
"confidence": 8
},
"AJ201": {
- "revision": 3,
- "explanation": "The second harmonic is twice the fundamental: $2 \\cdot 3.730 MHz = 7.460 MHz$.",
+ "revision": 4,
+ "explanation": "The second harmonic is twice the fundamental: $2 \\cdot 3.730\\,\\text{MHz} = 7.460\\,\\text{MHz}$.",
"source": "https://50ohm.de/NEA_nicht_sinus_signale.html#AJ201",
"confidence": 8
},
"AJ202": {
- "revision": 3,
- "explanation": "The third harmonic is three times the fundamental: $3 \\cdot 7.050 MHz = 21.150 MHz$.",
+ "revision": 4,
+ "explanation": "The third harmonic is three times the fundamental: $3 \\cdot 7.050\\,\\text{MHz} = 21.150\\,\\text{MHz}$.",
"source": "https://50ohm.de/NEA_nicht_sinus_signale.html#AJ202",
"confidence": 8
},
"AJ203": {
- "revision": 3,
- "explanation": "The third overtone is the fourth harmonic, so $4 \\cdot 7.20 MHz = 28.80 MHz$.",
+ "revision": 4,
+ "explanation": "The third overtone is the fourth harmonic, so $4 \\cdot 7.20\\,\\text{MHz} = 28.80\\,\\text{MHz}$.",
"source": "https://50ohm.de/NEA_unerwuenschte_aussendungen_3.html#AJ203",
"confidence": 8
},
"AJ204": {
- "revision": 2,
- "explanation": "The third harmonic is $3 \\cdot 29.5 MHz = 88.5 MHz$, which lies in the FM broadcast band.",
+ "revision": 3,
+ "explanation": "The third harmonic is $3 \\cdot 29.5\\,\\text{MHz} = 88.5\\,\\text{MHz}$, which lies in the FM broadcast band.",
"source": "https://50ohm.de/NEA_unerwuenschte_aussendungen_3.html#AJ204",
"confidence": 8
},
"AJ205": {
- "revision": 3,
- "explanation": "Odd harmonics are 1st, 3rd, 5th, ...; the second odd harmonic is the 3rd, so $3 \\cdot 144.690 MHz = 434.070 MHz$.",
+ "revision": 4,
+ "explanation": "Odd harmonics are 1st, 3rd, 5th, ...; the second odd harmonic is the 3rd, so $3 \\cdot 144.690\\,\\text{MHz} = 434.070\\,\\text{MHz}$.",
"source": "https://50ohm.de/NEA_nicht_sinus_signale.html#AJ205",
"confidence": 8
},
@@ -4194,20 +4194,20 @@
"confidence": 8
},
"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 simple field/power relation only holds in the far field, d > λ/(2π) (E = √(30Ω·P_EIRP)/d (Feldstärke im Fernfeld, S.15)).",
+ "revision": 5,
+ "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, $d > \\lambda/(2\\pi)$ ($E = \\sqrt{30\\,\\Omega\\cdot P_\\mathrm{EIRP}}/d$, 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: feed-line loss in dB subtracts from transmitter power before the antenna; a = 10·log10(P_in/P_out) (Pegel, S.15).",
+ "revision": 3,
+ "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; $a = 10\\cdot\\log_{10}(P_\\mathrm{in}/P_\\mathrm{out})$ (Pegel, S.15).",
"source": "https://50ohm.de/NEA_personenschutzabstand_3.html#AK104",
"confidence": 8
},
"AK105": {
- "revision": 3,
- "explanation": "Safety distance scales as $d \\propto \\sqrt{P_{EIRP}}/E$. A 6 dB pattern attenuation cuts EIRP in that direction by a factor of 4, so the distance shrinks by $\\sqrt{4}=2$: 20 m becomes 10 m.",
+ "revision": 4,
+ "explanation": "Safety distance scales as $d \\propto \\sqrt{P_\\mathrm{EIRP}}/E$. A 6 dB pattern attenuation cuts EIRP in that direction by a factor of 4, so the distance shrinks by $\\sqrt{4}=2$: 20 m becomes 10 m.",
"source": "https://50ohm.de/NEA_personenschutzabstand_richtantennen.html#AK105",
"confidence": 8
},
@@ -4218,8 +4218,8 @@
"confidence": 8
},
"AK107": {
- "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.",
+ "revision": 6,
+ "explanation": "Rearrange $E = \\sqrt{30\\,\\Omega \\cdot P_\\mathrm{EIRP}}/d$ for the radiated power: $P_\\mathrm{EIRP} = (E \\cdot d)^2/(30\\,\\Omega) = (28 \\cdot 5)^2/(30\\,\\Omega) = 140^2/(30\\,\\Omega) \\approx 653\\,\\text{W}$ 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 $P = P_\\mathrm{EIRP}/G = 653/6.53 \\approx 100\\,\\text{W}$. Hilfsmittel: rearrange $E = \\sqrt{30\\,\\Omega\\cdot P_\\mathrm{EIRP}}/d$ (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": 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.",
+ "revision": 6,
+ "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 $P_\\mathrm{EIRP} = 75\\,\\text{W} \\cdot 16.4 \\approx 1231\\,\\text{W}$. Then $d = \\sqrt{30\\,\\Omega \\cdot P_\\mathrm{EIRP}}/E = \\sqrt{30\\,\\Omega \\cdot 1231}/28 \\approx 192/28 \\approx 6.86\\,\\text{m}$. Hilfsmittel: rearrange $E = \\sqrt{30\\,\\Omega\\cdot P_\\mathrm{EIRP}}/d$ (Feldstärke im Fernfeld, S.15) for distance: $d = \\sqrt{30\\,\\Omega\\cdot P_\\mathrm{EIRP}}/E$.",
"source": "https://50ohm.de/NEA_naeherungsformel_2.html#AK110",
"confidence": 8
},
@@ -4254,20 +4254,20 @@
"confidence": 8
},
"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: 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).",
+ "revision": 8,
+ "explanation": "12.15 dBi is a linear gain of about 16.4; $\\sqrt{30\\,\\Omega \\cdot 250\\,\\text{W} \\cdot 16.4}/30\\,\\text{m}$ is about 11.7 V/m. Hilfsmittel: apply $E = \\sqrt{30\\,\\Omega\\cdot P_\\mathrm{EIRP}}/d$ (Feldstärke im Fernfeld, S.15) after converting 12,15 dBi to a factor ($G = 10^{g/(10\\,\\text{dB})}$, 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: apply E = √(30Ω·P_EIRP)/d (Feldstärke im Fernfeld, S.15) with the dipole gain (g_i = 2,15 dBi, S.15).",
+ "revision": 4,
+ "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 $E = \\sqrt{30\\,\\Omega\\cdot P_\\mathrm{EIRP}}/d$ (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: first ERP→EIRP (×1,64), then apply E = √(30Ω·P_EIRP)/d (Feldstärke im Fernfeld, S.15).",
+ "revision": 7,
+ "explanation": "Convert 100 W ERP to about 164 W EIRP, then $\\sqrt{30\\,\\Omega \\cdot 164\\,\\text{W}}/100\\,\\text{m}$ gives about 0.7 V/m. Hilfsmittel: first ERP→EIRP (×1,64), then apply $E = \\sqrt{30\\,\\Omega\\cdot P_\\mathrm{EIRP}}/d$ (Feldstärke im Fernfeld, S.15).",
"source": "https://50ohm.de/NEA_personenschutzabstand_3.html#AK115",
"confidence": 8
},
@@ -5364,62 +5364,62 @@
"confidence": 8
},
"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: apply g = 10·log10(P2/P1) (Pegel, S.15); the table (S.15) gives ×2 → 3 dB, ×10 → 10 dB.",
+ "revision": 5,
+ "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 $g = 10\\cdot\\log_{10}(P_2/P_1)$ (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: use the Zehnerpotenzen/SI-Präfix table (S.11): micro = 10⁻⁶.",
+ "revision": 4,
+ "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: use the Zehnerpotenzen/SI-Präfix table (S.11): milli = 10⁻³.",
+ "revision": 5,
+ "explanation": "Milli means $10^{-3}$. Move from amperes to milliamperes by multiplying by 1000: $0.042\\,\\text{A} = 42\\,\\text{mA} = 42 \\cdot 10^{-3}\\,\\text{A}$. 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: use the Zehnerpotenzen/SI-Präfix table (S.11) for the powers of ten (scientific notation).",
+ "revision": 5,
+ "explanation": "Scientific notation keeps one non-zero digit before the decimal point. Moving the decimal in $4{,}200{,}000$ six places gives $4.2 \\cdot 10^6\\,\\text{Hz}$, 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: use the Zehnerpotenzen/SI-Präfix table (S.11): milli = 10⁻³, micro = 10⁻⁶.",
+ "revision": 5,
+ "explanation": "A millivolt is $10^{-3}\\,\\text{V}$ and a microvolt is $10^{-6}\\,\\text{V}$. So $0.01\\,\\text{mV} = 0.01 \\cdot 10^{-3}\\,\\text{V} = 10 \\cdot 10^{-6}\\,\\text{V} = 10\\,\\mu\\text{V}$. 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: use the Zehnerpotenzen/SI-Präfix table (S.11): mega = 10⁶.",
+ "revision": 5,
+ "explanation": "Mega means $10^6$. Therefore $0.002\\,\\text{M}\\Omega = 0.002 \\cdot 10^6\\,\\Omega = 2000\\,\\Omega = 2\\,\\text{k}\\Omega$. Hilfsmittel: use the Zehnerpotenzen/SI-Präfix table (S.11): mega = $10^6$.",
"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: use the Zehnerpotenzen/SI-Präfix table (S.11): micro = 10⁻⁶.",
+ "revision": 5,
+ "explanation": "A microwatt is $10^{-6}\\,\\text{W}$. To convert watts to microwatts, divide by $10^{-6}$: $2 \\cdot 10^{-7}\\,\\text{W} / 10^{-6} = 0.2\\,\\mu\\text{W}$. 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: use the Zehnerpotenzen/SI-Präfix table (S.11): 10⁻¹ and milli = 10⁻³.",
+ "revision": 5,
+ "explanation": "$5 \\cdot 10^{-1}\\,\\text{W}$ is $0.5\\,\\text{W}$. Since $1\\,\\text{W} = 1000\\,\\text{mW}$, $0.5\\,\\text{W} = 500\\,\\text{mW}$. 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: use the Zehnerpotenzen/SI-Präfix table (S.11): micro = 10⁻⁶, nano = 10⁻⁹.",
+ "revision": 6,
+ "explanation": "Micro is $10^{-6}$ and nano is $10^{-9}$, so one microfarad is 1000 nanofarads. Thus $0.22\\,\\mu\\text{F} = 0.22 \\cdot 1000\\,\\text{nF} = 220\\,\\text{nF}$. 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: use the Zehnerpotenzen/SI-Präfix table (S.11): kilo = 10³, mega = 10⁶.",
+ "revision": 6,
+ "explanation": "Kilo is $10^3$ and mega is $10^6$, so converting kHz to MHz divides by 1000. $3750\\,\\text{kHz} = 3.750\\,\\text{MHz}$. Hilfsmittel: use the Zehnerpotenzen/SI-Präfix table (S.11): kilo = $10^3$, mega = $10^6$.",
"source": "https://50ohm.de/NEA_zehnerpotenzen.html#EA116",
"confidence": 8
},
@@ -5478,20 +5478,20 @@
"confidence": 8
},
"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: apply E = U/d (E-Feld im homogenen Feld, S.13); convert the gap to metres.",
+ "revision": 6,
+ "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: apply E = U/d (S.13); convert the gap to metres.",
+ "revision": 6,
+ "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: rearrange E = U/d → U_max = E·d (S.13); keep units consistent.",
+ "revision": 6,
+ "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 $E = U/d \\to U_\\mathrm{max} = E\\cdot d$ (S.13); keep units consistent.",
"source": "https://50ohm.de/NEA_e_feld.html#EB104",
"confidence": 8
},
@@ -5514,8 +5514,8 @@
"confidence": 8
},
"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: apply H = N·I/l_m with l_m = π·d (Magnetische Feldstärke, S.13).",
+ "revision": 5,
+ "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 $H = N\\cdot I/l_m$ with $l_m = \\pi d$ (Magnetische Feldstärke, S.13).",
"source": "https://50ohm.de/NEA_h_feld.html#EB203",
"confidence": 8
},
@@ -5598,104 +5598,104 @@
"confidence": 8
},
"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: apply c = f·λ (λ[m] ≈ 300/f[MHz]), S.17.",
+ "revision": 4,
+ "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 $c = f\\lambda$ ($\\lambda[\\text{m}] \\approx 300/f[\\text{MHz}]$), 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: apply c = f·λ (λ[m] ≈ 300/f[MHz]), S.17.",
+ "revision": 5,
+ "explanation": "Use $\\lambda = c/f$. For radio exam mental math, with $c \\approx 300$ million m/s and frequency in MHz, $\\lambda_m \\approx 300/f_{\\text{MHz}}$. At 21 MHz, $300/21 \\approx 14.29$ m. Hilfsmittel: apply $c = f\\lambda$ ($\\lambda[\\text{m}] \\approx 300/f[\\text{MHz}]$), 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: apply c = f·λ (λ[m] ≈ 300/f[MHz]), S.17.",
+ "revision": 5,
+ "explanation": "Wavelength and frequency are inversely related: $\\lambda = c/f$. With frequency in MHz, use $\\lambda_m \\approx 300/f_{\\text{MHz}}$; for 28.5 MHz this gives $300/28.5 \\approx 10.5$ m. Hilfsmittel: apply $c = f\\lambda$ ($\\lambda[\\text{m}] \\approx 300/f[\\text{MHz}]$), 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: rearrange c = f·λ (λ[m] ≈ 300/f[MHz]), S.17: f[MHz] ≈ 300/λ[m].",
+ "revision": 4,
+ "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 $c = f\\lambda$ ($\\lambda[\\text{m}] \\approx 300/f[\\text{MHz}]$), S.17: $f[\\text{MHz}] \\approx 300/\\lambda[\\text{m}]$.",
"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: apply f = c/λ (c = f·λ (λ[m] ≈ 300/f[MHz]), S.17).",
+ "revision": 4,
+ "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 $f = c/\\lambda$ ($c = f\\lambda$, $\\lambda[\\text{m}] \\approx 300/f[\\text{MHz}]$, 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: apply f = c/λ (c = f·λ (λ[m] ≈ 300/f[MHz]), S.17).",
+ "revision": 4,
+ "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 $f = c/\\lambda$ ($c = f\\lambda$, $\\lambda[\\text{m}] \\approx 300/f[\\text{MHz}]$, 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: apply Û = U_eff·√2 (Wechselspannung, S.12).",
+ "revision": 8,
+ "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 $U_\\mathrm{peak} = 230\\,\\text{V} \\cdot \\sqrt{2} \\approx 325\\,\\text{V}$. Hilfsmittel: apply $\\hat{U} = U_\\mathrm{eff}\\cdot\\sqrt{2}$ (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: Û = U_eff·√2 then U_SS = 2·Û (Wechselspannung, S.12).",
+ "revision": 8,
+ "explanation": "Peak-to-peak voltage is the full swing from negative peak to positive peak: $U_\\mathrm{pp} = 2 U_\\mathrm{peak}$. From 230 V RMS, $U_\\mathrm{peak} = 230\\sqrt{2} \\approx 325\\,\\text{V}$, so $U_\\mathrm{pp} \\approx 650\\,\\text{V}$. Hilfsmittel: $\\hat{U} = U_\\mathrm{eff}\\cdot\\sqrt{2}$ then $U_\\mathrm{SS} = 2\\hat{U}$ (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: Û = U_eff·√2 then U_SS = 2·Û (Wechselspannung, S.12).",
+ "revision": 7,
+ "explanation": "For sine waves, $U_\\mathrm{peak} = U_\\mathrm{RMS}\\sqrt{2}$ and $U_\\mathrm{pp} = 2U_\\mathrm{peak}$. With 12 V RMS, the peak is about 17 V and peak-to-peak is about 34 V. Hilfsmittel: $\\hat{U} = U_\\mathrm{eff}\\cdot\\sqrt{2}$ then $U_\\mathrm{SS} = 2\\hat{U}$ (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: rearrange Û = U_eff·√2 → U_eff = Û/√2 (Wechselspannung, S.12).",
+ "revision": 7,
+ "explanation": "For sine waves, RMS is peak divided by $\\sqrt{2}$ because RMS represents equal heating power. $12\\,\\text{V} / 1.414 \\approx 8.5\\,\\text{V}$. Hilfsmittel: rearrange $\\hat{U} = U_\\mathrm{eff}\\cdot\\sqrt{2} \\to U_\\mathrm{eff} = \\hat{U}/\\sqrt{2}$ (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: U_eff = Û/√2 (Û = U_eff·√2, Wechselspannung, S.12).",
+ "revision": 6,
+ "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: $U_\\mathrm{eff} = \\hat{U}/\\sqrt{2}$ ($\\hat{U} = U_\\mathrm{eff}\\cdot\\sqrt{2}$, 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: read U_SS off the grid; relate via U_SS = 2·Û, Û = U_eff·√2 (Wechselspannung, S.12).",
+ "revision": 6,
+ "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 $U_\\mathrm{SS}$ off the grid; relate via $U_\\mathrm{SS} = 2\\hat{U}$, $\\hat{U} = U_\\mathrm{eff}\\cdot\\sqrt{2}$ (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: U_SS = 2·Û (Wechselspannung, S.12).",
+ "revision": 6,
+ "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 $2 \\cdot 20\\,\\text{V} = 40\\,\\text{V}$. Hilfsmittel: $U_\\mathrm{SS} = 2\\hat{U}$ (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: apply f = 1/T (Wechselspannung, S.12).",
+ "revision": 9,
+ "explanation": "Frequency is cycles per second, so it is the reciprocal of period: $f = 1/T$. With $T = 50\\,\\mu\\text{s} = 50 \\cdot 10^{-6}\\,\\text{s}$, $f = 1/T = 20{,}000\\,\\text{Hz} = 20\\,\\text{kHz}$. 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: read T off the grid, then f = 1/T (Wechselspannung, S.12).",
+ "revision": 7,
+ "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 $12\\,\\mu\\text{s}$, so $f \\approx 1/(12 \\cdot 10^{-6}) = 83.3\\,\\text{kHz}$. 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: T = divisions × time/div, then f = 1/T (Wechselspannung, S.12).",
+ "revision": 7,
+ "explanation": "Oscilloscope timebase reading is divisions times time/div. Four divisions at 5 ms/div gives $T = 20\\,\\text{ms} = 0.020\\,\\text{s}$, so $f = 1/T = 50\\,\\text{Hz}$. Hilfsmittel: $T = \\text{divisions} \\times \\text{time/div}$, 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: read T off the grid, then f = 1/T (Wechselspannung, S.12).",
+ "revision": 8,
+ "explanation": "Read the period from the grid: $4 \\cdot 0.03\\,\\mu\\text{s} = 0.12\\,\\mu\\text{s}$. Then $f = 1/T = 1/(0.12 \\cdot 10^{-6}\\,\\text{s}) \\approx 8.33\\,\\text{MHz}$. Hilfsmittel: read T off the grid, then $f = 1/T$ (Wechselspannung, S.12).",
"source": "https://50ohm.de/NEA_oszilloskop_1.html#EB411",
"confidence": 8
},
@@ -5718,62 +5718,62 @@
"confidence": 8
},
"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: combine P = U·I with I = U/R to get P = U²/R, hence U = √(P·R) (Leistung, S.12).",
+ "revision": 4,
+ "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 $P = U\\cdot I$ with $I = U/R$ to get $P = U^2/R$, hence $U = \\sqrt{P\\cdot R}$ (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: from P = I²·R and P = U²/R: I = √(P/R), U = √(P·R) (Leistung, S.12).",
+ "revision": 4,
+ "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 $P = I^2\\cdot R$ and $P = U^2/R$: $I = \\sqrt{P/R}$, $U = \\sqrt{P\\cdot R}$ (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: rearrange P = U²/R and P = I²·R for R (Leistung, S.12).",
+ "revision": 4,
+ "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 $P = U^2/R$ and $P = I^2\\cdot R$ 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: apply P = U²/R (Leistung, S.12) with RMS voltage.",
+ "revision": 6,
+ "explanation": "For RF power in a resistive 50 Ohm load, use RMS voltage: $P = U^2/R$. With $U = 100\\,\\text{V}$ and $R = 50\\,\\Omega$, $P = 100^2/50 = 200\\,\\text{W}$. Hilfsmittel: apply $P = U^2/R$ (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: apply P = I²·R (Leistung, S.12) with RMS current.",
+ "revision": 6,
+ "explanation": "Use RMS current in the resistive power formula $P = I^2R$. With $I = 2\\,\\text{A}$ and $R = 50\\,\\Omega$, $P = 2^2 \\cdot 50 = 200\\,\\text{W}$. Hilfsmittel: apply $P = I^2\\cdot R$ (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: apply P = U²/R (Leistung, S.12).",
+ "revision": 5,
+ "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 $P = U^2/R$ (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 power limit gives U = √(P·R) (Leistung, S.12); compare with the voltage rating, lower wins.",
+ "revision": 5,
+ "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 $U = \\sqrt{P\\cdot R}$ (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 power limit gives U = √(P·R) (Leistung, S.12); compare with the voltage rating, lower wins.",
+ "revision": 5,
+ "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 $U = \\sqrt{P\\cdot R}$ (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: rearrange P = I²·R → I = √(P/R) (Leistung, S.12).",
+ "revision": 6,
+ "explanation": "Solve $P = I^2R$ for current: $I = \\sqrt{P/R}$. With $P = 23.0\\,\\text{W}$ and $R = 120\\,\\Omega$, $I = \\sqrt{23/120} \\approx 0.438\\,\\text{A} = 438\\,\\text{mA}$. Hilfsmittel: rearrange $P = I^2\\cdot R \\to I = \\sqrt{P/R}$ (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: first Û = U_SS/2 and U_eff = Û/√2 (Wechselspannung, S.12), then I = U/R (Ohmsches Gesetz, S.11).",
+ "revision": 5,
+ "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 $\\hat{U} = U_\\mathrm{SS}/2$ and $U_\\mathrm{eff} = \\hat{U}/\\sqrt{2}$ (Wechselspannung, S.12), then $I = U/R$ (Ohmsches Gesetz, S.11).",
"source": "https://50ohm.de/NEA_leistung_2.html#EB513",
"confidence": 8
},
@@ -5868,20 +5868,20 @@
"confidence": 8
},
"EC115": {
- "revision": 3,
- "explanation": "Three-digit SMD resistor codes use the first two digits as significant figures and the third as the zero count. `103` means $10 \\cdot 10^3 Ohm = 10,000 Ohm = 10 kOhm$.",
+ "revision": 5,
+ "explanation": "Three-digit SMD resistor codes use the first two digits as significant figures and the third as the zero count. `103` means $10 \\cdot 10^3\\,\\Omega = 10{,}000\\,\\Omega = 10\\,\\text{k}\\Omega$.",
"source": "https://50ohm.de/NEA_widerstand_smd.html#EC115",
"confidence": 8
},
"EC116": {
- "revision": 3,
- "explanation": "For SMD marking `221`, the significant digits are 22 and the multiplier digit 1 means one zero. So $22 \\cdot 10^1 Ohm = 220 Ohm$.",
+ "revision": 4,
+ "explanation": "For SMD marking `221`, the significant digits are 22 and the multiplier digit 1 means one zero. So $22 \\cdot 10^1\\,\\Omega = 220\\,\\Omega$.",
"source": "https://50ohm.de/NEA_widerstand_smd.html#EC116",
"confidence": 8
},
"EC117": {
- "revision": 3,
- "explanation": "The SMD code `223` means 22 with three zeros after it: $22 \\cdot 10^3 Ohm = 22,000 Ohm = 22 kOhm$.",
+ "revision": 5,
+ "explanation": "The SMD code `223` means 22 with three zeros after it: $22 \\cdot 10^3\\,\\Omega = 22{,}000\\,\\Omega = 22\\,\\text{k}\\Omega$.",
"source": "https://50ohm.de/NEA_widerstand_smd.html#EC117",
"confidence": 8
},
@@ -5952,14 +5952,14 @@
"confidence": 8
},
"EC305": {
- "revision": 3,
- "explanation": "Coil inductance follows $L = \\mu_0 \\mu_r N^2 A / l$. Squeezing the cylindrical coil shorter (smaller $l$) at unchanged turns raises $L$. A copper core would lower $L$ (eddy currents oppose the flux), and a shield can also reduce it — so only compressing the winding increases inductance.",
+ "revision": 4,
+ "explanation": "Coil inductance follows $L = \\mu_0\\,\\mu_r N^2 A / l$. Squeezing the cylindrical coil shorter (smaller $l$) at unchanged turns raises $L$. A copper core would lower $L$ (eddy currents oppose the flux), and a shield can also reduce it — so only compressing the winding increases inductance.",
"source": "https://50ohm.de/NEA_spule_1.html#EC305",
"confidence": 8
},
"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: L = μ0·μr·N²·A/l is ∝ 1/l (Zylinderspule, S.13); doubling length halves L.",
+ "revision": 6,
+ "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: $L = \\mu_0\\mu_r N^2 A/l$ is $\\propto 1/l$ (Zylinderspule, S.13); doubling length halves L.",
"source": "https://50ohm.de/NEA_spule_1.html#EC306",
"confidence": 8
},
@@ -5970,14 +5970,14 @@
"confidence": 8
},
"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: 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).",
+ "revision": 6,
+ "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 $230/15 \\approx 15.3\\,\\text{V}$, so about 15 V. Hilfsmittel: apply $U_2 = U_1\\cdot N_2/N_1$ (Übersetzungsverhältnis $\\ddot{u} = N_P/N_S = U_P/U_S = I_S/I_P = \\sqrt{Z_P/Z_S}$, 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: 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).",
+ "revision": 5,
+ "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 $U_2 = U_1\\cdot N_2/N_1$ (Übersetzungsverhältnis $\\ddot{u} = N_P/N_S = U_P/U_S = I_S/I_P = \\sqrt{Z_P/Z_S}$, S.13).",
"source": "https://50ohm.de/NEA_uebertrager_1.html#EC402",
"confidence": 8
},
@@ -6042,26 +6042,26 @@
"confidence": 8
},
"EC509": {
- "revision": 3,
- "explanation": "For a silicon diode to conduct, the anode must be about 0.7 V above the cathode. Here the anode-cathode difference is $1.3 V - 0.6 V = 0.7 V$, so it is forward-biased.",
+ "revision": 4,
+ "explanation": "For a silicon diode to conduct, the anode must be about 0.7 V above the cathode. Here the anode-cathode difference is $1.3\\,\\text{V} - 0.6\\,\\text{V} = 0.7\\,\\text{V}$, so it is forward-biased.",
"source": "https://50ohm.de/NEA_diode_1.html#EC509",
"confidence": 8
},
"EC510": {
- "revision": 3,
- "explanation": "Use voltage difference, not absolute voltage. The selected case has the anode at $0.3 V$ and cathode at $-0.4 V$, so $U_A-U_K = 0.7 V$ and the silicon diode conducts.",
+ "revision": 4,
+ "explanation": "Use voltage difference, not absolute voltage. The selected case has the anode at $0.3\\,\\text{V}$ and cathode at $-0.4\\,\\text{V}$, so $U_A-U_K = 0.7\\,\\text{V}$ and the silicon diode conducts.",
"source": "https://50ohm.de/NEA_diode_1.html#EC510",
"confidence": 8
},
"EC511": {
- "revision": 3,
- "explanation": "Forward bias depends on relative voltage. Even though both nodes are negative, $-1.3 V$ at the anode is $0.7 V$ higher than $-2.0 V$ at the cathode, so the silicon diode conducts.",
+ "revision": 4,
+ "explanation": "Forward bias depends on relative voltage. Even though both nodes are negative, $-1.3\\,\\text{V}$ at the anode is $0.7\\,\\text{V}$ higher than $-2.0\\,\\text{V}$ at the cathode, so the silicon diode conducts.",
"source": "https://50ohm.de/NEA_diode_1.html#EC511",
"confidence": 8
},
"EC512": {
- "revision": 3,
- "explanation": "The silicon-diode test is $U_A - U_K \\approx 0.7 V$. In the selected drawing, $-3.0 V - (-3.7 V) = 0.7 V$, so the anode is sufficiently above the cathode.",
+ "revision": 4,
+ "explanation": "The silicon-diode test is $U_A - U_K \\approx 0.7\\,\\text{V}$. In the selected drawing, $-3.0\\,\\text{V} - (-3.7\\,\\text{V}) = 0.7\\,\\text{V}$, so the anode is sufficiently above the cathode.",
"source": "https://50ohm.de/NEA_diode_1.html#EC512",
"confidence": 8
},
@@ -6084,8 +6084,8 @@
"confidence": 8
},
"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: first R = U/I with the LED drop subtracted (Ohmsches Gesetz, S.11), then P = U·I (Leistung, S.12).",
+ "revision": 5,
+ "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 $P = U\\cdot I$ (Leistung, S.12).",
"source": "https://50ohm.de/NEA_diode_1.html#EC516",
"confidence": 8
},
@@ -6114,8 +6114,8 @@
"confidence": 8
},
"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: apply R = U/I (Ohmsches Gesetz, S.11) to the resistor's drop and Zener current.",
+ "revision": 4,
+ "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
},
@@ -6180,8 +6180,8 @@
"confidence": 8
},
"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.",
+ "revision": 6,
+ "explanation": "A silicon BJT conducts once its base-emitter junction is forward-biased past the silicon threshold, about $0.6$-$0.7$ V, so $U_\\mathrm{BE} \\approx 0.6$ V (positive for an NPN). A negative or zero $U_\\mathrm{BE}$ leaves the junction off, so those options do not turn it on.",
"source": "https://50ohm.de/NEA_transistor_1.html#EC610",
"confidence": 8
},
@@ -6192,56 +6192,56 @@
"confidence": 8
},
"EC612": {
- "revision": 4,
- "explanation": "An NPN BJT conducts when its silicon base-emitter junction is forward biased, so $V_B - V_E \\approx +0.6 V$. The selected drawing has +2.0 V at the base and +1.4 V at the emitter, exactly the needed forward bias.",
+ "revision": 5,
+ "explanation": "An NPN BJT conducts when its silicon base-emitter junction is forward biased, so $V_B - V_E \\approx +0.6\\,\\text{V}$. The selected drawing has +2.0 V at the base and +1.4 V at the emitter, exactly the needed forward bias.",
"source": "https://50ohm.de/NEA_transistor_1.html#EC612",
"confidence": 8
},
"EC613": {
- "revision": 4,
- "explanation": "For an NPN transistor, compare base to emitter, not either point to ground: $V_{BE}=V_B-V_E$. Here $-5.6 V - (-6.2 V)=+0.6 V$, so the base-emitter junction is forward biased and collector current can flow.",
+ "revision": 6,
+ "explanation": "For an NPN transistor, compare base to emitter, not either point to ground: $V_\\mathrm{BE}=V_B-V_E$. Here $-5.6\\,\\text{V} - (-6.2\\,\\text{V})=+0.6\\,\\text{V}$, so the base-emitter junction is forward biased and collector current can flow.",
"source": "https://50ohm.de/NEA_transistor_1.html#EC613",
"confidence": 8
},
"EC614": {
- "revision": 4,
- "explanation": "A PNP BJT is the polarity mirror of an NPN: it conducts when the base is about 0.6 V lower than the emitter, so $V_B - V_E \\approx -0.6 V$. The selected drawing has -2.0 V at the base and -1.4 V at the emitter.",
+ "revision": 5,
+ "explanation": "A PNP BJT is the polarity mirror of an NPN: it conducts when the base is about 0.6 V lower than the emitter, so $V_B - V_E \\approx -0.6\\,\\text{V}$. The selected drawing has -2.0 V at the base and -1.4 V at the emitter.",
"source": "https://50ohm.de/NEA_transistor_1.html#EC614",
"confidence": 8
},
"EC615": {
- "revision": 4,
- "explanation": "For a PNP transistor the conducting condition is $V_{BE}\\approx -0.6 V$, meaning the base is 0.6 V below the emitter. Here $+5.6 V - +6.2 V = -0.6 V$, so the base-emitter junction is correctly forward biased.",
+ "revision": 6,
+ "explanation": "For a PNP transistor the conducting condition is $V_\\mathrm{BE}\\approx -0.6\\,\\text{V}$, meaning the base is 0.6 V below the emitter. Here $+5.6\\,\\text{V} - +6.2\\,\\text{V} = -0.6\\,\\text{V}$, so the base-emitter junction is correctly forward biased.",
"source": "https://50ohm.de/NEA_transistor_1.html#EC615",
"confidence": 8
},
"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: series voltages split as U1/U2 = R1/R2 (Spannungsteiler, S.12).",
+ "revision": 4,
+ "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 $U_1/U_2 = R_1/R_2$ (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: series voltages split as U1/U2 = R1/R2 (Spannungsteiler, S.12).",
+ "revision": 4,
+ "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 $U_1/U_2 = R_1/R_2$ (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: apply the divider U2 = U·R2/(R1+R2) (Spannungsteiler, S.12).",
+ "revision": 4,
+ "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 $U_2 = U\\cdot R_2/(R_1+R_2)$ (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: apply R_g = R1·R2/(R1+R2) (Reihen-/Parallelschaltung, S.12).",
+ "revision": 5,
+ "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: $100\\cdot400/(100+400)=40000/500=80\\,\\Omega$. Hilfsmittel: apply $R_g = R_1 R_2/(R_1+R_2)$ (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: apply R_g = R1·R2/(R1+R2) (Reihen-/Parallelschaltung, S.12).",
+ "revision": 5,
+ "explanation": "Parallel resistance is always lower than the smallest branch. For two branches use $R_g = R_1R_2/(R_1+R_2)$: $50\\cdot200/(50+200)=10000/250=40\\,\\Omega$. Hilfsmittel: apply $R_g = R_1 R_2/(R_1+R_2)$ (Reihen-/Parallelschaltung, S.12).",
"source": "https://50ohm.de/NEA_reihe_parallel_widerstand.html#ED105",
"confidence": 8
},
@@ -6258,44 +6258,44 @@
"confidence": 8
},
"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: series adds, two equal parallel resistors halve (Reihen-/Parallelschaltung, S.12).",
+ "revision": 4,
+ "explanation": "Reduce mixed resistor networks one obvious block at a time. Series resistors add directly: $500 + 500 = 1000\\,\\Omega$. That 1000 Ohm branch is parallel with another 1000 Ohm resistor, and two equal parallel resistors halve: $1000/2 = 500\\,\\Omega$. 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: series adds, two equal parallel resistors halve (Reihen-/Parallelschaltung, S.12).",
+ "revision": 4,
+ "explanation": "First combine the series path: $500\\,\\Omega + 1.5\\,\\text{k}\\Omega = 2\\,\\text{k}\\Omega$. That branch is parallel to another 2 kOhm branch, and equal parallel resistors give half the value, so $R_g = 1\\,\\text{k}\\Omega$. 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: two equal parallel resistors halve, then add the series part (Reihen-/Parallelschaltung, S.12).",
+ "revision": 4,
+ "explanation": "Parallel branches have the same voltage; for two equal 1 kOhm resistors the equivalent is $1\\,\\text{k}\\Omega / 2 = 500\\,\\Omega$. Add the remaining 500 Ohm series resistor and the total becomes $500 + 500 = 1000\\,\\Omega$. 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: two equal parallel resistors halve, then add the series part (Reihen-/Parallelschaltung, S.12).",
+ "revision": 4,
+ "explanation": "Start with the parallel part: two equal 2 kOhm resistors in parallel give $2\\,\\text{k}\\Omega / 2 = 1\\,\\text{k}\\Omega$. That equivalent is then in series with R1, so $1\\,\\text{k}\\Omega + 1\\,\\text{k}\\Omega = 2\\,\\text{k}\\Omega$. 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: parallel R_g = R1·R2/(R1+R2), then add the series part (Reihen-/Parallelschaltung, S.12).",
+ "revision": 6,
+ "explanation": "For two unequal parallel resistors use $R_g = R_1R_2/(R_1+R_2)$. Thus $3\\,\\text{k}\\Omega \\parallel 1.5\\,\\text{k}\\Omega = 4.5/4.5\\,\\text{k}\\Omega = 1\\,\\text{k}\\Omega$, and the series R1 adds another 1 kOhm for a total of 2 kOhm. Hilfsmittel: parallel $R_g = R_1 R_2/(R_1+R_2)$, 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: three in parallel via 1/R_g = Σ1/Ri, then add the series part (Reihen-/Parallelschaltung, S.12).",
+ "revision": 5,
+ "explanation": "Use reciprocal addition for three parallel resistors: $1/R = 1/10\\,\\text{k}\\Omega + 1/2.5\\,\\text{k}\\Omega + 1/500\\,\\Omega$, giving 400 Ohm. The remaining 600 Ohm is in series, so the total is $400 + 600 = 1000\\,\\Omega$. Hilfsmittel: three in parallel via $1/R_g = \\sum 1/R_i$, then add the series part (Reihen-/Parallelschaltung, S.12).",
"source": "https://50ohm.de/NEA_reihe_parallel_widerstand.html#ED113",
"confidence": 8
},
"ED114": {
- "revision": 3,
- "explanation": "Do not try to see the whole network at once; replace one series/parallel group by its equivalent, then repeat. Here $50+50=100 Ohm$, that 100 Ohm in parallel with 100 Ohm becomes 50 Ohm, and the remaining series values add to 250 Ohm.",
+ "revision": 4,
+ "explanation": "Do not try to see the whole network at once; replace one series/parallel group by its equivalent, then repeat. Here $50+50=100\\,\\Omega$, that 100 Ohm in parallel with 100 Ohm becomes 50 Ohm, and the remaining series values add to 250 Ohm.",
"source": "https://50ohm.de/NEA_reihe_parallel_widerstandsnetz_1.html#ED114",
"confidence": 8
},
@@ -6306,56 +6306,56 @@
"confidence": 8
},
"ED116": {
- "revision": 3,
- "explanation": "After each visible series/parallel block is replaced by its equivalent, the network no longer has branches: it is a series chain. The remaining values are 400 Ohm, 200 Ohm, 200 Ohm, and 150 Ohm, so $R_g = 950 Ohm$.",
+ "revision": 4,
+ "explanation": "After each visible series/parallel block is replaced by its equivalent, the network no longer has branches: it is a series chain. The remaining values are 400 Ohm, 200 Ohm, 200 Ohm, and 150 Ohm, so $R_g = 950\\,\\Omega$.",
"source": "https://50ohm.de/NEA_reihe_parallel_widerstandsnetz_1.html#ED116",
"confidence": 8
},
"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: parallel caps add: C_G = ΣCi (Kapazität, S.13).",
+ "revision": 4,
+ "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: $C_G = \\sum C_i$ (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: parallel caps add: C_G = ΣCi (Kapazität, S.13).",
+ "revision": 5,
+ "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: $C_G = \\sum C_i$ (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: series caps: 1/C_G = Σ1/Ci, equal ones → C/n (Kapazität, S.13).",
+ "revision": 5,
+ "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: $1/C_G = \\sum 1/C_i$, equal ones $\\to C/n$ (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: series caps via 1/C_G = Σ1/Ci (Kapazität, S.13).",
+ "revision": 5,
+ "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 $1/C_G = \\sum 1/C_i$ (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: series caps halve (1/C_G = Σ1/Ci), parallel caps add (Kapazität, S.13).",
+ "revision": 6,
+ "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 $10/2 = 5\\,\\text{nF}$; in parallel with $C_3 = 5\\,\\text{nF}$, the total is $10\\,\\text{nF}$. Hilfsmittel: series caps halve ($1/C_G = \\sum 1/C_i$), 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: parallel caps add, then series caps via 1/C_G = Σ1/Ci (Kapazität, S.13).",
+ "revision": 6,
+ "explanation": "First add the parallel capacitors: $C_2+C_3=1\\,\\mu\\text{F}+1\\,\\mu\\text{F}=2\\,\\mu\\text{F}$. That $2\\,\\mu\\text{F}$ equivalent is in series with $C_1 = 2\\,\\mu\\text{F}$; two equal capacitors in series halve, so $C_g=1.0\\,\\mu\\text{F}$. Hilfsmittel: parallel caps add, then series caps via $1/C_G = \\sum 1/C_i$ (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: parallel caps add, then equal series caps halve (Kapazität, S.13).",
+ "revision": 5,
+ "explanation": "Parallel capacitors add directly, so $C_2$ and $C_3$ become $4\\,\\text{nF} + 4\\,\\text{nF} = 8\\,\\text{nF}$. That $8\\,\\text{nF}$ equivalent is in series with $C_1 = 8\\,\\text{nF}$, and equal series capacitors halve, giving $C_g=4\\,\\text{nF}$. 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: parallel caps add, then equal series caps halve (Kapazität, S.13).",
+ "revision": 5,
+ "explanation": "Convert first so the units match: $100000\\,\\text{pF} = 100\\,\\text{nF}$. $C_2$ and $C_3$ are parallel, so they add to $200\\,\\text{nF}$; that is in series with $C_1 = 200\\,\\text{nF}$, and two equal series capacitances give $100\\,\\text{nF}$. Hilfsmittel: parallel caps add, then equal series caps halve (Kapazität, S.13).",
"source": "https://50ohm.de/NEA_reihe_parallel_kondensator.html#ED124",
"confidence": 8
},
@@ -6558,8 +6558,8 @@
"confidence": 8
},
"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 $21.250 MHz + 0.001 MHz = 21.251 MHz$.",
+ "revision": 5,
+ "explanation": "In upper sideband (USB), each audio component appears above the suppressed carrier by its audio frequency. A 1 kHz tone is $0.001\\,\\text{MHz}$, so $21.250\\,\\text{MHz} + 0.001\\,\\text{MHz} = 21.251\\,\\text{MHz}$.",
"source": "https://50ohm.de/NEA_ssb_2.html#EE203",
"confidence": 8
},
@@ -6726,32 +6726,32 @@
"confidence": 8
},
"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 difference product matches f_ZF = |f1 − f2| (S.14); the sum product f1 + f2 is mixer knowledge, not in the sheet.",
+ "revision": 5,
+ "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 $f_\\mathrm{ZF} = |f_1 - f_2|$ (S.14); the sum product $f_1 + f_2$ 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 difference product matches f_ZF = |f1 − f2| (S.14); the sum product f1 + f2 is mixer knowledge, not in the sheet.",
+ "revision": 5,
+ "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 $f_\\mathrm{ZF} = |f_1 - f_2|$ (S.14); the sum product $f_1 + f_2$ 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 difference product matches f_ZF = |f1 − f2| (S.14); the sum product f1 + f2 is mixer knowledge, not in the sheet.",
+ "revision": 5,
+ "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 $f_\\mathrm{ZF} = |f_1 - f_2|$ (S.14); the sum product $f_1 + f_2$ 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 difference product matches f_ZF = |f1 − f2| (S.14); the sum product f1 + f2 is mixer knowledge, not in the sheet.",
+ "revision": 5,
+ "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 $f_\\mathrm{ZF} = |f_1 - f_2|$ (S.14); the sum product $f_1 + f_2$ 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 difference product matches f_ZF = |f1 − f2| (S.14); the sum product f1 + f2 is mixer knowledge, not in the sheet.",
+ "revision": 6,
+ "explanation": "A mixer multiplies signals, producing first-order products at the sum and absolute difference: $f_\\mathrm{sum}=f_1+f_2$ and $f_\\mathrm{diff}=|f_1-f_2|$. For 145 MHz and 136 MHz these are 281 MHz and 9 MHz. Hilfsmittel: the difference product matches $f_\\mathrm{ZF} = |f_1 - f_2|$ (S.14); the sum product $f_1 + f_2$ is mixer knowledge, not in the sheet.",
"source": "https://50ohm.de/NEA_mischer.html#EF205",
"confidence": 8
},
@@ -6840,14 +6840,14 @@
"confidence": 8
},
"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 $145.2 MHz / 2 / 3 / 2 = 12.1 MHz$.",
+ "revision": 5,
+ "explanation": "A frequency multiplier chain multiplies stage by stage, so to find the oscillator frequency you work backward by dividing by each multiplier. Here $145.2\\,\\text{MHz} / 2 / 3 / 2 = 12.1\\,\\text{MHz}$.",
"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: $21.360 MHz / 3 / 2 = 3.560 MHz$.",
+ "revision": 5,
+ "explanation": "Frequency multipliers scale the input frequency by their factor. To recover the starting oscillator frequency, reverse the chain by division: $21.360\\,\\text{MHz} / 3 / 2 = 3.560\\,\\text{MHz}$.",
"source": "https://50ohm.de/NEA_frequenzvervielfacher_1.html#EF302",
"confidence": 8
},
@@ -7026,8 +7026,8 @@
"confidence": 8
},
"EG109": {
- "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).",
+ "revision": 5,
+ "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 $\\lambda = c/f$ ($c = f\\lambda$, $\\lambda[\\text{m}] \\approx 300/f[\\text{MHz}]$, 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
},
@@ -7182,8 +7182,8 @@
"confidence": 8
},
"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: add the 2,15 dB dipole offset: g_i = g_d + 2,15 dB (Antennen, Pegel, S.15).",
+ "revision": 5,
+ "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
},
@@ -7236,26 +7236,26 @@
"confidence": 8
},
"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: losses in dB add along the path; each is a = 10·log10(P_in/P_out) (Pegel, S.15).",
+ "revision": 5,
+ "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 $a = 10\\cdot\\log_{10}(P_\\mathrm{in}/P_\\mathrm{out})$ (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: apply a = 10·log10(P_in/P_out) (Pegel, S.15); ×2 power = 3 dB (table, S.15).",
+ "revision": 5,
+ "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 $a = 10\\cdot\\log_{10}(P_\\mathrm{in}/P_\\mathrm{out})$ (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: apply a = 10·log10(P_in/P_out) (Pegel, S.15); ×4 power = 6 dB (table, S.15).",
+ "revision": 5,
+ "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 $a = 10\\cdot\\log_{10}(P_\\mathrm{in}/P_\\mathrm{out})$ (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: apply a = 10·log10(P_in/P_out) (Pegel, S.15); ×10 power = 10 dB (table, S.15).",
+ "revision": 5,
+ "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 $a = 10\\cdot\\log_{10}(P_\\mathrm{in}/P_\\mathrm{out})$ (Pegel, S.15); ×10 power = 10 dB (table, S.15).",
"source": "https://50ohm.de/NEA_kabeldaempfung_1.html#EG310",
"confidence": 8
},
@@ -7296,8 +7296,8 @@
"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: first |r| = (s−1)/(s+1) (Reflexionsfaktor, S.17), then reflected power P_r = P_v·|r|² (S.17).",
+ "revision": 6,
+ "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 $P_r = P_v\\cdot|r|^2$ (S.17).",
"source": "https://50ohm.de/NEA_swr_2.html#EG401",
"confidence": 8
},
@@ -7350,62 +7350,62 @@
"confidence": 9
},
"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: 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).",
+ "revision": 6,
+ "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 $P_\\mathrm{ERP} = P_S\\cdot 10^{(g_d-a)/(10\\,\\text{dB})}$ and $P_\\mathrm{EIRP} = P_\\mathrm{ERP}\\cdot 1{,}64$ (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: first dBi→factor with G = 10^(g/10dB) (Pegel, S.15), then P_EIRP = P·G.",
+ "revision": 7,
+ "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 $G = 10^{g/(10\\,\\text{dB})}$ (Pegel, S.15), then $P_\\mathrm{EIRP} = P\\cdot G$.",
"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: first dBi→factor with G = 10^(g/10dB) (Pegel, S.15), then P_EIRP = P·G.",
+ "revision": 7,
+ "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 $G = 10^{g/(10\\,\\text{dB})}$ (Pegel, S.15), then $P_\\mathrm{EIRP} = P\\cdot G$.",
"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: net dB = gain − cable loss, then factor via G = 10^(g/10dB) (Pegel, S.15); ×10 = 10 dB.",
+ "revision": 7,
+ "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 $G = 10^{g/(10\\,\\text{dB})}$ (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: g_i = g_d + 2,15 dB and the equal cable loss cancel (Pegel, S.15); EIRP stays at the TX power.",
+ "revision": 5,
+ "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: 10 dB loss → ÷10 power, then ×1,64 for the dipole (g_i = 2,15 dB; Antennen, Pegel, S.15).",
+ "revision": 6,
+ "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: first dBd→dBi (g_i = g_d + 2,15 dB), subtract cable loss, then factor via G = 10^(g/10dB) (Pegel, S.15).",
+ "revision": 7,
+ "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 $G = 10^{g/(10\\,\\text{dB})}$ (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: first dBd→dBi (g_i = g_d + 2,15 dB), subtract cable loss, then factor via G = 10^(g/10dB) (Pegel, S.15).",
+ "revision": 7,
+ "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 $G = 10^{g/(10\\,\\text{dB})}$ (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: 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).",
+ "revision": 7,
+ "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 $G = 10^{g/(10\\,\\text{dB})}$ (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 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.",
+ "revision": 6,
+ "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 $G = 10^{g/(10\\,\\text{dB})}$ (Pegel, S.15), then back out $P = \\text{EIRP}/G$ from the 10 W limit.",
"source": "https://50ohm.de/NEA_aequivalente_isotrope_strahlungsleistung_eirp_2.html#EG511",
"confidence": 9
},
@@ -7608,14 +7608,14 @@
"confidence": 8
},
"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 $0.29 \\cdot 10 V = 2.9 V$.",
+ "revision": 5,
+ "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 $0.29 \\cdot 10\\,\\text{V} = 2.9\\,\\text{V}$.",
"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 $0.29 \\cdot 300 V \\approx 88 V$.",
+ "revision": 5,
+ "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 $0.29 \\cdot 300\\,\\text{V} \\approx 88\\,\\text{V}$.",
"source": "https://50ohm.de/NEA_zeigerinstrumente_ablesen.html#EI104",
"confidence": 8
},
@@ -7626,8 +7626,8 @@
"confidence": 8
},
"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: measure L and C and compute f₀ = 1/(2π·√(L·C)), S.14, or sweep with a VNA.",
+ "revision": 4,
+ "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 $f_0 = 1/(2\\pi\\sqrt{L\\cdot C})$, S.14, or sweep with a VNA.",
"source": "https://50ohm.de/NEA_vna_1.html#EI202",
"confidence": 8
},
@@ -7656,14 +7656,14 @@
"confidence": 8
},
"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 $T = 8 \\cdot 0.5 ms = 4 ms$.",
+ "revision": 6,
+ "explanation": "On an oscilloscope, time is read horizontally. One full sine period spans 8 divisions, and the timebase is 0.5 ms/div, so $T = 8 \\cdot 0.5\\,\\text{ms} = 4\\,\\text{ms}$.",
"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: apply f = 1/T (Wechselspannung, S.12).",
+ "revision": 7,
+ "explanation": "Frequency is the reciprocal of period: $f=1/T$. From the oscilloscope reading $T=4\\,\\text{ms} = 0.004\\,\\text{s}$, so $f=1/0.004\\,\\text{s} = 250\\,\\text{Hz}$. Hilfsmittel: apply $f = 1/T$ (Wechselspannung, S.12).",
"source": "https://50ohm.de/NEA_oszilloskop_1.html#EI302",
"confidence": 8
},
@@ -8016,8 +8016,8 @@
"confidence": 9
},
"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 far-field relation is invalid this close (valid only for d > λ/(2π); E = √(30Ω·P_EIRP)/d (Feldstärke im Fernfeld, S.15)).",
+ "revision": 4,
+ "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 $d > \\lambda/(2\\pi)$; $E = \\sqrt{30\\,\\Omega\\cdot P_\\mathrm{EIRP}}/d$, Feldstärke im Fernfeld, S.15).",
"source": "https://50ohm.de/NEA_naeherungsformel_1.html#EK105",
"confidence": 8
},
@@ -8034,8 +8034,8 @@
"confidence": 8
},
"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: 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)).",
+ "revision": 6,
+ "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 $d = \\sqrt{30\\,\\Omega \\cdot P_{\\text{EIRP}}} / E = \\sqrt{30\\,\\Omega \\cdot 653} / 28 \\approx 140 / 28 \\approx 5.0$ m. Hilfsmittel: convert dBd→dBi (+2,15 dB) and subtract cable loss for the EIRP factor (S.15), then $d = \\sqrt{30\\,\\Omega\\cdot P_\\mathrm{EIRP}}/E$ ($E = \\sqrt{30\\,\\Omega\\cdot P_\\mathrm{EIRP}}/d$, Feldstärke im Fernfeld, S.15).",
"source": "https://50ohm.de/NEA_naeherungsformel_1.html#EK108",
"confidence": 8
},
@@ -8160,38 +8160,38 @@
"confidence": 8
},
"NA207": {
- "revision": 2,
- "explanation": "One hertz means one cycle per second, so dimensionally $Hz = 1/s$.",
+ "revision": 4,
+ "explanation": "One hertz means one cycle per second, so dimensionally $\\text{Hz} = 1/\\text{s}$.",
"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: use the Zehnerpotenzen/SI-Präfix table (S.11): milli = 10⁻³.",
+ "revision": 4,
+ "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: use the Zehnerpotenzen/SI-Präfix table (S.11): milli = 10⁻³.",
+ "revision": 4,
+ "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: use the Zehnerpotenzen/SI-Präfix table (S.11): milli = 10⁻³.",
+ "revision": 4,
+ "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: use the Zehnerpotenzen/SI-Präfix table (S.11): milli = 10⁻³.",
+ "revision": 5,
+ "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: use the Zehnerpotenzen/SI-Präfix table (S.11): mega = 10⁶.",
+ "revision": 4,
+ "explanation": "Mega means $10^6$; $144000000$ Hz divided by $10^6$ is 144 MHz. Hilfsmittel: use the Zehnerpotenzen/SI-Präfix table (S.11): mega = $10^6$.",
"source": "https://50ohm.de/NEA_frequenz.html#NA212",
"confidence": 8
},
@@ -8244,14 +8244,14 @@
"confidence": 7
},
"NB204": {
- "revision": 4,
- "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).",
+ "revision": 6,
+ "explanation": "Series-connected cells add their voltages; six 1.5 V cells give $6 \\cdot 1.5\\,\\text{V} = 9\\,\\text{V}$. Hilfsmittel: series voltages add: $U_G = U_1 + U_2 + \\ldots$ (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: series voltages add: U_G = U1+U2 (Reihenschaltung, S.12).",
+ "revision": 5,
+ "explanation": "The voltmeter is connected across two 1.5 V cells in series, so it reads their sum: 3 V. Hilfsmittel: series voltages add: $U_G = U_1 + U_2$ (Reihenschaltung, S.12).",
"source": "https://50ohm.de/NEA_spannungsmessung.html#NB205",
"confidence": 7
},
@@ -8274,14 +8274,14 @@
"confidence": 8
},
"NB302": {
- "revision": 3,
- "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).",
+ "revision": 4,
+ "explanation": "Use $f = c/\\lambda$: $300000000 / 2.08$ is about 144 MHz. Hilfsmittel: apply $f = c/\\lambda$ ($c = f\\lambda$, $\\lambda[\\text{m}] \\approx 300/f[\\text{MHz}]$, 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: apply λ = c/f (c = f·λ (λ[m] ≈ 300/f[MHz]), S.17).",
+ "revision": 4,
+ "explanation": "Use $\\lambda = c/f$: $300000000 / 433500000$ is about 0.69 m. Hilfsmittel: apply $\\lambda = c/f$ ($c = f\\lambda$, $\\lambda[\\text{m}] \\approx 300/f[\\text{MHz}]$, S.17).",
"source": "https://50ohm.de/NEA_wellenlaenge.html#NB303",
"confidence": 8
},
@@ -8322,26 +8322,26 @@
"confidence": 7
},
"NB501": {
- "revision": 2,
- "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).",
+ "revision": 3,
+ "explanation": "Ohm's law relates voltage, current and resistance as $U = R \\cdot I$. Hilfsmittel: apply Ohm's law $U = R\\cdot I$ (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: rearrange Ohm's law → I = U/R (Ohmsches Gesetz, S.11).",
+ "revision": 3,
+ "explanation": "Rearranging Ohm's law gives current as voltage divided by resistance: $I = U/R$. Hilfsmittel: rearrange Ohm's law $\\to I = U/R$ (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: rearrange Ohm's law → R = U/I (Ohmsches Gesetz, S.11).",
+ "revision": 3,
+ "explanation": "Rearranging $U = R \\cdot I$ for resistance gives $R = U/I$. Hilfsmittel: rearrange Ohm's law $\\to R = U/I$ (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: apply Ohm's law U = R·I (Ohmsches Gesetz, S.11).",
+ "revision": 4,
+ "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 $U = R\\cdot I$ (Ohmsches Gesetz, S.11).",
"source": "https://50ohm.de/NEA_ohmsches_gesetz.html#NB504",
"confidence": 8
},
@@ -8352,38 +8352,38 @@
"confidence": 7
},
"NB601": {
- "revision": 3,
- "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).",
+ "revision": 5,
+ "explanation": "DC input power is $P = U \\cdot I$, so $13.8\\,\\text{V} \\cdot 1.5\\,\\text{A} = 20.7\\,\\text{W}$. Hilfsmittel: apply $P = U\\cdot I$ (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: apply P = U·I (Leistung, S.12).",
+ "revision": 4,
+ "explanation": "Power converted to heat is $P = U \\cdot I$; 50 V times 0.050 A gives 2.5 W. Hilfsmittel: apply $P = U\\cdot I$ (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: apply P = U·I (Leistung, S.12).",
+ "revision": 5,
+ "explanation": "20 mA is 0.020 A, and $3.2\\,\\text{V} \\cdot 0.020\\,\\text{A} = 0.064\\,\\text{W} = 64.0\\,\\text{mW}$. Hilfsmittel: apply $P = U\\cdot I$ (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: rearrange P = U·I → I = P/U (Leistung, S.12).",
+ "revision": 5,
+ "explanation": "From $P = U \\cdot I$, current is $I = P/U = 100\\,\\text{W} / 12\\,\\text{V} = 8.33\\,\\text{A}$. Hilfsmittel: rearrange $P = U\\cdot I \\to I = P/U$ (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: rearrange P = U·I → I = P/U (Leistung, S.12).",
+ "revision": 5,
+ "explanation": "A 3 W load at 12 V draws $I = P/U = 3/12 = 0.25\\,\\text{A}$, which is 250 mA. Hilfsmittel: rearrange $P = U\\cdot I \\to I = P/U$ (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: rearrange P = U·I → I = P/U (Leistung, S.12).",
+ "revision": 5,
+ "explanation": "A 48 W load at 12 V draws $I = P/U = 48/12 = 4\\,\\text{A}$. Hilfsmittel: rearrange $P = U\\cdot I \\to I = P/U$ (Leistung, S.12).",
"source": "https://50ohm.de/NEA_leistung.html#NB606",
"confidence": 8
},
@@ -9018,8 +9018,8 @@
"confidence": 7
},
"NG301": {
- "revision": 3,
- "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.",
+ "revision": 4,
+ "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
},
@@ -10074,20 +10074,20 @@
"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 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.",
+ "revision": 3,
+ "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 $P_\\mathrm{EIRP} = P_\\mathrm{ERP}\\cdot 1{,}64$ (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: 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).",
+ "revision": 4,
+ "explanation": "EIRP is transmitter power times antenna gain relative to isotropic: $5\\,\\text{W} \\cdot 2.5 = 12.5\\,\\text{W}$, which exceeds the 10 W class N limit. Hilfsmittel: EIRP = transmitter power × isotropic gain factor ($5\\,\\text{W} \\times 2{,}5 = 12{,}5\\,\\text{W}$); the 10 W EIRP class-N limit comes from $6{,}1\\,\\text{W}$ ERP $\\times 1{,}64$ (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: 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).",
+ "revision": 4,
+ "explanation": "EIRP is transmitter power times antenna gain relative to isotropic: $5\\,\\text{W} \\cdot 1.8 = 9\\,\\text{W}$, which stays below the 10 W class N limit. Hilfsmittel: EIRP = transmitter power × isotropic gain factor ($5\\,\\text{W} \\times 1{,}8 = 9\\,\\text{W}$); the 10 W EIRP class-N limit comes from $6{,}1\\,\\text{W}$ ERP $\\times 1{,}64$ (Anlage 1 S. 2 + conversion S. 15).",
"source": "https://50ohm.de/NEA_sendeleistung_klasse_n.html#VD726",
"confidence": 10
},