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2026-08-16 17:51:54 +02:00
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@@ -1415,21 +1415,21 @@
"confidence": 8
},
"AD426": {
"revision": 6,
"revision": 7,
"explanation": "A 16 dB power gain is a ratio of $10^{16/10} = 39.8$, so 1 W input becomes about 40 W output. <u>Hilfsmittel:</u> 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",
"source": "https://50ohm.de/NEA_dezibel_2.html#AD426",
"confidence": 8
},
"AD427": {
"revision": 6,
"revision": 7,
"explanation": "For equal impedances, voltage gain in dB is $20 \\log_{10}(U_2/U_1)$; $20 \\log_{10}(4\\,\\text{mV} / 1\\,\\text{mV}) = 20 \\log_{10}(4) = 12\\,\\text{dB}$. <u>Hilfsmittel:</u> apply the voltage form $g = 20\\cdot\\log_{10}(U_2/U_1)$ (Pegel, S.15).",
"source": "https://50ohm.de/NEA_verstaerkungsleistung.html#AD427",
"source": "https://50ohm.de/NEA_dezibel_2.html#AD427",
"confidence": 8
},
"AD428": {
"revision": 6,
"revision": 7,
"explanation": "Power gain in dB is $10 \\log_{10}(P_2/P_1)$; $10 \\log_{10}(38\\,\\text{W} / 2.5\\,\\text{W}) = 10 \\log_{10}(15.2) = 11.8\\,\\text{dB}$. <u>Hilfsmittel:</u> apply $g = 10\\cdot\\log_{10}(P_2/P_1)$ (Pegel, S.15).",
"source": "https://50ohm.de/NEA_verstaerkungsleistung.html#AD428",
"source": "https://50ohm.de/NEA_dezibel_2.html#AD428",
"confidence": 8
},
"AD429": {
@@ -1475,9 +1475,9 @@
"confidence": 7
},
"AD503": {
"revision": 3,
"revision": 4,
"explanation": "This is an envelope demodulator. The IF signal enters on the left, so X is not the IF output. Point X sits after an RC low-pass built with a high-capacitance electrolytic capacitor; the large $C$ pushes $f_g = \\frac{1}{2\\pi R C}$ very low, so neither the audio nor the oscillator signal gets through — only a slowly varying, near-DC voltage. That low-frequency voltage at X is used as a control (AGC) voltage. <u>Hilfsmittel:</u> $f_g = 1/(2\\pi R C)$ (RC-Tiefpass, S.14).",
"source": "https://50ohm.de/NEA_modulatoren.html#AD503",
"source": "https://50ohm.de/NEA_agc_2.html#AD503",
"confidence": 8,
"provenance": "50ohm-loesungsweg"
},
@@ -2401,9 +2401,9 @@
"confidence": 8
},
"AF401": {
"revision": 2,
"revision": 3,
"explanation": "HF amplifier efficiency is useful RF output power divided by the DC power taken from the supply.",
"source": "https://50ohm.de/NEA_leistungsvertaerker.html#AF401",
"source": "https://50ohm.de/NEA_verstaerker_wirkungsgrad.html#AF401",
"confidence": 8
},
"AF402": {
@@ -2566,9 +2566,9 @@
"provenance": "50ohm-loesungsweg"
},
"AF428": {
"revision": 4,
"revision": 5,
"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. <u>Hilfsmittel:</u> 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",
"source": "https://50ohm.de/NEA_dezibel_2.html#AF428",
"confidence": 7
},
"AF501": {
@@ -2945,14 +2945,14 @@
"confidence": 8
},
"AG114": {
"revision": 3,
"explanation": "In a 20/15/10 m trap dipole, the inner trap pair must stop the 15 m current, so it is tuned near 21.2 MHz.",
"revision": 4,
"explanation": "In a 20/15/10 m trap dipole, the outer trap pair a is tuned near the 15 m operating frequency of 21.2 MHz. At resonance it presents a high impedance and isolates the wire sections beyond it.",
"source": "https://50ohm.de/NEA_traps.html#AG114",
"confidence": 8
},
"AG115": {
"revision": 3,
"explanation": "The outer trap pair separates the 10 m section, so it is tuned near the 10 m operating frequency around 29 MHz.",
"revision": 4,
"explanation": "The inner trap pair b is tuned near the 10 m operating frequency of 29 MHz. Because the highest-frequency band needs the shortest active dipole section, its traps are closest to the feed point.",
"source": "https://50ohm.de/NEA_traps.html#AG115",
"confidence": 8
},
@@ -3041,27 +3041,27 @@
"confidence": 8
},
"AG203": {
"revision": 2,
"revision": 3,
"explanation": "The 28 MHz case is the highest listed harmonic, so it shows the largest number of current half-waves on the same dipole.",
"source": "https://50ohm.de/NEA_frequenzabhaengige_stromverteilung.html#AG203",
"source": "https://50ohm.de/NEA_strom_spannung_speisung_2.html#AG203",
"confidence": 8
},
"AG204": {
"revision": 2,
"explanation": "At 14 MHz the 20 m dipole is excited at the next lower shown harmonic pattern, with fewer current lobes than at 28 MHz.",
"source": "https://50ohm.de/NEA_frequenzabhaengige_stromverteilung.html#AG204",
"revision": 3,
"explanation": "The pictured 80 m dipole has a 3.5 MHz fundamental. At 14 MHz it is driven at the fourth harmonic, so the current distribution has four lobes.",
"source": "https://50ohm.de/NEA_strom_spannung_speisung_2.html#AG204",
"confidence": 8
},
"AG205": {
"revision": 2,
"explanation": "At 7 MHz the same 20 m wire is near one full wavelength overall, matching the intermediate current distribution.",
"source": "https://50ohm.de/NEA_frequenzabhaengige_stromverteilung.html#AG205",
"revision": 3,
"explanation": "The pictured 80 m dipole has a 3.5 MHz fundamental. At 7 MHz it is driven at the second harmonic, so the current distribution has two lobes.",
"source": "https://50ohm.de/NEA_strom_spannung_speisung_2.html#AG205",
"confidence": 8
},
"AG206": {
"revision": 2,
"explanation": "At 3.5 MHz the 20 m dipole is near its half-wave fundamental, so it has the simplest current distribution with the central maximum.",
"source": "https://50ohm.de/NEA_frequenzabhaengige_stromverteilung.html#AG206",
"revision": 3,
"explanation": "At its 3.5 MHz fundamental, the pictured 80 m dipole is approximately one half-wave long and has the simplest current distribution: one lobe with a current maximum at the center.",
"source": "https://50ohm.de/NEA_strom_spannung_speisung_2.html#AG206",
"confidence": 8
},
"AG207": {
@@ -3245,9 +3245,9 @@
"confidence": 8
},
"AG308": {
"revision": 3,
"revision": 4,
"explanation": "Over 60 m at 29 MHz the total loss must stay under 2 dB, so pick the lowest-loss cable — not the thinnest. Of the four, the 10.3 mm PE-foam cable has both the largest conductor and a low-loss foam dielectric, so it has the least loss per metre: the thin 4.95 mm RG58 and the thinner 7.3 mm foam cable have more conductor loss, and the equally-thick RG213 loses more because its solid PE dielectric is lossier than foam. Only the 10.3 mm foam cable therefore holds the 60 m run within 2 dB. <u>Hilfsmittel:</u> read dB/100 m at 29 MHz from the Kabeldämpfungsdiagramm Koaxialkabel (S. 22) and scale linearly with length.",
"source": "https://50ohm.de/NEA_kabeldaempfung_2.html",
"source": "https://50ohm.de/NEA_kabeldaempfung_2.html#AG308",
"confidence": 7
},
"AG309": {
@@ -3257,9 +3257,9 @@
"confidence": 8
},
"AG310": {
"revision": 3,
"revision": 4,
"explanation": "Coax loss falls as the cable gets thicker, so read the per-100 m attenuation at 5.7 GHz off the diagram and scale to the 8 m run ($8\\,\\text{m}$ is $0.08$ of $100\\,\\text{m}$, so the budget is $3\\,\\text{dB}/0.08 = 37.5\\,\\text{dB per }100\\,\\text{m}$). At 5.7 GHz the 12.7 mm <u>and</u> the fatter 16.4 mm PE-foam cables both stay under $3\\,\\text{dB}$ over $8\\,\\text{m}$; the thinner 10.3 mm and 7.3 mm types exceed it. The question asks for the <u>thinnest</u> cable that still meets the $3\\,\\text{dB}$ limit — not the only one that meets it — so the 12.7 mm cable is chosen over the needlessly thick 16.4 mm. <u>Hilfsmittel:</u> the Kabeldämpfungsdiagramm Koaxialkabel (S. 22) — read dB/100 m at the operating frequency and scale linearly with length.",
"source": "https://50ohm.de/NEA_kabeldaempfung_2.html",
"source": "https://50ohm.de/NEA_kabeldaempfung_2.html#AG310",
"confidence": 8
},
"AG311": {
@@ -3372,15 +3372,15 @@
"confidence": 8
},
"AG409": {
"revision": 2,
"revision": 3,
"explanation": "A quarter-wave line inverts impedance: a short circuit at one end appears as very high impedance at the other.",
"source": "https://50ohm.de/NEA_lecherleitung.html#AG409",
"source": "https://50ohm.de/NEA_impedanztransformation.html#AG409",
"confidence": 8
},
"AG410": {
"revision": 2,
"revision": 3,
"explanation": "The quarter-wave section transforms the far-end condition so point X is at a current maximum and nearly zero impedance.",
"source": "https://50ohm.de/NEA_lecherleitung.html#AG410",
"source": "https://50ohm.de/NEA_impedanztransformation.html#AG410",
"confidence": 8
},
"AG411": {
@@ -3897,9 +3897,9 @@
"confidence": 8
},
"AI403": {
"revision": 2,
"revision": 3,
"explanation": "For a purely resistive mismatch, SWR is the impedance ratio; $150/50 = 3$.",
"source": "https://50ohm.de/NEA_swr_meter_2.html#AI403",
"source": "https://50ohm.de/NEA_swr_3.html#AI403",
"confidence": 8
},
"AI501": {
@@ -4335,23 +4335,23 @@
"provenance": "50ohm-loesungsweg"
},
"AK102": {
"revision": 3,
"revision": 4,
"explanation": "In the far field the magnitudes of $E$ and $H$ are tied together by the wave impedance of the medium: $\\frac{E}{H} = Z$. For free space $Z_0 = \\sqrt{\\frac{\\mu_0}{\\varepsilon_0}} \\approx 120\\pi\\,\\Omega \\approx 377\\,\\Omega$ (air is practically identical). One field component therefore fixes the other — which is not possible in the near field.",
"source": "https://50ohm.de/NEA_personenschutzabstand_3.html#AK102",
"source": "https://50ohm.de/NEA_nahfeld.html#AK102",
"confidence": 8,
"provenance": "50ohm-loesungsweg"
},
"AK103": {
"revision": 7,
"revision": 8,
"explanation": "The far-field formula $d = \\frac{\\sqrt{30\\,\\Omega\\cdot P_\\mathrm{EIRP}}}{E}$ is valid for $d > \\frac{\\lambda}{2\\pi}$ for most antenna types (e.g. dipoles). In the radiating near field it generally gives a conservative over-estimate of the field, so it stays on the safe side — which is why it is allowed down to $\\frac{\\lambda}{2\\pi}$; only in the reactive near field ($d < \\frac{\\lambda}{2\\pi}$) is it not permitted. For electrically small or magnetically dominated antennas, though, the real near-field strength can exceed the predicted value. Where validity is not assured (such antennas, or shorter distances), determine the safety distance instead by field-strength measurement or by a near-field calculation (simulation), as the BNetzA BEMFV guidance prescribes.",
"source": "https://50ohm.de/NEA_nahfeld.html#AK103",
"source": "https://50ohm.de/NEA_naeherungsformel_2.html#AK103",
"confidence": 8,
"provenance": "50ohm-loesungsweg"
},
"AK104": {
"revision": 4,
"revision": 5,
"explanation": "The antenna-input power is the transmitter output reduced by the feedline loss. Write the loss as a factor $D = \\frac{P_\\mathrm{Ant}}{P_\\mathrm{Sender}}$ (with $0 < D < 1$), then $P_\\mathrm{Ant} = D\\cdot P_\\mathrm{Sender}$. The greater the cable loss, the smaller $D$ and the less power reaches the antenna.",
"source": "https://50ohm.de/NEA_personenschutzabstand_3.html#AK104",
"source": "https://50ohm.de/NEA_effektive_strahlungsleistung_erp_2.html#AK104",
"confidence": 8,
"provenance": "50ohm-loesungsweg"
},
@@ -4363,9 +4363,9 @@
"provenance": "50ohm-loesungsweg"
},
"AK106": {
"revision": 4,
"revision": 5,
"explanation": "A half-wave dipole has $g_i = 2.15\\,\\text{dBi}$, so the EIRP from the $100\\,\\text{W}$ transmitter power is $P_\\mathrm{EIRP} = 100\\,\\text{W}\\cdot 10^{2.15/10} \\approx 164\\,\\text{W}$. Then $d = \\frac{\\sqrt{30\\,\\Omega\\cdot 164\\,\\text{W}}}{28\\,\\text{V/m}} \\approx \\frac{70.2\\,\\text{V}}{28\\,\\text{V/m}} \\approx 2.50\\,\\text{m}$. The far-field condition $d > \\frac{\\lambda}{2\\pi} \\approx 1.59\\,\\text{m}$ holds. <u>Hilfsmittel:</u> $d = \\sqrt{30\\,\\Omega\\cdot P_\\mathrm{EIRP}}/E$ and $P_\\mathrm{EIRP} = P_S\\cdot 10^{(g_i - a)/(10\\,\\text{dB})}$, where $P_S$ is the transmitter power (Pegel / Antennen, S.15).",
"source": "https://50ohm.de/NEA_naeherungsformel_2.html#AK106",
"source": "https://50ohm.de/NEA_personenschutzabstand_3.html#AK106",
"confidence": 8,
"provenance": "50ohm-loesungsweg"
},
@@ -4377,37 +4377,37 @@
"provenance": "50ohm-loesungsweg"
},
"AK108": {
"revision": 4,
"revision": 5,
"explanation": "For a dipole $g_i = 2.15\\,\\text{dBi}$ with feedline loss $a = 0.5\\,\\text{dB}$, the EIRP from the $300\\,\\text{W}$ transmitter power is $P_\\mathrm{EIRP} = 300\\,\\text{W}\\cdot 10^{(2.15 - 0.5)/10} \\approx 439\\,\\text{W}$. Then $d = \\frac{\\sqrt{30\\,\\Omega\\cdot 439\\,\\text{W}}}{28\\,\\text{V/m}} \\approx \\frac{114.7\\,\\text{V}}{28\\,\\text{V/m}} \\approx 4.10\\,\\text{m}$. <u>Hilfsmittel:</u> $d = \\sqrt{30\\,\\Omega\\cdot P_\\mathrm{EIRP}}/E$ and $P_\\mathrm{EIRP} = P_S\\cdot 10^{(g_i - a)/(10\\,\\text{dB})}$, with $P_S$ the transmitter power (Pegel / Antennen, S.15).",
"source": "https://50ohm.de/NEA_naeherungsformel_2.html#AK108",
"source": "https://50ohm.de/NEA_personenschutzabstand_3.html#AK108",
"confidence": 8,
"provenance": "50ohm-loesungsweg"
},
"AK109": {
"revision": 4,
"revision": 5,
"explanation": "For a dipole $g_i = 2.15\\,\\text{dBi}$ with feedline loss $a = 0.5\\,\\text{dB}$, the EIRP from the $700\\,\\text{W}$ transmitter power is $P_\\mathrm{EIRP} = 700\\,\\text{W}\\cdot 10^{(2.15 - 0.5)/10} \\approx 1024\\,\\text{W}$. Then $d = \\frac{\\sqrt{30\\,\\Omega\\cdot 1024\\,\\text{W}}}{28\\,\\text{V/m}} \\approx \\frac{175.2\\,\\text{V}}{28\\,\\text{V/m}} \\approx 6.26\\,\\text{m}$. <u>Hilfsmittel:</u> $d = \\sqrt{30\\,\\Omega\\cdot P_\\mathrm{EIRP}}/E$ and $P_\\mathrm{EIRP} = P_S\\cdot 10^{(g_i - a)/(10\\,\\text{dB})}$, with $P_S$ the transmitter power (Pegel / Antennen, S.15).",
"source": "https://50ohm.de/NEA_naeherungsformel_2.html#AK109",
"source": "https://50ohm.de/NEA_personenschutzabstand_3.html#AK109",
"confidence": 8,
"provenance": "50ohm-loesungsweg"
},
"AK110": {
"revision": 8,
"revision": 9,
"explanation": "With $g_d = 11.5\\,\\text{dBd}$ and $a = 1.5\\,\\text{dB}$, the EIRP from the $75\\,\\text{W}$ transmitter power is $P_\\mathrm{EIRP} = 75\\,\\text{W}\\cdot 10^{(11.5 + 2.15 - 1.5)/10} \\approx 1230\\,\\text{W}$. Then $d = \\frac{\\sqrt{30\\,\\Omega\\cdot 1230\\,\\text{W}}}{28\\,\\text{V/m}} \\approx \\frac{192.1\\,\\text{V}}{28\\,\\text{V/m}} \\approx 6.86\\,\\text{m}$. <u>Hilfsmittel:</u> $d = \\sqrt{30\\,\\Omega\\cdot P_\\mathrm{EIRP}}/E$ and $P_\\mathrm{EIRP} = P_S\\cdot 10^{(g_d + 2.15 - a)/(10\\,\\text{dB})}$, with $P_S$ the transmitter power (Pegel / Antennen, S.15).",
"source": "https://50ohm.de/NEA_naeherungsformel_2.html#AK110",
"source": "https://50ohm.de/NEA_personenschutzabstand_3.html#AK110",
"confidence": 8,
"provenance": "50ohm-loesungsweg"
},
"AK111": {
"revision": 4,
"revision": 5,
"explanation": "With $g_d = 10.5\\,\\text{dBd}$ and $a = 1.5\\,\\text{dB}$, the EIRP from the $100\\,\\text{W}$ transmitter power is $P_\\mathrm{EIRP} = 100\\,\\text{W}\\cdot 10^{(10.5 + 2.15 - 1.5)/10} \\approx 1303\\,\\text{W}$. Then $d = \\frac{\\sqrt{30\\,\\Omega\\cdot 1303\\,\\text{W}}}{28\\,\\text{V/m}} \\approx \\frac{197.7\\,\\text{V}}{28\\,\\text{V/m}} \\approx 7.1\\,\\text{m}$. <u>Hilfsmittel:</u> $d = \\sqrt{30\\,\\Omega\\cdot P_\\mathrm{EIRP}}/E$ and $P_\\mathrm{EIRP} = P_S\\cdot 10^{(g_d + 2.15 - a)/(10\\,\\text{dB})}$, with $P_S$ the transmitter power (Pegel / Antennen, S.15).",
"source": "https://50ohm.de/NEA_naeherungsformel_2.html#AK111",
"source": "https://50ohm.de/NEA_personenschutzabstand_3.html#AK111",
"confidence": 8,
"provenance": "50ohm-loesungsweg"
},
"AK112": {
"revision": 4,
"revision": 5,
"explanation": "With $g_d = 18\\,\\text{dBd}$, $a = 2\\,\\text{dB}$ and the higher limit $E = 61\\,\\text{V/m}$, the EIRP from the $40\\,\\text{W}$ transmitter power is $P_\\mathrm{EIRP} = 40\\,\\text{W}\\cdot 10^{(18 + 2.15 - 2)/10} \\approx 2613\\,\\text{W}$. Then $d = \\frac{\\sqrt{30\\,\\Omega\\cdot 2613\\,\\text{W}}}{61\\,\\text{V/m}} \\approx \\frac{280\\,\\text{V}}{61\\,\\text{V/m}} \\approx 4.6\\,\\text{m}$. <u>Hilfsmittel:</u> $d = \\sqrt{30\\,\\Omega\\cdot P_\\mathrm{EIRP}}/E$ and $P_\\mathrm{EIRP} = P_S\\cdot 10^{(g_d + 2.15 - a)/(10\\,\\text{dB})}$, with $P_S$ the transmitter power (Pegel / Antennen, S.15).",
"source": "https://50ohm.de/NEA_naeherungsformel_2.html#AK112",
"source": "https://50ohm.de/NEA_personenschutzabstand_3.html#AK112",
"confidence": 8,
"provenance": "50ohm-loesungsweg"
},
@@ -7373,8 +7373,8 @@
"confidence": 8
},
"EG302": {
"revision": 3,
"explanation": "Good coaxial cable keeps the RF field trapped between inner conductor and shield, so in normal (common-mode-free) operation it radiates almost nothing. That makes it the right choice for HF links between station devices; open balanced feeders deliberately radiate from both wires and would couple into nearby equipment.",
"revision": 4,
"explanation": "Good coaxial cable confines the wanted RF field between its inner conductor and shield, so in normal common-mode-free operation it produces very little external field. Its shielding therefore makes it the appropriate choice for HF connections between station devices. Balanced open-wire line can also have low radiation when correctly operated, but it has no surrounding shield and is less suitable for interconnecting equipment in the station.",
"source": "https://50ohm.de/NEA_uebertragungsleitungen_2.html#EG302",
"confidence": 8
},
@@ -8149,8 +8149,8 @@
"confidence": 8
},
"EJ218": {
"revision": 3,
"explanation": "For FT8, JS8, PSK31 and similar modes, set the audio (NF) drive low enough that the ALC does <u>not</u> engage at all. ALC action on these constant-envelope digital signals causes distortion and splatter, so the clean operating point is just below the ALC threshold — not at maximum.",
"revision": 4,
"explanation": "For FT8, JS8, PSK31 and similar modes, set the audio (NF) drive low enough that the ALC does <u>not</u> engage at all. ALC action can distort the digitally generated SSB waveform and cause splatter, so the clean operating point is just below the ALC threshold — not at maximum.",
"source": "https://50ohm.de/NEA_digimod_uebersteuerung.html#EJ218",
"confidence": 8
},