diff --git a/explanations.json b/explanations.json
index d21c572..06c9a5d 100644
--- a/explanations.json
+++ b/explanations.json
@@ -1350,8 +1350,8 @@
"confidence": 7
},
"AD504": {
- "revision": 1,
- "explanation": "A tuned circuit offset from the IF converts FM frequency deviations into amplitude changes, which a diode detector can then recover as audio; that is a slope discriminator.",
+ "revision": 2,
+ "explanation": "A tuned circuit offset from the IF converts FM frequency deviation, German Hub/Frequenzhub, into amplitude changes. A diode detector can then recover those amplitude changes as audio; that method is a slope discriminator.",
"source": "https://50ohm.de/EA_demodulator.html",
"confidence": 7
},
@@ -1380,10 +1380,10 @@
"confidence": 7
},
"AD509": {
- "revision": 1,
- "explanation": "For FM the audio amplitude determines frequency deviation; antiparallel diodes and the level control limit and set that deviation, i.e. the FM deviation or hub.",
+ "revision": 2,
+ "explanation": "For FM, German Hub or Frequenzhub means frequency deviation: the maximum shift of the carrier above and below its centre frequency. The audio amplitude sets that deviation; the antiparallel diodes and level control limit it so the FM signal does not become too wide.",
"source": "https://50ohm.de/A_modulatoren.html",
- "confidence": 7
+ "confidence": 8
},
"AD510": {
"revision": 1,
@@ -1668,14 +1668,14 @@
"confidence": 7
},
"AE301": {
- "revision": 1,
- "explanation": "In FM, the modulating signal frequency sets how often the RF carrier frequency is moved back and forth; the modulation amplitude sets how far it moves.",
+ "revision": 2,
+ "explanation": "In FM, the modulating signal frequency sets how often the RF carrier is moved back and forth. The modulation amplitude sets how far it moves; that frequency swing is called deviation, German Hub or Frequenzhub.",
"source": "https://50ohm.de/A_fm_3.html",
"confidence": 8
},
"AE302": {
- "revision": 1,
- "explanation": "Impulse noise mainly changes amplitude; FM carries the information in frequency deviation, so amplitude disturbances have less effect than with AM or SSB.",
+ "revision": 2,
+ "explanation": "Impulse noise mainly changes amplitude. FM carries the information in frequency deviation, German Hub/Frequenzhub, so a limiter can remove much of the amplitude disturbance before the FM demodulator.",
"source": "https://50ohm.de/A_fm_3.html",
"confidence": 8
},
@@ -1686,32 +1686,32 @@
"confidence": 8
},
"AE304": {
- "revision": 1,
- "explanation": "Carson's rule shows FM bandwidth increases with both deviation and the highest modulation frequency, so too high an audio frequency makes the RF bandwidth too large.",
+ "revision": 2,
+ "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.",
"source": "https://50ohm.de/A_fm_3.html",
"confidence": 8
},
"AE305": {
- "revision": 1,
- "explanation": "For FM speech, larger deviation represents a larger audio amplitude after demodulation, so the received audio becomes louder.",
+ "revision": 2,
+ "explanation": "For FM speech, audio amplitude is represented by deviation, German Hub/Frequenzhub. A larger Hub means the carrier swings farther from centre, and after demodulation that corresponds to louder received audio.",
"source": "https://50ohm.de/A_fm_3.html",
"confidence": 8
},
"AE306": {
- "revision": 1,
- "explanation": "Excessive FM deviation widens the transmitted spectrum, so the signal can spill into adjacent channels and cause interference.",
+ "revision": 2,
+ "explanation": "Excessive FM deviation, German Hub/Frequenzhub, means the carrier swings too far from its centre frequency. That widens the transmitted spectrum and can spill into adjacent channels, causing interference.",
"source": "https://50ohm.de/A_fm_3.html",
"confidence": 8
},
"AE307": {
- "revision": 1,
- "explanation": "Stronger FM modulator drive increases deviation; higher deviation increases the occupied RF bandwidth.",
+ "revision": 2,
+ "explanation": "Stronger FM modulator drive increases deviation, German Hub/Frequenzhub: the carrier swings farther away from centre. Larger Hub increases occupied RF bandwidth, so overdriving the modulator makes the signal too wide.",
"source": "https://50ohm.de/A_fm_3.html",
"confidence": 8
},
"AE308": {
- "revision": 1,
- "explanation": "Carson's rule gives $B = 2 x (deviation + highest modulation frequency)$, so $2 x (2.5 kHz + 2.7 kHz) = 10.4 kHz$.",
+ "revision": 2,
+ "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 $f_{mod,max}=2.7 kHz$, $B=2(2.5+2.7)=10.4 kHz$.",
"source": "https://50ohm.de/A_fm_3.html",
"confidence": 8
},
@@ -1722,20 +1722,20 @@
"confidence": 8
},
"AE310": {
- "revision": 1,
- "explanation": "Narrowband FM in a 12.5 kHz channel uses a typical peak deviation of about 2.5 kHz, leaving room for modulation sidebands and channel spacing.",
+ "revision": 2,
+ "explanation": "Narrowband FM in a 12.5 kHz channel uses a typical peak deviation, German Hub/Frequenzhub, of about 2.5 kHz. That leaves room for the modulation sidebands inside the channel spacing.",
"source": "https://50ohm.de/A_fm_3.html",
"confidence": 8
},
"AE311": {
- "revision": 1,
- "explanation": "Rearrange Carson's rule: $f_mod = B/2 - deviation = 10 kHz/2 - 2.5 kHz = 2.5 kHz$.",
+ "revision": 2,
+ "explanation": "Rearrange Carson's rule $B \\approx 2(\\Delta f + f_{mod})$. Here $\\Delta f$ is deviation, German Hub/Frequenzhub, so $f_{mod}=B/2-\\Delta f=10 kHz/2-2.5 kHz=2.5 kHz$.",
"source": "https://50ohm.de/A_fm_3.html",
"confidence": 8
},
"AE312": {
- "revision": 1,
- "explanation": "Rearrange Carson's rule: deviation $= B/2 - f_mod = 10 kHz/2 - 2.7 kHz = 2.3 kHz$.",
+ "revision": 2,
+ "explanation": "Rearrange Carson's rule $B \\approx 2(\\Delta f + f_{mod})$. The deviation $\\Delta f$, German Hub/Frequenzhub, is $B/2-f_{mod}=10 kHz/2-2.7 kHz=2.3 kHz$.",
"source": "https://50ohm.de/A_fm_3.html",
"confidence": 8
},
@@ -2148,8 +2148,8 @@
"confidence": 8
},
"AF226": {
- "revision": 1,
- "explanation": "In FM, information is in frequency deviation, so a limiter removes amplitude variations and suppresses AM noise before demodulation.",
+ "revision": 2,
+ "explanation": "In FM, information is in frequency deviation, German Hub/Frequenzhub, not in RF amplitude. A limiter can therefore remove amplitude variations and suppress AM noise before the FM demodulator recovers the audio.",
"source": "https://50ohm.de/A_slide_a_empfaenger.html",
"confidence": 8
},
@@ -5328,2780 +5328,2780 @@
"confidence": 8
},
"EA101": {
- "revision": 1,
- "explanation": "Capacitance is charge stored per voltage, $C = Q/U$, and its named SI-derived unit is the farad.",
+ "revision": 2,
+ "explanation": "Capacitance is the charge stored per volt, $C = Q/U$, so its named SI-derived unit is the farad (F) — one coulomb per volt. Ohm ($\\Omega$) is resistance, henry (H) is inductance, and ampere-hours measure charge, so none of those fit capacitance.",
"source": "https://50ohm.de/EA_kondensator_1.html",
"confidence": 8
},
"EA102": {
- "revision": 1,
- "explanation": "Inductance describes magnetic flux linkage per current; the named SI-derived unit for it is the henry.",
+ "revision": 2,
+ "explanation": "Inductance describes how much magnetic flux linkage a conductor or coil produces per ampere: $L = \\Psi/I$. A larger inductance stores more magnetic-field energy for the same current, and its named SI-derived unit is the henry (H).",
"source": "https://50ohm.de/EA_spule_1.html",
"confidence": 8
},
"EA103": {
- "revision": 1,
- "explanation": "For a plate field, $E = U/d$, so the unit is volts divided by metres: V/m.",
+ "revision": 2,
+ "explanation": "Electric field strength is voltage per distance: $E = U/d$. That is why the unit is volts per metre (V/m), not just volts; it tells how steeply electric potential changes through space.",
"source": "https://50ohm.de/EA_e_feld.html",
"confidence": 8
},
"EA104": {
- "revision": 1,
- "explanation": "For a long straight conductor, $H = I/(2\\pi r)$, so magnetic field strength has amperes divided by metres: A/m.",
+ "revision": 2,
+ "explanation": "Magnetic field strength $H$ is given in amperes per metre (A/m). The straight-wire case shows why: $H = I/(2\\pi r)$ is a current (amperes) spread over a circumference (metres). Keep $H$ (field strength, A/m) distinct from flux density $B$ (tesla); option D, H/m, is the unit of permeability instead.",
"source": "https://50ohm.de/EA_h_feld.html",
"confidence": 8
},
"EA105": {
- "revision": 2,
- "explanation": "Bandwidth is a frequency interval, so it is measured in hertz just like frequency.",
+ "revision": 3,
+ "explanation": "Bandwidth is not a separate physical kind of quantity; it is the width of a frequency range. Since both endpoints are frequencies, their difference is also measured in hertz: $B = f_2 - f_1$.",
"source": "https://50ohm.de/E_slide_e_modulation.html",
- "confidence": 7
+ "confidence": 8
},
"EA106": {
- "revision": 1,
- "explanation": "A data rate counts transferred bits per unit time, so the usual unit is bit/s rather than baud or hertz.",
+ "revision": 2,
+ "explanation": "A data-transmission rate counts how many bits cross per second, so the unit is bit/s. The trap is baud (Bd), which counts symbols per second and equals bit/s only when each symbol carries exactly one bit; hertz is for frequency and decibel for ratios.",
"source": "https://50ohm.de/NEA_datenuebertragungsdrate.html",
"confidence": 8
},
"EA107": {
- "revision": 1,
- "explanation": "Power ratios in dB use $10\\log_{10}(P_2/P_1)$; doubling power gives $10\\log_{10}(2) \\approx 3$ dB.",
+ "revision": 2,
+ "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 $20\\log_{10}$.)",
"source": "https://50ohm.de/E_dezibel_1.html",
"confidence": 8
},
"EA108": {
- "revision": 1,
- "explanation": "$0.00042$ A equals $420 \\cdot 10^{-6}$ A because moving the decimal six places expresses the value in micro-units.",
+ "revision": 2,
+ "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.",
"source": "https://50ohm.de/NEA_zehnerpotenzen.html",
"confidence": 8
},
"EA109": {
- "revision": 1,
- "explanation": "$0.042$ A equals $42 \\cdot 10^{-3}$ A because milli means $10^{-3}$.",
+ "revision": 2,
+ "explanation": "Milli means $10^{-3}$. Move from amperes to milliamperes by multiplying by 1000: $0.042 A = 42 mA = 42 \\cdot 10^{-3} A$.",
"source": "https://50ohm.de/NEA_zehnerpotenzen.html",
"confidence": 8
},
"EA110": {
- "revision": 1,
- "explanation": "$4,200,000$ Hz is $4.2 \\cdot 10^6$ Hz in scientific notation.",
+ "revision": 2,
+ "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.",
"source": "https://50ohm.de/NEA_zehnerpotenzen.html",
"confidence": 8
},
"EA111": {
- "revision": 1,
- "explanation": "$0.01$ mV is $0.01 \\cdot 10^{-3}$ V, which is $10 \\cdot 10^{-6}$ V.",
+ "revision": 2,
+ "explanation": "A millivolt is $10^{-3} V$ and a microvolt is $10^{-6} V$. So $0.01 mV = 0.01 \\cdot 10^{-3} V = 10 \\cdot 10^{-6} V = 10 µV$.",
"source": "https://50ohm.de/NEA_zehnerpotenzen.html",
"confidence": 8
},
"EA112": {
- "revision": 1,
- "explanation": "$0.002$ MOhm is $0.002 \\cdot 10^6$ ohm, which is $2 \\cdot 10^3$ ohm.",
+ "revision": 2,
+ "explanation": "Mega means $10^6$. Therefore $0.002 MOhm = 0.002 \\cdot 10^6 Ohm = 2000 Ohm = 2 kOhm$.",
"source": "https://50ohm.de/NEA_zehnerpotenzen.html",
"confidence": 8
},
"EA113": {
- "revision": 1,
- "explanation": "$2 \\cdot 10^{-7}$ W divided by $10^{-6}$ W/µW gives $0.2$ µW.",
+ "revision": 2,
+ "explanation": "A microwatt is $10^{-6} W$. To convert watts to microwatts, divide by $10^{-6}$: $2 \\cdot 10^{-7} W / 10^{-6} = 0.2 µW$.",
"source": "https://50ohm.de/NEA_zehnerpotenzen.html",
"confidence": 8
},
"EA114": {
- "revision": 1,
- "explanation": "$5 \\cdot 10^{-1}$ W is $0.5$ W; multiplying by 1000 converts that to 500 mW.",
+ "revision": 2,
+ "explanation": "$5 \\cdot 10^{-1} W$ is $0.5 W$. Since $1 W = 1000 mW$, $0.5 W = 500 mW$.",
"source": "https://50ohm.de/NEA_zehnerpotenzen.html",
"confidence": 8
},
"EA115": {
- "revision": 1,
- "explanation": "Micro is $10^{-6}$ and nano is $10^{-9}$, so $0.22$ µF is $0.22 \\cdot 1000 = 220$ nF.",
+ "revision": 2,
+ "explanation": "Micro is $10^{-6}$ and nano is $10^{-9}$, so one microfarad is 1000 nanofarads. Thus $0.22 µF = 0.22 \\cdot 1000 nF = 220 nF$.",
"source": "https://50ohm.de/NEA_zehnerpotenzen.html",
"confidence": 8
},
"EA116": {
- "revision": 1,
- "explanation": "Kilo to mega divides by 1000, so 3750 kHz is 3.750 MHz.",
+ "revision": 2,
+ "explanation": "Kilo is $10^3$ and mega is $10^6$, so converting kHz to MHz divides by 1000. $3750 kHz = 3.750 MHz$.",
"source": "https://50ohm.de/NEA_zehnerpotenzen.html",
"confidence": 8
},
"EA201": {
- "revision": 1,
- "explanation": "Digital circuits can represent two robust electrical states as 0 and 1, so binary maps naturally to switching devices such as transistors.",
+ "revision": 2,
+ "explanation": "Binary needs only two symbols, 0 and 1, which map directly onto two robust electrical states — a switching element such as a transistor is simply off or on (low or high voltage). Multi-level analog states (option C) are far harder to keep reliable across temperature and noise, so digital logic builds everything from clean two-state switching.",
"source": "https://50ohm.de/EA_binaer.html",
"confidence": 8
},
"EA202": {
- "revision": 1,
- "explanation": "Each bit doubles the number of possible states; with 3 bits there are $2^3 = 8$ states.",
+ "revision": 2,
+ "explanation": "A bit has two possible states. With independent bits, the possibilities multiply, so $n$ bits can represent $2^n$ combinations. For 3 bits: $2^3 = 8$.",
"source": "https://50ohm.de/EA_binaer.html",
"confidence": 8
},
"EA203": {
- "revision": 1,
- "explanation": "Each bit doubles the number of possible states; with 4 bits there are $2^4 = 16$ states.",
+ "revision": 2,
+ "explanation": "Every additional bit doubles the number of representable values. Four bits therefore give $2^4 = 16$ combinations, from binary 0000 to 1111.",
"source": "https://50ohm.de/EA_binaer.html",
"confidence": 8
},
"EA204": {
- "revision": 1,
- "explanation": "A five-bit binary number has $2^5$ possible combinations, so it can represent 32 values.",
+ "revision": 2,
+ "explanation": "For binary numbers, count values with $2^n$. A five-bit word has $2^5 = 32$ possible patterns, so it can represent 32 values.",
"source": "https://50ohm.de/EA_binaer.html",
"confidence": 8
},
"EA205": {
- "revision": 1,
- "explanation": "$01001110_2 = 64 + 8 + 4 + 2 = 78$; the leading zero only pads the width.",
+ "revision": 2,
+ "explanation": "Read binary by place values: 128, 64, 32, 16, 8, 4, 2, 1. In $01001110_2$, the set bits are $64 + 8 + 4 + 2 = 78$; the leading zero just pads the byte.",
"source": "https://50ohm.de/EA_binaer.html",
"confidence": 8
},
"EA206": {
- "revision": 1,
- "explanation": "$10001110_2 = 128 + 8 + 4 + 2 = 142$.",
+ "revision": 2,
+ "explanation": "Use binary place values. $10001110_2$ has bits set at 128, 8, 4 and 2, so the decimal value is $128 + 8 + 4 + 2 = 142$.",
"source": "https://50ohm.de/EA_binaer.html",
"confidence": 8
},
"EA207": {
- "revision": 1,
- "explanation": "$10011100_2 = 128 + 16 + 8 + 4 = 156$.",
+ "revision": 2,
+ "explanation": "$10011100_2$ uses the 128, 16, 8 and 4 places. Adding those weights gives $128 + 16 + 8 + 4 = 156$.",
"source": "https://50ohm.de/EA_binaer.html",
"confidence": 8
},
"EA208": {
- "revision": 1,
- "explanation": "$11111000_2 = 128 + 64 + 32 + 16 + 8 = 248$.",
+ "revision": 2,
+ "explanation": "$11111000_2$ has the five upper places set and the three lower places clear. $128 + 64 + 32 + 16 + 8 = 248$.",
"source": "https://50ohm.de/EA_binaer.html",
"confidence": 8
},
"EB101": {
- "revision": 2,
- "explanation": "Between large parallel plates the field lines are almost straight, parallel, and evenly spaced, so the approximation is a homogeneous electric field.",
+ "revision": 3,
+ "explanation": "Well inside a large parallel-plate capacitor the field lines run straight, parallel, and evenly spaced — a homogeneous (uniform) electric field, the same strength and direction everywhere. (Fringing curves the field only at the plate edges.) The field between charged plates is electric, not magnetic.",
"source": "https://50ohm.de/NEA_slide_nea_em_feld.html",
"confidence": 8
},
"EB102": {
- "revision": 2,
- "explanation": "For a plate capacitor, $E = U/d$. With $d = 0.6$ cm $= 0.006$ m, $E = 9/0.006 = 1500$ V/m.",
+ "revision": 3,
+ "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.",
"source": "https://50ohm.de/NEA_slide_nea_em_feld.html",
"confidence": 8
},
"EB103": {
- "revision": 2,
- "explanation": "Use $E = U/d$ and convert $0.15$ mm to $1.5 \\cdot 10^{-4}$ m. Thus $300/(1.5 \\cdot 10^{-4}) = 2.0 \\cdot 10^6$ V/m = 2000 kV/m.",
+ "revision": 3,
+ "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.",
"source": "https://50ohm.de/NEA_slide_nea_em_feld.html",
"confidence": 8
},
"EB104": {
- "revision": 2,
- "explanation": "Breakdown strength is an electric field strength, so $U = E \\cdot d$. $400$ kV/cm across $0.15$ mm gives $400 \\cdot 0.015 = 6$ kV.",
+ "revision": 3,
+ "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.",
"source": "https://50ohm.de/NEA_slide_nea_em_feld.html",
"confidence": 8
},
"EB105": {
- "revision": 2,
- "explanation": "At a vertical antenna the electric field lines run between the conductor and the surrounding reference/ground; the concentric loops around the conductor are the magnetic field, not the marked vertical electric lines.",
+ "revision": 3,
+ "explanation": "At a vertical antenna, the electric field (E-field) is drawn along the voltage path between the conductor and the surrounding ground/reference. The closed concentric loops around the conductor are the magnetic field (H-field), caused by RF current, so the marked vertical/open field lines are the electric field.",
"source": "https://50ohm.de/NEA_slide_nea_em_feld.html",
- "confidence": 7
+ "confidence": 8
},
"EB201": {
- "revision": 1,
- "explanation": "A current through a straight conductor creates magnetic field lines that close around the conductor; in the simple straight-wire case they are concentric circles.",
+ "revision": 2,
+ "explanation": "A current in a straight conductor produces a magnetic field whose lines close on themselves, encircling the wire as concentric circles (right-hand rule). They are magnetic, not electric, and circular rather than radial/star-shaped.",
"source": "https://50ohm.de/EA_h_feld.html",
"confidence": 8
},
"EB202": {
- "revision": 2,
- "explanation": "A long current-carrying solenoid concentrates nearly parallel magnetic field lines inside the winding, so its interior field is approximately homogeneous and magnetic.",
+ "revision": 3,
+ "explanation": "Inside a long solenoid (Zylinderspule) carrying DC, the turns' fields add to give nearly straight, parallel, evenly spaced lines along the axis — an approximately homogeneous (uniform) magnetic field. It is magnetic (current-driven) and uniform, like the electric analogue inside a plate capacitor.",
"source": "https://50ohm.de/NEA_slide_nea_em_feld.html",
"confidence": 8
},
"EB203": {
- "revision": 1,
- "explanation": "For a toroidal core, $H = NI/l_m$ with $l_m = \\pi d$. Here $H = 6 \\cdot 2.5/(\\pi \\cdot 0.026) \\approx 183.6$ A/m.",
+ "revision": 2,
+ "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 $100\\times$.",
"source": "https://50ohm.de/NEA_slide_nea_em_feld.html?print-pdf=&showNotes=true",
"confidence": 8
},
"EB204": {
- "revision": 1,
- "explanation": "Iron is ferromagnetic at room temperature; copper and aluminium are not ferromagnetic, and chromium is not the standard room-temperature ferromagnet used here.",
+ "revision": 2,
+ "explanation": "Of these, only iron (Eisen) is ferromagnetic at room temperature — strongly magnetisable, ideal for inductor and transformer cores. Copper and aluminium are non-magnetic, and chromium is not ferromagnetic at room temperature, so they cannot serve as a core.",
"source": "https://50ohm.de/E_spule_1.html",
"confidence": 8
},
"EB205": {
- "revision": 4,
- "explanation": "Eddy currents induced in the conductive copper or aluminium core oppose the changing RF flux, partially cancelling the field inside the core and shrinking the effective magnetic cross-section, so inductance drops. The 'field cannot penetrate the core' wording is a simplification of this eddy-current screening — a memorable shorthand rather than a literal account.",
+ "revision": 5,
+ "explanation": "At RF, a conductive copper or aluminium core develops eddy currents. By Lenz's law these currents create an opposing magnetic field, so the changing RF flux is screened from the core interior and the effective magnetic cross-section becomes smaller. Less linked flux for the same current means lower inductance.",
"source": "https://50ohm.de/NEA_spule_1.html",
- "confidence": 7
+ "confidence": 8
},
"EB206": {
- "revision": 2,
- "explanation": "Around a vertical current-carrying antenna conductor, the closed concentric loops are magnetic field lines. The electric field lines are the open/vertical ones tied to the conductor and ground reference.",
+ "revision": 3,
+ "explanation": "A current-carrying antenna conductor produces a magnetic field (H-field) in closed loops around the conductor, following the right-hand rule. The electric field (E-field) is not the closed loop; it is associated with voltage between conductor and reference/ground.",
"source": "https://50ohm.de/NEA_slide_nea_em_feld.html",
- "confidence": 7
+ "confidence": 8
},
"EB301": {
- "revision": 2,
- "explanation": "Radio radiation needs time-varying fields. A time-varying current in a conductor, such as an antenna, produces coupled electric and magnetic field components.",
+ "revision": 3,
+ "explanation": "Radiation needs changing fields. A time-varying current in a conductor (e.g. the RF current in an antenna) produces a time-varying magnetic field, which in turn induces an electric field — the coupled pair is the electromagnetic field. A steady DC current gives only a static magnetic field that does not radiate.",
"source": "https://50ohm.de/NEA_slide_nea_em_feld.html",
"confidence": 8
},
"EB302": {
- "revision": 2,
- "explanation": "An electromagnetic wave propagates because changing electric and magnetic fields continually sustain each other; neither field travels independently in the far-field wave model.",
+ "revision": 3,
+ "explanation": "An electromagnetic wave is self-sustaining: a changing electric field induces a magnetic field and the changing magnetic field induces an electric field, so the two regenerate each other as the wave travels (Maxwell's equations). Neither field propagates alone in the far field — they are inseparable partners.",
"source": "https://50ohm.de/NEA_slide_nea_em_feld.html",
"confidence": 8
},
"EB303": {
- "revision": 1,
- "explanation": "In free-space far-field propagation the electric and magnetic field vectors are transverse to each other, so their angle is $90^\\circ$.",
+ "revision": 2,
+ "explanation": "In the far field of free-space propagation the wave is transverse (TEM): the $E$ and $H$ vectors lie at right angles to each other and to the direction of travel. So the angle between the electric and magnetic field components is $90^\\circ$.",
"source": "https://50ohm.de/NEA_fernfeld.html",
"confidence": 8
},
"EB304": {
- "revision": 1,
- "explanation": "In an undisturbed far field, the $E$ field, $H$ field, and propagation direction form a mutually perpendicular triad.",
+ "revision": 2,
+ "explanation": "An undisturbed far field is a TEM wave: the $E$-field, the $H$-field, and the propagation direction form a mutually perpendicular (right-angled) triad, with $E$ and $H$ in phase. The propagation direction is fixed by $E \\times H$ (the Poynting vector), so it is not free as the wrong options claim.",
"source": "https://50ohm.de/NEA_fernfeld.html",
"confidence": 8
},
"EB305": {
- "revision": 2,
- "explanation": "Electromagnetic-wave polarization is defined by the orientation or motion of the electric-field vector, not by the magnetic field or travel direction.",
+ "revision": 3,
+ "explanation": "By convention the polarisation of a wave is the direction of its electric-field vector ($E$). A vertical $E$-field is vertical polarisation, horizontal is horizontal. It is defined by the E-field, not the magnetic field, the travel direction, or the near field.",
"source": "https://50ohm.de/NEA_polarisation_2.html",
"confidence": 8
},
"EB306": {
- "revision": 2,
- "explanation": "Polarization follows the electric-field vector in the drawing. Here that vector lies horizontally, so the wave is horizontally polarized.",
+ "revision": 3,
+ "explanation": "Polarization is defined by the direction of the electric-field vector, not by the magnetic field or travel direction. In this drawing the E-field lies horizontally, so the electromagnetic wave is horizontally polarized.",
"source": "https://50ohm.de/NEA_polarisation_2.html",
- "confidence": 7
+ "confidence": 8
},
"EB307": {
- "revision": 2,
- "explanation": "Polarization is read from the electric-field vector. In this figure the electric field is vertical, so the wave is vertically polarized.",
+ "revision": 3,
+ "explanation": "For radio waves, polarization follows the electric field (E-field). The E-field in the figure is vertical, so the wave is vertically polarized; the magnetic field is at right angles to it and does not name the polarization.",
"source": "https://50ohm.de/NEA_polarisation_2.html",
- "confidence": 7
+ "confidence": 8
},
"EB308": {
- "revision": 2,
- "explanation": "When the electric-field direction rotates during propagation rather than staying along one fixed line, the wave is circularly polarized.",
+ "revision": 3,
+ "explanation": "Circular polarization means the electric-field vector rotates as the wave propagates instead of staying vertical or horizontal. A receiving antenna then sees a rotating field rather than one fixed linear direction.",
"source": "https://50ohm.de/NEA_polarisation_2.html",
- "confidence": 7
+ "confidence": 8
},
"EB309": {
- "revision": 2,
- "explanation": "For a linear antenna, the transmitted wave's polarization follows the orientation of the radiating element in the main direction. The shown elements are horizontal, so the signal is horizontally polarized.",
+ "revision": 3,
+ "explanation": "A linear antenna launches an electric field in the same orientation as the radiating element in its main direction. Since the shown elements are horizontal, the emitted wave is horizontally polarized.",
"source": "https://50ohm.de/NEA_polarisation_2.html",
- "confidence": 7
+ "confidence": 8
},
"EB310": {
- "revision": 2,
- "explanation": "A linearly radiating element gives polarization in the same orientation as the electric field it launches. The shown main-direction field is vertical, so the signal is vertically polarized.",
+ "revision": 3,
+ "explanation": "The polarization of a linearly polarized wave is the orientation of its electric field. In the shown main radiation direction that field is vertical, so the transmitted signal is vertically polarized.",
"source": "https://50ohm.de/NEA_polarisation_2.html",
- "confidence": 7
+ "confidence": 8
},
"EB311": {
- "revision": 1,
- "explanation": "Use $\\lambda = c/f$ with $c \\approx 300$ Mm/s. $300/1.84 \\approx 163$, so 1.84 MHz corresponds to about 163 m.",
+ "revision": 2,
+ "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.)",
"source": "https://50ohm.de/NE_wellenlaenge_2.html",
"confidence": 8
},
"EB312": {
- "revision": 1,
- "explanation": "Using $\\lambda \\approx 300/f_{MHz}$, $300/21 \\approx 14.29$ m.",
+ "revision": 2,
+ "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.",
"source": "https://50ohm.de/NE_wellenlaenge_2.html",
"confidence": 8
},
"EB313": {
- "revision": 1,
- "explanation": "Using $\\lambda \\approx 300/f_{MHz}$, $300/28.5 \\approx 10.5$ m.",
+ "revision": 2,
+ "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.",
"source": "https://50ohm.de/NE_wellenlaenge_2.html",
"confidence": 8
},
"EB314": {
- "revision": 1,
- "explanation": "Rearrange $\\lambda = c/f$ to $f \\approx 300/\\lambda$ in MHz for metres. $300/80.0 = 3.75$ MHz.",
+ "revision": 2,
+ "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.",
"source": "https://50ohm.de/NE_wellenlaenge_2.html",
"confidence": 8
},
"EB315": {
- "revision": 1,
- "explanation": "A wavelength of 30 mm is 0.03 m. $f = c/\\lambda \\approx 3 \\cdot 10^8 / 0.03 = 1 \\cdot 10^{10}$ Hz = 10 GHz.",
+ "revision": 2,
+ "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.",
"source": "https://50ohm.de/NE_wellenlaenge_2.html",
"confidence": 8
},
"EB316": {
- "revision": 1,
- "explanation": "A wavelength of 10 cm is 0.1 m. $f = c/\\lambda \\approx 3 \\cdot 10^8 / 0.1 = 3 \\cdot 10^9$ Hz = 3 GHz.",
+ "revision": 2,
+ "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.)",
"source": "https://50ohm.de/NE_wellenlaenge_2.html",
"confidence": 8
},
"EB401": {
- "revision": 2,
- "explanation": "For a sine wave, the peak value is RMS times sqrt(2). Mains 230 V is an RMS value, so 230 * 1.414 is about 325 V.",
+ "revision": 3,
+ "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 $U_{peak} = 230 V \\cdot \\sqrt{2} \\approx 325 V$.",
"source": "https://50ohm.de/E_slide_e_strom_spannung_widerstand_leistung_energie.html",
"confidence": 8
},
"EB402": {
- "revision": 2,
- "explanation": "Peak-to-peak voltage is twice the peak value. From 230 V RMS, the peak is about 325 V, so peak-to-peak is about 650 V, rounded here to 651 V.",
+ "revision": 3,
+ "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 $U_{pp} \\approx 650 V$.",
"source": "https://50ohm.de/E_slide_e_strom_spannung_widerstand_leistung_energie.html",
"confidence": 8
},
"EB403": {
- "revision": 2,
- "explanation": "For a sine wave, peak voltage is RMS times sqrt(2): 12 V * 1.414 is about 17 V. Peak-to-peak is twice that, about 34 V.",
+ "revision": 3,
+ "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.",
"source": "https://50ohm.de/E_slide_e_strom_spannung_widerstand_leistung_energie.html",
"confidence": 8
},
"EB404": {
- "revision": 2,
- "explanation": "For a sine wave, RMS is peak divided by sqrt(2). 12 V / 1.414 is about 8.5 V.",
+ "revision": 3,
+ "explanation": "For sine waves, RMS is peak divided by $\\sqrt{2}$ because RMS represents equal heating power. $12 V / 1.414 \\approx 8.5 V$.",
"source": "https://50ohm.de/E_slide_e_strom_spannung_widerstand_leistung_energie.html",
"confidence": 8
},
"EB405": {
- "revision": 2,
- "explanation": "A DC voltage that gives the same heating power as a sine wave is the RMS value. For a 1 V sine peak, RMS is 1/sqrt(2), about 0.7 V in either polarity.",
+ "revision": 3,
+ "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.",
"source": "https://50ohm.de/E_slide_e_strom_spannung_widerstand_leistung_energie.html",
"confidence": 8
},
"EB406": {
- "revision": 2,
- "explanation": "Peak-to-peak voltage is the vertical distance from the lowest trough to the highest crest on the screen. Reading the divisions in the shown trace gives 12 V.",
+ "revision": 3,
+ "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.",
"source": "https://50ohm.de/E_slide_e_strom_spannung_widerstand_leistung_energie.html",
- "confidence": 7
+ "confidence": 8
},
"EB407": {
- "revision": 2,
- "explanation": "The peak-to-peak value is twice the peak value shown in the diagram. A 20 V peak therefore gives 40 V peak-to-peak.",
+ "revision": 3,
+ "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 $2 \\cdot 20 V = 40 V$.",
"source": "https://50ohm.de/E_slide_e_strom_spannung_widerstand_leistung_energie.html",
- "confidence": 7
+ "confidence": 8
},
"EB408": {
- "revision": 2,
- "explanation": "Frequency is the reciprocal of period: f = 1/T. With T = 50 microseconds, f = 1/(50e-6 s) = 20000 Hz = 20 kHz.",
+ "revision": 3,
+ "explanation": "Frequency is cycles per second, so it is the reciprocal of period: $f = 1/T$. With $T = 50 µs = 50 \\cdot 10^{-6} s$, $f = 1/T = 20,000 Hz = 20 kHz$.",
"source": "https://50ohm.de/E_slide_e_strom_spannung_widerstand_leistung_energie.html",
"confidence": 8
},
"EB409": {
- "revision": 2,
- "explanation": "Read one period from the oscilloscope grid, then use f = 1/T. The trace period is about 12 microseconds, so f is about 83.3 kHz.",
+ "revision": 3,
+ "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 $f \\approx 1/(12 \\cdot 10^{-6}) = 83.3 kHz$.",
"source": "https://50ohm.de/E_slide_e_strom_spannung_widerstand_leistung_energie.html",
- "confidence": 7
+ "confidence": 8
},
"EB410": {
- "revision": 2,
- "explanation": "The oscilloscope trace spans 4 divisions at 5 ms/div, so T = 20 ms. f = 1/0.020 s = 50 Hz.",
+ "revision": 3,
+ "explanation": "Oscilloscope timebase reading is divisions times time/div. Four divisions at 5 ms/div gives $T = 20 ms = 0.020 s$, so $f = 1/T = 50 Hz$.",
"source": "https://50ohm.de/E_slide_e_strom_spannung_widerstand_leistung_energie.html",
- "confidence": 7
+ "confidence": 8
},
"EB411": {
- "revision": 2,
- "explanation": "The trace period is 4 divisions at 0.03 microseconds/div, so T = 0.12 microseconds. f = 1/T is about 8.33 MHz.",
+ "revision": 3,
+ "explanation": "Read the period from the grid: $4 \\cdot 0.03 µs = 0.12 µs$. Then $f = 1/T = 1/(0.12 \\cdot 10^{-6} s) \\approx 8.33 MHz$.",
"source": "https://50ohm.de/NE_oszilloskop_1.html",
- "confidence": 7
+ "confidence": 8
},
"EB501": {
- "revision": 1,
- "explanation": "PEP is defined at the crest of the modulation envelope: it is the average power over one RF cycle at that highest envelope point under normal operating conditions.",
+ "revision": 2,
+ "explanation": "PEP (peak envelope power, Spitzenleistung) is the average power over one RF cycle taken at the highest crest of the modulation envelope, into a real (resistive) load, under normal operation. The key is 'one RF cycle at the envelope peak' — averaging over a long interval instead would give mean power, and the dipole-gain option describes ERP.",
"source": "https://life.itu.int/radioclub/rr/art1.pdf",
"confidence": 9
},
"EB502": {
- "revision": 1,
- "explanation": "Mean transmitter power is averaged over a time interval long enough compared with the lowest modulation frequency period, so it describes the longer-term power delivered to the antenna feed line.",
+ "revision": 2,
+ "explanation": "Mean power (mittlere Leistung) is averaged over a time long compared with the period of the lowest modulation frequency — i.e. the long-term average delivered to the feed line. Contrast PEP, which is measured over a single RF cycle at the envelope peak; the gain-product option describes ERP.",
"source": "https://life.itu.int/radioclub/rr/art1.pdf",
"confidence": 9
},
"EB503": {
- "revision": 1,
- "explanation": "For a purely ohmic load, AC power formulas keep the same form when voltage and current are RMS values. Peak values would overstate the heating power.",
+ "revision": 2,
+ "explanation": "Yes — the DC power formulas ($P = UI = U^2/R = I^2 R$) apply unchanged to AC across a purely resistive load, provided you use RMS (Effektivwert) values. RMS is defined precisely so that it produces the same heating as the equivalent DC; peak values would overstate the power by up to a factor of two.",
"source": "https://50ohm.de/EA_leistung_2.html",
"confidence": 8
},
"EB504": {
- "revision": 1,
- "explanation": "Combine P = U * I with Ohm's law I = U/R to get P = U^2/R. Solving for voltage gives U = sqrt(P * R).",
+ "revision": 2,
+ "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.)",
"source": "https://50ohm.de/EA_leistung_2.html",
"confidence": 8
},
"EB505": {
- "revision": 1,
- "explanation": "From P = I^2 * R, current is I = sqrt(P/R). From P = U^2/R, voltage is U = sqrt(P * R).",
+ "revision": 2,
+ "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 $U = \\sqrt{PR}$.",
"source": "https://50ohm.de/EA_leistung_2.html",
"confidence": 8
},
"EB506": {
- "revision": 1,
- "explanation": "Rearrange P = U^2/R to get R = U^2/P, and rearrange P = I^2 * R to get R = P/I^2.",
+ "revision": 2,
+ "explanation": "These are just Ohm's law plus power rearranged. From $P = U^2/R$, $R = U^2/P$; from $P = I^2R$, $R = P/I^2$.",
"source": "https://50ohm.de/EA_leistung_2.html",
"confidence": 8
},
"EB507": {
- "revision": 1,
- "explanation": "Use RMS voltage in P = U^2/R for the 50 Ohm load. 100^2/50 = 10000/50 = 200 W.",
+ "revision": 2,
+ "explanation": "For RF power in a resistive 50 Ohm load, use RMS voltage: $P = U^2/R$. With $U = 100 V$ and $R = 50 Ohm$, $P = 100^2/50 = 200 W$.",
"source": "https://50ohm.de/EA_leistung_2.html",
"confidence": 8
},
"EB508": {
- "revision": 1,
- "explanation": "Use P = I^2 * R with RMS current. 2^2 * 50 = 4 * 50 = 200 W.",
+ "revision": 2,
+ "explanation": "Use RMS current in the resistive power formula $P = I^2R$. With $I = 2 A$ and $R = 50 Ohm$, $P = 2^2 \\cdot 50 = 200 W$.",
"source": "https://50ohm.de/EA_leistung_2.html",
"confidence": 8
},
"EB509": {
- "revision": 1,
- "explanation": "The resistor power is P = U^2/R. With 10 V across 100 Ohm, P = 100/100 = 1.00 W, so the rating must be at least that.",
+ "revision": 2,
+ "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.",
"source": "https://50ohm.de/EA_leistung_2.html",
"confidence": 8
},
"EB510": {
- "revision": 1,
- "explanation": "Check both limits. The power limit gives U = sqrt(P * R) = sqrt(1 W * 10000 Ohm) = 100 V, which is below the 700 V voltage limit.",
+ "revision": 2,
+ "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.",
"source": "https://50ohm.de/EA_leistung_2.html",
"confidence": 8
},
"EB511": {
- "revision": 1,
- "explanation": "The power limit gives U = sqrt(P * R) = sqrt(6 W * 100000 Ohm) about 775 V. That is below the 1000 V voltage rating, so power is the limiting factor.",
+ "revision": 2,
+ "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.",
"source": "https://50ohm.de/EA_leistung_2.html",
"confidence": 8
},
"EB512": {
- "revision": 1,
- "explanation": "From P = I^2 * R, I = sqrt(P/R). sqrt(23.0/120) is about 0.438 A, or 438 mA.",
+ "revision": 2,
+ "explanation": "Solve $P = I^2R$ for current: $I = \\sqrt{P/R}$. With $P = 23.0 W$ and $R = 120 Ohm$, $I = \\sqrt{23/120} \\approx 0.438 A = 438 mA$.",
"source": "https://50ohm.de/EA_leistung_2.html",
"confidence": 8
},
"EB513": {
- "revision": 1,
- "explanation": "A 25 V peak-to-peak sine has a 12.5 V peak, so RMS voltage is 12.5/sqrt(2) about 8.84 V. Through 1000 Ohm, that is about 8.8 mA RMS.",
+ "revision": 2,
+ "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.",
"source": "https://50ohm.de/EA_leistung_2.html",
"confidence": 8
},
"EB514": {
- "revision": 1,
- "explanation": "With 11 equal resistors in parallel, each resistor can still dissipate 5 W. The total rating is 11 * 5 W = 55 W.",
+ "revision": 2,
+ "explanation": "Parallel resistors each dissipate independently, so power ratings add: $11 \\cdot 5\\ \\text{W} = 55$ W. (As a check, eleven $560\\ \\Omega$ in parallel give $560/11 \\approx 51\\ \\Omega \\approx 50\\ \\Omega$, the wanted dummy-load value.) The total handles much more than a single $5$ W resistor.",
"source": "https://50ohm.de/EA_leistung_2.html",
"confidence": 8
},
"EC101": {
- "revision": 1,
- "explanation": "Wirewound resistors can dissipate high power, but the winding adds inductance, so they are best suited to DC and low-frequency high-load use.",
+ "revision": 2,
+ "explanation": "Wirewound resistors (Drahtwiderstände) dissipate large power well, but the winding acts as a coil and adds inductance. That stray inductance is harmless at DC and low frequencies but spoils RF behaviour — so they suit high-power, low-frequency loads, not HF.",
"source": "https://50ohm.de/E_widerstand_materialien.html",
"confidence": 8
},
"EC102": {
- "revision": 1,
- "explanation": "Metal-film resistors are made for tight value tolerance and low temperature dependence, which is why they are used as precision resistors.",
+ "revision": 2,
+ "explanation": "Metal-film resistors (Metallschichtwiderstände) are made with tight tolerance and a very low temperature coefficient, so their value barely drifts with manufacturing spread or heat. Those two properties are exactly what a precision resistor needs; wirewound is too inductive and LDRs change with light.",
"source": "https://50ohm.de/E_widerstand_materialien.html",
"confidence": 8
},
"EC103": {
- "revision": 1,
- "explanation": "Metal-oxide film resistors are relatively low-inductance and stable at higher frequencies, unlike wirewound parts whose winding behaves like an inductor.",
+ "revision": 2,
+ "explanation": "Metal-oxide film resistors (Metalloxidschichtwiderstände) are essentially non-inductive — no coil winding — so their impedance stays resistive above $30$ MHz. Wirewound parts behave like inductors at RF, which is why they are avoided there despite their power rating.",
"source": "https://50ohm.de/E_widerstand_materialien.html",
"confidence": 8
},
"EC104": {
- "revision": 1,
- "explanation": "A VHF/UHF dummy load should behave like a pure resistance. Low stray inductance and capacitance keep the impedance near 50 Ohm as frequency rises.",
+ "revision": 2,
+ "explanation": "A dummy load (künstliche Antenne) must look like a pure $50\\ \\Omega$ resistance, presenting an SWR near 1 instead of radiating. Low self-inductance and self-capacitance keep the impedance resistive as frequency climbs into VHF/UHF; any reactance would skew it away from $50\\ \\Omega$.",
"source": "https://50ohm.de/E_widerstand_materialien.html",
"confidence": 8
},
"EC105": {
- "revision": 1,
- "explanation": "Ten 500 Ohm resistors in parallel give 500/10 = 50 Ohm. Carbon-film parts avoid the wirewound inductance that would spoil a dummy load at RF.",
+ "revision": 2,
+ "explanation": "Ten $500\\ \\Omega$ resistors in parallel give $500/10 = 50\\ \\Omega$, the wanted load value. Spreading the power over ten carbon-film parts also avoids the winding inductance of a single wirewound $50\\ \\Omega$ resistor, which would ruin the match at $28$ MHz.",
"source": "https://50ohm.de/E_widerstand_materialien.html",
"confidence": 8
},
"EC106": {
- "revision": 1,
- "explanation": "Ten equal 500 Ohm resistors in parallel give 50 Ohm, and unwound carbon-film parts keep parasitic inductance low enough for this RF dummy-load use.",
+ "revision": 2,
+ "explanation": "Ten unwound $500\\ \\Omega$ carbon-film resistors in parallel give $500/10 = 50\\ \\Omega$, while staying low-inductance enough to behave resistively at $50$ MHz. A single wirewound $50\\ \\Omega$ part, or wound carbon parts, would add reactance and spoil the load.",
"source": "https://50ohm.de/E_widerstand_materialien.html",
"confidence": 8
},
"EC107": {
- "revision": 1,
- "explanation": "For VHF dummy loads, unwound metal-oxide resistors are preferred because they can be made low-inductance and thermally robust.",
+ "revision": 2,
+ "explanation": "For a VHF dummy load you want resistors that combine a clean resistive impedance with good power handling: unwound (non-inductive) metal-oxide resistors do both. High-power wirewound parts are too inductive at VHF, and reactive ('Blind') components by definition are not pure resistors.",
"source": "https://50ohm.de/E_widerstand_materialien.html",
"confidence": 8
},
"EC108": {
- "revision": 1,
- "explanation": "NTC thermistors have a deliberately temperature-dependent resistance, making them useful as temperature sensors.",
+ "revision": 2,
+ "explanation": "An NTC thermistor is a temperature-dependent resistor with a Negative Temperature Coefficient: as temperature rises, resistance falls. That predictable resistance change lets the circuit infer temperature from a resistance or voltage-divider measurement.",
"source": "https://50ohm.de/E_widerstand_ntc_ptc.html",
"confidence": 8
},
"EC109": {
- "revision": 1,
- "explanation": "The symbol shows a temperature-dependent resistor whose resistance falls as temperature rises; that negative temperature coefficient is an NTC thermistor.",
+ "revision": 2,
+ "explanation": "NTC means negative temperature coefficient: as temperature rises, resistance falls. The matching symbol is therefore the temperature-dependent resistor whose resistance trend goes downward with heat.",
"source": "https://50ohm.de/E_widerstand_ntc_ptc.html",
- "confidence": 7
+ "confidence": 8
},
"EC110": {
- "revision": 1,
- "explanation": "An NTC symbol indicates temperature dependence with resistance decreasing as temperature increases. In the shown choices, that is the symbol with the temperature arrow up and resistance/conductance indication downward as described on 50ohm.",
+ "revision": 2,
+ "explanation": "For an NTC thermistor, hotter means lower resistance. In the symbol choices, look for the temperature indication rising while the resistance indication falls; that is the negative coefficient.",
"source": "https://50ohm.de/E_widerstand_ntc_ptc.html",
- "confidence": 7
+ "confidence": 8
},
"EC111": {
- "revision": 1,
- "explanation": "A PTC thermistor has a positive temperature coefficient: as temperature rises, resistance rises. The matching symbol is the one with both temperature and resistance trend upward.",
+ "revision": 2,
+ "explanation": "PTC means positive temperature coefficient: resistance increases with temperature. The right symbol is the one where heating and resistance go in the same upward direction.",
"source": "https://50ohm.de/E_widerstand_ntc_ptc.html",
- "confidence": 7
+ "confidence": 8
},
"EC112": {
- "revision": 1,
- "explanation": "A 10 percent tolerance on 5.6 kOhm is 0.56 kOhm. The possible range is 5.6 - 0.56 to 5.6 + 0.56 kOhm, or 5040 to 6160 Ohm.",
+ "revision": 2,
+ "explanation": "A $10\\,\\%$ tolerance means $\\pm 0.10 \\cdot 5.6\\ \\text{k}\\Omega = \\pm 0.56\\ \\text{k}\\Omega = \\pm 560\\ \\Omega$. So the true value lies in $5600 \\pm 560\\ \\Omega$, i.e. $5040$ to $6160\\ \\Omega$. The wider/narrower options use the wrong tolerance band.",
"source": "https://50ohm.de/NE_widerstand_toleranz.html",
"confidence": 8
},
"EC113": {
- "revision": 1,
- "explanation": "Green-blue-red is 56 times 100, so the nominal value is 5600 Ohm. Silver means 10 percent tolerance, giving 5040 to 6160 Ohm.",
+ "revision": 2,
+ "explanation": "Read the colour bands (Farbringe): green $= 5$, blue $= 6$, red $= \\times 100$, so the nominal value is $56 \\cdot 100 = 5600\\ \\Omega$. The silver band is $\\pm 10\\,\\%$, i.e. $\\pm 560\\ \\Omega$, giving $5040$ to $6160\\ \\Omega$ — the same maths as a stated $10\\,\\%$ tolerance.",
"source": "https://50ohm.de/NE_widerstand_toleranz.html",
"confidence": 8
},
"EC114": {
- "revision": 1,
- "explanation": "Common three-digit SMD resistor marking uses the first digits as significant figures and the last digit as the power of ten multiplier.",
+ "revision": 2,
+ "explanation": "Three-digit SMD codes work like the colour code in numbers: the first digits are significant figures and the last digit is the power-of-ten multiplier (e.g. $472 = 47 \\cdot 10^2 = 4700\\ \\Omega$). SMD chips are too small for colour rings, so a printed-number scheme is used.",
"source": "https://50ohm.de/E_widerstand_smd.html",
"confidence": 8
},
"EC115": {
- "revision": 1,
- "explanation": "The marking 103 means 10 followed by 3 zeros: 10 * 10^3 Ohm = 10000 Ohm = 10 kOhm.",
+ "revision": 2,
+ "explanation": "Three-digit SMD resistor codes use the first two digits as significant figures and the third as the zero count. `103` means $10 \\cdot 10^3 Ohm = 10,000 Ohm = 10 kOhm$.",
"source": "https://50ohm.de/E_widerstand_smd.html",
"confidence": 8
},
"EC116": {
- "revision": 1,
- "explanation": "The marking 221 means 22 followed by one zero: 22 * 10^1 Ohm = 220 Ohm.",
+ "revision": 2,
+ "explanation": "For SMD marking `221`, the significant digits are 22 and the multiplier digit 1 means one zero. So $22 \\cdot 10^1 Ohm = 220 Ohm$.",
"source": "https://50ohm.de/E_widerstand_smd.html",
"confidence": 8
},
"EC117": {
- "revision": 1,
- "explanation": "The marking 223 means 22 followed by three zeros: 22 * 10^3 Ohm = 22000 Ohm = 22 kOhm.",
+ "revision": 2,
+ "explanation": "The SMD code `223` means 22 with three zeros after it: $22 \\cdot 10^3 Ohm = 22,000 Ohm = 22 kOhm$.",
"source": "https://50ohm.de/E_widerstand_smd.html",
"confidence": 8
},
"EC201": {
- "revision": 1,
- "explanation": "An initially discharged capacitor charges quickly at first, then the voltage rise flattens as it approaches the supply voltage. That is the rising exponential charging curve.",
+ "revision": 2,
+ "explanation": "A capacitor voltage cannot jump instantly; it charges through the resistor with an RC time constant $\\tau = RC$. Current is largest at first, then falls as the capacitor voltage approaches the supply, producing a rising exponential curve.",
"source": "https://50ohm.de/EA_kondensator_1.html",
- "confidence": 7
+ "confidence": 8
},
"EC202": {
- "revision": 1,
- "explanation": "A capacitor's AC reactance is inversely proportional to frequency. As frequency increases, an ideal capacitor's opposition to AC decreases.",
+ "revision": 2,
+ "explanation": "A capacitor's AC reactance is $X_C = 1/(2\\pi f C)$ — inversely proportional to frequency. So as frequency rises, an ideal capacitor's opposition to AC steadily falls: it increasingly looks like a short circuit to high frequencies, which is exactly how coupling and bypass capacitors work.",
"source": "https://50ohm.de/EA_kondensator_1.html",
"confidence": 8
},
"EC203": {
- "revision": 1,
- "explanation": "For a plate capacitor, capacitance is proportional to plate area and dielectric constant, and inversely proportional to plate spacing. A larger spacing therefore reduces capacitance.",
+ "revision": 2,
+ "explanation": "Plate-capacitor capacitance is $C = \\varepsilon_0 \\varepsilon_r A / d$ — proportional to plate area $A$ and dielectric constant $\\varepsilon_r$, inversely proportional to the gap $d$. Increasing the spacing $d$ is therefore the way to reduce capacitance; larger area or higher $\\varepsilon_r$ would increase it, and applied voltage does not change $C$ at all.",
"source": "https://50ohm.de/EA_kondensator_1.html",
"confidence": 8
},
"EC204": {
- "revision": 1,
- "explanation": "Increasing the plate distance puts the same plates farther apart, so the plate capacitor's capacitance falls.",
+ "revision": 2,
+ "explanation": "For a plate capacitor, $C=\\varepsilon A/d$: capacitance rises with plate area and permittivity, but falls when plate spacing $d$ increases. Moving the same plates farther apart weakens their electric-field coupling, so capacitance decreases.",
"source": "https://50ohm.de/EA_kondensator_1.html",
"confidence": 8
},
"EC205": {
- "revision": 1,
- "explanation": "Ideal plate-capacitor capacitance depends on geometry and dielectric material, not on the applied voltage.",
+ "revision": 2,
+ "explanation": "For an ideal plate capacitor, $C=\\varepsilon A/d$. Geometry and dielectric material set the capacitance; the applied voltage only changes stored charge by $Q=CU$, not the capacitance itself.",
"source": "https://50ohm.de/EA_kondensator_1.html",
"confidence": 8
},
"EC206": {
- "revision": 1,
- "explanation": "A variable capacitor with rotor plates moving between fixed stator plates is a Drehkondensator; rotation changes the overlapping plate area and thus the capacitance.",
+ "revision": 2,
+ "explanation": "A component whose rotor plates turn on an insulated shaft between fixed stator plates is a variable/tuning capacitor (Drehkondensator). Rotating the rotor changes how much plate area overlaps, and since $C \\propto A$, that varies the capacitance — the classic way to tune an LC resonant circuit.",
"source": "https://50ohm.de/EA_kondensator_1.html",
"confidence": 8
},
"EC207": {
- "revision": 1,
- "explanation": "Electrolytic capacitors are polarized because their oxide dielectric depends on the correct DC polarity; reversed polarity can damage them.",
+ "revision": 2,
+ "explanation": "Electrolytic capacitors are polarized because the thin oxide dielectric is formed and maintained only with the correct DC polarity. Reverse voltage can destroy that oxide, causing high leakage, heating, gas pressure, or failure.",
"source": "https://50ohm.de/EA_kondensator_1.html",
"confidence": 8
},
"EC301": {
- "revision": 1,
- "explanation": "After DC is applied through a resistor, an inductor initially opposes the current change, so the voltage across it starts high and then decays toward zero.",
+ "revision": 2,
+ "explanation": "An inductor resists changes in current: $u_L = L \\cdot di/dt$. Right after switch-on, current is changing fastest and the inductor voltage is high; once steady DC flows, $di/dt$ becomes zero and the inductor voltage decays toward zero.",
"source": "https://50ohm.de/EA_spule_1.html",
- "confidence": 7
+ "confidence": 8
},
"EC302": {
- "revision": 1,
- "explanation": "The coil initially limits current because it opposes the sudden current change, so the lamp fed through the plain resistor lights first.",
+ "revision": 2,
+ "explanation": "An inductor opposes any change in current via its self-induced voltage, $u_L = L \\cdot di/dt$. At switch-on the iron-cored, many-turn coil briefly blocks the current, so lamp 2 ramps up slowly while lamp 1 — fed through a plain resistor — lights first. The coil only delays the current; it does not stop it, so lamp 2 follows shortly after.",
"source": "https://50ohm.de/EA_spule_1.html",
"confidence": 8
},
"EC303": {
- "revision": 1,
- "explanation": "An ideal inductor's AC reactance is proportional to frequency, so its opposition to AC rises as frequency increases.",
+ "revision": 2,
+ "explanation": "An ideal inductor's reactance is $X_L = 2\\pi f L$ — directly proportional to frequency. So its opposition to AC rises with frequency: an inductor increasingly blocks high frequencies, the mirror image of a capacitor. This is why RF chokes (Drosseln) work.",
"source": "https://50ohm.de/EA_spule_1.html",
"confidence": 8
},
"EC304": {
- "revision": 1,
- "explanation": "Any current-carrying conductor has a magnetic field and therefore some inductance, even if it is only a straight piece of wire.",
+ "revision": 2,
+ "explanation": "Every conductor carrying current sets up a magnetic field and therefore stores some magnetic energy, which is exactly what inductance measures — so even a straight piece of wire has inductance (a few nH per cm). No curvature or winding is required; at VHF/UHF that lead inductance already matters.",
"source": "https://50ohm.de/E_spule_1.html",
"confidence": 8
},
"EC305": {
- "revision": 1,
- "explanation": "For the same winding, shortening the coil length increases inductance. Compressing the cylindrical coil in the length direction therefore raises L.",
+ "revision": 2,
+ "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/EA_spule_1.html",
"confidence": 8
},
"EC306": {
- "revision": 1,
- "explanation": "For the same turns and cross-section, inductance is inversely proportional to coil length. Doubling the length halves 12 microhenry to 6 microhenry.",
+ "revision": 2,
+ "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: $12\\ \\mu\\text{H} / 2 = 6\\ \\mu\\text{H}$.",
"source": "https://50ohm.de/EA_spule_1.html",
"confidence": 8
},
"EC307": {
- "revision": 1,
- "explanation": "Inductance is proportional to the square of the number of turns. Doubling the turns multiplies 12 microhenry by 4, giving 48 microhenry.",
+ "revision": 2,
+ "explanation": "Inductance grows with the square of the turns: $L \\propto N^2$. Doubling the turns (same winding length) multiplies $L$ by $2^2 = 4$, so $12\\ \\mu\\text{H} \\cdot 4 = 48\\ \\mu\\text{H}$. That $N^2$ dependence is why adding a few turns changes $L$ sharply.",
"source": "https://50ohm.de/EA_spule_1.html",
"confidence": 8
},
"EC401": {
- "revision": 1,
- "explanation": "A 15:1 transformer ratio steps the 230 V primary down by 15. 230/15 is about 15.3 V, so the secondary is about 15 V.",
+ "revision": 2,
+ "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.",
"source": "https://50ohm.de/E_uebertrager_1.html",
"confidence": 8
},
"EC402": {
- "revision": 1,
- "explanation": "If the primary has five times as many turns as the secondary, the secondary voltage is one fifth of the primary voltage. 230/5 = 46 V.",
+ "revision": 2,
+ "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.",
"source": "https://50ohm.de/E_uebertrager_1.html",
"confidence": 8
},
"EC403": {
- "revision": 1,
- "explanation": "Turns ratio follows voltage ratio: 230/11.5 = 20. The secondary therefore has 600/20 = 30 turns.",
+ "revision": 2,
+ "explanation": "For an ideal transformer, voltage ratio equals turns ratio: $U_1/U_2 = N_1/N_2$. Here $230/11.5 = 20$, so the secondary has $N_2 = 600/20 = 30$ turns.",
"source": "https://50ohm.de/E_uebertrager_1.html",
"confidence": 8
},
"EC404": {
- "revision": 1,
- "explanation": "The secondary voltage is four times the primary voltage, so the secondary must have four times the turns. 150 * 4 = 600 turns.",
+ "revision": 2,
+ "explanation": "Turns follow the voltage ratio, $N_2/N_1 = U_2/U_1$. Here $U_2/U_1 = 180/45 = 4$, so the secondary needs four times the primary turns: $N_2 = 150 \\cdot 4 = 600$ turns. A step-up in voltage means a step-up in turns.",
"source": "https://50ohm.de/E_uebertrager_1.html",
"confidence": 8
},
"EC501": {
- "revision": 1,
- "explanation": "In reverse bias a normal diode blocks current except for a tiny leakage current, so it behaves like a high resistance.",
+ "revision": 2,
+ "explanation": "A diode conducts only above its forward threshold; biased the other way (reverse, Sperrrichtung) it passes just a tiny leakage current. Very little current for an applied voltage means $R = U/I$ is very large — the diode looks like a high resistance, effectively an open switch.",
"source": "https://50ohm.de/EA_diode_1.html",
"confidence": 8
},
"EC502": {
- "revision": 1,
- "explanation": "A diode conducts mainly in one direction, so it can pass one half-cycle polarity and block the other; that is the basic rectifier function.",
+ "revision": 2,
+ "explanation": "A diode conducts in only one direction, so it passes one polarity of an AC half-cycle and blocks the other. That one-way action is exactly what rectification (Gleichrichtung) needs — turning AC into pulsating DC in power supplies and detectors. It cannot amplify or store, so the other options are out.",
"source": "https://50ohm.de/EA_diode_1.html",
"confidence": 8
},
"EC503": {
- "revision": 1,
- "explanation": "Germanium diodes have a lower forward threshold, roughly 0.2 to 0.4 V, while silicon diodes are typically around 0.6 to 0.8 V.",
+ "revision": 2,
+ "explanation": "Forward threshold (Schwellspannung) depends on the semiconductor: germanium conducts from about $0.2$-$0.4$ V, silicon from about $0.6$-$0.8$ V. The lower germanium drop is why crystal/detector receivers favour Ge or Schottky diodes for weak signals. Option B simply swaps the two materials.",
"source": "https://50ohm.de/EA_diode_1.html",
"confidence": 8
},
"EC504": {
- "revision": 1,
- "explanation": "A Schottky diode uses a metal-semiconductor junction, giving a low forward voltage and very fast switching compared with ordinary pn diodes.",
+ "revision": 2,
+ "explanation": "A Schottky diode uses a metal-semiconductor junction instead of a pn junction. That gives a low forward voltage (around $0.3$ V) and — because it has almost no minority-carrier storage charge — extremely fast switching, making it ideal for HF mixers, detectors, and switch-mode rectifiers.",
"source": "https://50ohm.de/EA_diode_1.html",
"confidence": 8
},
"EC505": {
- "revision": 1,
- "explanation": "On the shown characteristic plot, curve 1 starts conducting at the lowest forward voltage near 0.2 V, which matches a Schottky diode.",
+ "revision": 2,
+ "explanation": "A Schottky diode has the lowest forward voltage because it uses a metal-semiconductor junction rather than a pn junction. On the characteristic plot, the curve that starts conducting at the lowest voltage is therefore Schottky.",
"source": "https://50ohm.de/EA_diode_1.html",
- "confidence": 7
+ "confidence": 8
},
"EC506": {
- "revision": 1,
- "explanation": "Curve 2 begins conducting around 0.2 to 0.4 V, the typical forward-threshold range for a germanium diode.",
+ "revision": 2,
+ "explanation": "Germanium pn diodes conduct at a lower forward voltage than silicon, typically around 0.2-0.4 V. So the curve above Schottky but below silicon is the germanium diode.",
"source": "https://50ohm.de/EA_diode_1.html",
- "confidence": 7
+ "confidence": 8
},
"EC507": {
- "revision": 1,
- "explanation": "Curve 3 starts its steep rise around 0.6 to 0.8 V, matching the usual silicon-diode forward threshold.",
+ "revision": 2,
+ "explanation": "A normal silicon diode needs about 0.6-0.8 V forward bias before current rises steeply. The curve with that knee voltage is the silicon diode.",
"source": "https://50ohm.de/EA_diode_1.html",
- "confidence": 7
+ "confidence": 8
},
"EC508": {
- "revision": 1,
- "explanation": "Curve 4 has the highest forward threshold in the plot, around the LED range, so it represents a light-emitting diode.",
+ "revision": 2,
+ "explanation": "An LED has a higher forward voltage because electron-hole recombination releases photon energy. On the plot, the curve with the highest forward threshold is the light-emitting diode.",
"source": "https://50ohm.de/EA_diode_1.html",
- "confidence": 7
+ "confidence": 8
},
"EC509": {
- "revision": 1,
- "explanation": "A silicon diode conducts when its anode is about 0.7 V more positive than its cathode. In the selected drawing, the right/anode side is 1.3 V and the left/cathode side is 0.6 V.",
+ "revision": 2,
+ "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.",
"source": "https://50ohm.de/EA_diode_1.html",
- "confidence": 7
+ "confidence": 8
},
"EC510": {
- "revision": 1,
- "explanation": "Use the silicon-diode rule: anode about 0.7 V above cathode. The selected drawing has 0.3 V on the anode side and -0.4 V on the cathode side.",
+ "revision": 2,
+ "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.",
"source": "https://50ohm.de/EA_diode_1.html",
- "confidence": 7
+ "confidence": 8
},
"EC511": {
- "revision": 1,
- "explanation": "Forward conduction depends on voltage difference, not whether the node voltages are positive. Here the anode is -1.3 V and the cathode is -2.0 V, so the anode is 0.7 V higher.",
+ "revision": 2,
+ "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.",
"source": "https://50ohm.de/EA_diode_1.html",
- "confidence": 7
+ "confidence": 8
},
"EC512": {
- "revision": 1,
- "explanation": "The conducting silicon-diode case is the one with the anode about 0.7 V above the cathode. In the selected drawing, -3.0 V is 0.7 V higher than -3.7 V.",
+ "revision": 2,
+ "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.",
"source": "https://50ohm.de/EA_diode_1.html",
- "confidence": 7
+ "confidence": 8
},
"EC513": {
- "revision": 1,
- "explanation": "A silicon diode becomes forward-biased when the anode is about 0.7 V above the cathode. 5.7 V at the anode and 5.0 V at the cathode meets that condition.",
+ "revision": 2,
+ "explanation": "A silicon diode conducts when the anode is about $0.7$ V above the cathode (forward bias). Check each pair by anode minus cathode: $5.7\\ \\text{V} - 5.0\\ \\text{V} = +0.7$ V, which just meets the threshold. The others give $-0.7$ V, $-0.1$ V, or $-0.7$ V — reverse-biased or below threshold.",
"source": "https://50ohm.de/EA_diode_1.html",
"confidence": 8
},
"EC514": {
- "revision": 1,
- "explanation": "The circuit is a current-limited LED: the resistor sets the LED current and the diode symbol with outgoing arrows indicates light emission.",
+ "revision": 2,
+ "explanation": "The outgoing arrows identify an LED (light-emitting diode). The series resistor is essential because the diode voltage is roughly fixed once conducting; the resistor limits current to a safe value.",
"source": "https://50ohm.de/EA_diode_1.html",
- "confidence": 7
+ "confidence": 8
},
"EC515": {
- "revision": 1,
- "explanation": "The resistor must drop the remaining voltage: 5.0 V - 1.4 V = 3.6 V. At 20 mA, R = 3.6/0.020 = 180 Ohm.",
+ "revision": 2,
+ "explanation": "The series resistor (Vorwiderstand) drops the supply voltage that the LED does not: $5.0\\ \\text{V} - 1.4\\ \\text{V} = 3.6$ V. By Ohm's law at the wanted current, $R = U/I = 3.6\\ \\text{V} / 0.020\\ \\text{A} = 180\\ \\Omega$. The LED itself sets the current via this resistor, since its own voltage is nearly fixed.",
"source": "https://50ohm.de/EA_diode_1.html",
"confidence": 8
},
"EC516": {
- "revision": 1,
- "explanation": "The resistor drops 5.5 V - 1.75 V = 3.75 V. With 25 mA, R = 3.75/0.025 = 150 Ohm and P = 3.75 * 0.025 about 0.094 W, so 0.1 W is needed.",
+ "revision": 2,
+ "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.",
"source": "https://50ohm.de/EA_diode_1.html",
"confidence": 8
},
"EC517": {
- "revision": 1,
- "explanation": "The bent cathode bar is the distinctive schematic mark for a Zener diode, used in reverse breakdown operation.",
+ "revision": 2,
+ "explanation": "A Zener diode is drawn like a diode with a bent cathode bar. That symbol hints at its special use: reverse breakdown at a defined voltage for voltage limiting or stabilisation.",
"source": "https://50ohm.de/EA_diode_1.html",
- "confidence": 7
+ "confidence": 8
},
"EC518": {
- "revision": 1,
- "explanation": "A Zener diode is designed to operate in reverse breakdown at a defined voltage, making it useful for voltage stabilization.",
+ "revision": 2,
+ "explanation": "A Zener diode (Z-Diode) is run in reverse breakdown, where it holds an almost constant voltage over a wide current range. That fixed-voltage behaviour makes it a simple voltage reference/stabiliser (Spannungsstabilisierung) — it pins a voltage, not a current.",
"source": "https://50ohm.de/EA_diode_1.html",
"confidence": 8
},
"EC519": {
- "revision": 1,
- "explanation": "The shown circuit puts a Zener diode across the output after a series resistor, the standard simple shunt voltage stabilizer arrangement.",
+ "revision": 2,
+ "explanation": "A simple Zener stabiliser is a shunt regulator: the series resistor limits current, and the reverse-biased Zener sits across the output to clamp the voltage near its Zener voltage.",
"source": "https://50ohm.de/EA_diode_1.html",
- "confidence": 7
+ "confidence": 8
},
"EC520": {
- "revision": 1,
- "explanation": "For positive output stabilization, the Zener diode is placed after the series resistor and reverse-biased across the output. That lets it clamp the output near its Zener voltage.",
+ "revision": 2,
+ "explanation": "For a positive shunt regulator, the Zener diode is reverse-biased across the output after a series resistor. When voltage tries to rise above the Zener voltage, the diode conducts more and clamps it.",
"source": "https://50ohm.de/EA_diode_1.html",
- "confidence": 7
+ "confidence": 8
},
"EC521": {
- "revision": 1,
- "explanation": "The resistor drops 13.8 V - 5 V = 8.8 V at 30 mA. R = 8.8/0.030 = 293 Ohm approximately.",
+ "revision": 2,
+ "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.)",
"source": "https://50ohm.de/EA_diode_1.html",
"confidence": 8
},
"EC522": {
- "revision": 1,
- "explanation": "The series resistor carries both Zener and load current: 25 mA + 20 mA = 45 mA. With a 4.7 V output, R = (13.8 - 4.7)/0.045 about 202 Ohm.",
+ "revision": 2,
+ "explanation": "The series resistor carries both the Zener current and the load current: $25\\ \\text{mA} + 20\\ \\text{mA} = 45$ mA. From the schematic's $13.8$ V supply and $4.7$ V stabilised output it drops $13.8 - 4.7 = 9.1$ V, so $R = 9.1\\ \\text{V} / 0.045\\ \\text{A} \\approx 202\\ \\Omega$. The resistor must be sized for the worst case where the load draws nothing and all $45$ mA flows in the Zener.",
"source": "https://50ohm.de/EA_diode_1.html",
"confidence": 8
},
"EC601": {
- "revision": 1,
- "explanation": "A transistor can be biased to switch fully on/off, to operate linearly as an amplifier, or in some cases to behave as a controllable resistance.",
+ "revision": 2,
+ "explanation": "A transistor can be driven into saturation or cut-off to act as a switch, biased in its linear region to act as an amplifier, or used as a voltage-controlled resistance. That versatility is unique to the transistor here — a transformer, capacitor, or diode cannot amplify.",
"source": "https://50ohm.de/NEA_transistor_1.html",
"confidence": 8
},
"EC602": {
- "revision": 1,
- "explanation": "A transistor is built from semiconductor regions; the usual bipolar types use alternating n- and p-doped semiconductor zones.",
+ "revision": 2,
+ "explanation": "A transistor is a semiconductor device (Halbleiterbauelement): a bipolar type is built from three alternating n- and p-doped zones (NPN or PNP). Its controllable conduction comes precisely from the physics of those doped semiconductor junctions, not from being an insulator or a simple conductor.",
"source": "https://50ohm.de/NEA_transistor_1.html",
"confidence": 8
},
"EC603": {
- "revision": 1,
- "explanation": "In the practical current-control model, a small base current controls a much larger collector current; their ratio is the current gain.",
+ "revision": 2,
+ "explanation": "Current gain (Stromverstärkung) means a small base current steers a much larger collector current, $I_C = \\beta \\cdot I_B$, with $\\beta$ (hFE) often around 100-300. The base acts as the control input; the collector-emitter path carries the controlled output current.",
"source": "https://50ohm.de/NEA_transistor_1.html",
"confidence": 8
},
"EC604": {
- "revision": 1,
- "explanation": "Bipolar junction transistors are the NPN and PNP types. FET names such as MOS-FET or junction-FET belong to field-effect transistors, not bipolar transistors.",
+ "revision": 2,
+ "explanation": "Bipolar junction transistors (BJTs) come in just two polarities: NPN and PNP, named for their doped-layer order. All the FET names — MOSFET, dual-gate MOS, junction-FET (JFET) — are field-effect transistors, a different family that is voltage- rather than current-controlled.",
"source": "https://50ohm.de/NEA_transistor_1.html",
"confidence": 8
},
"EC605": {
- "revision": 2,
- "explanation": "A bipolar transistor symbol has base, collector, and emitter terminals, with an emitter arrow; FET symbols use gate, drain, and source structures instead.",
+ "revision": 3,
+ "explanation": "A bipolar junction transistor (BJT) symbol is identified by its three terminals base, collector, and emitter, plus the arrow on the emitter. Field-effect transistors (FETs) use gate, drain, and source instead, so the terminal names alone already separate the symbol families.",
"source": "https://50ohm.de/NEA_slide_nea_bauelemente.html",
- "confidence": 7
+ "confidence": 8
},
"EC606": {
- "revision": 2,
- "explanation": "In an NPN transistor symbol the emitter arrow points outward, matching the common mnemonic 'NPN: Not Pointing iN'.",
+ "revision": 3,
+ "explanation": "In an NPN BJT, conventional emitter current leaves the transistor symbol, so the emitter arrow points outward. The common English mnemonic is 'NPN: Not Pointing iN'.",
"source": "https://50ohm.de/NEA_slide_nea_bauelemente.html",
- "confidence": 7
+ "confidence": 8
},
"EC607": {
- "revision": 2,
- "explanation": "In a PNP transistor symbol the emitter arrow points inward toward the transistor body; 50ohm gives the mnemonic 'PNP: Pfeil Nach Platte'.",
+ "revision": 3,
+ "explanation": "In a PNP BJT, conventional emitter current enters the transistor symbol, so the emitter arrow points inward toward the base region. 50ohm's German memory aid is 'PNP: Pfeil Nach Platte'.",
"source": "https://50ohm.de/NEA_slide_nea_bauelemente.html",
- "confidence": 7
+ "confidence": 8
},
"EC608": {
- "revision": 1,
- "explanation": "The three terminals of a bipolar transistor are emitter, base, and collector. Drain, source, and gate are FET terminal names.",
+ "revision": 2,
+ "explanation": "A bipolar transistor's three terminals are emitter, base, and collector (Emitter, Basis, Kollektor). Drain, source, and gate are the terminals of a field-effect transistor instead — mixing the two naming sets is the trap in the other options.",
"source": "https://50ohm.de/NEA_transistor_1.html",
"confidence": 8
},
"EC609": {
- "revision": 2,
- "explanation": "The shown NPN symbol has the collector at terminal 1, the base at terminal 2, and the emitter with the arrow at terminal 3.",
+ "revision": 3,
+ "explanation": "Read a BJT symbol by finding the base line first and then the emitter arrow. In the shown NPN symbol, terminal 2 is the base, terminal 3 is the emitter because it carries the outward arrow, and terminal 1 is therefore the collector.",
"source": "https://50ohm.de/NEA_slide_nea_bauelemente.html",
- "confidence": 7
+ "confidence": 8
},
"EC610": {
- "revision": 3,
- "explanation": "A silicon BJT conducts when its forward-biased base-emitter junction reaches its threshold; for silicon that is about 0.6 to 0.7 V.",
+ "revision": 4,
+ "explanation": "A silicon BJT conducts once its base-emitter junction is forward-biased past the silicon threshold, about $0.6$-$0.7$ V, so $U_{BE} \\approx 0.6$ V (positive for an NPN). A negative or zero $U_{BE}$ leaves the junction off, so those options do not turn it on.",
"source": "https://50ohm.de/NEA_slide_nea_bauelemente.html",
"confidence": 8
},
"EC611": {
- "revision": 1,
- "explanation": "The emitter current is the sum of collector current and base current, so in the conducting state the emitter carries the largest current.",
+ "revision": 2,
+ "explanation": "Kirchhoff's current law at the transistor gives $I_E = I_C + I_B$: the emitter current is the sum of the other two. Since base current is tiny and collector current large, the emitter necessarily carries the most current of the three terminals.",
"source": "https://50ohm.de/NEA_transistor_1.html",
"confidence": 8
},
"EC612": {
- "revision": 2,
- "explanation": "For an NPN transistor, collector current flows when the base is about 0.6 V above the emitter. The selected drawing has +2.0 V at the base and +1.4 V at the emitter.",
+ "revision": 3,
+ "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.",
"source": "https://50ohm.de/NEA_slide_nea_bauelemente.html",
- "confidence": 7
+ "confidence": 8
},
"EC613": {
- "revision": 2,
- "explanation": "Only the voltage difference matters: for NPN, the base must be about 0.6 V above the emitter. Here -5.6 V is 0.6 V above -6.2 V, so the transistor conducts.",
+ "revision": 3,
+ "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.",
"source": "https://50ohm.de/NEA_slide_nea_bauelemente.html",
- "confidence": 7
+ "confidence": 8
},
"EC614": {
- "revision": 2,
- "explanation": "For a PNP transistor, collector current flows when the base is about 0.6 V below the emitter. The selected drawing has -2.0 V at the base and -1.4 V at the emitter.",
+ "revision": 3,
+ "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.",
"source": "https://50ohm.de/NEA_slide_nea_bauelemente.html",
- "confidence": 7
+ "confidence": 8
},
"EC615": {
- "revision": 2,
- "explanation": "For PNP, the base-emitter voltage is negative when conducting: the base is about 0.6 V lower than the emitter. Here +5.6 V at the base and +6.2 V at the emitter satisfy that.",
+ "revision": 3,
+ "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.",
"source": "https://50ohm.de/NEA_slide_nea_bauelemente.html",
- "confidence": 7
+ "confidence": 8
},
"ED101": {
- "revision": 1,
- "explanation": "In a series voltage divider, voltage drops in the same ratio as resistance. If R1 is 5 times R2, then U1 is 5 times U2.",
+ "revision": 2,
+ "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.",
"source": "https://50ohm.de/NE_spannungsteiler_1.html",
"confidence": 8
},
"ED102": {
- "revision": 1,
- "explanation": "In a series voltage divider, U1/U2 = R1/R2. If R1 is one sixth of R2, then U1 is one sixth of U2.",
+ "revision": 2,
+ "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.",
"source": "https://50ohm.de/NE_spannungsteiler_1.html",
"confidence": 8
},
"ED103": {
- "revision": 1,
- "explanation": "Use the divider rule: U2 = U * R2/(R1 + R2). With 9 V, 10 kOhm, and 20 kOhm, U2 = 9 * 20/30 = 6.0 V.",
+ "revision": 2,
+ "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.",
"source": "https://50ohm.de/NE_spannungsteiler_1.html",
"confidence": 8
},
"ED104": {
- "revision": 1,
- "explanation": "For two parallel resistors, Rg = R1*R2/(R1+R2). 100*400/(100+400) = 40000/500 = 80 Ohm.",
+ "revision": 2,
+ "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: $100\\cdot400/(100+400)=40000/500=80 Ohm$.",
"source": "https://50ohm.de/E_reihe_parallel_widerstand.html",
"confidence": 8
},
"ED105": {
- "revision": 1,
- "explanation": "For two parallel resistors, Rg = R1*R2/(R1+R2). 50*200/(50+200) = 10000/250 = 40 Ohm.",
+ "revision": 2,
+ "explanation": "Parallel resistance is always lower than the smallest branch. For two branches use $R_g = R_1R_2/(R_1+R_2)$: $50\\cdot200/(50+200)=10000/250=40 Ohm$.",
"source": "https://50ohm.de/E_reihe_parallel_widerstand.html",
"confidence": 8
},
"ED106": {
- "revision": 1,
- "explanation": "For n equal resistors in parallel, Rg = R/n. Therefore each resistor is R = n*Rg = 3*1.7 kOhm = 5.1 kOhm.",
+ "revision": 2,
+ "explanation": "For $n$ equal resistors in parallel the total is $R_g = R/n$, so a single resistor is $R = n \\cdot R_g$. Here $R = 3 \\cdot 1.7\\ \\text{k}\\Omega = 5.1\\ \\text{k}\\Omega$. (Parallel resistance is always smaller than the smallest resistor, which it is.)",
"source": "https://50ohm.de/E_reihe_parallel_widerstand.html",
"confidence": 8
},
"ED107": {
- "revision": 1,
- "explanation": "With three equal resistors, the allowed total power is the sum of the individual ratings when the load is shared. That gives 3*1 W = 3 W in both series and parallel arrangements.",
+ "revision": 2,
+ "explanation": "Power ratings add when several resistors share the load, regardless of how they are wired: the assembly dissipates $3 \\cdot 1\\ \\text{W} = 3$ W in both series and parallel. Series or parallel changes the total resistance, not the total heat the three parts can shed.",
"source": "https://50ohm.de/E_reihe_parallel_widerstand.html",
"confidence": 8
},
"ED108": {
- "revision": 1,
- "explanation": "R1 and R2 are in series, giving 500 + 500 = 1000 Ohm. That 1000 Ohm branch is in parallel with R3 = 1000 Ohm, so the total is 500 Ohm.",
+ "revision": 2,
+ "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: $1000/2 = 500 Ohm$.",
"source": "https://50ohm.de/E_reihe_parallel_widerstand.html",
- "confidence": 7
+ "confidence": 8
},
"ED109": {
- "revision": 1,
- "explanation": "R1 and R2 first add in series: 500 Ohm + 1.5 kOhm = 2 kOhm. That is in parallel with R3 = 2 kOhm, so the result is 1 kOhm.",
+ "revision": 2,
+ "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 $R_g = 1 kOhm$.",
"source": "https://50ohm.de/E_reihe_parallel_widerstand.html",
- "confidence": 7
+ "confidence": 8
},
"ED110": {
- "revision": 1,
- "explanation": "The two 1 kOhm resistors are parallel, so they reduce to 500 Ohm. In series with the remaining 500 Ohm, the total is 1 kOhm.",
+ "revision": 2,
+ "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 $500 + 500 = 1000 Ohm$.",
"source": "https://50ohm.de/E_reihe_parallel_widerstand.html",
- "confidence": 7
+ "confidence": 8
},
"ED111": {
- "revision": 1,
- "explanation": "R2 and R3 are both 2 kOhm in parallel, giving 1 kOhm. Adding the series R1 of 1 kOhm gives 2 kOhm total.",
+ "revision": 2,
+ "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 $1 kOhm + 1 kOhm = 2 kOhm$.",
"source": "https://50ohm.de/E_reihe_parallel_widerstand.html",
- "confidence": 7
+ "confidence": 8
},
"ED112": {
- "revision": 1,
- "explanation": "R2 and R3 are parallel: 3 kOhm || 1.5 kOhm = 1 kOhm. Add the series R1 of 1 kOhm to get 2 kOhm.",
+ "revision": 2,
+ "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.",
"source": "https://50ohm.de/E_reihe_parallel_widerstand.html",
- "confidence": 7
+ "confidence": 8
},
"ED113": {
- "revision": 1,
- "explanation": "R1, R2, and R3 are parallel: 10 kOhm || 2.5 kOhm || 500 Ohm = 400 Ohm. Adding the series 600 Ohm resistor gives 1 kOhm.",
+ "revision": 2,
+ "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 $400 + 600 = 1000 Ohm$.",
"source": "https://50ohm.de/E_reihe_parallel_widerstand.html",
- "confidence": 7
+ "confidence": 8
},
"ED114": {
- "revision": 1,
- "explanation": "Reduce the obvious groups step by step: 50 Ohm + 50 Ohm gives 100 Ohm, 100 Ohm in parallel with 100 Ohm gives 50 Ohm, then the remaining series parts total 250 Ohm.",
+ "revision": 2,
+ "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.",
"source": "https://50ohm.de/NE_reihe_parallel_widerstandsnetz_1.html",
- "confidence": 7
+ "confidence": 8
},
"ED115": {
- "revision": 1,
- "explanation": "Combine the clear series and parallel subgroups in stages; the network reduces to a final series sum of 550 Ohm.",
+ "revision": 2,
+ "explanation": "The method is staged simplification: series groups add directly, parallel groups use reciprocal addition or the equal-resistor shortcut. After replacing the drawn subgroups by their equivalents, the remaining series chain sums to 550 Ohm.",
"source": "https://50ohm.de/NE_reihe_parallel_widerstandsnetz_1.html",
- "confidence": 7
+ "confidence": 8
},
"ED116": {
- "revision": 1,
- "explanation": "After reducing the drawn subgroups, the remaining series values are 400 Ohm, 200 Ohm, 200 Ohm, and 150 Ohm. Their sum is 950 Ohm.",
+ "revision": 2,
+ "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 $R_g = 950 Ohm$.",
"source": "https://50ohm.de/NE_reihe_parallel_widerstandsnetz_1.html",
- "confidence": 7
+ "confidence": 8
},
"ED117": {
- "revision": 1,
- "explanation": "Parallel capacitances add directly. 0.1 uF = 100 nF and 50000 pF = 50 nF, so 100 + 150 + 50 = 300 nF = 0.3 uF.",
+ "revision": 2,
+ "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 $= 100 + 150 + 50 = 300\\ \\text{nF} = 0.3\\ \\mu\\text{F}$.",
"source": "https://50ohm.de/NEA_reihe_parallel_kondensator.html",
"confidence": 8
},
"ED118": {
- "revision": 1,
- "explanation": "Parallel capacitors add directly after unit conversion: 22 nF + 0.033 uF (33 nF) + 15000 pF (15 nF) = 70 nF = 0.070 uF.",
+ "revision": 2,
+ "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.",
"source": "https://50ohm.de/NEA_reihe_parallel_kondensator.html",
"confidence": 8
},
"ED119": {
- "revision": 1,
- "explanation": "For equal capacitors in series, Cg = C/n. Three 0.33 uF capacitors therefore give 0.33/3 = 0.110 uF.",
+ "revision": 2,
+ "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.",
"source": "https://50ohm.de/NEA_reihe_parallel_kondensator.html",
"confidence": 8
},
"ED120": {
- "revision": 1,
- "explanation": "Convert 200000 nF to 200 uF. The series formula gives 1/Cg = 1/100 + 1/200 + 1/200, so Cg = 50 uF.",
+ "revision": 2,
+ "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 $C_g = 50\\ \\mu\\text{F}$.",
"source": "https://50ohm.de/NEA_reihe_parallel_kondensator.html",
"confidence": 8
},
"ED121": {
- "revision": 1,
- "explanation": "C1 and C2 are equal 10 nF capacitors in series, so their equivalent is 5 nF. In parallel with C3 = 5 nF, the total is 10 nF.",
+ "revision": 2,
+ "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.",
"source": "https://50ohm.de/NEA_reihe_parallel_kondensator.html",
- "confidence": 7
+ "confidence": 8
},
"ED122": {
- "revision": 1,
- "explanation": "C2 and C3 are parallel, giving 1 uF + 1 uF = 2 uF. That is in series with C1 = 2 uF, so two equal 2 uF capacitors in series give 1.0 uF.",
+ "revision": 2,
+ "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 $C_g=1.0 uF$.",
"source": "https://50ohm.de/NEA_reihe_parallel_kondensator.html",
- "confidence": 7
+ "confidence": 8
},
"ED123": {
- "revision": 1,
- "explanation": "C2 and C3 are parallel, so 4 nF + 4 nF = 8 nF. That 8 nF equivalent is in series with C1 = 8 nF, giving 4 nF.",
+ "revision": 2,
+ "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 $C_g=4 nF$.",
"source": "https://50ohm.de/NEA_reihe_parallel_kondensator.html",
- "confidence": 7
+ "confidence": 8
},
"ED124": {
- "revision": 1,
- "explanation": "Convert C3 = 100000 pF to 100 nF. C2 and C3 are parallel, giving 200 nF; that is in series with C1 = 200 nF, so the total is 100 nF.",
+ "revision": 2,
+ "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.",
"source": "https://50ohm.de/NEA_reihe_parallel_kondensator.html",
- "confidence": 7
+ "confidence": 8
},
"ED201": {
- "revision": 1,
- "explanation": "The graph passes low frequencies and attenuates frequencies above the cutoff, which is the defining response of a low-pass filter.",
+ "revision": 2,
+ "explanation": "A Tiefpass is a low-pass filter: it lets low frequencies through and attenuates frequencies above the cutoff. On the graph that looks like high transmission on the left, then a roll-off as frequency increases.",
"source": "https://50ohm.de/EA_schwingkreis_1.html",
- "confidence": 7
+ "confidence": 8
},
"ED202": {
- "revision": 1,
- "explanation": "The graph attenuates low frequencies and passes higher frequencies after the cutoff, so it is a high-pass response.",
+ "revision": 2,
+ "explanation": "A Hochpass is a high-pass filter: it blocks or attenuates low frequencies and passes frequencies above the cutoff. On the graph the response starts low on the left and rises toward the right.",
"source": "https://50ohm.de/EA_schwingkreis_1.html",
- "confidence": 7
+ "confidence": 8
},
"ED203": {
- "revision": 1,
- "explanation": "The curve passes only a middle frequency range and attenuates both low and high frequencies, which is a band-pass response.",
+ "revision": 2,
+ "explanation": "A Bandpass (band-pass filter) passes only a chosen frequency range around its centre/resonant frequency. Frequencies below and above that range are attenuated, so the graph has a passband peak or plateau in the middle.",
"source": "https://50ohm.de/EA_schwingkreis_1.html",
- "confidence": 7
+ "confidence": 8
},
"ED204": {
- "revision": 1,
- "explanation": "The curve passes frequencies on both sides but rejects a middle range around resonance, so it is a band-stop response.",
+ "revision": 2,
+ "explanation": "A Bandsperre is a band-stop or notch filter: it passes frequencies below and above a selected range but rejects the middle range around resonance. The recognition cue is the dip in the response curve.",
"source": "https://50ohm.de/EA_schwingkreis_1.html",
- "confidence": 7
+ "confidence": 8
},
"ED205": {
- "revision": 1,
- "explanation": "A series resonant circuit has minimum impedance at resonance because inductive and capacitive reactance cancel, giving the V-shaped impedance curve.",
+ "revision": 2,
+ "explanation": "The V-shaped impedance curve — high either side, dipping to a sharp minimum at resonance — is the signature of a series resonant circuit. At $f_0$ the inductive and capacitive reactances cancel ($X_L = X_C$), leaving only the small series resistance, so impedance is minimum.",
"source": "https://50ohm.de/EA_schwingkreis_1.html",
"confidence": 8
},
"ED206": {
- "revision": 1,
- "explanation": "A parallel resonant circuit has maximum impedance at resonance, producing the peaked impedance curve shown.",
+ "revision": 2,
+ "explanation": "In a parallel resonant circuit, inductor and capacitor exchange energy while their branch currents largely cancel at resonance. The input impedance becomes maximum at $f_0=1/(2\\pi\\sqrt{LC})$, producing the peaked impedance curve.",
"source": "https://50ohm.de/EA_schwingkreis_1.html",
"confidence": 8
},
"ED207": {
- "revision": 1,
- "explanation": "At resonance a parallel LC circuit presents a very high impedance; away from resonance one branch becomes comparatively low impedance.",
+ "revision": 2,
+ "explanation": "At resonance a parallel LC circuit (Parallelschwingkreis) presents maximum, high-ohmic impedance: the equal and opposite branch currents circulate internally and largely cancel at the terminals, drawing little current from the source. This is why a parallel tank is used as a selective load in tuned amplifiers.",
"source": "https://50ohm.de/EA_schwingkreis_1.html",
"confidence": 8
},
"ED208": {
- "revision": 1,
- "explanation": "The circuit takes the output after a series resistor with a capacitor shunting high frequencies to ground, so low frequencies pass and high frequencies are attenuated: a low-pass filter.",
+ "revision": 2,
+ "explanation": "This RC network is a Tiefpass (low-pass filter). The capacitor reactance is $X_C=1/(2\\pi fC)$, so at low frequency the capacitor is nearly open and the output remains high; at high frequency it becomes a low-impedance path to ground and shunts the signal away.",
"source": "https://50ohm.de/EA_schwingkreis_1.html",
- "confidence": 7
+ "confidence": 8
},
"ED209": {
- "revision": 1,
- "explanation": "A series inductor followed by a shunt capacitor passes low frequencies: the inductor is low impedance at low frequency and the capacitor shunts high frequency components.",
+ "revision": 2,
+ "explanation": "This LC network is a low-pass filter. The inductor has $X_L=2\\pi fL$, so it is easy for low frequencies and increasingly blocks high frequencies; the shunt capacitor has lower impedance at high frequency and bypasses those components to ground.",
"source": "https://50ohm.de/EA_schwingkreis_1.html",
- "confidence": 7
+ "confidence": 8
},
"ED210": {
- "revision": 1,
- "explanation": "For microphone audio low-pass filtering, the selected RC network uses capacitors to bypass high-frequency components while the wanted lower audio range remains at the output.",
+ "revision": 2,
+ "explanation": "A microphone low-pass keeps speech audio below the cutoff and removes unwanted high-frequency components. In an RC low-pass, $X_C=1/(2\\pi fC)$ falls as frequency rises, so high-frequency audio/RF components are bypassed while the wanted speech range remains at the output.",
"source": "https://50ohm.de/EA_schwingkreis_1.html",
- "confidence": 7
+ "confidence": 8
},
"ED211": {
- "revision": 1,
- "explanation": "A series capacitor followed by a resistor load is a high-pass: the capacitor blocks low-frequency components and passes higher-frequency components more easily.",
+ "revision": 2,
+ "explanation": "A series capacitor followed by a load resistor is a Hochpass (high-pass filter). Because $X_C=1/(2\\pi fC)$ is large at low frequency, low-frequency components are blocked; as frequency rises the capacitor impedance falls and more signal reaches the output.",
"source": "https://50ohm.de/EA_schwingkreis_1.html",
- "confidence": 7
+ "confidence": 8
},
"ED212": {
- "revision": 1,
- "explanation": "With a series capacitor and a shunt inductor, low frequencies are shunted through the inductor while higher frequencies pass through the capacitor path, so the circuit is a high-pass.",
+ "revision": 2,
+ "explanation": "This LC circuit is a high-pass filter. At low frequency the series capacitor has high impedance and the shunt inductor has low impedance, so low-frequency energy is blocked and bypassed; at high frequency the capacitor passes and the inductor no longer shorts the output.",
"source": "https://50ohm.de/EA_schwingkreis_1.html",
- "confidence": 7
+ "confidence": 8
},
"ED213": {
- "revision": 1,
- "explanation": "The selected LC ladder has a series capacitor path with shunt inductors, the high-pass pattern: low frequencies are bypassed, higher frequencies are passed.",
+ "revision": 2,
+ "explanation": "The LC ladder has the classic high-pass pattern: capacitors in the series signal path and inductors as shunt paths. Low frequencies see blocked series capacitors and easy shunt inductors; high frequencies pass through the capacitors while the inductors become high impedance.",
"source": "https://50ohm.de/EA_schwingkreis_1.html",
- "confidence": 7
+ "confidence": 8
},
"ED214": {
- "revision": 1,
- "explanation": "A parallel resonant circuit placed in series with the signal path has high impedance at resonance and blocks that frequency, so it is a Sperrkreis.",
+ "revision": 2,
+ "explanation": "A Sperrkreis is a blocking or trap circuit. A parallel LC circuit has very high impedance at resonance, where $X_L=X_C$ and $f_0=1/(2\\pi\\sqrt{LC})$; placed in series with the signal path, that high impedance blocks the resonant frequency.",
"source": "https://50ohm.de/EA_schwingkreis_1.html",
- "confidence": 7
+ "confidence": 8
},
"ED215": {
- "revision": 1,
- "explanation": "A series resonant LC branch connected across the signal path has low impedance at resonance and diverts that frequency away from the output, so it is a Saugkreis.",
+ "revision": 2,
+ "explanation": "A Saugkreis is a shunt notch or absorption trap. A series LC branch has minimum impedance at resonance, $f_0=1/(2\\pi\\sqrt{LC})$; connected across the signal path, it 'sucks out' that frequency by diverting it away from the output.",
"source": "https://50ohm.de/EA_schwingkreis_1.html",
- "confidence": 7
+ "confidence": 8
},
"ED216": {
- "revision": 1,
- "explanation": "HF filters need low-loss, high-Q capacitors with small parasitic effects; ceramic and air capacitors are preferred over electrolytics.",
+ "revision": 2,
+ "explanation": "HF filters need low-loss, high-$Q$ capacitors with stable value and minimal parasitics, so ceramic (Class 1) or air-dielectric types are preferred. Electrolytics (aluminium, tantalum) are polarised, lossy, and inductive — fine for smoothing DC, useless as RF filter elements.",
"source": "https://50ohm.de/EA_kondensator_1.html",
"confidence": 8
},
"ED301": {
- "revision": 1,
- "explanation": "A useful DC supply should keep its output voltage nearly constant under load; otherwise the connected radio stages see supply-voltage changes.",
+ "revision": 2,
+ "explanation": "A good DC supply should hold its output voltage nearly constant as the load current changes (good Spannungskonstanz / low output resistance). A radio's stages misbehave if the rail sags or carries ripple under load, so high voltage stability — not a sagging or AC-laden output — is the wanted property.",
"source": "https://50ohm.de/EA_spannungsquelle.html",
"confidence": 8
},
"ED302": {
- "revision": 1,
- "explanation": "Switch-mode supplies convert power at high switching frequency, allowing small transformers and heat sinks, so they are efficient, light, and compact.",
+ "revision": 2,
+ "explanation": "A switch-mode supply (Schaltnetzteil) processes power by switching at tens of kHz or more, which shrinks the transformer and lets it run efficiently. The result is high efficiency, low weight, and small volume — its drawback is switching noise, not size or efficiency.",
"source": "https://50ohm.de/NEA_schaltnetzteil_1.html",
"confidence": 8
},
"ED303": {
- "revision": 1,
- "explanation": "The high-frequency switching action can generate RF interference unless the supply is well filtered and shielded.",
+ "revision": 2,
+ "explanation": "A switching power supply chops current at high frequency, creating fast edges rich in harmonics. Those components can leave through cables or radiation unless the supply has proper filtering, layout, and shielding.",
"source": "https://50ohm.de/NEA_schaltnetzteil_1.html",
"confidence": 8
},
"ED304": {
- "revision": 1,
- "explanation": "The circuit is a single-diode half-wave rectifier. The load voltage contains only the conducting half-cycles, with the opposite half-cycles blocked by the diode.",
+ "revision": 2,
+ "explanation": "This is a single-diode half-wave rectifier. The diode conducts only on the half-cycle where it is forward biased, so current flows through the load in one direction; the opposite half-cycle is blocked, leaving pulsating DC.",
"source": "https://50ohm.de/EA_gleichrichter_1.html",
- "confidence": 7
+ "confidence": 8
},
"ED401": {
- "revision": 1,
- "explanation": "Power gain means the output signal power is greater than the input signal power. That extra power must come from an external supply, not from the input signal alone.",
+ "revision": 2,
+ "explanation": "Power gain means the output signal power exceeds the input signal power. Energy is conserved, so that extra power cannot come from the small input signal — it is drawn from an external DC supply, which the amplifier modulates under control of the input. No supply, no power gain.",
"source": "https://50ohm.de/NE_verstaerker.html",
"confidence": 8
},
"ED402": {
- "revision": 1,
- "explanation": "The shown transistor audio-stage topology is an NF amplifier; it is meant for low-frequency/audio signal amplification, not RF or IF selection.",
+ "revision": 2,
+ "explanation": "NF means Niederfrequenz, i.e. audio/low frequency. The shown transistor stage has the topology of a small-signal audio amplifier: it biases the transistor and couples an audio signal through capacitors, rather than using tuned RF/IF resonant circuits for frequency selection.",
"source": "https://50ohm.de/NE_verstaerker.html",
- "confidence": 7
+ "confidence": 8
},
"ED403": {
- "revision": 1,
- "explanation": "An HF power amplifier raises the transmitter's RF signal level to the desired output power before feeding the antenna system.",
+ "revision": 2,
+ "explanation": "An HF power amplifier (Leistungsverstärker, the PA or 'Endstufe') boosts the transmitter's already-formed RF signal up to the wanted output power before the antenna. It only raises level — modulation, mixing, and filtering happen in earlier stages.",
"source": "https://50ohm.de/NE_verstaerker.html",
"confidence": 8
},
"ED501": {
- "revision": 1,
- "explanation": "An LC oscillator uses a tuned circuit made from an inductor L and capacitor C; that resonant circuit sets the oscillation frequency.",
+ "revision": 2,
+ "explanation": "An LC oscillator gets its frequency from a resonant (tuned) circuit of an inductor $L$ and capacitor $C$: $f_0 = 1/(2\\pi\\sqrt{LC})$. The amplifier merely sustains the oscillation; the LC tank sets the frequency. (A crystal-controlled version would be a quartz oscillator instead.)",
"source": "https://50ohm.de/E_oszillatoren.html",
"confidence": 8
},
"ED502": {
- "revision": 1,
- "explanation": "The LC resonance frequency is inversely related to the square root of capacitance. If C increases, the oscillator frequency decreases.",
+ "revision": 2,
+ "explanation": "From $f_0 = 1/(2\\pi\\sqrt{LC})$, frequency falls as $C$ rises. So if warming raises the capacitor's value, the oscillator frequency drifts lower — a typical cause of thermal frequency drift in simple VFOs.",
"source": "https://50ohm.de/E_oszillatoren.html",
"confidence": 8
},
"ED503": {
- "revision": 1,
- "explanation": "The LC resonance frequency rises when capacitance falls, because frequency is inversely related to sqrt(L*C).",
+ "revision": 2,
+ "explanation": "LC resonance follows $f_0=1/(2\\pi\\sqrt{LC})$. If capacitance $C$ is made smaller while $L$ stays the same, the denominator shrinks, so the oscillator frequency rises.",
"source": "https://50ohm.de/E_oszillatoren.html",
"confidence": 8
},
"ED504": {
- "revision": 1,
- "explanation": "The LC resonance frequency is inversely related to the square root of inductance. If L increases, the oscillator frequency decreases.",
+ "revision": 2,
+ "explanation": "Same resonance law, $f_0 = 1/(2\\pi\\sqrt{LC})$: frequency falls when $L$ increases. A coil whose inductance grows with temperature therefore pulls the oscillator frequency down.",
"source": "https://50ohm.de/E_oszillatoren.html",
"confidence": 8
},
"ED505": {
- "revision": 1,
- "explanation": "When inductance decreases, the LC product becomes smaller, so the resonant frequency becomes higher.",
+ "revision": 2,
+ "explanation": "The oscillator frequency of an LC circuit is $f_0=1/(2\\pi\\sqrt{LC})$. Decreasing inductance $L$ makes the $LC$ product smaller, so the square-root denominator gets smaller and the frequency increases.",
"source": "https://50ohm.de/E_oszillatoren.html",
"confidence": 8
},
"ED506": {
- "revision": 1,
- "explanation": "In a crystal oscillator, the quartz crystal is the frequency-determining resonator.",
+ "revision": 2,
+ "explanation": "In a crystal oscillator, the quartz crystal is the frequency-determining resonator. Its mechanical resonance has very high Q and stable dimensions, so the oscillator frequency is far more stable than a simple adjustable LC circuit.",
"source": "https://50ohm.de/E_oszillatoren.html",
"confidence": 8
},
"ED507": {
- "revision": 1,
- "explanation": "A quartz crystal's resonance changes much less with temperature and component tolerances than a simple LC circuit, so crystal oscillators are more frequency-stable.",
+ "revision": 2,
+ "explanation": "A quartz crystal has an extremely high $Q$ and a resonance that barely shifts with temperature or component tolerance, far more stable than an LC tank. So crystal oscillators win on frequency stability — at the cost of a very narrow tuning range, which is the trade-off the other options get backwards.",
"source": "https://50ohm.de/E_oszillatoren.html",
"confidence": 8
},
"EE101": {
- "revision": 1,
- "explanation": "An unmodulated carrier is a steady sine wave with constant amplitude, frequency, and phase; the selected diagram shows that unchanged carrier.",
+ "revision": 2,
+ "explanation": "An unmodulated carrier is just the RF sine wave before information is added. Its amplitude, frequency, and phase stay constant, so the correct diagram is the steady, unchanged sinusoid without envelope changes or frequency variation.",
"source": "https://50ohm.de/E_unmodulierter_traeger.html",
- "confidence": 7
+ "confidence": 8
},
"EE201": {
- "revision": 1,
- "explanation": "AM carries both sidebands plus carrier, while SSB suppresses the carrier and one sideband. Therefore SSB needs less than half the bandwidth of AM.",
+ "revision": 2,
+ "explanation": "Full AM transmits a carrier plus both sidebands; SSB removes the carrier and one sideband, sending only the information-bearing sideband. With one sideband instead of two-plus-carrier, SSB occupies less than half the bandwidth of AM — and puts all its power into useful signal.",
"source": "https://50ohm.de/E_ssb_2.html",
"confidence": 8
},
"EE202": {
- "revision": 1,
- "explanation": "In SSB, only one translated sideband is transmitted, so the occupied RF bandwidth is essentially the same as the audio/NF bandwidth being sent.",
+ "revision": 2,
+ "explanation": "SSB sends just one frequency-translated sideband, a copy of the audio (NF) spectrum shifted up to RF. So the occupied RF bandwidth equals the audio bandwidth — about $2.4$-$3$ kHz for speech — not double or half it.",
"source": "https://50ohm.de/E_ssb_2.html",
"confidence": 8
},
"EE203": {
- "revision": 1,
- "explanation": "USB places the audio component above the carrier. 21.250 MHz + 0.001 MHz = 21.251 MHz.",
+ "revision": 2,
+ "explanation": "In upper sideband (USB), each audio component appears above the suppressed carrier by its audio frequency. A 1 kHz tone is $0.001 MHz$, so $21.250 MHz + 0.001 MHz = 21.251 MHz$.",
"source": "https://50ohm.de/E_ssb_2.html",
"confidence": 8
},
"EE204": {
- "revision": 1,
- "explanation": "LSB places the audio component below the carrier and suppresses the carrier in ideal SSB. 3.650 MHz - 0.002 MHz = 3.648 MHz.",
+ "revision": 2,
+ "explanation": "Ideal SSB suppresses the carrier and keeps one sideband. LSB (lower sideband) places the audio below the carrier: $3.650\\ \\text{MHz} - 0.002\\ \\text{MHz} = 3.648$ MHz. Only that single component is transmitted, so options listing the carrier or upper sideband are wrong.",
"source": "https://50ohm.de/E_ssb_2.html",
"confidence": 8
},
"EE205": {
- "revision": 2,
- "explanation": "For SSB voice, RF output follows the audio drive level. Reducing the NF/audio amplitude reduces the modulated transmitter output power.",
+ "revision": 3,
+ "explanation": "In SSB the RF output is produced directly from the audio drive — no audio, no output. Reducing the audio (NF) amplitude therefore lowers the transmitter's output power. Speaking louder or widening the audio would raise power or bandwidth, not reduce output.",
"source": "https://50ohm.de/E_slide_e_modulation.html",
"confidence": 8
},
"EE206": {
- "revision": 2,
- "explanation": "Too little microphone gain gives too little audio drive to the SSB modulator, so the transmitter produces low output power.",
+ "revision": 3,
+ "explanation": "SSB output power tracks the audio drive into the modulator. If the microphone gain is set too low, the modulator is under-driven and the transmitter produces too little output power — the fix is more gain, up to (but not past) the point of distortion.",
"source": "https://50ohm.de/E_slide_e_modulation.html",
"confidence": 8
},
"EE207": {
- "revision": 1,
- "explanation": "CW keys one carrier rather than transmitting a full speech spectrum, so its occupied bandwidth is smaller than both SSB voice and AM voice.",
+ "revision": 2,
+ "explanation": "CW (Morse) simply keys a single carrier on and off; it carries no voice spectrum, so its occupied bandwidth (a few tens of Hz, set by the keying speed) is smaller than both SSB and AM speech. That narrowness is why CW gets through when voice cannot.",
"source": "https://50ohm.de/E_ssb_2.html",
"confidence": 8
},
"EE301": {
- "revision": 1,
- "explanation": "The shown waveform keeps amplitude essentially constant while the instantaneous carrier frequency changes, which identifies frequency modulation.",
+ "revision": 2,
+ "explanation": "Frequency modulation (FM) carries information by varying the instantaneous carrier frequency around the centre frequency. The German term Hub or Frequenzhub means frequency deviation: how far the carrier swings above and below centre. In the diagram the clue is changing wave spacing/period, not changing height.",
"source": "https://50ohm.de/EA_fm_2.html",
- "confidence": 7
+ "confidence": 8
},
"EE302": {
- "revision": 1,
- "explanation": "FM carries information in frequency deviation rather than amplitude, so amplitude noise has less direct effect than it does in SSB.",
+ "revision": 2,
+ "explanation": "FM carries information in frequency deviation, German Hub/Frequenzhub, rather than in RF amplitude. Because the receiver can limit amplitude variations before demodulation, amplitude noise has less direct effect than it does in AM or SSB.",
"source": "https://50ohm.de/EA_fm_2.html",
"confidence": 8
},
"EE303": {
- "revision": 1,
- "explanation": "Vehicle electrical noise often appears as amplitude disturbance. FM is least affected because the receiver can limit amplitude and use frequency deviation instead.",
+ "revision": 2,
+ "explanation": "Vehicle electrical noise often appears as unwanted amplitude spikes. FM is least affected because the wanted information is in frequency deviation, German Hub/Frequenzhub, and the receiver can limit amplitude before recovering the audio.",
"source": "https://50ohm.de/EA_fm_2.html",
"confidence": 8
},
"EE304": {
- "revision": 1,
- "explanation": "In FM, a larger frequency deviation spreads the signal over a wider range of frequencies, increasing RF bandwidth.",
+ "revision": 2,
+ "explanation": "In FM, Hub or Frequenzhub means frequency deviation: the carrier is shifted farther from its centre frequency as modulation increases. A larger deviation spreads the signal over a wider RF range, so occupied bandwidth increases.",
"source": "https://50ohm.de/EA_fm_2.html",
"confidence": 8
},
"EE305": {
- "revision": 1,
- "explanation": "Excessive FM bandwidth is reduced by lowering the deviation setting, because deviation directly determines how far the carrier moves from center frequency.",
+ "revision": 2,
+ "explanation": "Excessive FM bandwidth is reduced by lowering the deviation setting. Deviation is the German Hub/Frequenzhub: the maximum frequency swing away from centre. Smaller Hub means the carrier moves less, so the transmitted signal occupies less bandwidth.",
"source": "https://50ohm.de/EA_fm_2.html",
"confidence": 8
},
"EE306": {
- "revision": 1,
- "explanation": "In FM, loudness is represented by the size of the carrier-frequency deviation, not by RF amplitude.",
+ "revision": 2,
+ "explanation": "In FM, speech loudness is represented by frequency deviation, German Hub/Frequenzhub. Louder audio pushes the instantaneous carrier farther above and below the centre frequency, while RF amplitude ideally stays constant; that is why FM receivers can use amplitude limiting.",
"source": "https://50ohm.de/EA_fm_2.html",
"confidence": 8
},
"EE401": {
- "revision": 1,
- "explanation": "Bandwidth is occupied frequency range measured in hertz. Data rate is the amount of information transferred per time, measured in bit/s.",
+ "revision": 2,
+ "explanation": "Keep the two ideas separate: bandwidth is the frequency range a signal occupies, measured in hertz; data rate is the information per time, measured in bit/s. A wide channel does not guarantee a high bit rate, and symbol rate (baud) is a third, distinct quantity.",
"source": "https://50ohm.de/NEA_datenuebertragungsdrate.html",
"confidence": 8
},
"EE402": {
- "revision": 1,
- "explanation": "SSB translates the audio-frequency digimode signal to RF while preserving its narrow bandwidth, which is why modes such as FT8 or BPSK31 are sent through an SSB transmitter path.",
+ "revision": 2,
+ "explanation": "Narrowband digimodes such as FT8 or BPSK31 are fed in as audio tones and sent through the rig in SSB. SSB just shifts the audio block up to RF unchanged, preserving the signal's narrow bandwidth; FM or AM would widen it and break the mode.",
"source": "https://50ohm.de/NEA_digimode_ssb.html",
"confidence": 8
},
"EE403": {
- "revision": 1,
- "explanation": "With SSB, the RF bandwidth follows the bandwidth of the audio/NF signal. A 50 Hz audio signal therefore occupies about 50 Hz RF bandwidth.",
+ "revision": 2,
+ "explanation": "In SSB the RF bandwidth equals the audio (NF) bandwidth, because SSB is a pure frequency translation of the audio with no extra sidebands. So a $50$ Hz audio signal occupies about $50$ Hz of RF — narrow in, narrow out.",
"source": "https://50ohm.de/NEA_digimode_ssb.html",
"confidence": 8
},
"EE404": {
- "revision": 1,
- "explanation": "A 2.4 kHz SSB passband can contain several much narrower digimode signals at different audio frequencies, and software can decode one or more of them.",
+ "revision": 2,
+ "explanation": "A $2.4$ kHz SSB passband is wide enough to hold many far narrower digimode signals sitting at different audio pitches. The computer demodulates the whole passband, so depending on the mode it can decode one or several of them at once — modern FT8/PSK decoders routinely show many simultaneously.",
"source": "https://50ohm.de/NEA_digimode_ssb.html",
"confidence": 8
},
"EE405": {
- "revision": 2,
- "explanation": "Reporter networks collect received digimode spots by callsign. Sending a suitable signal such as WSPR and then searching the reporting platform shows where it was received.",
+ "revision": 3,
+ "explanation": "Use a reporting network: transmit with a mode designed for it (e.g. WSPR, or CW into the Reverse Beacon Network), then look up your callsign on the matching internet platform to see who copied you and where. The spots are gathered automatically by listening stations — no email handshake or made-up 'AUTO RSVP' procedure exists.",
"source": "https://50ohm.de/NEA_slide_nea_digitale_uebertragungsverfahren.html",
"confidence": 8
},
"EE406": {
- "revision": 1,
- "explanation": "ASK changes the carrier amplitude between symbol states while frequency stays recognisably the same; the selected diagram shows those amplitude changes.",
+ "revision": 2,
+ "explanation": "ASK means Amplitude Shift Keying. The digital symbol is represented by changing carrier amplitude, often between a large amplitude and a small or zero amplitude, while the carrier frequency itself remains recognisably the same.",
"source": "https://50ohm.de/EA_ask_fsk_afsk.html",
- "confidence": 7
+ "confidence": 8
},
"EE407": {
- "revision": 1,
- "explanation": "FSK changes between different carrier frequencies while keeping amplitude essentially constant; the selected diagram shows changing period/frequency.",
+ "revision": 2,
+ "explanation": "FSK means Frequency Shift Keying. The digital symbol is represented by switching between carrier frequencies, so the period/wave spacing changes while the amplitude remains essentially constant.",
"source": "https://50ohm.de/EA_ask_fsk_afsk.html",
- "confidence": 7
+ "confidence": 8
},
"EE408": {
- "revision": 1,
- "explanation": "In AFSK, frequency-shift keying is first generated as an audio-frequency signal, which then modulates an RF transmitter such as an FM, AM, or SSB rig.",
+ "revision": 2,
+ "explanation": "AFSK (audio frequency-shift keying) generates the FSK as an audio tone pair first, then feeds that audio into an ordinary transmitter (commonly FM) to put it on RF. The keying lives in the audio stage, which is why any SSB/FM rig with an audio input can send it — distinct from direct FSK that shifts the RF carrier itself.",
"source": "https://50ohm.de/NEA_afsk.html",
"confidence": 8
},
"EE409": {
- "revision": 2,
- "explanation": "TDMA separates users by time slots: signals take rapid turns on the same frequency rather than transmitting continuously at once.",
+ "revision": 3,
+ "explanation": "TDMA (time-division multiple access) gives each signal its own brief time slot and lets them take rapid turns on the same frequency. They are not truly simultaneous — interleaving in time is what keeps them apart.",
"source": "https://50ohm.de/NEA_vielfachzugriff.html",
"confidence": 8
},
"EE410": {
- "revision": 2,
- "explanation": "FDMA separates simultaneous signals by frequency, so users transmit at the same time but on different frequency channels.",
+ "revision": 3,
+ "explanation": "FDMA (frequency-division multiple access) gives each signal its own frequency channel, so all users transmit at the same time but on different frequencies. Separation is by frequency, the complement of TDMA's separation by time.",
"source": "https://50ohm.de/NEA_vielfachzugriff.html",
"confidence": 8
},
"EE411": {
- "revision": 2,
- "explanation": "CDMA lets signals share time and frequency by applying different spreading codes that the receiver uses to separate them.",
+ "revision": 3,
+ "explanation": "CDMA (code-division multiple access) lets all signals share the same time and frequency band; each uses a different spreading code, and the receiver correlates with the wanted code to pull its signal out of the others. Separation is by code, not time or frequency.",
"source": "https://50ohm.de/NEA_vielfachzugriff.html",
"confidence": 8
},
"EE412": {
- "revision": 1,
- "explanation": "In a packet-switched network, packets can be forwarded through intermediate stations or routers when the two endpoints cannot reach each other directly.",
+ "revision": 2,
+ "explanation": "In a packet-switched network, two stations that cannot hear each other directly communicate by having intermediate stations relay (forward) their packets hop by hop toward the destination — the principle behind digipeaters and routed mesh networks like HAMNET.",
"source": "https://50ohm.de/NEA_paketvermittelte_netzwerke.html",
"confidence": 8
},
"EE413": {
- "revision": 1,
- "explanation": "The IP address plus subnet mask defines which addresses are on the same local subnet and are reachable directly without routing.",
+ "revision": 2,
+ "explanation": "The IP address together with the subnet mask defines the local subnet — the block of addresses an interface can reach directly, without going through a router. Anything outside that range must be sent via a gateway. It does not encode ports, bandwidth, or hop counts.",
"source": "https://50ohm.de/NEA_paketvermittelte_netzwerke.html",
"confidence": 8
},
"EE414": {
- "revision": 1,
- "explanation": "IP is a network protocol, not something limited to the public Internet, so it can also be used in amateur-radio networks such as HAMNET.",
+ "revision": 2,
+ "explanation": "IP is a general networking protocol, not something confined to the public Internet, so it works fine over amateur links — German amateurs run it on HAMNET. The callsign is not hidden in a subnet mask, and using IP does not expose ham bands to Internet users.",
"source": "https://50ohm.de/NEA_paketvermittelte_netzwerke.html",
"confidence": 8
},
"EE415": {
- "revision": 1,
- "explanation": "SSTV sends still pictures slowly, while ATV is amateur television with moving pictures and much larger bandwidth needs.",
+ "revision": 2,
+ "explanation": "SSTV (slow-scan TV) sends still images line by line over a narrow voice-bandwidth channel; ATV (amateur TV) sends full moving pictures and therefore needs a much wider bandwidth (only practical on UHF and up). Stills-vs-motion is the defining difference; both can be colour.",
"source": "https://50ohm.de/NEA_digimode_ssb.html",
"confidence": 8
},
"EF101": {
- "revision": 1,
- "explanation": "The circuit is a detector receiver: the tuned circuit selects the station and the diode recovers the audio envelope without an active oscillator or amplifier.",
+ "revision": 2,
+ "explanation": "This is a detector receiver (crystal set style). The LC tuned circuit selects one AM station, and the diode rectifies the RF envelope so headphones can reproduce the audio; there is no local oscillator, mixer, IF stage, or active RF/audio amplification.",
"source": "https://50ohm.de/NE_detektorempf%C3%A4nger.html",
- "confidence": 7
+ "confidence": 8
},
"EF102": {
- "revision": 1,
- "explanation": "A superhet converts received signals to a fixed IF, so fixed filters can provide much better selectivity than a tuned-radio-frequency receiver.",
+ "revision": 2,
+ "explanation": "A superheterodyne (Überlagerungsempfänger) mixes every signal down to one fixed intermediate frequency, so a single well-designed IF filter sets the selectivity for all bands. That gives far better adjacent-channel rejection (Trennschärfe) than a tuned-radio-frequency (Geradeaus) receiver, whose filters must retune with frequency.",
"source": "https://50ohm.de/E_ueberlagerungsempfaenger_einfachsuper_1.html",
"confidence": 8
},
"EF201": {
- "revision": 1,
- "explanation": "A mixer mainly produces the sum and absolute difference: 31.7 MHz + 21 MHz = 52.7 MHz and |31.7 MHz - 21 MHz| = 10.7 MHz.",
+ "revision": 2,
+ "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.)",
"source": "https://50ohm.de/E_mischer.html",
"confidence": 8
},
"EF202": {
- "revision": 1,
- "explanation": "Mixer products are the sum and absolute difference: 38.7 MHz + 28 MHz = 66.7 MHz and |38.7 MHz - 28 MHz| = 10.7 MHz.",
+ "revision": 2,
+ "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.",
"source": "https://50ohm.de/E_mischer.html",
"confidence": 8
},
"EF203": {
- "revision": 1,
- "explanation": "The desired mixer products are sum and difference, so 30 MHz and 39 MHz produce 69 MHz and 9 MHz.",
+ "revision": 2,
+ "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.",
"source": "https://50ohm.de/E_mischer.html",
"confidence": 8
},
"EF204": {
- "revision": 1,
- "explanation": "A mixer gives sum and absolute difference: 145 MHz + 136 MHz = 281 MHz and |145 MHz - 136 MHz| = 9 MHz.",
+ "revision": 2,
+ "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.",
"source": "https://50ohm.de/E_mischer.html",
"confidence": 8
},
"EF205": {
- "revision": 1,
- "explanation": "The wanted first-order mixer products are the sum and difference, here 281 MHz and 9 MHz.",
+ "revision": 2,
+ "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.",
"source": "https://50ohm.de/E_mischer.html",
"confidence": 8
},
"EF206": {
- "revision": 1,
- "explanation": "Mixers generate many RF products, so good shielding is needed to keep unwanted signals from being radiated or coupled into other stages.",
+ "revision": 2,
+ "explanation": "A mixer deliberately creates many sum/difference products and harmonics, most of them unwanted. Good shielding (a screened enclosure) keeps those products from radiating or coupling into other stages. Loose VFO coupling or removing the earth would make leakage worse, not better.",
"source": "https://50ohm.de/E_mischer.html",
"confidence": 8
},
"EF207": {
- "revision": 1,
- "explanation": "An oscillator should be enclosed in a grounded metal shield so its RF energy is not unintentionally radiated.",
+ "revision": 2,
+ "explanation": "An oscillator radiates RF that could leak out as a spurious emission or reach other stages, so it belongs inside a grounded metal screen (Abschirmung). Shielding plus good supply decoupling contains its energy; leaving it unshielded or unfiltered does the opposite.",
"source": "https://50ohm.de/NE_oszillatoren.html",
"confidence": 8
},
"EF208": {
- "revision": 1,
- "explanation": "In direct conversion the IF is audio, so the local oscillator must be very close to the received RF frequency.",
+ "revision": 2,
+ "explanation": "A direct-conversion (Direktüberlagerung) receiver mixes straight down to audio, i.e. an IF of essentially zero. For the difference frequency to land at audio, the local oscillator must sit right next to the received frequency — only a few kHz away.",
"source": "https://50ohm.de/E_ueberlagerungsempfaenger_einfachsuper_1.html",
"confidence": 8
},
"EF209": {
- "revision": 1,
- "explanation": "A BFO inserts the missing carrier needed to demodulate CW or SSB, making those signals audible.",
+ "revision": 2,
+ "explanation": "A BFO is a beat frequency oscillator. CW and SSB do not arrive as ordinary AM audio by themselves; the BFO supplies a local carrier so the detector can mix it with the received signal and produce an audible beat or speech audio.",
"source": "https://50ohm.de/NEA_bfo_1.html",
"confidence": 8
},
"EF210": {
- "revision": 1,
- "explanation": "Narrow receiver bandwidth rejects nearby unwanted signals, which is exactly high selectivity.",
+ "revision": 2,
+ "explanation": "Selectivity is a receiver's ability to accept the wanted frequency while rejecting nearby signals. A narrower IF/audio bandwidth passes less adjacent-channel energy, so high selectivity means nearby stations interfere less.",
"source": "https://50ohm.de/E_trennschaerfe_1.html",
"confidence": 8
},
"EF211": {
- "revision": 1,
- "explanation": "AGC changes receiver gain as the RF input varies, keeping the demodulated audio level more constant.",
+ "revision": 2,
+ "explanation": "AGC means Automatic Gain Control. It reduces receiver gain when strong signals arrive and increases gain again for weak signals, keeping the demodulated audio level more constant and preventing overload.",
"source": "https://50ohm.de/NE_agc_1.html",
"confidence": 8
},
"EF212": {
- "revision": 1,
- "explanation": "AGC stands for Automatic Gain Control, the automatic receiver gain regulation used to reduce level swings.",
+ "revision": 2,
+ "explanation": "AGC stands for Automatic Gain Control. It is a feedback system in the receiver that automatically changes gain according to signal strength, reducing large audio level swings between weak and strong stations.",
"source": "https://50ohm.de/NE_agc_1.html",
"confidence": 8
},
"EF213": {
- "revision": 1,
- "explanation": "Noise Reduction tries to distinguish wanted signal from noise and suppress the noise component in the received signal.",
+ "revision": 2,
+ "explanation": "Receiver noise reduction tries to tell wanted signal from random noise and suppress the noise part of the received audio, improving readability. It works on the demodulated signal content, not on the power supply or the IF dynamic range.",
"source": "https://50ohm.de/NE_noise_reduction.html",
"confidence": 8
},
"EF214": {
- "revision": 1,
- "explanation": "A noise blanker blanks short impulse disturbances, unlike a notch filter or AGC which target different problems.",
+ "revision": 2,
+ "explanation": "A noise blanker is designed for short impulsive noise, such as ignition clicks. It detects the pulse and briefly gates or blanks the receiver path, whereas a notch filter removes one narrow tone and AGC only changes gain.",
"source": "https://50ohm.de/NE_noise_reduction.html",
"confidence": 8
},
"EF215": {
- "revision": 1,
- "explanation": "A notch filter is a narrow rejection filter, so it can suppress interference at one small frequency range while leaving the rest mostly unchanged.",
+ "revision": 2,
+ "explanation": "A notch filter is a very narrow band-stop: it rejects one small slice of frequency (e.g. an interfering carrier or heterodyne whistle) while passing everything else. Low-, high-, and band-pass filters all pass a broad range instead, so they cannot remove a single narrow tone cleanly.",
"source": "https://50ohm.de/NE_notchfilter.html",
"confidence": 8
},
"EF216": {
- "revision": 1,
- "explanation": "A notch response is recognized by a narrow dip in an otherwise passed band; the correct diagram shows that sharp rejection notch.",
+ "revision": 2,
+ "explanation": "A notch filter is a very narrow band-stop filter. The response should pass most nearby frequencies but have one sharp rejection dip at the interfering tone, so the correct diagram is the otherwise flat passband with a narrow notch.",
"source": "https://50ohm.de/NE_notchfilter.html",
- "confidence": 7
+ "confidence": 8
},
"EF217": {
- "revision": 1,
- "explanation": "An attenuator reduces the RF input level before the receiver front end, preventing overload from strong signals.",
+ "revision": 2,
+ "explanation": "An attenuator is a deliberate RF loss inserted before the receiver front end. It lowers all incoming signals, which can improve real reception when strong local signals would otherwise overload the mixer or RF amplifier.",
"source": "https://50ohm.de/NE_vorverstaerker_daempfungsglied.html",
"confidence": 8
},
"EF218": {
- "revision": 1,
- "explanation": "A UHF preamplifier should be at the antenna so it amplifies the signal before feed-line loss degrades the noise figure.",
+ "revision": 2,
+ "explanation": "Put a UHF preamplifier right at the antenna. Amplifying the weak signal before it suffers feed-line loss sets a low overall noise figure (Friis): loss ahead of the first amplifier adds directly to the noise figure, so preamplifying after the cable would be far worse.",
"source": "https://50ohm.de/NE_vorverstaerker_daempfungsglied.html",
"confidence": 8
},
"EF219": {
- "revision": 1,
- "explanation": "A 9600-port bypasses audio filtering and takes receive data directly after the FM demodulator, which is point 4 in the shown chain.",
+ "revision": 2,
+ "explanation": "9600-baud packet/FM data needs the discriminator or demodulator signal before speech audio shaping filters distort it. Therefore the receive 9600-port is taken directly after the FM demodulator, at point 4 in the shown chain.",
"source": "https://50ohm.de/NEA_9600_port.html",
- "confidence": 7
+ "confidence": 8
},
"EF301": {
- "revision": 1,
- "explanation": "The multiplier chain is reversed by division: 145.2 MHz / 2 / 3 / 2 = 12.1 MHz.",
+ "revision": 2,
+ "explanation": "A frequency multiplier chain multiplies stage by stage, so to find the oscillator frequency you work backward by dividing by each multiplier. Here $145.2 MHz / 2 / 3 / 2 = 12.1 MHz$.",
"source": "https://50ohm.de/NE_frequenzvervielfacher_1.html",
"confidence": 8
},
"EF302": {
- "revision": 1,
- "explanation": "Work backward through the multipliers in the diagram: 21.360 MHz / 3 / 2 = 3.560 MHz.",
+ "revision": 2,
+ "explanation": "Frequency multipliers scale the input frequency by their factor. To recover the starting oscillator frequency, reverse the chain by division: $21.360 MHz / 3 / 2 = 3.560 MHz$.",
"source": "https://50ohm.de/NE_frequenzvervielfacher_1.html",
"confidence": 8
},
"EF303": {
- "revision": 1,
- "explanation": "Work forward through the multiplier chain: 3.51 MHz x 2 x 2 = 14.04 MHz at output a.",
+ "revision": 2,
+ "explanation": "Trace the multiplier chain: the VFO at $3.51$ MHz passes two doublers, $\\times 2 \\times 2 = \\times 4$, giving $3.51 \\cdot 4 = 14.04$ MHz at output a. Frequency multipliers move a low, stable VFO up to the wanted band.",
"source": "https://50ohm.de/NE_frequenzvervielfacher_1.html",
"confidence": 8
},
"EF304": {
- "revision": 1,
- "explanation": "Temperature changes alter oscillator L/C values gradually, so a VFO under changing temperature slowly drifts in frequency.",
+ "revision": 2,
+ "explanation": "A VFO's frequency is set by its $L$ and $C$, whose values drift slowly as temperature changes. The result is a slow frequency drift (not amplitude jumps) — which is why stable oscillators use temperature compensation or are locked to a reference.",
"source": "https://50ohm.de/NE_oszillatoren.html",
"confidence": 8
},
"EF305": {
- "revision": 1,
- "explanation": "ALC protects the transmit chain from overdrive by reducing the signal amplitude before the power amplifier when level is too high.",
+ "revision": 2,
+ "explanation": "ALC (automatic level control) guards the transmit chain: when the audio (NF) drive is too strong, it reduces the signal amplitude ahead of the power amplifier to prevent overdrive, distortion, and splatter. It acts in the transmit path, not on the receiver gain.",
"source": "https://50ohm.de/NEA_alc.html",
"confidence": 8
},
"EF306": {
- "revision": 2,
- "explanation": "A dynamic compressor raises quiet speech parts relative to loud ones, compressing the speech dynamic range.",
+ "revision": 3,
+ "explanation": "A dynamic compressor reduces speech dynamic range: loud syllables are limited or reduced, while quiet syllables are brought up relative to them. In radio this can raise average talk power without increasing peaks as much.",
"source": "https://50ohm.de/NE_slide_ne_modulation.html",
"confidence": 8
},
"EF307": {
- "revision": 1,
- "explanation": "A speech microphone amplifier should pass roughly 300 Hz to 3 kHz and reject lower and higher frequencies, matching the band-pass graph.",
+ "revision": 2,
+ "explanation": "A speech microphone amplifier is an audio band-pass stage: it should pass the intelligible voice range, roughly 300 Hz to 3 kHz, while rejecting rumble below it and hiss/RF products above it. That corresponds to the band-pass graph.",
"source": "https://50ohm.de/NE_verstaerker.html",
- "confidence": 7
+ "confidence": 8
},
"EF308": {
- "revision": 1,
- "explanation": "Intelligible SSB speech needs only about 2.5 to 3 kHz of audio bandwidth, so about 2.5 kHz is the minimum matching answer.",
+ "revision": 2,
+ "explanation": "Speech intelligibility needs roughly the $300$ Hz-$2.7$ kHz band, so an audio amplifier for SSB voice needs at least about $2.5$ kHz of bandwidth. More than that wastes bandwidth; much less (1 kHz) would make speech hard to understand.",
"source": "https://50ohm.de/NE_verstaerker.html",
"confidence": 8
},
"EF309": {
- "revision": 1,
- "explanation": "For 9600-baud FM data the signal should bypass speech audio filters and enter directly at the FM modulator, point 2 in the transmitter diagram.",
+ "revision": 2,
+ "explanation": "For 9600-baud FM data, normal microphone filtering and pre-emphasis would distort the baseband waveform. The data signal therefore enters directly at the FM modulator input, point 2 in the transmitter diagram.",
"source": "https://50ohm.de/NEA_9600_port.html",
- "confidence": 7
+ "confidence": 8
},
"EF310": {
- "revision": 1,
- "explanation": "An SSB speech filter only needs the voice sideband width; practical SSB generation commonly uses about 2.4 kHz.",
+ "revision": 2,
+ "explanation": "An SSB voice signal only needs the width of one speech sideband, so the sideband-selecting filter is about $2.4$ kHz wide. The other values are unrelated ($455$ kHz and $10.7$ MHz are IF centre frequencies, not bandwidths; $800$ Hz is too narrow for voice).",
"source": "https://50ohm.de/E_ssb_2.html",
"confidence": 8
},
"EF401": {
- "revision": 1,
- "explanation": "Transmitter output power is measured directly at the transmitter output before tuners, filters, feed lines, or other accessories change it.",
+ "revision": 2,
+ "explanation": "Transmitter output power is the power measured right at the transmitter output, before any tuner, filter, or feed line alters it. It is not a forward-minus-reflected figure or a near-field antenna measurement — just the raw power leaving the rig.",
"source": "https://50ohm.de/E_senderausgangsleistung.html",
"confidence": 8
},
"EF402": {
- "revision": 1,
- "explanation": "For SSB the relevant PEP is measured at the transmitter output using a steady one- or two-tone drive, not with an unmodulated carrier at the antenna.",
+ "revision": 2,
+ "explanation": "SSB PEP (peak envelope power) is measured directly at the transmitter output while it is driven with a steady one- or two-tone test signal, which produces a constant envelope peak you can read. An unmodulated carrier gives no SSB output (no audio, no signal), so it cannot be used.",
"source": "https://50ohm.de/E_senderausgangsleistung.html",
"confidence": 8
},
"EF403": {
- "revision": 1,
- "explanation": "SSB carries information in signal amplitude and phase, so its final stage must be linear to avoid distorting the waveform.",
+ "revision": 2,
+ "explanation": "An SSB signal carries its information in amplitude and phase variations, so the final stage must be a linear amplifier (Klasse A/AB) to reproduce the envelope faithfully. A limiter or non-linear stage would distort the envelope and splatter into neighbouring channels; a multiplier would destroy the signal entirely.",
"source": "https://50ohm.de/EA_verstaerker.html",
"confidence": 8
},
"EF404": {
- "revision": 1,
- "explanation": "Changing the final amplifier bias can change linearity, so the transmitter must then be checked for harmonic output.",
+ "revision": 2,
+ "explanation": "Re-adjusting the final amplifier's operating point (bias) changes its linearity and can create new harmonics, so the transmitter must be re-checked for harmonic emissions afterwards. Checking only when interference is heard would be too late — verify after any change to the PA.",
"source": "https://50ohm.de/NE_unerwuenschte_aussendungen_2.html",
"confidence": 8
},
"EF405": {
- "revision": 1,
- "explanation": "The transmitter supply should be well decoupled against RF so RF energy cannot couple into the power wiring or other stages.",
+ "revision": 2,
+ "explanation": "The transmitter's DC supply leads should be well decoupled against RF (chokes and bypass capacitors), so stray RF cannot ride back onto the power wiring or feed between stages and cause instability or interference. Making the supply high-impedance or low-capacitance to ground would worsen, not improve, RF behaviour.",
"source": "https://50ohm.de/EA_verstaerker.html",
"confidence": 8
},
"EF501": {
- "revision": 1,
- "explanation": "A transverter converts both directions: on receive it downconverts the higher band to the transceiver band, and on transmit it upconverts the transceiver signal.",
+ "revision": 2,
+ "explanation": "A transverter shifts a band in both directions: on receive it converts the higher band (e.g. $70$ cm) down into the transceiver's band ($10$ m), and on transmit it converts the $10$ m signal back up to $70$ cm. It only moves frequency — it does not change the modulation type or digital protocol.",
"source": "https://50ohm.de/NE_transverter_1.html",
"confidence": 8
},
"EF502": {
- "revision": 1,
- "explanation": "A transverter changes bands by mixing the input signal with a local oscillator and filtering the wanted product.",
+ "revision": 2,
+ "explanation": "A transverter changes band by mixing: it beats the signal against a local oscillator and keeps the wanted sum or difference product. Multiplication, division, or feedback would not translate a whole modulated band intact the way heterodyne mixing does.",
"source": "https://50ohm.de/NE_transverter_1.html",
"confidence": 8
},
"EF503": {
- "revision": 1,
- "explanation": "The block diagram shows receive and transmit frequency conversion around a VHF transceiver, which is a transverter for the 2 m band.",
+ "revision": 2,
+ "explanation": "A transverter performs bidirectional frequency conversion around an existing transceiver: on receive it downconverts the target band, and on transmit it upconverts the transceiver output. The shown receive/transmit conversion around a VHF transceiver identifies it as a 2 m transverter setup.",
"source": "https://50ohm.de/NE_transverter_1.html",
- "confidence": 7
+ "confidence": 8
},
"EF504": {
- "revision": 1,
- "explanation": "The diagram upconverts a VHF transmit signal to the 13 cm range, so it is a 13 cm converter placed before a VHF transmitter.",
+ "revision": 2,
+ "explanation": "The diagram shows only transmit upconversion: a VHF signal is mixed with a local oscillator and filtered to produce a signal in the 13 cm band. That makes it a transmit converter for 13 cm, not a complete bidirectional transverter.",
"source": "https://50ohm.de/NE_transverter_1.html",
- "confidence": 7
+ "confidence": 8
},
"EF505": {
- "revision": 1,
- "explanation": "In a GHz transverter the oscillator is multiplied, so any oscillator frequency error is multiplied too and can be too large for SSB satellite operation.",
+ "revision": 2,
+ "explanation": "For a $2.4$ GHz uplink the local oscillator is multiplied up to the transmit frequency, and multiplication scales any oscillator error by the same factor. A small drift at the XO becomes a large drift at GHz — easily more than SSB's few-hundred-Hz tolerance — so the LO must be temperature-stabilised or locked to a high-grade reference.",
"source": "https://50ohm.de/NE_transverter_1.html",
"confidence": 8
},
"EF601": {
- "revision": 1,
- "explanation": "Digital signal processing first converts the analog input with an A/D converter and later reconstructs an analog output with a D/A converter.",
+ "revision": 2,
+ "explanation": "In a DSP chain the analogue input must first be sampled by an A/D converter (block 1, ADC) so the maths can run on numbers; after processing, a D/A converter (block 2, DAC) rebuilds an analogue output. The order is fixed by the signal flow: digitise, process, reconstruct.",
"source": "https://50ohm.de/NEA_digitale_signalverarbeitung_einleitung.html",
"confidence": 8
},
"EF602": {
- "revision": 1,
- "explanation": "A digital filter can only process digital samples, so the analog input signal must first be digitized by A/D conversion.",
+ "revision": 2,
+ "explanation": "A digital filter operates on numerical samples, not on a continuous waveform, so the analogue input must first be digitised (A/D conversion) before it can be filtered. Demodulation or noise/harmonic removal are not prerequisites — sampling is.",
"source": "https://50ohm.de/NEA_digitale_signalverarbeitung_einleitung.html",
"confidence": 8
},
"EF603": {
- "revision": 1,
- "explanation": "SDR means Software Defined Radio: at least part of the receiver or transceiver signal processing is implemented in software.",
+ "revision": 2,
+ "explanation": "SDR means Software-Defined Radio: at least part of the signal processing (filtering, demodulation, often the whole IF chain) is done in software rather than fixed hardware. It is still radio over the air, just with the heavy lifting moved into code.",
"source": "https://50ohm.de/NEA_digitale_signalverarbeitung_einleitung.html",
"confidence": 8
},
"EG101": {
- "revision": 1,
- "explanation": "A loop made from three equal wire sections forms a triangle, so it is the delta-loop form of a full-wave loop antenna.",
+ "revision": 2,
+ "explanation": "A loop built from three equal wire sections forms a triangle, i.e. a delta loop (Delta-Loop) — a full-wave loop bent into three sides. A quad loop is four-sided, a W3DZZ is a trap dipole, and a beam is a multi-element directional array, so none of those match a three-sided loop.",
"source": "https://50ohm.de/EA_antennenformen_2.html",
"confidence": 8
},
"EG102": {
- "revision": 1,
- "explanation": "A wire antenna can have many lengths if an appropriate matching network is used; resonance and feed impedance change with length.",
+ "revision": 2,
+ "explanation": "A wire HF antenna can in principle be any length: with a suitable matching network (Antennentuner) you can present $50\\ \\Omega$ to the transmitter even when the wire is not resonant. The wire's length sets its resonance and feed-point impedance, not whether it can be used at all — so the rigid $\\lambda/2$ or $\\lambda/4$ options are wrong.",
"source": "https://50ohm.de/NE_antenne_laenge_resonanz.html",
"confidence": 8
},
"EG103": {
- "revision": 1,
- "explanation": "The diagram shows a wire fed from one end through a simple matching unit, which is an end-fed antenna with a basic matching network.",
+ "revision": 2,
+ "explanation": "The diagram shows an end-fed antenna: the wire is fed at one end rather than in the middle. Because an end-fed half-wave wire has high feed impedance, it needs a matching unit to transform that impedance toward the transmitter/feed-line impedance.",
"source": "https://50ohm.de/E_antennenformen_2.html",
- "confidence": 7
+ "confidence": 8
},
"EG104": {
- "revision": 1,
- "explanation": "The shown end-fed wire with a tuned Fuchs matching circuit is the characteristic Fuchs antenna arrangement.",
+ "revision": 2,
+ "explanation": "A Fuchs antenna is a form of end-fed half-wave antenna using a tuned matching circuit, often a parallel resonant circuit, at the feed end. The tuned circuit transforms the high end impedance so the transmitter can feed the wire efficiently.",
"source": "https://50ohm.de/E_antennenformen_2.html",
- "confidence": 7
+ "confidence": 8
},
"EG105": {
- "revision": 1,
- "explanation": "A magnetic loop is small compared with wavelength, about lambda/10 circumference, and its near field is dominated by a strong magnetic component.",
+ "revision": 2,
+ "explanation": "A magnetic loop is electrically small, often around $\\lambda/10$ circumference or less. Its current is high and voltage is comparatively low, so the near field is dominated by the magnetic H-field; this is why it behaves differently from a full-size resonant wire antenna.",
"source": "https://50ohm.de/EA_antennenformen_2.html",
"confidence": 8
},
"EG106": {
- "revision": 1,
- "explanation": "Common HF transmitting antennas include long wire, Yagi-Uda, dipole, Windom, and delta-loop; horn, patch, and parabolic antennas are mainly higher-frequency forms.",
+ "revision": 2,
+ "explanation": "Typical HF (Kurzwelle) transmitting antennas are wire/boom types: long wire, Yagi-Uda, dipole, Windom, and delta loop. Horn, patch, slot, and parabolic dish antennas need dimensions of several wavelengths, which is only practical at VHF/UHF and microwave — so any answer list containing them is wrong for HF.",
"source": "https://50ohm.de/EA_antennenformen_2.html",
"confidence": 8
},
"EG107": {
- "revision": 2,
- "explanation": "For 80 m HF operation, dipoles, delta loops, and W3DZZ trap dipoles are practical wire antennas; parabolic, cross-Yagi, and trap-sleeve forms are not suitable choices here.",
+ "revision": 3,
+ "explanation": "For the 80 m band ($\\lambda \\approx 80$ m) you need physically large wire antennas: a dipole, a delta loop, and the W3DZZ (a trap dipole for multiband HF) all fit. A parabolic dish or cross-Yagi would be enormous at 80 m, and a sleeve/Sperrtopf antenna is a VHF/UHF form, so the lists containing those are out.",
"source": "https://50ohm.de/NE_slide_ne_antennen_uebertragungsleitungen.html",
"confidence": 8
},
"EG108": {
- "revision": 1,
- "explanation": "A 5/8-wave vertical is chosen because its length gives better antenna gain than a quarter-wave mobile vertical.",
+ "revision": 2,
+ "explanation": "A $5/8\\,\\lambda$ vertical concentrates more of its radiation toward the horizon than a $\\lambda/4$ whip, giving roughly 1-3 dB more gain in useful low-angle directions — ideal for mobile VHF/UHF where you want range along the ground. The other options (power handling, mounting, noise) are not the reason it is preferred.",
"source": "https://50ohm.de/E_antennenformen_2.html",
"confidence": 8
},
"EG109": {
- "revision": 1,
- "explanation": "The wavelength is 300 / 28.5 = 10.53 m, and 5/8 of that is about 6.58 m.",
+ "revision": 2,
+ "explanation": "First the wavelength: $\\lambda = 300 / f_{\\text{MHz}} = 300/28.5 \\approx 10.53$ m. The electrical length of a $5/8\\,\\lambda$ radiator is $0.625 \\cdot 10.53 \\approx 6.58$ m. (This is the electrical length; a real whip is trimmed a few percent shorter for the velocity factor.)",
"source": "https://50ohm.de/NE_antenne_laenge_resonanz.html",
"confidence": 8
},
"EG110": {
- "revision": 1,
- "explanation": "A folded dipole is essentially a flattened full-wave loop, so the total wire length is one wavelength.",
+ "revision": 2,
+ "explanation": "A folded dipole (Faltdipol) is a half-wave dipole whose two parallel conductors are joined at the ends, forming a thin loop. Going out along the top and back along the bottom totals two half-waves, so the wire used is one full wavelength ($\\lambda$).",
"source": "https://50ohm.de/NE_antenne_laenge_resonanz.html",
"confidence": 8
},
"EG111": {
- "revision": 1,
- "explanation": "A simple Yagi-Uda has the longer reflector behind the driven element and a shorter director in front, giving the order reflector, driven element, director.",
+ "revision": 2,
+ "explanation": "In a Yagi-Uda antenna, only the driven element is fed directly. The reflector is slightly longer and sits behind it; the director is slightly shorter and sits in front, so the order along the boom is reflector, driven element, director.",
"source": "https://50ohm.de/NE_yagi_uda_2.html",
- "confidence": 7
+ "confidence": 8
},
"EG112": {
- "revision": 2,
- "explanation": "For a directional HF antenna, placing it high and far from neighboring equipment reduces field strength at the neighbor and therefore coupling risk.",
+ "revision": 3,
+ "explanation": "Coupling into a neighbour's equipment grows with the field strength reaching it, which falls with distance and with how much the main beam misses their house. Mounting the directional antenna as high and as far away as possible therefore minimises the field at the neighbour — the opposite of placing it low, close, or aimed over their roof.",
"source": "https://50ohm.de/NE_slide_ne_antennen_uebertragungsleitungen.html",
"confidence": 8
},
"EG113": {
- "revision": 1,
- "explanation": "Microwave dish antennas use a paraboloid reflector plus a feed antenna; the feed illuminates the reflector that forms the narrow beam.",
+ "revision": 2,
+ "explanation": "A sharply focused microwave antenna is usually a parabolic dish: a paraboloid reflector (Spiegel) plus a small feed antenna (Erreger/Feed) at its focus. The feed illuminates the dish, which reflects the energy into a narrow parallel beam. An isotropic radiator is only a theoretical reference, not a real feed.",
"source": "https://50ohm.de/EA_parabolspiegel_1.html",
"confidence": 8
},
"EG114": {
- "revision": 1,
- "explanation": "Dish gain improves when the reflector is many wavelengths across; at least about five wavelengths is the suitable choice for high gain.",
+ "revision": 2,
+ "explanation": "Dish gain rises with aperture measured in wavelengths — bigger relative to $\\lambda$ means a narrower beam and more gain. A reflector of at least about five wavelengths diameter is the practical minimum for high gain; one or two wavelengths is too small to form a tight beam.",
"source": "https://50ohm.de/EA_parabolspiegel_1.html",
"confidence": 8
},
"EG201": {
- "revision": 1,
- "explanation": "The shortening factor compares wave speed on the line or wire with wave speed in vacuum, so it is the velocity ratio.",
+ "revision": 2,
+ "explanation": "The velocity factor (Verkürzungsfaktor) is the ratio of wave propagation speed on the line or wire to its speed in vacuum, always $< 1$. Because the wave travels slower, a resonant element is physically shorter than the free-space $\\lambda/2$ — hence the name shortening factor.",
"source": "https://50ohm.de/E_verkuerzungsfaktor_1.html",
"confidence": 8
},
"EG202": {
- "revision": 1,
- "explanation": "For wire antennas the usual shortening correction is about 0.95, meaning about 95 percent of the free-space calculated length.",
+ "revision": 2,
+ "explanation": "A real wire antenna is electrically a little longer than its physical length because of end effects and the wave speed near the conductor. The usual shortening factor is about 0.95, so physical length is roughly $0.95 \\cdot \\lambda/2$ for a half-wave wire.",
"source": "https://50ohm.de/E_verkuerzungsfaktor_1.html",
"confidence": 8
},
"EG203": {
- "revision": 1,
- "explanation": "At a dipole end charge and voltage are high while current goes to zero, so the ends are voltage maxima and current nodes.",
+ "revision": 2,
+ "explanation": "A dipole's ends are open circuits: charge piles up there, so voltage is maximum (voltage antinode, Spannungsbauch) while the current must fall to zero (current node, Stromknoten). Current can only be large where it has somewhere to flow, i.e. toward the centre, not at the open tips.",
"source": "https://50ohm.de/NEA_strom_spannung_speisung_1.html",
"confidence": 8
},
"EG204": {
- "revision": 1,
- "explanation": "Current feeding means high current and low voltage at the feed point, a current maximum and voltage node, which gives low impedance.",
+ "revision": 2,
+ "explanation": "Current feeding (Stromspeisung) means the feed point sits at a current antinode (Strombauch) and a voltage node (Spannungsknoten). With high current and low voltage there, the impedance $Z = U/I$ is low — a centre-fed half-wave dipole is the classic example at about $50$-$75\\ \\Omega$.",
"source": "https://50ohm.de/NE_strom_spannung_speisung_1.html",
"confidence": 8
},
"EG205": {
- "revision": 1,
- "explanation": "Voltage feeding is the opposite case: high voltage and nearly zero current at the feed point, so the feed point is high impedance.",
+ "revision": 2,
+ "explanation": "Voltage feeding (Spannungsspeisung) is the opposite of current feeding: the feed point is a voltage antinode (Spannungsbauch) and a current node (Stromknoten). High voltage with near-zero current makes $Z = U/I$ very high — e.g. feeding a half-wave element at its end.",
"source": "https://50ohm.de/NE_strom_spannung_speisung_1.html",
"confidence": 8
},
"EG206": {
- "revision": 1,
- "explanation": "A half-wave dipole fed in the middle has its current maximum at the center, so it is current-fed on its fundamental frequency.",
+ "revision": 2,
+ "explanation": "On its fundamental a half-wave dipole carries a half-sine current distribution: maximum in the middle, zero at the ends. Feeding it at the centre therefore taps a current antinode (Strombauch), which is by definition current feeding (low impedance).",
"source": "https://50ohm.de/NE_strom_spannung_speisung_1.html",
"confidence": 8
},
"EG207": {
- "revision": 1,
- "explanation": "A center-fed half-wave dipole high above ground has a free-space feed impedance near 73 Ohm, so the rounded answer is 75 Ohm.",
+ "revision": 2,
+ "explanation": "A center-fed half-wave dipole in free space has a feed-point resistance of about $73\\ \\Omega$; raised at least a wavelength above ground it stays near that value, which the exam rounds to $75\\ \\Omega$. That is also why $75\\ \\Omega$ coax is a natural match for a simple dipole.",
"source": "https://50ohm.de/E_fusspunktimpedanz_1.html",
"confidence": 8
},
"EG208": {
- "revision": 1,
- "explanation": "Ground interaction changes a center-fed half-wave dipole impedance with height, typically over about 40 to 90 Ohm.",
+ "revision": 2,
+ "explanation": "The free-space figure ($\\approx 73\\ \\Omega$) shifts when the dipole is near ground: reflections add in or out of phase depending on height, so the feed resistance swings over roughly $40$-$90\\ \\Omega$ at practical heights. That spread is why a dipole still works acceptably on either $50$ or $75\\ \\Omega$ coax.",
"source": "https://50ohm.de/E_fusspunktimpedanz_1.html",
"confidence": 8
},
"EG209": {
- "revision": 1,
- "explanation": "A straight center-fed half-wave dipole is in the same practical impedance range as the height-dependent value, about 40 to 90 Ohm.",
+ "revision": 2,
+ "explanation": "A straight, center-fed half-wave dipole has the same height-dependent feed resistance as above: about $40$-$90\\ \\Omega$ over realistic installation heights, centred on the free-space $\\approx 73\\ \\Omega$. The wider answers (240-300 $\\Omega$) belong to a folded dipole, not a plain one.",
"source": "https://50ohm.de/E_fusspunktimpedanz_1.html",
"confidence": 8
},
"EG210": {
- "revision": 1,
- "explanation": "A folded dipole approximately quadruples the feed impedance of a normal dipole, giving about 240 to 300 Ohm.",
+ "revision": 2,
+ "explanation": "A folded dipole (Faltdipol) splits the antenna current between its two parallel conductors. Halving the current at the same power raises the feed-point resistance by a factor of about four ($2^2$), turning the dipole's $\\approx 70\\ \\Omega$ into roughly $240$-$300\\ \\Omega$ — the classic match for $300\\ \\Omega$ ribbon feeder.",
"source": "https://50ohm.de/E_fusspunktimpedanz_1.html",
"confidence": 8
},
"EG211": {
- "revision": 2,
- "explanation": "A quarter-wave vertical against a counterpoise has a feed impedance near 35 Ohm with horizontal radials; sloping the radials downward raises it toward 50 Ohm, so the practical range is about 30 to 50 Ohm.",
+ "revision": 3,
+ "explanation": "A quarter-wave ground-plane (Groundplane) works against its radials rather than a second arm. With horizontal radials the feed resistance is near $35\\ \\Omega$; drooping the radials downward raises it toward $50\\ \\Omega$, so the practical range is about $30$-$50\\ \\Omega$ — a good direct match to $50\\ \\Omega$ coax.",
"source": "https://50ohm.de/E_fusspunktimpedanz_1.html",
"confidence": 8
},
"EG212": {
- "revision": 1,
- "explanation": "In a Yagi-Uda antenna the feed is applied to the driven element, called the Strahler; reflector and directors are parasitic elements.",
+ "revision": 2,
+ "explanation": "In a Yagi-Uda only one element is actually fed — the driven element (Strahler). The reflector and directors are parasitic: they carry current induced by the driven element and reshape the pattern, but no feeder connects to them.",
"source": "https://50ohm.de/NE_yagi_uda_2.html",
"confidence": 8
},
"EG213": {
- "revision": 1,
- "explanation": "A ground-plane is unbalanced because the radial side is at earth or counterpoise potential; dipoles, folded dipoles, and Yagis are balanced antenna forms.",
+ "revision": 2,
+ "explanation": "A ground-plane is unbalanced (asymmetric): one side is the radiator, the other is the radial/counterpoise system near earth potential, so the two halves are not mirror images. A folded dipole, a long Yagi, and a center-fed $\\lambda/2$ dipole are all symmetric about their feed point.",
"source": "https://50ohm.de/EA_antennenformen_2.html",
"confidence": 8
},
"EG214": {
- "revision": 1,
- "explanation": "A half-wave dipole pattern has two equal broad lobes perpendicular to the wire, matching the symmetric two-lobed diagram.",
+ "revision": 2,
+ "explanation": "A half-wave dipole radiates broadside: strongest at right angles to the wire and weakest off the wire ends. From above this gives two equal lobes, so the symmetric two-lobed diagram is the dipole pattern.",
"source": "https://50ohm.de/NE_antennenformen_2.html",
- "confidence": 7
+ "confidence": 8
},
"EG215": {
- "revision": 1,
- "explanation": "The shown two-lobed pattern perpendicular to the wire is the typical radiation pattern of a half-wave dipole.",
+ "revision": 2,
+ "explanation": "The clue is maximum radiation perpendicular to the conductor and nulls toward the conductor ends. That is the horizontal pattern of a half-wave dipole, often drawn as a figure-eight from above.",
"source": "https://50ohm.de/NE_antennenformen_2.html",
- "confidence": 7
+ "confidence": 8
},
"EG216": {
- "revision": 1,
- "explanation": "The nearly circular horizontal pattern around the vertical radiator is typical of a ground-plane antenna viewed from above.",
+ "revision": 2,
+ "explanation": "A vertical ground-plane antenna radiates roughly equally in all horizontal directions, so its top-view azimuth pattern is nearly circular. Its useful directivity is mainly in elevation, not around the compass.",
"source": "https://50ohm.de/EA_antennenformen_2.html",
- "confidence": 7
+ "confidence": 8
},
"EG217": {
- "revision": 1,
- "explanation": "A large forward lobe with a smaller rear lobe indicates directional gain, so the diagram is for a directional antenna.",
+ "revision": 2,
+ "explanation": "A directional antenna concentrates radiation into a main lobe instead of radiating equally in all directions. The large forward lobe, with smaller rear or side lobes, is the visual cue for directional gain.",
"source": "https://50ohm.de/EA_antennenformen_2.html",
- "confidence": 7
+ "confidence": 8
},
"EG218": {
- "revision": 1,
- "explanation": "A Yagi-Uda radiation pattern has a strong main lobe toward the directors and smaller rear or side lobes, matching the shown diagram.",
+ "revision": 2,
+ "explanation": "A Yagi-Uda antenna uses parasitic reflector/director elements to reinforce radiation in one direction. Its pattern therefore has a strong main lobe toward the directors and much smaller rear/side lobes.",
"source": "https://50ohm.de/NE_yagi_uda_2.html",
- "confidence": 7
+ "confidence": 8
},
"EG219": {
- "revision": 1,
- "explanation": "A vertical half-wave antenna radiates mainly perpendicular to the vertical element, giving a low elevation or flat radiation angle.",
+ "revision": 2,
+ "explanation": "A vertical half-wave antenna radiates mainly perpendicular to the vertical conductor. Since the conductor is vertical, the strongest radiation leaves at low elevation angles, which is useful for DX because low-angle rays make longer ionospheric hops.",
"source": "https://50ohm.de/E_antennenformen_2.html",
"confidence": 8
},
"EG220": {
- "revision": 1,
- "explanation": "The suffix dBi means gain in dB relative to an isotropic radiator, the ideal antenna radiating equally in all directions.",
+ "revision": 2,
+ "explanation": "The suffix dBi means gain in decibels relative to an isotropic radiator (Isotropstrahler) — the ideal point source that radiates equally in every direction. It is a pure reference: no real antenna is isotropic, but it gives an absolute baseline for comparing gains.",
"source": "https://50ohm.de/NE_antennengewinn.html",
"confidence": 8
},
"EG221": {
- "revision": 1,
- "explanation": "dBd is referenced to a half-wave dipole, which is 2.15 dB above isotropic; 5 dBd + 2.15 dB = 7.15 dBi.",
+ "revision": 2,
+ "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.",
"source": "https://50ohm.de/NE_antennengewinn.html",
"confidence": 8
},
"EG222": {
- "revision": 1,
- "explanation": "Antenna polarization is defined by the electric field orientation in the main radiation direction relative to the earth surface.",
+ "revision": 2,
+ "explanation": "Polarisation is defined by the orientation of the electric field component in the main beam direction, relative to the Earth's surface — horizontal E-field is horizontal polarisation, vertical E-field is vertical. It is referenced to the ground, not to compass north, and to the E-field, not the magnetic field.",
"source": "https://50ohm.de/E_polarisation_2.html",
"confidence": 8
},
"EG223": {
- "revision": 2,
- "explanation": "Putting the transmitting antenna outdoors reduces coupling into house wiring and nearby electrical installations.",
+ "revision": 3,
+ "explanation": "An indoor transmitting antenna puts strong RF fields close to house wiring and consumer electronics, so those leads can act as unintended receiving antennas. Moving the antenna outdoors and away from installations reduces coupling and interference risk.",
"source": "https://50ohm.de/NE_slide_ne_antennen_uebertragungsleitungen.html",
"confidence": 8
},
"EG301": {
- "revision": 1,
- "explanation": "A line's characteristic impedance is set by its conductor geometry and dielectric; in the HF range it is roughly constant and does not depend on the load connected at the end.",
+ "revision": 2,
+ "explanation": "The characteristic impedance (Wellenwiderstand) is fixed by the line's geometry and dielectric — conductor spacing and insulation — not by what is connected at the far end. Across the HF range it is essentially constant and independent of the load, which is why a $50\\ \\Omega$ cable stays $50\\ \\Omega$ regardless of the antenna on it.",
"source": "https://50ohm.de/E_uebertragungsleitungen_2.html",
"confidence": 8
},
"EG302": {
- "revision": 1,
- "explanation": "Good coaxial cable confines the RF field inside the shield in normal use, reducing unwanted radiation between station devices.",
+ "revision": 2,
+ "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.",
"source": "https://50ohm.de/E_uebertragungsleitungen_2.html",
"confidence": 8
},
"EG303": {
- "revision": 1,
- "explanation": "N connectors are designed for 50 Ohm operation into the GHz range and are suitable for higher power and voltage than SMA or BNC in this comparison.",
+ "revision": 2,
+ "explanation": "The N connector (N-Stecker) holds a defined $50\\ \\Omega$ impedance well into the GHz range and has the highest voltage rating of the four, so it handles high power best. SMA is small and fine to microwave but limited in power; BNC tops out lower; the UHF (PL) connector has no constant impedance and is unsuitable above a few hundred MHz.",
"source": "https://50ohm.de/E_uebertragungsleitungen_2.html",
"confidence": 8
},
"EG304": {
- "revision": 1,
- "explanation": "A feed line is unbalanced when the two conductors are not equivalent, as in coax where the inner conductor and shield have different shapes and potentials.",
+ "revision": 2,
+ "explanation": "A feeder is unbalanced (unsymmetrisch) when its two conductors are not geometrically equivalent — as in coax, where a central inner conductor sits inside an outer shield at a different potential. A balanced line uses two identical, symmetric conductors instead. Reflection, resonance, and length do not define balance.",
"source": "https://50ohm.de/E_uebertragungsleitungen_2.html",
"confidence": 8
},
"EG305": {
- "revision": 1,
- "explanation": "Open parallel-wire feed line avoids much dielectric loss and can withstand high voltages better than coaxial cable.",
+ "revision": 2,
+ "explanation": "Open parallel-wire line (Hühnerleiter/Paralleldraht) runs mostly through air rather than a lossy solid dielectric, so it has markedly lower loss than coax and withstands high voltages well — ideal for feeding a high-SWR multiband antenna. Its drawback is that it radiates if unbalanced, the opposite of the shielding option B claims.",
"source": "https://50ohm.de/E_uebertragungsleitungen_2.html",
"confidence": 8
},
"EG306": {
- "revision": 1,
- "explanation": "Running RF feed lines directly beside mains leads can couple RF into the power wiring, so a shared cable duct can worsen interference risk.",
+ "revision": 2,
+ "explanation": "Bundling RF feeders next to mains leads in one duct lets RF couple from the cable into the power wiring, carrying interference out onto the supply network. The realistic hazard is this conducted coupling/EMC problem, not flashover (voltages are modest) — neat cable management can quietly make interference worse.",
"source": "https://50ohm.de/E_uebertragungsleitungen_2.html",
"confidence": 8
},
"EG307": {
- "revision": 1,
- "explanation": "Cable losses in dB are added as positive attenuation values; the shown station layout sums to 5 dB of cable loss.",
+ "revision": 2,
+ "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.",
"source": "https://50ohm.de/EA_kabeldaempfung_1.html",
- "confidence": 7
+ "confidence": 8
},
"EG308": {
- "revision": 1,
- "explanation": "With SWR 1 there is no reflection; 100 W reduced to 50 W is a factor of 2 loss, which corresponds to 3 dB attenuation.",
+ "revision": 2,
+ "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.)",
"source": "https://50ohm.de/EA_kabeldaempfung_1.html",
"confidence": 8
},
"EG309": {
- "revision": 1,
- "explanation": "Only one quarter of the power remains, so the loss factor is 4; a power factor of 4 is about 6 dB.",
+ "revision": 2,
+ "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.)",
"source": "https://50ohm.de/EA_kabeldaempfung_1.html",
"confidence": 8
},
"EG310": {
- "revision": 1,
- "explanation": "Only one tenth of the power remains, so the loss factor is 10; a power factor of 10 is 10 dB.",
+ "revision": 2,
+ "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.",
"source": "https://50ohm.de/EA_kabeldaempfung_1.html",
"confidence": 8
},
"EG311": {
- "revision": 1,
- "explanation": "Cable attenuation scales with length for the same cable and frequency: 20 dB per 100 m times 20/100 gives 4 dB.",
+ "revision": 2,
+ "explanation": "For one cable at one frequency, attenuation in decibels is proportional to length. Scale directly: $20\\ \\text{dB} \\cdot (20\\ \\text{m} / 100\\ \\text{m}) = 4$ dB. Because dB already represents the loss logarithmically, you scale the dB value linearly with length — no powers needed.",
"source": "https://50ohm.de/E_kabeldaempfung_1.html",
"confidence": 8
},
"EG312": {
- "revision": 1,
- "explanation": "The cable-loss chart gives RG58 at 145 MHz as about 20 dB per 100 m, and the question length is exactly 100 m.",
+ "revision": 2,
+ "explanation": "Read the cable-loss chart for RG58 (full-PE, $4.95$ mm) at $145$ MHz: about $20$ dB per $100$ m. The cable here is exactly $100$ m long, so the attenuation is that chart value, $20$ dB. (Thin RG58 is quite lossy at VHF — a good reason to use thicker cable on $2$ m.)",
"source": "https://50ohm.de/E_kabeldaempfung_1.html",
"confidence": 8
},
"EG313": {
- "revision": 1,
- "explanation": "RG58 is about 20 dB per 100 m at 145 MHz; for 15 m the attenuation is 20 x 15/100 = 3 dB.",
+ "revision": 2,
+ "explanation": "From the chart, RG58 at $145$ MHz is about $20$ dB per $100$ m. Attenuation scales with length, so $15$ m gives $20 \\cdot 15/100 = 3$ dB.",
"source": "https://50ohm.de/E_kabeldaempfung_1.html",
"confidence": 8
},
"EG314": {
- "revision": 1,
- "explanation": "The chart value for RG174 at 145 MHz is about 40 dB per 100 m; for 50 m this is 40 x 50/100 = 20 dB.",
+ "revision": 2,
+ "explanation": "From the chart, the thin RG174 ($2.8$ mm) at $145$ MHz is about $40$ dB per $100$ m — twice as lossy as RG58. Scaled to $50$ m: $40 \\cdot 50/100 = 20$ dB. Thinner cable means higher loss.",
"source": "https://50ohm.de/E_kabeldaempfung_1.html",
"confidence": 8
},
"EG315": {
- "revision": 1,
- "explanation": "The chart gives about 7 dB per 100 m for the 12.7 mm PE-foam cable at 435 MHz; 40 m gives 7 x 40/100 = 2.8 dB.",
+ "revision": 2,
+ "explanation": "From the chart, the thick $12.7$ mm PE-foam cable at $435$ MHz is about $7$ dB per $100$ m. For $40$ m: $7 \\cdot 40/100 = 2.8$ dB. Larger diameter and foam dielectric keep the loss low even at UHF.",
"source": "https://50ohm.de/E_kabeldaempfung_1.html",
"confidence": 8
},
"EG316": {
- "revision": 1,
- "explanation": "The chart gives about 20.5 dB per 100 m for the 10.3 mm PE-foam cable at 1296 MHz; 40 m gives about 8.2 dB.",
+ "revision": 2,
+ "explanation": "From the chart, the $10.3$ mm PE-foam cable at $1296$ MHz is about $20.5$ dB per $100$ m — loss climbs steeply with frequency. For $40$ m: $20.5 \\cdot 40/100 \\approx 8.2$ dB.",
"source": "https://50ohm.de/E_kabeldaempfung_1.html",
"confidence": 8
},
"EG401": {
- "revision": 2,
- "explanation": "For SWR 3 the reflection coefficient is (3 - 1)/(3 + 1) = 0.5, so reflected power is 0.5 squared = 25 percent of 100 W, i.e. 25 W.",
+ "revision": 3,
+ "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.",
"source": "https://50ohm.de/NEA_swr.html",
"confidence": 8
},
"EG402": {
- "revision": 2,
- "explanation": "SWR 3 gives voltage reflection coefficient 0.5; power reflection is 0.5 squared, so 25 percent of forward power is reflected.",
+ "revision": 3,
+ "explanation": "Convert SWR to the voltage reflection coefficient $\\Gamma = (S-1)/(S+1) = (3-1)/(3+1) = 0.5$, then square it for power: $\\Gamma^2 = 0.25 = 25\\,\\%$ of the forward power is reflected. Power always goes with $\\Gamma^2$, not $\\Gamma$.",
"source": "https://50ohm.de/NEA_swr.html",
"confidence": 8
},
"EG403": {
- "revision": 2,
- "explanation": "If SWR 3 reflects 25 percent of the forward power, the remaining 75 percent is delivered to the load.",
+ "revision": 3,
+ "explanation": "If $\\Gamma^2 = ((3-1)/(3+1))^2 = 0.25 = 25\\,\\%$ of the forward power is reflected, the rest reaches the load. Energy is conserved, so the delivered fraction is $1 - 0.25 = 75\\,\\%$.",
"source": "https://50ohm.de/NEA_swr.html",
"confidence": 8
},
"EG404": {
- "revision": 1,
- "explanation": "The current on the outside of the coax shield is the common-mode or mantle current, called Mantelstrom in the diagram.",
+ "revision": 2,
+ "explanation": "In normal coax operation, RF current flows on the inner conductor and the inside of the shield. Current on the outside of the shield is common-mode current, called Mantelstrom or mantle current; it can make the feed line radiate as part of the antenna.",
"source": "https://50ohm.de/NE_mantelwellen_1.html",
- "confidence": 7
+ "confidence": 8
},
"EG405": {
- "revision": 1,
- "explanation": "Mantle waves make the coax shield radiate or receive as part of the antenna, which can disturb other devices and worsen the station's own reception.",
+ "revision": 2,
+ "explanation": "Mantle waves (German Mantelwellen, common-mode currents on the coax outer braid) make the feed line itself radiate and receive as part of the antenna. That stray radiation can disturb nearby devices (EMC) and let household noise back into the receiver, worsening your own reception.",
"source": "https://50ohm.de/NE_mantelwellen_1.html",
"confidence": 8
},
"EG406": {
- "revision": 1,
- "explanation": "A balanced dipole fed directly with unbalanced coax can drive common-mode current on the shield, distorting the radiation pattern and creating mantle waves.",
+ "revision": 2,
+ "explanation": "A dipole is a symmetric (balanced) load, but coax is asymmetric (unbalanced). Feeding one directly with the other unbalances the currents, driving a common-mode current (Mantelwelle) on the braid. That braid current distorts the radiation pattern and radiates where it should not — which is exactly why a balun belongs at the feed point.",
"source": "https://50ohm.de/NE_mantelwellen_1.html",
"confidence": 8
},
"EG407": {
- "revision": 2,
- "explanation": "A balun connects a balanced antenna such as a dipole to an unbalanced feed line such as coax while suppressing common-mode current.",
+ "revision": 3,
+ "explanation": "A balun (balanced-unbalanced transformer, Symmetrierglied) joins a balanced antenna such as a dipole to an unbalanced feeder such as coax. By forcing equal and opposite currents at the feed point it suppresses common-mode current (Mantelwellen) on the braid.",
"source": "https://50ohm.de/NE_mantelwellen_1.html",
"confidence": 8
},
"EG408": {
- "revision": 2,
- "explanation": "Coax turns on a ferrite core form a common-mode choke, increasing impedance for mantle currents and therefore damping mantle waves.",
+ "revision": 3,
+ "explanation": "Coax turns on a ferrite core form a common-mode choke. The wanted differential-mode signal still travels inside the coax, but common-mode mantle current on the shield outside sees high impedance and is strongly reduced.",
"source": "https://50ohm.de/NE_mantelwellen_1.html",
- "confidence": 7
+ "confidence": 8
},
"EG501": {
- "revision": 2,
- "explanation": "EIRP is antenna input power multiplied by antenna gain in the chosen direction, with the gain referenced to an isotropic radiator.",
+ "revision": 3,
+ "explanation": "EIRP (equivalent isotropically radiated power) is the power actually fed to the antenna multiplied by the antenna's gain in the chosen direction, with that gain referenced to an isotropic radiator. The traps: it uses average (not peak-envelope) power, and the reference is isotropic — referencing a dipole instead would give ERP.",
"source": "https://life.itu.int/radioclub/rr/art1.pdf",
"confidence": 9
},
"EG502": {
- "revision": 2,
- "explanation": "First subtract losses from transmitter power to get power at the antenna, then multiply by antenna gain referenced to an isotropic radiator.",
+ "revision": 3,
+ "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).",
"source": "https://50ohm.de/NE_aequivalente_isotrope_strahlungsleistung_eirp_1.html",
"confidence": 8
},
"EG503": {
- "revision": 2,
- "explanation": "26 dBi is a gain factor of about 10^2.6 = 398; 0.25 W times 398 is about 100 W EIRP.",
+ "revision": 3,
+ "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 $= \\times 100 \\cdot \\times 4 = \\times 400$.",
"source": "https://50ohm.de/NE_aequivalente_isotrope_strahlungsleistung_eirp_1.html",
"confidence": 8
},
"EG504": {
- "revision": 2,
- "explanation": "36 dBi is a gain factor of about 10^3.6 = 3981; 5 W times 3981 is about 20000 W EIRP.",
+ "revision": 3,
+ "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.",
"source": "https://50ohm.de/NE_aequivalente_isotrope_strahlungsleistung_eirp_1.html",
"confidence": 8
},
"EG505": {
- "revision": 2,
- "explanation": "The net isotropic gain is 11 dBi - 1 dB = 10 dB, a factor of 10, so 100 W becomes 1000 W EIRP.",
+ "revision": 3,
+ "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.",
"source": "https://50ohm.de/NE_aequivalente_isotrope_strahlungsleistung_eirp_1.html",
"confidence": 8
},
"EG506": {
- "revision": 2,
- "explanation": "A dipole has 2.15 dBi gain, factor 1.64, and the cable loss is also factor 1.64; they cancel, leaving 75 W EIRP.",
+ "revision": 3,
+ "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.",
"source": "https://50ohm.de/NE_aequivalente_isotrope_strahlungsleistung_eirp_1.html",
"confidence": 8
},
"EG507": {
- "revision": 2,
- "explanation": "10 dB cable loss reduces 100 W to 10 W at the dipole; dipole gain is factor 1.64 relative to isotropic, giving 16.4 W EIRP.",
+ "revision": 3,
+ "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.",
"source": "https://50ohm.de/NE_aequivalente_isotrope_strahlungsleistung_eirp_1.html",
"confidence": 8
},
"EG508": {
- "revision": 2,
- "explanation": "5 dBd equals 7.15 dBi; after 2 dB cable loss the net gain is 5.15 dB, factor about 3.28, so 5 W becomes 16.4 W EIRP.",
+ "revision": 3,
+ "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.",
"source": "https://50ohm.de/NE_aequivalente_isotrope_strahlungsleistung_eirp_1.html",
"confidence": 8
},
"EG509": {
- "revision": 2,
- "explanation": "11 dBd equals 13.15 dBi; minus 1 dB cable loss gives 12.15 dB, factor about 16.4, and 0.6 W times that is about 9.8 W.",
+ "revision": 3,
+ "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.",
"source": "https://50ohm.de/NE_aequivalente_isotrope_strahlungsleistung_eirp_1.html",
"confidence": 8
},
"EG510": {
- "revision": 2,
- "explanation": "0 dBd equals 2.15 dBi; after 1.5 dB cable loss the net gain is 0.65 dB, factor about 1.17, so 8.5 W becomes about 9.9 W.",
+ "revision": 3,
+ "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.",
"source": "https://50ohm.de/NE_aequivalente_isotrope_strahlungsleistung_eirp_1.html",
"confidence": 8
},
"EG511": {
- "revision": 1,
- "explanation": "BEMFV notification starts at 10 W EIRP. A 5.15 dBi antenna has factor about 3.28, so transmitter power must be at most about 10 / 3.28 = 3 W.",
+ "revision": 2,
+ "explanation": "The BEMFV $\\S\\,9$ notification (Anzeige) for a fixed station is required once radiated power reaches $10$ W EIRP, so stay at or below it. Convert the gain: $5.15\\ \\text{dBi} \\to 10^{0.515} \\approx 3.28$. Working back, $P_{\\text{max}} = 10\\ \\text{W} / 3.28 \\approx 3$ W of transmitter power keeps you under the limit.",
"source": "https://www.gesetze-im-internet.de/bemfv/__9.html",
"confidence": 9
},
"EH101": {
- "revision": 1,
- "explanation": "HF long-distance propagation uses sky waves refracted by ionized, electrically charged regions of the ionosphere.",
+ "revision": 2,
+ "explanation": "HF long-distance propagation mainly uses sky waves, German Raumwellen. Ionized layers of the ionosphere contain free electrons that gradually bend/refract HF waves back toward Earth, enabling paths far beyond line of sight.",
"source": "https://50ohm.de/E_ionosphaere_2.html",
"confidence": 8
},
"EH102": {
- "revision": 1,
- "explanation": "The important HF DX regions are mainly the F regions, which lie roughly from 130 km up to about 450 km altitude.",
+ "revision": 2,
+ "explanation": "For HF DX, the important ionospheric regions are mainly the F regions, roughly 130 km to 450 km high. They are high enough that a refracted ray returns hundreds or thousands of kilometres away, while the lower D and E regions mainly affect absorption or shorter hops.",
"source": "https://50ohm.de/E_ionosphaere_2.html",
"confidence": 8
},
"EH103": {
- "revision": 1,
- "explanation": "The F2 region persists high in the ionosphere and is the main refracting region for long-distance HF communication.",
+ "revision": 2,
+ "explanation": "The F2 region is the highest and most persistent ionospheric region used for normal HF DX. Because it is high and often strongly ionized, it can bend HF waves back over long distances and supports multi-hop worldwide propagation.",
"source": "https://50ohm.de/E_ionosphaere_2.html",
"confidence": 8
},
"EH104": {
- "revision": 1,
- "explanation": "At night the D region absorption largely disappears, and 80 m DX is then mainly enabled by F2-region refraction.",
+ "revision": 2,
+ "explanation": "On 80 m, daytime sky waves are often absorbed before they reach the useful refracting layer. At night the D region largely recombines and absorption drops, so the signal can reach the F2 region and return as DX.",
"source": "https://50ohm.de/E_ionosphaere_2.html",
"confidence": 8
},
"EH105": {
- "revision": 1,
- "explanation": "The D region is strongly ionized by daylight and absorbs lower HF, especially 80 m and 160 m, causing strong daytime attenuation.",
+ "revision": 2,
+ "explanation": "The D region is the lowest ionospheric region and is produced mainly by daylight. It is not useful as a DX reflector; instead collisions in this dense lower atmosphere turn RF energy into heat, absorbing low HF especially on 160 m and 80 m during the day.",
"source": "https://50ohm.de/E_ionosphaere_2.html",
"confidence": 8
},
"EH106": {
- "revision": 1,
- "explanation": "Sporadic-E occurs as unusually ionized patches in the E region and can support upper-HF to VHF propagation in summer.",
+ "revision": 2,
+ "explanation": "Sporadic-E is not the normal smooth E layer. It is patchy, dense ionization in the E region, often seasonal, that can refract much higher frequencies than usual, producing surprisingly short-to-medium paths on 10 m and sometimes VHF.",
"source": "https://50ohm.de/E_ionosphaere_2.html",
"confidence": 8
},
"EH107": {
- "revision": 1,
- "explanation": "Solar activity follows the sunspot cycle, whose average period is about 11 years.",
+ "revision": 2,
+ "explanation": "Solar activity follows the sunspot cycle, averaging about 11 years. More sunspots generally mean more solar UV/X-ray radiation, stronger ionization of the upper ionosphere, and better higher-HF propagation.",
"source": "https://50ohm.de/E_ionosphaere_2.html",
"confidence": 8
},
"EH201": {
- "revision": 1,
- "explanation": "The dead zone is between the end of ground-wave coverage and the first point where the sky wave returns to earth.",
+ "revision": 2,
+ "explanation": "The dead zone, German tote Zone, is a geometry problem: the ground wave (Bodenwelle) has already become too weak, but the first sky-wave hop (Raumwelle) has not yet returned to Earth. Stations in that gap hear little or nothing even though stations farther away may be readable.",
"source": "https://50ohm.de/E_tote_zone_1.html",
"confidence": 8
},
"EH202": {
- "revision": 1,
- "explanation": "Where ground wave and sky wave overlap, phase differences can make the received field strength vary, producing fading.",
+ "revision": 2,
+ "explanation": "Where ground wave (Bodenwelle) and sky wave (Raumwelle) both reach the receiver, they are two versions of the same signal with different path lengths and phases. Their vector sum changes with ionospheric motion and frequency, causing reinforcement or cancellation: fading.",
"source": "https://50ohm.de/NE_fading.html",
"confidence": 8
},
"EH203": {
- "revision": 1,
- "explanation": "Signal weakening from overlap and interference of ground and sky waves is called fading.",
+ "revision": 2,
+ "explanation": "Fading is signal-strength variation caused by multiple received paths adding with changing phase. If ground wave and sky wave overlap, they can sometimes reinforce and sometimes cancel, making the signal rise and fall.",
"source": "https://50ohm.de/NE_fading.html",
"confidence": 8
},
"EH204": {
- "revision": 1,
- "explanation": "MUF means Maximum Usable Frequency, the highest frequency still refracted back for the wanted path.",
+ "revision": 2,
+ "explanation": "MUF means Maximum Usable Frequency. It is the highest frequency that the ionosphere can still refract back to Earth for a specific path; above the MUF the wave escapes instead of returning.",
"source": "https://50ohm.de/NE_muf_luf_1.html",
"confidence": 8
},
"EH205": {
- "revision": 1,
- "explanation": "At sunspot maximum solar UV and X-ray output are high, increasing ionization especially in the F region.",
+ "revision": 2,
+ "explanation": "At sunspot maximum, the Sun emits more UV and X-ray radiation that ionizes the upper atmosphere. The F region then has more free electrons, so it can refract higher HF frequencies that would otherwise pass through into space.",
"source": "https://50ohm.de/E_ionosphaere_2.html",
"confidence": 8
},
"EH206": {
- "revision": 1,
- "explanation": "More free electrons in the F2 region allow higher frequencies to be refracted back, so the MUF rises.",
+ "revision": 2,
+ "explanation": "The MUF rises when the F2 region has a higher electron density. Physically, more free electrons change the refractive index more strongly, so even shorter-wavelength, higher-frequency HF waves can be bent back to Earth.",
"source": "https://50ohm.de/NE_muf_luf_1.html",
"confidence": 8
},
"EH207": {
- "revision": 1,
- "explanation": "To use frequencies above the current MUF, the refracting region needs stronger ionization so it can bend those higher frequencies back.",
+ "revision": 2,
+ "explanation": "If you want to use a frequency above the current MUF, the ionosphere must become more ionized or the path geometry must improve. Otherwise the wave is not bent enough and continues through the F region instead of returning.",
"source": "https://50ohm.de/NE_muf_luf_1.html",
"confidence": 8
},
"EH208": {
- "revision": 1,
- "explanation": "Skip distance depends strongly on takeoff angle: a flatter radiation angle produces a longer hop, while a steeper angle returns sooner.",
+ "revision": 2,
+ "explanation": "Skip distance, German Sprungdistanz, is mainly controlled by launch angle and layer height. A low takeoff angle hits the ionosphere far away and returns far away; a steep angle goes up and comes back sooner. This is why antenna elevation pattern matters for DX.",
"source": "https://50ohm.de/E_sprungdistanz_1.html",
"confidence": 8
},
"EH209": {
- "revision": 1,
- "explanation": "LUF is limited mainly by absorption, and lower HF absorption is controlled by the ionization level of the D region.",
+ "revision": 2,
+ "explanation": "LUF means Lowest Usable Frequency, but it is not set by refraction like MUF. It is set mainly by absorption: if D-region loss is too high, lower-frequency signals are absorbed before completing the path, so the usable frequency must be raised.",
"source": "https://50ohm.de/NE_muf_luf_1.html",
"confidence": 8
},
"EH210": {
- "revision": 1,
- "explanation": "During the day the D region absorbs low HF strongly, so 160 m and 80 m sky-wave signals are weak for worldwide communication.",
+ "revision": 2,
+ "explanation": "During daylight, the D region absorbs low-HF sky waves strongly. That makes 160 m and 80 m poor for daytime worldwide sky-wave communication even though those bands can be excellent after dark when D-region absorption collapses.",
"source": "https://50ohm.de/E_ionosphaere_2.html",
"confidence": 8
},
"EH211": {
- "revision": 1,
- "explanation": "On 160 m in daytime, D-region absorption prevents useful sky-wave propagation, so propagation is mainly by ground wave.",
+ "revision": 2,
+ "explanation": "On 160 m in daytime, the ionosphere is often more of an absorber than a reflector. The D region eats the sky wave, so useful coverage is mainly by ground wave along the Earth's surface, usually over much shorter ranges.",
"source": "https://50ohm.de/E_bodenwelle.html",
"confidence": 8
},
"EH212": {
- "revision": 1,
- "explanation": "HF ground waves follow the earth beyond the optical horizon, but their attenuation increases at higher frequencies.",
+ "revision": 2,
+ "explanation": "A ground wave, German Bodenwelle, travels along the Earth's surface rather than being refracted by the ionosphere. It can reach beyond the optical horizon, especially at lower frequencies, but ground losses increase with frequency so higher-HF ground-wave range is limited.",
"source": "https://50ohm.de/E_bodenwelle.html",
"confidence": 8
},
"EH213": {
- "revision": 1,
- "explanation": "The greyline is the twilight zone near sunrise and sunset where D-region absorption is reduced while higher-layer refraction can remain useful.",
+ "revision": 2,
+ "explanation": "Greyline propagation uses the sunrise/sunset transition. The D region, which causes much HF absorption, weakens quickly in twilight, while the F region can remain ionized enough to refract signals; that combination can briefly improve DX paths.",
"source": "https://50ohm.de/NE_greyline.html",
"confidence": 8
},
"EH214": {
- "revision": 1,
- "explanation": "A solar flare can abruptly increase D-region ionization and absorb HF sky waves; this shortwave fadeout is the Moegel-Dellinger effect.",
+ "revision": 2,
+ "explanation": "A strong solar flare sends extra X-ray radiation to the sunlit side of Earth, abruptly increasing D-region ionization. Because the D region is absorptive for HF, sky-wave signals can disappear suddenly: the Moegel-Dellinger shortwave fadeout.",
"source": "https://50ohm.de/E_moegel_dellinger_effekt.html",
"confidence": 8
},
"EH215": {
- "revision": 1,
- "explanation": "The Moegel-Dellinger effect causes a temporary loss or severe impairment of HF sky-wave propagation.",
+ "revision": 2,
+ "explanation": "The Moegel-Dellinger effect is a shortwave fadeout after a strong solar flare. Extra X-ray radiation suddenly increases D-region ionization, causing heavy HF absorption and temporary loss or severe impairment of sky-wave propagation.",
"source": "https://50ohm.de/E_moegel_dellinger_effekt.html",
"confidence": 8
},
"EH216": {
- "revision": 1,
- "explanation": "Long path means the signal travels in the direction opposite the shortest bearing to the other station, around the longer side of the earth.",
+ "revision": 2,
+ "explanation": "Short path is the smaller great-circle route to the other station; long path is the same great circle in the opposite direction around the larger side of Earth. Long path can work when illumination, greyline, or ionospheric conditions are better that way.",
"source": "https://50ohm.de/EA_langer_kurzer_weg_1.html",
"confidence": 8
},
"EH217": {
- "revision": 1,
- "explanation": "For Germany to VK, long path points away from the direct route and reaches Australia via the opposite direction, over South America.",
+ "revision": 2,
+ "explanation": "VK is Australia. From Germany, the short path points generally east/southeast; the long path is the opposite great-circle direction, around the other side of Earth, so the route goes roughly over South America before reaching Australia.",
"source": "https://50ohm.de/EA_langer_kurzer_weg_1.html",
"confidence": 8
},
"EH218": {
- "revision": 1,
- "explanation": "Short-skip paths under 1000 km on 10 m are produced by refraction in localized sporadic-E ionization patches.",
+ "revision": 2,
+ "explanation": "A normal F-layer 10 m hop is usually much longer. Short-skip contacts under 1000 km on 10 m point to sporadic-E, where a lower E-region ionization patch returns the signal after a shorter geometric path.",
"source": "https://50ohm.de/NE_sporadic_e_2.html",
"confidence": 8
},
"EH219": {
- "revision": 1,
- "explanation": "At sunspot maximum the F region is strongly ionized, so the 10 m band can support worldwide daytime contacts even with low power.",
+ "revision": 2,
+ "explanation": "The 10 m band needs a high MUF for normal worldwide F-layer propagation. Around sunspot maximum the F region is strongly ionized, so MUF often rises above 28 MHz and even low-power daytime DX can become possible.",
"source": "https://50ohm.de/E_ionosphaere_2.html",
"confidence": 8
},
"EH301": {
- "revision": 1,
- "explanation": "The troposphere is the lower atmospheric layer where weather processes occur.",
+ "revision": 2,
+ "explanation": "The troposphere is the lowest atmospheric layer, where weather, temperature gradients, and humidity changes occur. For VHF/UHF radio this matters because refractive-index changes in this layer can bend, duct, or scatter signals.",
"source": "https://50ohm.de/NE_troposphaere_2.html",
"confidence": 8
},
"EH302": {
- "revision": 1,
- "explanation": "VHF/UHF over-horizon propagation can occur when waves are bent, reflected, or scattered by tropospheric regions with different temperature and density.",
+ "revision": 2,
+ "explanation": "VHF/UHF normally behaves close to line-of-sight, but the troposphere can extend range. Temperature, humidity, and pressure gradients change refractive index, so signals may bend, scatter, or get trapped in ducts beyond the optical horizon.",
"source": "https://50ohm.de/NE_troposphaere_2.html",
"confidence": 8
},
"EH303": {
- "revision": 1,
- "explanation": "VHF long-distance contacts mainly use tropospheric propagation effects rather than HF-style ionospheric sky-wave propagation.",
+ "revision": 2,
+ "explanation": "VHF long-distance contacts usually rely on tropospheric effects, not normal HF-style F-layer sky-wave propagation. Temperature inversions, ducts, and scatter in the lower atmosphere can bend or guide VHF/UHF signals beyond line of sight.",
"source": "https://50ohm.de/NE_troposphaere_2.html",
"confidence": 8
},
"EH304": {
- "revision": 1,
- "explanation": "Sporadic-E is refraction by locally limited, unusually highly ionized regions inside the E layer.",
+ "revision": 2,
+ "explanation": "Sporadic-E propagation comes from patchy, unusually dense ionization clouds in the E layer. These localized regions can refract upper-HF and sometimes VHF signals back to Earth over medium-distance paths.",
"source": "https://50ohm.de/NE_sporadic_e_2.html",
"confidence": 8
},
"EH305": {
- "revision": 2,
- "explanation": "Aurora makes CW tone quality rough and unstable, so the report uses R and S plus A for Aurora instead of a normal tone rating.",
+ "revision": 3,
+ "explanation": "Auroral propagation scatters VHF signals from disturbed, moving ionized regions near the polar aurora. The rapid motion spreads and roughens the CW tone, so reports use readability and strength plus A for Aurora instead of a normal tone-quality number.",
"source": "https://50ohm.de/E_slide_e_wellenausbreitung.html",
"confidence": 8
},
"EI101": {
- "revision": 1,
- "explanation": "Voltage is measured across a component, so the meter is connected in parallel; high input resistance prevents the meter from loading the circuit.",
+ "revision": 2,
+ "explanation": "A voltmeter measures the voltage across a component, so it is wired in parallel with it. To avoid drawing significant current and disturbing the circuit, it must be high-impedance (hochohmig) — ideally infinite. (An ammeter is the opposite: in series and low-impedance.)",
"source": "https://50ohm.de/E_strom_spannung_messung_2.html",
"confidence": 8
},
"EI102": {
- "revision": 1,
- "explanation": "To use Ohm's law for a resistor, current must be measured in series through it and voltage in parallel across it.",
+ "revision": 2,
+ "explanation": "To verify Ohm's law, measure the same current that flows through the resistor and the voltage across that resistor. Therefore the ammeter goes in series with the resistor, while the voltmeter goes in parallel across it; then $R = U/I$ is meaningful.",
"source": "https://50ohm.de/E_strom_spannung_messung_2.html",
- "confidence": 7
+ "confidence": 8
},
"EI103": {
- "revision": 1,
- "explanation": "The pointer is at 29 percent of full scale; on the 10 V range that is 0.29 x 10 V = 2.9 V.",
+ "revision": 2,
+ "explanation": "For an analog meter, first read the pointer as a fraction of full scale, then multiply by the selected range. The pointer is at about 29 percent of full scale, so on the 10 V range the value is $0.29 \\cdot 10 V = 2.9 V$.",
"source": "https://50ohm.de/NEA_zeigerinstrumente_ablesen.html",
- "confidence": 7
+ "confidence": 8
},
"EI104": {
- "revision": 1,
- "explanation": "On the 300 V range the same pointer position corresponds to about 29 percent of full scale, so 0.29 x 300 V is about 88 V.",
+ "revision": 2,
+ "explanation": "The pointer fraction is independent of the selected range. With the pointer at about 29 percent of full scale and the range set to 300 V, the reading is $0.29 \\cdot 300 V \\approx 88 V$.",
"source": "https://50ohm.de/NEA_zeigerinstrumente_ablesen.html",
- "confidence": 7
+ "confidence": 8
},
"EI201": {
- "revision": 1,
- "explanation": "A VNA measures frequency-dependent impedance and reflection behavior, so it is suited to finding resonances and impedances of tuned circuits and antennas.",
+ "revision": 2,
+ "explanation": "A vector network analyser measures complex (magnitude and phase) impedance and reflection versus frequency, so it directly reveals resonant frequencies and feed-point impedances of tuned circuits and antennas. It is not a time-domain scope, a spectral-purity checker, or an earth-resistance meter.",
"source": "https://50ohm.de/NEA_vna_1.html",
"confidence": 8
},
"EI202": {
- "revision": 1,
- "explanation": "Resonance can be calculated from measured L and C or found directly by sweeping the circuit with a VNA.",
+ "revision": 2,
+ "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.",
"source": "https://50ohm.de/NEA_vna_1.html",
"confidence": 8
},
"EI203": {
- "revision": 1,
- "explanation": "A vector network analyzer directly measures complex impedance, including resistance, reactance, and reflection/SWR quantities.",
+ "revision": 2,
+ "explanation": "A vector network analyzer (VNA) measures magnitude and phase, so it can derive complex impedance, not just signal level. That includes resistance, reactance, reflection coefficient, return loss, and SWR over frequency.",
"source": "https://50ohm.de/NEA_vna_1.html",
"confidence": 8
},
"EI204": {
- "revision": 1,
- "explanation": "Impedance measurement is a core VNA use because it compares voltage/current or incident/reflected waves over frequency.",
+ "revision": 2,
+ "explanation": "Measuring impedance is a core VNA function: it excites the device over frequency and compares incident and reflected waves to derive the complex impedance. Data rates, transmit power, and harmonics are jobs for other instruments.",
"source": "https://50ohm.de/NEA_vna_1.html",
"confidence": 8
},
"EI205": {
- "revision": 1,
- "explanation": "A VNA must be calibrated with the measurement setup so its reference plane and systematic errors are corrected before use.",
+ "revision": 2,
+ "explanation": "A VNA must be calibrated at the intended reference plane, usually with open/short/load standards. Calibration removes systematic errors from cables, adapters, and the instrument so the displayed impedance belongs to the device under test.",
"source": "https://50ohm.de/NEA_vna_1.html",
"confidence": 8
},
"EI206": {
- "revision": 1,
- "explanation": "Open and short should reflect almost all power, giving very high SWR, while a matched load should show SWR near 1.",
+ "revision": 2,
+ "explanation": "Sanity-check the VNA with the three standard terminations: a matched load should read SWR near 1 (almost no reflection), while open circuit and short circuit reflect everything and read SWR toward infinity. Those three known states confirm the instrument and its calibration before you trust antenna readings.",
"source": "https://50ohm.de/NEA_vna_1.html",
"confidence": 8
},
"EI301": {
- "revision": 2,
- "explanation": "The displayed sine period spans 8 divisions; at 0.5 ms per division the period is 8 x 0.5 ms = 4 ms.",
+ "revision": 3,
+ "explanation": "On an oscilloscope, time is read horizontally. One full sine period spans 8 divisions, and the timebase is 0.5 ms/div, so $T = 8 \\cdot 0.5 ms = 4 ms$.",
"source": "https://50ohm.de/NE_oszilloskop_1.html",
- "confidence": 7
+ "confidence": 8
},
"EI302": {
- "revision": 2,
- "explanation": "The period is 4 ms, so frequency is 1 / 0.004 s = 250 Hz.",
+ "revision": 3,
+ "explanation": "Frequency is the reciprocal of period: $f=1/T$. From the oscilloscope reading $T=4 ms = 0.004 s$, so $f=1/0.004 s = 250 Hz$.",
"source": "https://50ohm.de/NE_oszilloskop_1.html",
"confidence": 8
},
"EI303": {
- "revision": 2,
- "explanation": "Pulse duration is read from the middle of the rising edge to the middle of the falling edge; the shown interval is 200 microseconds.",
+ "revision": 3,
+ "explanation": "Pulse duration is conventionally measured between the 50 percent points of the rising and falling edges, not from the first visible slope to the last. Reading those midpoint crossings on the oscilloscope gives 200 microseconds.",
"source": "https://50ohm.de/E_slide_e_strom_spannung_widerstand_leistung_energie.html",
- "confidence": 7
+ "confidence": 8
},
"EI304": {
- "revision": 2,
- "explanation": "Audio distortion changes the waveform shape, and an oscilloscope displays waveform shape directly.",
+ "revision": 3,
+ "explanation": "Audio distortion changes the time-domain waveform: clipping flattens peaks, nonlinear stages bend the sine shape, and hum/noise adds visible components. An oscilloscope displays that waveform directly, so it is the right tool for seeing distortion.",
"source": "https://50ohm.de/E_slide_e_strom_spannung_widerstand_leistung_energie.html",
"confidence": 8
},
"EI401": {
- "revision": 2,
- "explanation": "An SWR meter measures the match between feed line and load, so in transmitter use it indicates antenna-system matching.",
+ "revision": 3,
+ "explanation": "An SWR meter compares forward and reflected power on the feed line, which reveals how well the antenna system is matched (Antennenanpassung). A bad match sends power back; a good match (SWR near 1) does not. It does not measure harmonics, bandwidth, or efficiency.",
"source": "https://50ohm.de/NE_slide_ne_antennen_uebertragungsleitungen.html",
"confidence": 8
},
"EI402": {
- "revision": 2,
- "explanation": "The instrument for showing the match between a UHF transmitter and its feed line is an SWR meter.",
+ "revision": 3,
+ "explanation": "To show the match between a UHF transmitter and its feed line you use an SWR meter, which senses forward versus reflected waves. An ohmmeter cannot measure RF impedance, and the other options are not measuring instruments for this.",
"source": "https://50ohm.de/NE_slide_ne_antennen_uebertragungsleitungen.html",
"confidence": 8
},
"EI403": {
- "revision": 2,
- "explanation": "In transmit operation SWR is measured with an SWR bridge that compares forward and reflected power on the line.",
+ "revision": 3,
+ "explanation": "On transmit, SWR is read with an SWR bridge (SWR-Messbrücke / directional coupler) that samples forward and reflected power while the signal is live. A simple current or voltage reading at the line ends cannot separate forward from reflected waves, and an absorption wavemeter measures frequency, not SWR.",
"source": "https://50ohm.de/NE_slide_ne_antennen_uebertragungsleitungen.html",
"confidence": 8
},
"EI404": {
- "revision": 2,
- "explanation": "To judge the antenna itself, the SWR meter should be as close to the antenna as possible, between antenna cable and antenna.",
+ "revision": 3,
+ "explanation": "Insert the SWR meter as close to the antenna as possible — between antenna cable and antenna — so it reports the antenna's own match. Placed at the transmitter end, cable loss masks the reflected wave and flatters the reading, and a tuner in between would hide the real antenna SWR.",
"source": "https://50ohm.de/NE_slide_ne_antennen_uebertragungsleitungen.html",
"confidence": 8
},
"EI405": {
- "revision": 2,
- "explanation": "To check whether the whole antenna system is well matched to the transmitter, the SWR meter belongs at the transmitter output, point 1.",
+ "revision": 3,
+ "explanation": "If you want to know what match the transmitter actually sees, place the SWR meter at the transmitter output. Then the meter includes the combined effect of feed line, tuner, and antenna as seen by the transmitter; in the diagram that is point 1.",
"source": "https://50ohm.de/NE_slide_ne_antennen_uebertragungsleitungen.html",
- "confidence": 7
+ "confidence": 8
},
"EI501": {
- "revision": 1,
- "explanation": "An unmodulated RF signal has a single stable frequency, which a frequency counter can count directly.",
+ "revision": 2,
+ "explanation": "A frequency counter counts zero crossings or cycles during a gate time. An unmodulated RF carrier has one stable frequency, so the counter can measure it directly; modulated or unstable signals are harder to count cleanly.",
"source": "https://50ohm.de/NE_frequenzmessung_1.html",
"confidence": 8
},
"EI502": {
- "revision": 1,
- "explanation": "The marked digit is in the 10^3 Hz position of the counter display, so its place value is one kilohertz.",
+ "revision": 2,
+ "explanation": "A frequency counter display is read by decimal place value. The marked digit is in the $10^3$ Hz position, so changing that digit changes the reading by 1000 Hz, which is one kilohertz.",
"source": "https://50ohm.de/NE_frequenzmessung_1.html",
- "confidence": 7
+ "confidence": 8
},
"EI503": {
- "revision": 1,
- "explanation": "In this display the marked digit is in the 10 Hz position, so its place value is ten hertz.",
+ "revision": 2,
+ "explanation": "Read the marked counter digit by its decimal place. In this display the marked position is $10^1$ Hz, so one count in that digit corresponds to 10 Hz.",
"source": "https://50ohm.de/NE_frequenzmessung_1.html",
- "confidence": 7
+ "confidence": 8
},
"EI504": {
- "revision": 1,
- "explanation": "A 10:1 prescaler divides the input by 10 before counting, so the real frequency is 10 x 14.5625 MHz = 145.625 MHz.",
+ "revision": 2,
+ "explanation": "A 10:1 prescaler (Frequenzteiler) divides the input frequency by ten before the counter sees it, so the display reads one tenth of the true value. Multiply back: $10 \\cdot 14.5625\\ \\text{MHz} = 145.625$ MHz. Prescalers are used precisely so a slower counter can measure high (VHF) frequencies.",
"source": "https://50ohm.de/NE_frequenzmessung_1.html",
"confidence": 8
},
"EJ101": {
- "revision": 1,
- "explanation": "Conducted RF interference enters equipment through attached leads such as mains, antenna, or speaker cables; that is Einströmung.",
+ "revision": 2,
+ "explanation": "Einströmung (conducted ingress) is RF getting into a device through its attached wires — mains lead, speaker, or signal cables acting as receiving antennas. It is conducted in via cables, distinct from radiated pickup through a poorly shielded case.",
"source": "https://50ohm.de/E_stoerungen_elektronischer_geraete_1.html",
"confidence": 8
},
"EJ102": {
- "revision": 1,
- "explanation": "Radiated RF entering through poor enclosure shielding is Einstrahlung, distinct from conducted entry via cables.",
+ "revision": 2,
+ "explanation": "Einstrahlung means radiated RF coupling directly into equipment, usually through inadequate enclosure shielding or circuit layout. That is different from Einströmung, where RF is conducted into the device through attached leads.",
"source": "https://50ohm.de/E_stoerungen_elektronischer_geraete_1.html",
"confidence": 8
},
"EJ103": {
- "revision": 1,
- "explanation": "Even a clean wanted signal can overload nearby receiver stages or otherwise influence them, so the issue is overload or disturbing influence, not spurious emission.",
+ "revision": 2,
+ "explanation": "A perfectly clean transmitted signal can still overload the front end of a nearby receiver tuned elsewhere, desensitising it or producing spurious responses inside that receiver. This is overload/disturbing influence (Übersteuerung) — the fault is the victim's selectivity, not any unwanted emission from your transmitter.",
"source": "https://50ohm.de/E_stoerungen_elektronischer_geraete_1.html",
"confidence": 8
},
"EJ104": {
- "revision": 1,
- "explanation": "Lower transmitter power lowers field strength and coupling risk, so use only the minimum needed for satisfactory communication.",
+ "revision": 2,
+ "explanation": "Interference scales with the field strength you radiate, so the standard good-practice rule is to use only the minimum power needed for satisfactory communication. Running full legal power 'because it is allowed' needlessly raises the interference probability.",
"source": "https://50ohm.de/E_stoerungen_elektronischer_geraete_1.html",
"confidence": 8
},
"EJ105": {
- "revision": 1,
- "explanation": "In dense residential areas during TV viewing hours, the practical interference-reduction step is to transmit with no more power than needed for reliable communication.",
+ "revision": 2,
+ "explanation": "In a built-up area during peak TV-viewing hours, the same minimum-power principle applies: transmit with no more power than reliable communication requires. Lowering the antenna or insisting on a high-gain beam are not the recommended courtesy measure — reducing power is.",
"source": "https://50ohm.de/NE_stoerungen_elektronischer_geraete_1.html",
"confidence": 8
},
"EJ106": {
- "revision": 1,
- "explanation": "A high-gain 432 MHz antenna pointed at a TV receive antenna can create a very strong local signal and overload the TV receiver input.",
+ "revision": 2,
+ "explanation": "A high-gain $432$ MHz beam aimed straight at a TV antenna delivers a very strong local field into the TV's input, overloading its front end (Übersteuerung). The TV receiver, not your transmitter, is the limiting part — its input cannot handle the huge nearby signal even though your emission is clean.",
"source": "https://50ohm.de/E_stoerungen_elektronischer_geraete_1.html",
"confidence": 8
},
"EJ107": {
- "revision": 1,
- "explanation": "Receiver overload drives input stages out of their normal range, reducing effective sensitivity or even blocking reception.",
+ "revision": 2,
+ "explanation": "Receiver overload happens when a strong signal drives the RF input, mixer, or amplifier outside its linear range. The receiver then creates distortion or desensitization, so effective sensitivity drops or reception can be blocked.",
"source": "https://50ohm.de/E_stoerungen_elektronischer_geraete_1.html",
"confidence": 8
},
"EJ108": {
- "revision": 1,
- "explanation": "A nearly closed metal enclosure provides RF shielding by enclosing the circuitry in a conductive shell.",
+ "revision": 2,
+ "explanation": "A conductive metal enclosure acts as RF shielding by giving induced RF currents a path around the electronics instead of through them. It must be nearly closed and well bonded, because gaps, long leads, and slots can leak RF.",
"source": "https://50ohm.de/E_stoerungen_elektronischer_geraete_1.html",
"confidence": 8
},
"EJ109": {
- "revision": 1,
- "explanation": "A parallel nearby HF antenna can inductively or capacitively couple RF current into the 230 V mains wiring.",
+ "revision": 2,
+ "explanation": "An HF antenna running close and parallel to a $230$ V mains line couples to it like a transformer/capacitor, inducing RF currents into the power wiring. From there the RF spreads and can disturb other devices. The effect is RF ingress into the mains, not mains overvoltage or harmonics.",
"source": "https://50ohm.de/E_stoerungen_elektronischer_geraete_1.html",
"confidence": 8
},
"EJ110": {
- "revision": 1,
- "explanation": "Running the 80 m wire at right angles to the row of houses avoids long parallel coupling to building wiring and neighboring installations.",
+ "revision": 2,
+ "explanation": "In a terraced house, run the $80$ m wire at right angles to the row so it does not lie long and parallel to the neighbours' wiring and antennas — parallel runs maximise coupling, perpendicular minimises it. Placing it by the shared chimney next to the TV antenna would do the opposite.",
"source": "https://50ohm.de/E_standortwahl.html",
"confidence": 8
},
"EJ111": {
- "revision": 1,
- "explanation": "A separate RF earth for transmitting antennas helps keep RF currents out of house wiring and therefore lowers in-house interference risk.",
+ "revision": 2,
+ "explanation": "Give transmitting antennas their own separate RF earth so antenna return currents flow there rather than through the house wiring and protective-earth network. Keeping RF out of the building's wiring is what lowers in-house interference; a water pipe or the mains PE is not a proper RF earth.",
"source": "https://50ohm.de/E_stoerungen_elektronischer_geraete_1.html",
"confidence": 8
},
"EJ112": {
- "revision": 1,
- "explanation": "LED lamps with mains-connected electronics can be susceptible to RF influence, unlike simple thermal or motor loads in the alternatives.",
+ "revision": 2,
+ "explanation": "The susceptible device is the mains-powered LED lamp: its built-in switch-mode electronics can rectify and respond to strong RF. The purely thermal iron (bimetal) and the plain commutator/AC motors have no sensitive electronics to be disturbed.",
"source": "https://50ohm.de/E_stoerungen_elektronischer_geraete_1.html",
"confidence": 8
},
"EJ113": {
- "revision": 1,
- "explanation": "Strong RF can be rectified by nonlinear semiconductor junctions in an audio power stage, producing audible noise even when the stereo is nominally off.",
+ "revision": 2,
+ "explanation": "Strong RF picked up on speaker or mains leads can be rectified at non-linear semiconductor junctions in the audio output stage, turning your modulation into audible sound even with the set switched off. Demodulation by an unintended detector is the classic audio-rectification (BCI) mechanism.",
"source": "https://50ohm.de/E_stoerungen_elektronischer_geraete_1.html",
"confidence": 8
},
"EJ114": {
- "revision": 1,
- "explanation": "If RF is entering the audio power stage through speaker leads, shielding those leads reduces the conducted RF path.",
+ "revision": 2,
+ "explanation": "The RF is entering through the speaker leads into the audio power stage, so screen that path: shielded speaker cable (plus a common-mode choke) blocks the conducted RF. Filtering the coax or the receiver mains lead would not help, because those are not the entry route here.",
"source": "https://50ohm.de/E_slide_e_sender.html?print-pdf=&showNotes=true",
"confidence": 8
},
"EJ115": {
- "revision": 1,
- "explanation": "A shielded intercom cable reduces RF pickup on the wiring that otherwise conducts the interfering signal into the door-phone electronics.",
+ "revision": 2,
+ "explanation": "The interfering RF is conducted in on the intercom's wiring, so a shielded connecting cable (ideally with a ferrite choke) reduces the pickup that carries RF into the door-phone electronics. Changing cable length or conductor material does not address the coupling.",
"source": "https://50ohm.de/E_slide_e_sender.html?print-pdf=&showNotes=true",
"confidence": 8
},
"EJ116": {
- "revision": 1,
- "explanation": "A DVB-T2 input should pass UHF TV frequencies while rejecting the much lower 28 MHz amateur signal, so a high-pass filter is appropriate.",
+ "revision": 2,
+ "explanation": "The wanted DVB-T2 signals are UHF, while the interferer is the much lower $28$ MHz amateur signal. A high-pass filter at the TV antenna input passes the high UHF channels but strongly attenuates $28$ MHz, removing the overload. A low-pass would do the reverse.",
"source": "https://50ohm.de/E_stoerungen_elektronischer_geraete_1.html",
"confidence": 8
},
"EJ117": {
- "revision": 1,
- "explanation": "For HF interference in a TV antenna lead, use the high-pass filter: it rejects low HF while passing the higher TV bands.",
+ "revision": 2,
+ "explanation": "TV antenna signals are at much higher frequencies than HF amateur signals. A high-pass filter passes the wanted TV band but rejects lower-frequency HF energy conducted on the antenna lead, so it reduces the interference path.",
"source": "https://50ohm.de/E_stoerungen_elektronischer_geraete_1.html",
- "confidence": 7
+ "confidence": 8
},
"EJ118": {
- "revision": 1,
- "explanation": "A mantle-wave choke raises impedance for common-mode RF on the outside of the coax shield, suppressing those RF interference currents.",
+ "revision": 2,
+ "explanation": "A common-mode (mantle-wave) choke (Mantelwellendrossel) on the coax presents a high impedance only to common-mode RF current flowing on the outside of the braid, choking it off, while the wanted differential signal inside passes freely. It targets RF common-mode interference, not mains hum or low-frequency noise.",
"source": "https://50ohm.de/E_slide_e_sender.html?print-pdf=&showNotes=true",
"confidence": 8
},
"EJ119": {
- "revision": 1,
- "explanation": "If 144 MHz RF is induced as common-mode current on the broadcast receiver coax, a mantle-wave choke before the receiver reduces that current.",
+ "revision": 2,
+ "explanation": "The $144$ MHz energy is riding as common-mode current on the outside of the broadcast receiver's coax. A common-mode choke (Mantelwellendrossel) fitted just before the receiver raises the impedance to that current and suppresses it — without disconnecting the screen or the transmitter earth, which would be unsafe and ineffective.",
"source": "https://50ohm.de/E_slide_e_sender.html?print-pdf=&showNotes=true",
"confidence": 8
},
"EJ120": {
- "revision": 1,
- "explanation": "Intermodulation creates phantom signals from two or more strong signals; removing one participating signal removes the product.",
+ "revision": 2,
+ "explanation": "Intermodulation occurs in a nonlinear device when two or more strong signals mix and create new frequencies such as $2f_1-f_2$ or $f_1+f_2$. If one participating signal is removed, that particular phantom product disappears.",
"source": "https://50ohm.de/E_slide_e_sender.html?print-pdf=&showNotes=true",
"confidence": 8
},
"EJ121": {
- "revision": 1,
- "explanation": "Corroded metal contacts are nonlinear and can rectify or mix nearby transmitter signals, creating unwanted products that disturb TV reception.",
+ "revision": 2,
+ "explanation": "A corroded joint forms a non-linear (rectifying) metal contact. Combined with a strong nearby transmitter signal it acts as an unintended mixer/diode, generating intermodulation products that fall on and disturb TV channels — the classic 'rusty-bolt effect'. The fix is to remake the corroded connection.",
"source": "https://50ohm.de/E_stoerungen_elektronischer_geraete_1.html",
"confidence": 8
},
"EJ122": {
- "revision": 1,
- "explanation": "The first useful step is to check whether the disturbance actually coincides in time with your transmissions.",
+ "revision": 2,
+ "explanation": "Before changing equipment or blaming a transmitter, establish correlation. If the disturbance appears only during your transmissions and disappears when you stop, you have useful evidence; if not, the source may be unrelated.",
"source": "https://50ohm.de/E_stoerungen_elektronischer_geraete_1.html",
"confidence": 8
},
"EJ123": {
- "revision": 2,
- "explanation": "An indoor antenna sits inside the noisy local RF environment and gets little wanted signal; an outdoor antenna can be placed away from the interferer and offers more gain and directivity, improving the wanted-to-interferer ratio.",
+ "revision": 3,
+ "explanation": "An indoor antenna sits in the neighbour's noisy near field and captures little wanted signal, so the interference-to-signal ratio is poor. An outdoor antenna can be sited away from the interferer, higher and directional, raising wanted signal and lowering pickup of the local transmitter — the most effective remedy here.",
"source": "https://50ohm.de/E_slide_e_sender.html?print-pdf=&showNotes=true",
"confidence": 8
},
"EJ124": {
- "revision": 1,
- "explanation": "After cooperative mitigation attempts fail, the proper next step is to ask the responsible Bundesnetzagentur field office to examine the situation.",
+ "revision": 2,
+ "explanation": "Once cooperative, technically reasonable mitigation has genuinely failed, the correct next step is to ask the responsible BNetzA field office (Außenstelle der Bundesnetzagentur) to investigate. Opening and earthing the neighbour's TV yourself is neither permitted nor a proper diagnosis.",
"source": "https://50ohm.de/E_stoerungen_elektronischer_geraete_1.html",
"confidence": 8
},
"EJ201": {
- "revision": 1,
- "explanation": "A pure sine wave contains only one frequency component; non-sinusoidal carriers contain harmonics that can cause interference.",
+ "revision": 2,
+ "explanation": "A pure sine wave contains only its single fundamental frequency; any distorted (square, triangle) waveform is mathematically a fundamental plus harmonics (Fourier). To avoid radiating harmonics, the carrier should therefore be as sinusoidal as possible — clean shape means clean spectrum.",
"source": "https://50ohm.de/NE_unerwuenschte_aussendungen_2.html",
"confidence": 8
},
"EJ202": {
- "revision": 1,
- "explanation": "Harmonics are unwanted multiples of the wanted RF frequency, so an harmonic filter is used to reduce them.",
+ "revision": 2,
+ "explanation": "Harmonics (Oberwellen) are unwanted whole-number multiples of the operating frequency, all above it. A harmonic filter (a low-pass, the 'Oberwellenfilter') removes them while passing the fundamental. An IF or adjacent-channel filter addresses different problems.",
"source": "https://50ohm.de/NE_unerwuenschte_aussendungen_2.html",
"confidence": 8
},
"EJ203": {
- "revision": 1,
- "explanation": "A low-pass filter passes the wanted fundamental output while attenuating higher-frequency harmonics.",
+ "revision": 2,
+ "explanation": "Harmonics are multiples above the wanted transmit frequency, such as $2f$ and $3f$. A low-pass filter passes the fundamental output while attenuating those higher-frequency harmonic components before they reach the antenna.",
"source": "https://50ohm.de/NE_unerwuenschte_aussendungen_2.html",
"confidence": 8
},
"EJ204": {
- "revision": 1,
- "explanation": "Between transmitter and antenna, a low-pass filter is best for reducing harmonic radiation above the operating frequency.",
+ "revision": 2,
+ "explanation": "Harmonics lie at $2f$, $3f$, … above the operating frequency, so a low-pass filter between transmitter and antenna passes the fundamental and attenuates everything higher. That is the standard harmonic-suppression filter; a high-pass would remove the wanted signal instead.",
"source": "https://50ohm.de/NE_unerwuenschte_aussendungen_2.html",
"confidence": 8
},
"EJ205": {
- "revision": 1,
- "explanation": "A UHF transmitter's harmonics are at still higher frequencies, so a following low-pass filter attenuates them while passing the wanted UHF signal.",
+ "revision": 2,
+ "explanation": "A UHF transmitter's harmonics sit at even higher frequencies, so a low-pass filter after the transmitter passes the wanted UHF output and rejects the harmonics above it. Filters placed before the transmitter, or high-pass/notch types, cannot do this job.",
"source": "https://50ohm.de/NE_unerwuenschte_aussendungen_2.html",
"confidence": 8
},
"EJ206": {
- "revision": 1,
- "explanation": "The correct circuit is the low-pass output filter, with series inductors and shunt capacitors arranged to pass the fundamental and shunt harmonics.",
+ "revision": 2,
+ "explanation": "A transmitter harmonic filter is normally a low-pass output filter: it passes the wanted fundamental frequency but attenuates higher harmonics. The circuit cue is series inductors plus shunt capacitors, where high-frequency energy is increasingly blocked and bypassed to ground.",
"source": "https://50ohm.de/NE_unerwuenschte_aussendungen_2.html",
- "confidence": 7
+ "confidence": 8
},
"EJ207": {
- "revision": 1,
- "explanation": "A harmonic-reduction filter should pass the HF operating range and roll off higher frequencies, i.e. a low-pass characteristic.",
+ "revision": 2,
+ "explanation": "Harmonics are integer multiples of the transmit frequency, so they lie above the wanted fundamental. A harmonic-reduction filter for a transmitter should therefore have a low-pass characteristic: pass the operating range, roll off above it.",
"source": "https://50ohm.de/NE_unerwuenschte_aussendungen_2.html",
- "confidence": 7
+ "confidence": 8
},
"EJ208": {
- "revision": 1,
- "explanation": "For an HF multiband transmitter, the output filter should pass all HF bands while attenuating frequencies above them, so the wide low-pass curve is best.",
+ "revision": 2,
+ "explanation": "For an HF multiband transmitter, the filter must pass the whole wanted HF operating range but attenuate harmonics above it. That is the wide low-pass response: broad passband across HF, then strong attenuation at higher frequencies.",
"source": "https://50ohm.de/NE_unerwuenschte_aussendungen_2.html",
- "confidence": 7
+ "confidence": 8
},
"EJ209": {
- "revision": 1,
- "explanation": "Unwanted-emission power is assessed at the transmitter output including normally used inline devices such as the SWR meter and any low-pass filter.",
+ "revision": 2,
+ "explanation": "Unwanted-emission power is measured at the transmitter output including the devices normally in line during operation — the SWR meter and any low-pass/harmonic filter — because those affect what actually reaches the antenna. The compliance figure must reflect the real transmitting configuration.",
"source": "https://50ohm.de/NE_unerwuenschte_aussendungen_2.html",
"confidence": 8
},
"EJ210": {
- "revision": 1,
- "explanation": "Keeping SSB occupied bandwidth to at most about 2.7 kHz limits spillover onto adjacent frequencies.",
+ "revision": 2,
+ "explanation": "In SSB, RF occupied bandwidth follows the speech/audio bandwidth. Keeping the transmitted sideband to about 2.7 kHz avoids unnecessary high-audio components and reduces spillover into adjacent channels.",
"source": "https://50ohm.de/E_ssb_2.html",
"confidence": 8
},
"EJ211": {
- "revision": 1,
- "explanation": "SSB speech audio above about 3 kHz would widen the RF sideband unnecessarily, increasing adjacent-channel interference risk.",
+ "revision": 2,
+ "explanation": "In SSB the RF sideband width equals the audio bandwidth, so limiting the highest transmitted audio (NF) frequency to under about $3$ kHz keeps the occupied bandwidth tight and limits splatter onto adjacent frequencies. Allowing $5$ or $10$ kHz of audio would needlessly widen the signal.",
"source": "https://50ohm.de/E_ssb_2.html",
"confidence": 8
},
"EJ212": {
- "revision": 1,
- "explanation": "For FM AFSK, occupied bandwidth rises with frequency deviation, so lowering audio drive or deviation reduces the transmitted bandwidth.",
+ "revision": 2,
+ "explanation": "For FM AFSK, occupied bandwidth rises with frequency deviation. In German this deviation is Hub or Frequenzhub: the carrier swing caused by the audio tones. Lowering audio drive reduces the Hub and therefore narrows the transmitted FM signal.",
"source": "https://50ohm.de/EA_fm_2.html",
"confidence": 8
},
"EJ213": {
- "revision": 1,
- "explanation": "Overdriving a power amplifier makes it nonlinear, creating distortion products and a high level of unwanted emissions.",
+ "revision": 2,
+ "explanation": "A power amplifier should operate in its intended linear range for linear modes. Overdrive pushes it into compression or clipping, creating harmonics and intermodulation products, which appear as unwanted emissions.",
"source": "https://50ohm.de/NE_unerwuenschte_aussendungen_2.html",
"confidence": 8
},
"EJ214": {
- "revision": 1,
- "explanation": "An overdriven SSB linear amplifier produces intermodulation products that spread into neighboring frequencies.",
+ "revision": 2,
+ "explanation": "SSB needs linear amplification because the envelope carries information. If the linear amplifier is overdriven, nonlinear mixing of the speech components creates intermodulation products, heard as splatter on neighboring frequencies.",
"source": "https://50ohm.de/NE_unerwuenschte_aussendungen_2.html",
"confidence": 8
},
"EJ215": {
- "revision": 1,
- "explanation": "Too much microphone gain overdrives the SSB transmit chain and creates splatter affecting nearby stations.",
+ "revision": 2,
+ "explanation": "Excess microphone gain overdrives the SSB modulator or later amplifier stages. The result is flat-topping and intermodulation, so the transmitted signal becomes wider than necessary and causes splatter near the operating frequency.",
"source": "https://50ohm.de/NE_unerwuenschte_aussendungen_2.html",
"confidence": 8
},
"EJ216": {
- "revision": 1,
- "explanation": "Poor frequency stability can make the transmitter drift, potentially moving the emission outside authorized band limits.",
+ "revision": 2,
+ "explanation": "Frequency stability is the transmitter's ability to stay on its assigned frequency despite temperature, supply, or time changes. If it drifts, the emission can move into another channel or even outside the authorized amateur band.",
"source": "https://50ohm.de/NE_unerwuenschte_aussendungen_2.html",
"confidence": 8
},
"EJ217": {
- "revision": 1,
- "explanation": "If ALC acts during SSB digital modes, it can distort the audio/RF envelope and create unwanted products on neighboring frequencies.",
+ "revision": 2,
+ "explanation": "If ALC clamps the peaks of a digital SSB signal (RTTY, FT8, Olivia), it distorts the envelope and generates intermodulation products that splatter onto neighbouring frequencies. The damage shows up as interference on adjacent channels, the hallmark of an over-driven digimode signal.",
"source": "https://50ohm.de/NEA_digimode_ssb.html",
"confidence": 8
},
"EJ218": {
- "revision": 1,
- "explanation": "The audio drive for FT8, JS8, PSK31, and similar modes should be low enough that ALC does not engage, avoiding distortion and splatter.",
+ "revision": 2,
+ "explanation": "For FT8, JS8, PSK31 and similar modes, set the audio (NF) drive low enough that the ALC does not engage at all. ALC action on these constant-envelope digital signals causes distortion and splatter, so the clean operating point is just below the ALC threshold — not at maximum.",
"source": "https://50ohm.de/NEA_digimode_ssb.html",
"confidence": 8
},
"EJ219": {
- "revision": 1,
- "explanation": "If ALC is causing interference in SSB digital operation, reduce the audio input level so the transmitter is no longer driven into ALC action.",
+ "revision": 2,
+ "explanation": "When ALC action is causing splatter in a digital mode, the fix is to reduce the audio (NF) input level until the ALC no longer engages, restoring a clean envelope. Raising power, disabling the harmonic filter, or using RIT would not address the over-drive cause.",
"source": "https://50ohm.de/NEA_digimode_ssb.html",
"confidence": 8
},
"EK101": {
- "revision": 1,
- "explanation": "RF energy absorption in the human body depends on frequency, including penetration depth and resonance effects, so exposure limits are frequency-dependent.",
+ "revision": 2,
+ "explanation": "RF energy absorption in the human body depends on frequency. Penetration depth, body-size resonance, and tissue heating efficiency change with wavelength, so safety limits are frequency-dependent rather than one fixed field value for all bands.",
"source": "https://50ohm.de/E_personenschutzabstand_grenzwerte.html",
"confidence": 8
},
"EK102": {
- "revision": 1,
- "explanation": "The 26th BImSchV uses different time references: Annex 1b values are RMS-averaged over 6 minutes, Annex 1a values are short-term RMS values, and Annex 3 uses instantaneous peak limits for pulsed fields.",
+ "revision": 2,
+ "explanation": "The 26th BImSchV applies different time references by annex: Annex 1b limits are RMS-averaged over $6$ minutes, Annex 1a limits are taken as a short-term RMS value, and Annex 3 (pulsed fields) uses the instantaneous peak value. The trap option swaps the $6$-minute window for $3$ minutes.",
"source": "https://www.gesetze-im-internet.de/bimschv_26/BJNR196600996.html",
"confidence": 9
},
"EK103": {
- "revision": 1,
- "explanation": "For active body aids, the relevant protection criterion is the maximum instantaneous field value, not a 3- or 6-minute average.",
+ "revision": 2,
+ "explanation": "For people with active implants (aktive Körperhilfen, e.g. pacemakers) the protective criterion is the maximum instantaneous field value, not a time average — a single peak could disrupt the device, so averaging over $3$ or $6$ minutes would not be safe.",
"source": "https://50ohm.de/E_personenschutzabstand_grenzwerte.html",
"confidence": 8
},
"EK104": {
- "revision": 1,
- "explanation": "13 dBd is 15.15 dBi, a factor about 32.7; 6 W therefore gives about 196 W EIRP, well above the 10 W EIRP amateur-station threshold requiring proof/notification duties.",
+ "revision": 2,
+ "explanation": "Yes. Convert the gain: $13\\ \\text{dBd} = 15.15$ dBi $\\to 10^{1.515} \\approx 32.7$, so EIRP $= 6\\ \\text{W} \\cdot 32.7 \\approx 196$ W. That is far above the $10$ W EIRP threshold at which a fixed station must demonstrate compliance with the person-protection limits — so proof is required regardless of mode or duty cycle.",
"source": "https://www.gesetze-im-internet.de/bemfv/__8.html",
"confidence": 9
},
"EK105": {
- "revision": 1,
- "explanation": "At 80 m the 3.65 m result lies in the reactive near field, where the far-field approximation is invalid, so measurement, simulation, or near-field calculation is needed.",
+ "revision": 2,
+ "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.",
"source": "https://50ohm.de/E_naeherungsformel_1.html",
"confidence": 8
},
"EK106": {
- "revision": 1,
- "explanation": "The far-field approximation is invalid below roughly lambda/(2*pi); for 160 m this is about 25.5 m and for 80 m about 12.7 m.",
+ "revision": 2,
+ "explanation": "The reactive near-field boundary is roughly $r \\approx \\lambda/(2\\pi)$. Below that distance the E-field and H-field can vary independently and the far-field approximation is unreliable; for 160 m this is about 25.5 m, and for 80 m about 12.7 m.",
"source": "https://50ohm.de/E_naeherungsformel_1.html",
"confidence": 8
},
"EK107": {
- "revision": 2,
- "explanation": "When the safety distance is calculated from the antenna field, the distance must be maintained from every radiating point of the antenna, not only the feed point.",
+ "revision": 3,
+ "explanation": "With a far-field-based calculation, the safety distance (Sicherheitsabstand) must be kept from every radiating point of the antenna, since any part can radiate. Measuring only from the feed point or mast attachment would underestimate the exposure near the antenna's ends.",
"source": "https://50ohm.de/NEA_slide_nea_personenschutzabstand.html",
"confidence": 8
},
"EK108": {
- "revision": 1,
- "explanation": "Convert 7.5 dBd to 9.65 dBi, subtract 1.5 dB cable loss for 8.15 dB net gain, then use d = sqrt(30 x EIRP) / 28 V/m; the result is about 5.0 m.",
+ "revision": 2,
+ "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.",
"source": "https://50ohm.de/E_naeherungsformel_1.html",
"confidence": 8
},
"EK201": {
- "revision": 1,
- "explanation": "Microwave antennas can concentrate high fields in a narrow beam, so people should not stay in the direct beam path of transmitting antennas.",
+ "revision": 2,
+ "explanation": "Microwave dish and horn antennas concentrate their power into a narrow, intense beam, so the field on the beam axis can far exceed the exposure limits close in. The key safety rule is to avoid being in the direct beam path of transmitting antennas; shielding hats or 'EMC clothing' are not real protection, and duty cycle is not the point.",
"source": "https://50ohm.de/NE_strahlengang_aufenthalt.html",
"confidence": 8
},
"EK202": {
- "revision": 1,
- "explanation": "Transmitting antennas can have high RF voltages on their conductors; touching them can cause burns and other RF voltage injuries.",
+ "revision": 2,
+ "explanation": "A transmitting antenna carries high RF voltages on its conductors, especially at voltage antinodes. Touching it can give RF burns and other injuries — and unlike DC, the skin effect does not protect you. Earthing for lightning does not remove the RF voltage present while transmitting.",
"source": "https://50ohm.de/E_slide_e_sicherheit.html?print-pdf=&showNotes=true",
"confidence": 8
},
"EK203": {
- "revision": 1,
- "explanation": "Power-supply capacitors can remain charged after the mains plug is removed, so opening disconnected equipment can still expose you to electric shock.",
+ "revision": 2,
+ "explanation": "Even with the mains plug pulled, the smoothing/filter capacitors in the power supply can stay charged to a few hundred volts and deliver a dangerous shock. Always treat a just-opened supply as live and discharge the large capacitors before touching the circuit.",
"source": "https://50ohm.de/NEA_slide_nea_sicherheit.html?print-pdf=&showNotes=true",
"confidence": 8
},
"EK204": {
- "revision": 1,
- "explanation": "A fuse is a safety component matched to current and trip speed; after repair it must be replaced with the same current rating and same fast characteristic.",
+ "revision": 2,
+ "explanation": "A fuse is a calibrated safety device: replace it with the same current rating and the same tripping characteristic. A 20 A fast-blow ('flink') must be replaced by another 20 A fast-blow — a slower characteristic or a wire bridge would let fault current run too long and could start a fire.",
"source": "https://50ohm.de/E_sicherungen.html",
"confidence": 8
},
"EK205": {
- "revision": 1,
- "explanation": "For a 3-core mains cable the standard colors are PE green-yellow, live conductor brown, and neutral blue.",
+ "revision": 2,
+ "explanation": "German/IEC mains colour code: protective earth (Schutzleiter) is green-yellow, the live conductor (Außenleiter) is brown, and neutral (Neutralleiter) is blue. In the requested order PE, L, N that is green-yellow, brown, blue.",
"source": "https://50ohm.de/E_spannungsquelle.html",
"confidence": 8
},
"EK206": {
- "revision": 1,
- "explanation": "Ungrounded wire antennas can accumulate static charge from weather such as rain or hail, creating a safety hazard.",
+ "revision": 2,
+ "explanation": "An ungrounded wire antenna is an isolated metal collector: precipitation such as rain or hail deposits electric charge on it, which can build up to a dangerous static voltage with no path to drain. That charge hazard, not transmit voltage or solar storms, is the special concern.",
"source": "https://50ohm.de/E_slide_e_sicherheit.html?print-pdf=&showNotes=true",
"confidence": 8
},
"EK207": {
- "revision": 1,
- "explanation": "High-value bleed resistors drain static charge to the station earth while their high resistance avoids significantly affecting RF operation.",
+ "revision": 2,
+ "explanation": "Fit a high-value bleed resistor (hochohmiger Ableitwiderstand) from the antenna to station earth. Its high resistance quietly drains static charge to ground at DC, yet looks like a near-open circuit at RF, so it does not load or detune the antenna. A low-value resistor would short the signal.",
"source": "https://50ohm.de/NE_slide_ne_sicherheit.html?print-pdf=&showNotes=true",
"confidence": 8
},
"EK208": {
- "revision": 1,
- "explanation": "Bonding all antenna coax shields together and to the main earthing bar prevents dangerous potential differences between coax systems.",
+ "revision": 2,
+ "explanation": "Bond the shields of all antenna coax cables together and to the main earthing bar (Haupterdungsschiene). Tying them to one common potential prevents dangerous voltage differences between cable systems during a fault or lightning surge — equipotential bonding is the person-protection measure here.",
"source": "https://50ohm.de/NE_slide_ne_sicherheit.html",
"confidence": 8
},
"EK209": {
- "revision": 2,
- "explanation": "An existing building earthing system per VDE 0855-300 may be used for antenna earthing; no separate electrode or special approval is required.",
+ "revision": 3,
+ "explanation": "Per VDE 0855-300 any existing building earthing system may be used for antenna earthing — you do not need a separate electrode or special BNetzA approval. Using the common building earth actually improves safety by keeping everything at one potential.",
"source": "https://50ohm.de/NE_blitzerdung.html",
"confidence": 8
},
"EK210": {
- "revision": 1,
- "explanation": "VDE 0855-300 requires a solid earthing conductor, with example minimum cross sections of 16 mm2 copper, 25 mm2 aluminium, or 50 mm2 steel.",
+ "revision": 2,
+ "explanation": "VDE 0855-300 expects a robust solid earthing conductor; the example minimum cross-sections are $16\\ \\text{mm}^2$ copper, $25\\ \\text{mm}^2$ aluminium, or $50\\ \\text{mm}^2$ steel. The larger sections for aluminium and steel compensate for their poorer conductivity, so they can carry a lightning surge without melting.",
"source": "https://50ohm.de/NE_blitzerdung.html",
"confidence": 8
},
"EK211": {
- "revision": 1,
- "explanation": "Connecting an antenna mast to an existing lightning protection system changes that system and must be included in the lightning protection concept by a qualified specialist.",
+ "revision": 2,
+ "explanation": "Tying an antenna mast into a building's lightning-protection system alters that system, so it may only be done if a qualified lightning-protection specialist provides for it in the lightning-protection concept (Blitzschutzkonzept). It is neither always mandatory nor always forbidden — it depends on that professional assessment.",
"source": "https://50ohm.de/NE_blitzerdung.html",
"confidence": 8
},
@@ -8670,14 +8670,14 @@
"confidence": 7
},
"NE303": {
- "revision": 2,
- "explanation": "FM information is carried by frequency deviation, so the RF amplitude is ideally unaffected by microphone audio.",
+ "revision": 3,
+ "explanation": "FM information is carried by frequency deviation, German Hub/Frequenzhub: the carrier swings above and below its centre frequency. The RF amplitude is ideally unaffected by microphone audio.",
"source": "https://50ohm.de/NEA_fm.html",
"confidence": 7
},
"NE304": {
- "revision": 2,
- "explanation": "In ideal FM the transmitter output power is essentially constant; speaking louder changes deviation, not the set RF power.",
+ "revision": 3,
+ "explanation": "In ideal FM the transmitter output power is essentially constant. Speaking louder increases frequency deviation, German Hub/Frequenzhub, but it does not increase the set RF carrier power.",
"source": "https://50ohm.de/NEA_fm.html",
"confidence": 7
},
@@ -8688,10 +8688,10 @@
"confidence": 7
},
"NE306": {
- "revision": 2,
- "explanation": "Too much FM deviation usually comes from excessive audio level, so speaking more quietly reduces the modulation hub.",
+ "revision": 3,
+ "explanation": "Hub or Frequenzhub means FM frequency deviation: the carrier's maximum swing away from its centre frequency. Too much deviation usually comes from excessive audio level, so speaking more quietly reduces the Hub and narrows the FM signal.",
"source": "https://50ohm.de/NEA_fm.html",
- "confidence": 7
+ "confidence": 8
},
"NE307": {
"revision": 2,