From 3179f546811e6ee7212bbb083eb681faab2d322e Mon Sep 17 00:00:00 2001 From: Renat Nurgaliyev Date: Wed, 24 Jun 2026 13:09:52 +0200 Subject: [PATCH] Pull explanations from 50ohm.de when available --- EXPLANATIONS.md | 24 +- amateurfunk_anki.py | 131 ++++- explanations.json | 1190 ++++++++++++++++++++++---------------- test_amateurfunk_anki.py | 114 ++++ 4 files changed, 932 insertions(+), 527 deletions(-) diff --git a/EXPLANATIONS.md b/EXPLANATIONS.md index e9e63dd..450b8b0 100644 --- a/EXPLANATIONS.md +++ b/EXPLANATIONS.md @@ -27,7 +27,8 @@ entry is purely additive — no regenerate ceremony beyond ### Per-entry schema -Every entry MUST have exactly these four fields: +Every entry MUST have these four required fields, and MAY carry the one +optional field below: | Field | Type | Constraint | |---------------|---------|--------------------------------------------| @@ -35,13 +36,22 @@ Every entry MUST have exactly these four fields: | `explanation` | string | Non-empty. **English.** Correct & helpful, WHY-focused | | `source` | string | Non-empty. URL or citation like `AFuV §16(2)` | | `confidence` | integer | `1..10` inclusive. See scale in §5 | +| `provenance` | string | *Optional.* Only allowed value: `"50ohm-loesungsweg"` | -Extra keys are rejected by `load_explanations()` — the build fails -with `unknown fields [...]` listing them. The loader is similarly -strict about types: a JSON `true` will not satisfy the integer -contract for `revision` or `confidence`. If you need to track -editorial metadata that isn't shown on the card, propose a schema -change rather than smuggling fields in. +`provenance` records **how the text was produced**, which `source` (a +citation) does not. Set it to `"50ohm-loesungsweg"` only on entries +that are genuinely a translation/condensation of a 50ohm.de worked +solution (`contents/solutions/.md` in `DARC-e-V/50ohm-contents-dl`). +The build uses it — and *not* the `source` domain — to decide whether +to show the CC BY 4.0 derivative-work credit on the card (CC BY +requires naming the author team and indicating modification). Do **not** +add it just because an entry cites a 50ohm.de study page; merely citing +a page is not a derivative work. Any other value, or any other extra +field, is rejected by `load_explanations()` with +`unknown fields [...]` / `provenance must be one of [...]`. The loader +is also strict about types: a JSON `true` will not satisfy the integer +contract for `revision` or `confidence`. To track other editorial +metadata, propose a schema change rather than smuggling fields in. Top-level keys (the question numbers) must also match the catalog exactly. An entry keyed on a number that no live question carries diff --git a/amateurfunk_anki.py b/amateurfunk_anki.py index 3dcb55b..01d1007 100644 --- a/amateurfunk_anki.py +++ b/amateurfunk_anki.py @@ -63,6 +63,17 @@ DEFAULT_EXPLANATIONS_PATH = Path("explanations.json") # EXPLANATIONS.md for the full editorial contract. EXPLANATION_FIELDS = ("revision", "explanation", "source", "confidence") +# Optional, validated provenance marker. A `source` URL is only a +# citation; it does NOT establish that the explanation is a derivative +# work. `provenance` records *how the text was produced* so the CC BY +# derivative-work credit is shown only for records that genuinely are +# one. The single recognised value marks an explanation translated and +# condensed from a 50ohm.de worked solution +# (contents/solutions/.md in DARC-e-V/50ohm-contents-dl). +PROVENANCE_50OHM_SOLUTION = "50ohm-loesungsweg" +ALLOWED_PROVENANCE = frozenset({PROVENANCE_50OHM_SOLUTION}) +OPTIONAL_EXPLANATION_FIELDS = ("provenance",) + # Confidence values strictly below this threshold are surfaced on # the card as a "low confidence" badge — a learner-facing warning # that the explanation may be incomplete or weakly sourced. Matches @@ -70,6 +81,49 @@ EXPLANATION_FIELDS = ("revision", "explanation", "source", "confidence") # below 7 is review-worthy"). LOW_CONFIDENCE_THRESHOLD = 7 +# ---- Attribution (licensing) ---------------------------------------------- +# Two sources are redistributed in these decks and each carries an +# attribution obligation: +# * the questions/figures come from the BNetzA question catalog under +# Datenlizenz Deutschland – Namensnennung 2.0 (DL-DE→BY-2.0); and +# * the English explanations are translated and condensed from the +# worked solutions on 50ohm.de, which is CC BY 4.0 and requires both +# naming the author team and indicating that changes were made. +# DECK_ATTRIBUTION goes in every deck's description (the whole-deck +# notice); EXPLANATION_CC_BY_CREDIT is the per-card line appended to +# each 50ohm-derived explanation so the "modified" indication travels +# with the modified text itself. +EXPLANATION_CC_BY_CREDIT = ( + "Explanation translated & condensed from the worked solution on " + '50ohm.de — ' + "© 50ohm.de-Autorenteam (AJW-Referat, DARC e. V.), " + 'CC BY 4.0. ' + "Question © Bundesnetzagentur, " + 'DL-DE→BY-2.0.' +) + +DECK_ATTRIBUTION = ( + "Amateurfunk-Lernkarten / amateur-radio exam flashcards.

" + "Questions & figures: official question catalog " + ""Prüfungsfragen zum Erwerb von " + "Amateurfunkprüfungsbescheinigungen", Bundesnetzagentur, " + "3. Auflage, März 2024 " + "(https://www.bundesnetzagentur.de/amateurfunk). Licensed under " + "Datenlizenz Deutschland – Namensnennung – Version 2.0 " + "(https://www.govdata.de/dl-de/by-2-0).

" + "Explanations: written for this project. Where an explanation " + "is marked as derived from a 50ohm.de worked solution " + "("Lösungsweg"), that text is © 50ohm.de-Autorenteam, " + "koordiniert durch das AJW-Referat des DARC e. V., licensed under " + "CC BY 4.0 (https://creativecommons.org/licenses/by/4.0/), and has " + "been modified — translated from German and shortened. Other " + "explanations cite their own sources.

" + "Exam-aid references cite the BNetzA "Hilfsmittel" " + "formula sheet.

" + "This deck is an independent project and is not endorsed by the " + "Bundesnetzagentur or by the DARC e. V." +) + # Exit codes. The builder is much simpler than the fetcher — there is # no "operator must resolve local state" case here, so two codes are # enough. @@ -246,12 +300,15 @@ def load_explanations(path): def _validate_explanation(key, value): """Raise `AnkiBuildError` if one explanation entry is malformed. - Strict shape check: the four documented fields are required, no - extras allowed, and integer fields reject `bool` (which would - pass `isinstance(..., int)` because `bool` subclasses `int` in - Python). Extras are rejected so a stray `"note"` or `"author"` - field can't accumulate silently and drift from the documented - schema in EXPLANATIONS.md. + Strict shape check: the four documented fields are required and the + optional `provenance` field is allowed (and, if present, must be a + string from `ALLOWED_PROVENANCE`); any *other* extra field is + rejected, so a stray `"note"` or `"author"` field can't accumulate + silently and drift from the documented schema in EXPLANATIONS.md. + Integer fields reject `bool` (which would pass `isinstance(..., int)` + because `bool` subclasses `int` in Python), and every malformed + value — including an unhashable `provenance` — surfaces as an + `AnkiBuildError`, never a raw `TypeError`. """ if not isinstance(value, dict): raise AnkiBuildError( @@ -262,11 +319,24 @@ def _validate_explanation(key, value): raise AnkiBuildError( f"explanation {key!r} missing required fields: {missing}" ) - extra = sorted(set(value) - set(EXPLANATION_FIELDS)) + extra = sorted( + set(value) - set(EXPLANATION_FIELDS) - set(OPTIONAL_EXPLANATION_FIELDS) + ) if extra: raise AnkiBuildError( f"explanation {key!r}: unknown fields {extra}" ) + if "provenance" in value and ( + not isinstance(value["provenance"], str) + or value["provenance"] not in ALLOWED_PROVENANCE + ): + # The isinstance check must come first: an unhashable JSON value + # (a list or object) would make the `not in` set test raise a raw + # TypeError instead of the documented AnkiBuildError. + raise AnkiBuildError( + f"explanation {key!r}: provenance must be one of " + f"{sorted(ALLOWED_PROVENANCE)}" + ) if type(value["revision"]) is not int or value["revision"] < 1: raise AnkiBuildError( f"explanation {key!r}: revision must be a positive integer" @@ -619,11 +689,23 @@ def render_explanation(explanation): 'title="This explanation is weakly sourced or incomplete">' 'low confidence' ) + # CC BY 4.0 requires naming the author and indicating modifications. + # Only explanations actually derived from a 50ohm.de worked solution + # carry this credit — gated on the explicit `provenance` marker, NOT + # on the `source` URL (a citation is not provenance; many records + # merely cite a 50ohm study page without being a derivative work). + credit = "" + if explanation.get("provenance") == PROVENANCE_50OHM_SOLUTION: + credit = ( + '
' + f'{EXPLANATION_CC_BY_CREDIT}
' + ) return ( '
' f'
Explanation{badge}
' f'
{body}
' f'
Source: {source_html}
' + f'{credit}' '
' ) @@ -1218,8 +1300,15 @@ def insert_collection_metadata(conn, decks, cur_deck_id, model_id, now): package can ship a deck tree, not just one deck. `cur_deck_id` is the deck made current in `col.conf`. """ + # The attribution notice goes on the root (current) deck only, so it + # shows once for the package rather than on every sub-deck. decks_json = { - str(deck_id): deck_json(deck_id, deck_name, now) + str(deck_id): deck_json( + deck_id, + deck_name, + now, + desc=DECK_ATTRIBUTION if deck_id == cur_deck_id else "", + ) for deck_name, deck_id in decks.items() } conn.execute( @@ -1278,11 +1367,14 @@ def collection_conf(deck_id): } -def deck_json(deck_id, deck_name, now): +def deck_json(deck_id, deck_name, now, desc=""): """Return one deck entry for `col.decks`. The `Amateurfunk::Foo` naming gives the user a single top-level Amateurfunk node in Anki's deck browser with three child decks. + `desc` is the deck description shown in Anki's deck overview; we use + it on the root deck to carry the source-attribution notice (see + DECK_ATTRIBUTION). """ return { "id": deck_id, @@ -1297,7 +1389,7 @@ def deck_json(deck_id, deck_name, now): "browserCollapsed": False, "dyn": 0, "conf": 1, - "desc": "", + "desc": desc, "extendNew": 0, "extendRev": 0, } @@ -1502,6 +1594,17 @@ CARD_CSS = """ color: #1a73e8; text-decoration: none; } +.af-explanation-credit { + font-family: Arial, sans-serif; + font-style: normal; + font-size: 11px; + color: #999; + margin-top: 0.35rem; +} +.af-explanation-credit a { + color: #1a73e8; + text-decoration: none; +} .nightMode .af-explanation, .card.nightMode .af-explanation { border-top-color: #555; @@ -1519,6 +1622,14 @@ CARD_CSS = """ .card.nightMode .af-explanation-source a { color: #8ab4f8; } +.nightMode .af-explanation-credit, +.card.nightMode .af-explanation-credit { + color: #999; +} +.nightMode .af-explanation-credit a, +.card.nightMode .af-explanation-credit a { + color: #8ab4f8; +} .nightMode .af-explanation-low-confidence, .card.nightMode .af-explanation-low-confidence { background: #4a3510; diff --git a/explanations.json b/explanations.json index d558abb..6863ba9 100644 --- a/explanations.json +++ b/explanations.json @@ -1,111 +1,129 @@ { "AA101": { - "revision": 2, - "explanation": "Impedance is the AC form of resistance, including resistive and reactive parts, so it is measured in ohms just like resistance.", + "revision": 3, + "explanation": "Impedance is AC (complex) resistance; like ordinary resistance it is measured in the ohm, $\\Omega$. The other units belong to different quantities — the henry to inductance, the farad to capacitance, the siemens to conductance.", "source": "https://50ohm.de/NEA_spule_2.html#AA101", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AA102": { - "revision": 2, - "explanation": "Charge is current integrated over time; one coulomb is one ampere-second, so As is the practical unit here.", + "revision": 3, + "explanation": "Charge is current times time, so its unit is the ampere-second: $1\\,\\text{As} = 1\\,\\text{A}\\cdot 1\\,\\text{s} = 1\\,\\text{C}$ (coulomb). Writing it as $\\text{As}$ rather than the coulomb makes the unit check in $Q = I\\cdot t$ obvious.", "source": "https://50ohm.de/NEA_ladung_energie.html#AA102", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AA103": { - "revision": 2, - "explanation": "Energy is power over time, so it can be expressed as joules in SI terms or as watt-hours in practical electrical use.", + "revision": 3, + "explanation": "Energy is power times time. One joule is one watt acting for one second: $1\\,\\text{J} = 1\\,\\text{Ws}$, and $1\\,\\text{Wh} = 3600\\,\\text{J}$ — the kWh on an electricity bill is energy. The volt is a voltage and the watt a power (energy per time), neither is an energy.", "source": "https://50ohm.de/NEA_ladung_energie.html#AA103", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AA104": { - "revision": 2, - "explanation": "Symbol rate counts transmitted symbols per second, and that rate is measured in baud.", + "revision": 3, + "explanation": "Symbol rate counts transmitted symbols per second; its unit is the baud, $1\\,\\text{Bd} = 1$ symbol per second. Because one symbol can encode several bits, baud and bit/s are not the same. Hertz is a frequency and the bit an amount of information.", "source": "https://50ohm.de/NEA_mehrwertige_verfahren.html#AA104", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AA105": { - "revision": 5, - "explanation": "For power ratios, gain in dB is $10\\log_{10}(P_2/P_1)$; $10\\log_{10}(40)$ is about 16 dB. Hilfsmittel: apply $g = 10\\cdot\\log_{10}(P_2/P_1)$ (Pegel, S.15); the table has no 16 dB row — combine +10 dB (×10) and +6 dB (×4) → ×40.", + "revision": 6, + "explanation": "A power ratio in dB: $g = 10\\cdot\\log_{10}(40) \\approx 16\\,\\text{dB}$. Without a calculator, factor it — $40 = 10\\cdot 4$, and since logs of factors add, $10\\,\\text{dB} + 6\\,\\text{dB} = 16\\,\\text{dB}$. Hilfsmittel: $g = 10\\cdot\\log_{10}(P_2/P_1)$ (Verstärkung, S.15); the dB↔ratio table has no 16 dB row, so combine the +10 dB (×10) and +6 dB (×4) rows.", "source": "https://50ohm.de/NEA_dezibel_2.html#AA105", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AA106": { - "revision": 4, - "explanation": "A 16 dB power gain is approximately 40 times; with 1 W drive the output is about 40 W, below the 100 W maximum. Hilfsmittel: reverse $g = 10\\cdot\\log_{10}(P_2/P_1)$ (Pegel, S.15); 16 dB is not a table row — it is +10 dB (×10) plus +6 dB (×4) = ×40.", + "revision": 5, + "explanation": "Invert the gain formula: $P_2 = P_1\\cdot 10^{g/(10\\,\\text{dB})} = 1\\,\\text{W}\\cdot 10^{16/10} \\approx 40\\,\\text{W}$. Equivalently $16\\,\\text{dB} = 10\\,\\text{dB} + 6\\,\\text{dB} = \\times 10\\cdot\\times 4 = \\times 40$. The 100 W rating is only the ceiling, not reached here. Hilfsmittel: $g = 10\\cdot\\log_{10}(P_2/P_1)$ (Verstärkung, S.15).", "source": "https://50ohm.de/NEA_dezibel_2.html#AA106", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AA107": { - "revision": 4, - "explanation": "1 W is 0 dBW, and a 10 dB amplifier raises the power level by 10 dB, giving 10 dBW. Hilfsmittel: apply $p = 10\\cdot\\log_{10}(P/1\\,\\text{W})$ dBW (Pegel, S.15); +10 dB simply adds 10 to the dBW level.", + "revision": 5, + "explanation": "A level needs a reference; dBW is referenced to 1 W, so $1\\,\\text{W}\\to 0\\,\\text{dBW}$ (because $\\log_{10}1 = 0$). Adding the $10\\,\\text{dB}$ stage gain gives $0\\,\\text{dBW} + 10\\,\\text{dB} = 10\\,\\text{dBW}$. Hilfsmittel: $p = 10\\cdot\\log_{10}(P/1\\,\\text{W})\\,\\text{dBW}$ (Pegel, S.15).", "source": "https://50ohm.de/NEA_dezibel_2.html#AA107", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AA108": { - "revision": 4, - "explanation": "dBW is referenced to 1 W, so 20 dBW means $10^{20/10}\\,\\text{W} = 100\\,\\text{W} = 10^2\\,\\text{W}$. Hilfsmittel: invert $p = 10\\cdot\\log_{10}(P/1\\,\\text{W})$ dBW (Pegel, S.15): $P = 1\\,\\text{W}\\cdot 10^{p/10}$.", + "revision": 5, + "explanation": "Invert the dBW level: $P = 1\\,\\text{W}\\cdot 10^{p/(10\\,\\text{dB})} = 1\\,\\text{W}\\cdot 10^{20/10} = 10^2\\,\\text{W}$. Every 10 dB is a factor of ten, so 20 dB is ×100. Hilfsmittel: $p = 10\\cdot\\log_{10}(P/1\\,\\text{W})\\,\\text{dBW}$ (Pegel, S.15).", "source": "https://50ohm.de/NEA_dezibel_2.html#AA108", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AA109": { - "revision": 5, - "explanation": "The amplifier output is 10 W; in dBm that is $10\\,\\text{W} = 10000\\,\\text{mW}$, and $10\\log_{10}(10000) = 40\\,\\text{dBm}$. Hilfsmittel: apply $p = 10\\cdot\\log_{10}(P/1\\,\\text{mW})$ dBm (Pegel, S.15) after converting 10 W to 10000 mW.", + "revision": 6, + "explanation": "dBm is referenced to 1 mW. The sender's $1\\,\\text{W} = 1000\\,\\text{mW}\\to 30\\,\\text{dBm}$ (since $10\\cdot\\log_{10}10^3 = 30$). Adding the $10\\,\\text{dB}$ stage gain gives $40\\,\\text{dBm}$. Hilfsmittel: $p = 10\\cdot\\log_{10}(P/1\\,\\text{mW})\\,\\text{dBm}$ (Pegel, S.15).", "source": "https://50ohm.de/NEA_dezibel_2.html#AA109", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AA110": { - "revision": 4, - "explanation": "dBm is referenced to 1 mW: 0 dBm is 1 mW, +3 dB is about double or 2 mW, and +20 dB is 100 times or 100 mW. Hilfsmittel: use $p = 10\\cdot\\log_{10}(P/1\\,\\text{mW})$ dBm (Pegel, S.15); the dB↔ratio table (S.15) gives the +3/+20 dB factors.", + "revision": 5, + "explanation": "Invert the dBm level: $P = 1\\,\\text{mW}\\cdot 10^{p/(10\\,\\text{dB})}$. So $0\\,\\text{dBm}\\to 1\\,\\text{mW}$; $3\\,\\text{dB}$ is a doubling, so $3\\,\\text{dBm}\\to 2\\,\\text{mW}$; and $20\\,\\text{dB} = \\times 100$, so $20\\,\\text{dBm}\\to 100\\,\\text{mW}$. Hilfsmittel: $p = 10\\cdot\\log_{10}(P/1\\,\\text{mW})\\,\\text{dBm}$ (Pegel, S.15).", "source": "https://50ohm.de/NEA_dezibel_2.html#AA110", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AA111": { - "revision": 5, - "explanation": "For voltage ratios use $20\\log_{10}(U_2/U_1)$; $20\\log_{10}(15)$ is about 23.5 dB. Hilfsmittel: apply the voltage form $g = 20\\cdot\\log_{10}(U_2/U_1)$ (Pegel, S.15).", + "revision": 6, + "explanation": "A voltage ratio uses the factor 20, not 10 (because $P\\propto U^2$): $g = 20\\cdot\\log_{10}(U_2/U_1) = 20\\cdot\\log_{10}(15) \\approx 23.5\\,\\text{dB}$. Hilfsmittel: $g = 20\\cdot\\log_{10}(U_2/U_1)$ (Verstärkung, S.15).", "source": "https://50ohm.de/NEA_dezibel_2.html#AA111", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AA112": { - "revision": 4, - "explanation": "120 dB relative to 1 microvolt per meter is a voltage ratio of $10^{120/20} = 10^6$, so the field is 1 V/m. Hilfsmittel: invert $g = 20\\cdot\\log_{10}(U_2/U_1)$ (Pegel, S.15): $\\text{ratio} = 10^{120/20} = 10^6$.", + "revision": 5, + "explanation": "The prefix on $\\text{dB}\\mu\\text{V/m}$ sets the reference at $1\\,\\mu\\text{V/m}$, and field strength is a voltage-like quantity, so the factor is 20: $E = 1\\,\\mu\\text{V/m}\\cdot 10^{120/(20\\,\\text{dB})} = 1\\,\\mu\\text{V/m}\\cdot 10^6 = 1\\,\\text{V/m}$. Hilfsmittel: voltage-like levels use $20\\cdot\\log_{10}$ (Pegel, S.15).", "source": "https://50ohm.de/NEA_dezibel_2.html#AA112", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AA113": { - "revision": 4, - "explanation": "Each S-step is 6 dB; S4 to S7 is three steps, so $3 \\times 6\\,\\text{dB} = 18\\,\\text{dB}$.", + "revision": 5, + "explanation": "One S-unit is defined as $6\\,\\text{dB}$. S7 is three S-units above S4, so the difference is $3\\cdot 6\\,\\text{dB} = 18\\,\\text{dB}$. (The 6 dB per S-unit is a memory item, not in the Formelsammlung.)", "source": "https://50ohm.de/NEA_s_meter.html#AA113", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AA114": { - "revision": 3, - "explanation": "From S9+20 dB down to S9 removes 20 dB, and from S9 to S8 removes another 6 dB, totaling 26 dB.", + "revision": 4, + "explanation": "Above S9, signal strength is quoted as S9+$x\\,\\text{dB}$. Going from S9+20 dB back to S9 drops $20\\,\\text{dB}$; from S9 to S8 is one S-unit, $6\\,\\text{dB}$; together the input voltage falls by $26\\,\\text{dB}$. (S-unit = 6 dB is a memory item.)", "source": "https://50ohm.de/NEA_s_meter.html#AA114", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AA115": { - "revision": 4, - "explanation": "1 ppm is one part in one million; $435\\,\\text{MHz}$ divided by $10^6$ is $435\\,\\text{Hz}$.", + "revision": 5, + "explanation": "1 ppm means one part in $10^6$. Since $1\\,\\text{MHz} = 10^6\\,\\text{Hz}$, one ppm of a frequency expressed in MHz is simply the same number of Hz: $435\\,\\text{MHz}\\cdot 10^{-6} = 435\\,\\text{Hz}$.", "source": "https://50ohm.de/NEA_frequenzgenauigkeit.html#AA115", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AA116": { - "revision": 4, - "explanation": "10 ppm at 14.200000 MHz is $14.2\\,\\text{MHz} \\times 10/10^6 = 142\\,\\text{Hz}$, so the possible frequency is 14.200000 MHz plus or minus 0.000142 MHz.", + "revision": 5, + "explanation": "10 ppm is $10/10^6$ of the frequency: $\\Delta f = 10\\cdot\\frac{14.2\\,\\text{MHz}}{10^6} = 142\\,\\text{Hz}$. The true frequency therefore lies within $\\pm 142\\,\\text{Hz}$ of the display, i.e. between $14.199858\\,\\text{MHz}$ and $14.200142\\,\\text{MHz}$.", "source": "https://50ohm.de/NEA_frequenzgenauigkeit.html#AA116", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AB101": { - "revision": 5, - "explanation": "Use $R = \\rho l/A$ with copper $\\rho = 0.018\\,\\Omega\\,\\text{mm}^2/\\text{m}$ and $A = \\pi(0.1\\,\\text{mm})^2 = 0.0314\\,\\text{mm}^2$; $0.018 \\cdot 1.8 / 0.0314$ is about 1.02 ohm.", + "revision": 6, + "explanation": "Wire resistance is $R = \\frac{\\rho\\,l}{A}$ with cross-section $A = \\frac{d^2\\pi}{4} = \\frac{(0.2\\,\\text{mm})^2\\pi}{4} \\approx 0.0314\\,\\text{mm}^2$. With copper's $\\rho \\approx 0.018\\,\\Omega\\text{mm}^2/\\text{m}$ and $l = 1.8\\,\\text{m}$: $R = \\frac{0.018\\,\\Omega\\text{mm}^2/\\text{m}\\cdot 1.8\\,\\text{m}}{0.0314\\,\\text{mm}^2} \\approx 1.0\\,\\Omega$. Hilfsmittel: $R = \\rho\\,l/A_\\mathrm{Dr}$ and $A_\\mathrm{Dr} = d^2\\pi/4$ (Widerstand von Drähten, S.11); copper $\\rho$ from the material table (S.21).", "source": "https://50ohm.de/NEA_leiterwiderstand.html#AB101", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AB102": { - "revision": 5, - "explanation": "Rearrange $R = \\rho l/A$ to $l = RA/\\rho$; $1.5\\,\\Omega \\cdot 0.5\\,\\text{mm}^2 / 0.018\\,\\Omega\\,\\text{mm}^2/\\text{m}$ is about 41.7 m.", + "revision": 6, + "explanation": "Rearrange $R = \\frac{\\rho\\,l}{A}$ for length: $l = \\frac{R\\,A}{\\rho} = \\frac{1.5\\,\\Omega\\cdot 0.5\\,\\text{mm}^2}{0.018\\,\\Omega\\text{mm}^2/\\text{m}} \\approx 41.7\\,\\text{m}$. Hilfsmittel: $R = \\rho\\,l/A_\\mathrm{Dr}$ (Widerstand von Drähten, S.11); copper $\\rho$ from the material table (S.21).", "source": "https://50ohm.de/NEA_leiterwiderstand.html#AB102", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AB103": { "revision": 2, @@ -114,40 +132,46 @@ "confidence": 8 }, "AB104": { - "revision": 2, - "explanation": "Semiconductors such as silicon are poor conductors when pure, but heat or small amounts of dopant atoms can provide mobile charge carriers.", + "revision": 3, + "explanation": "Pure silicon at room temperature has very few free carriers and behaves almost like an insulator. Adding tiny amounts of dopant (e.g. boron or phosphorus), or raising the temperature, frees up charge carriers and turns it into a conductor. This switchable conductivity between insulator and conductor is what makes a material a semiconductor.", "source": "https://50ohm.de/NEA_halbleiter_2.html#AB104", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AB105": { - "revision": 2, - "explanation": "Doping means deliberately adding atoms with different valence to a semiconductor so extra electrons or holes become available as charge carriers.", + "revision": 3, + "explanation": "Doping is the deliberate introduction of small amounts of chemically different foreign atoms into a semiconductor base to supply free charge carriers. Pentavalent atoms (e.g. phosphorus) create an electron surplus (n-type); trivalent atoms (e.g. boron) create holes (p-type).", "source": "https://50ohm.de/NEA_halbleiter_2.html#AB105", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AB106": { - "revision": 2, - "explanation": "N-type material is doped to have extra mobile electrons; electrons are the majority carriers.", + "revision": 3, + "explanation": "N-type material has a surplus of free, mobile electrons. It is produced by doping with pentavalent atoms (e.g. phosphorus), each of which contributes one extra electron that is available as a charge carrier.", "source": "https://50ohm.de/NEA_halbleiter_2.html#AB106", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AB107": { - "revision": 2, - "explanation": "P-type material is doped to create mobile holes; holes are the majority carriers.", + "revision": 3, + "explanation": "P-type material has a surplus of holes (electron vacancies), produced by doping with trivalent atoms (e.g. boron). The holes behave like positive mobile charge carriers and carry the current through the material.", "source": "https://50ohm.de/NEA_halbleiter_2.html#AB107", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AB108": { - "revision": 2, - "explanation": "At the PN junction, electrons diffuse from the N side and recombine with holes on the P side, leaving a depleted insulating region at the boundary.", + "revision": 3, + "explanation": "At the PN junction, free electrons diffuse from the n-side into the p-side and recombine with the holes there. This cancels carriers near the boundary, leaving behind fixed (immobile) ionised charges and forming a carrier-depleted barrier — the depletion (space-charge) layer. It acts as an insulator and initially blocks further charge transport.", "source": "https://50ohm.de/NEA_halbleiter_2.html#AB108", - "confidence": 7 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AB109": { - "revision": 2, - "explanation": "The shown polarity reverse-biases the diode, pulling majority carriers away from the junction, so the depletion region widens.", + "revision": 3, + "explanation": "Here the p-region sits at $0\\,\\text{V}$ and the n-region at $+5\\,\\text{V}$, so the diode is reverse-biased. The reverse voltage pulls the free carriers away from the junction, so the depletion zone widens.", "source": "https://50ohm.de/NEA_halbleiter_2.html#AB109", - "confidence": 7 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AB201": { "revision": 2, @@ -156,10 +180,11 @@ "confidence": 8 }, "AB202": { - "revision": 2, - "explanation": "Maximum power transfer occurs when the load resistance equals the source internal resistance.", + "revision": 3, + "explanation": "Maximum power reaches the load when the load resistance equals the source's internal resistance: $R_L = R_i$ (power matching). Larger or smaller $R_L$ delivers less. (Matching condition — a memory item, not a Formelsammlung formula.)", "source": "https://50ohm.de/NEA_innenwiderstand.html#AB202", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AB203": { "revision": 2, @@ -174,82 +199,95 @@ "confidence": 8 }, "AB205": { - "revision": 3, - "explanation": "The load current is $4.8\\,\\text{V} / 1.2\\,\\Omega = 4\\,\\text{A}$ and the source drops 0.2 V internally, so $R_i = 0.2\\,\\text{V} / 4\\,\\text{A} = 0.05\\,\\Omega$.", + "revision": 4, + "explanation": "Find the load current first: $I = \\frac{U}{R} = \\frac{4.8\\,\\text{V}}{1.2\\,\\Omega} = 4\\,\\text{A}$. The terminal voltage sagged by $U_i = 5.0\\,\\text{V} - 4.8\\,\\text{V} = 0.2\\,\\text{V}$, all dropped across $R_i$. So $R_i = \\frac{U_i}{I} = \\frac{0.2\\,\\text{V}}{4\\,\\text{A}} = 0.05\\,\\Omega$. Hilfsmittel: $R_i = \\Delta U/\\Delta I$ (Innenwiderstand, S.11).", "source": "https://50ohm.de/NEA_innenwiderstand.html#AB205", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AB206": { - "revision": 3, - "explanation": "The internal voltage drop is $13.5\\,\\text{V} - 12.4\\,\\text{V} = 1.1\\,\\text{V}$; dividing by 0.9 A gives about 1.22 ohm.", + "revision": 4, + "explanation": "The terminal voltage sags by $U_i = 13.5\\,\\text{V} - 12.4\\,\\text{V} = 1.1\\,\\text{V}$, dropped across the internal resistance at the delivered current $I = 0.9\\,\\text{A}$: $R_i = \\frac{U_i}{I} = \\frac{1.1\\,\\text{V}}{0.9\\,\\text{A}} \\approx 1.22\\,\\Omega$. Hilfsmittel: $R_i = \\Delta U/\\Delta I$ (Innenwiderstand, S.11).", "source": "https://50ohm.de/NEA_innenwiderstand.html#AB206", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AB207": { - "revision": 3, - "explanation": "The terminal voltage falls by 0.5 V at 2 A, so $R_i = 0.5\\,\\text{V} / 2\\,\\text{A} = 0.25\\,\\Omega$.", + "revision": 4, + "explanation": "The sag from the open-circuit voltage is $U_i = 13.5\\,\\text{V} - 13\\,\\text{V} = 0.5\\,\\text{V}$, dropped across $R_i$ at $I = 2\\,\\text{A}$: $R_i = \\frac{U_i}{I} = \\frac{0.5\\,\\text{V}}{2\\,\\text{A}} = 0.25\\,\\Omega$. Hilfsmittel: $R_i = \\Delta U/\\Delta I$ (Innenwiderstand, S.11).", "source": "https://50ohm.de/NEA_innenwiderstand.html#AB207", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AB208": { - "revision": 4, - "explanation": "The voltage drop is 0.2 V at 20 A, so $R_i = 0.2\\,\\text{V} / 20\\,\\text{A} = 0.01\\,\\Omega = 10\\,\\text{m}\\Omega$. Hilfsmittel: apply $R_i = \\Delta U/\\Delta I$ (Innenwiderstand, S.11).", + "revision": 5, + "explanation": "The sag is $U_i = 13.8\\,\\text{V} - 13.6\\,\\text{V} = 0.2\\,\\text{V}$, dropped across $R_i$ at $I = 20\\,\\text{A}$: $R_i = \\frac{U_i}{I} = \\frac{0.2\\,\\text{V}}{20\\,\\text{A}} = 0.01\\,\\Omega = 10\\,\\text{m}\\Omega$. Hilfsmittel: $R_i = \\Delta U/\\Delta I$ (Innenwiderstand, S.11).", "source": "https://50ohm.de/NEA_innenwiderstand.html#AB208", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AB209": { - "revision": 4, - "explanation": "The six 2 V cells are in series, so voltages add to 12 V while the ampere-hour capacity remains that of one cell, 10 Ah. Hilfsmittel: series voltages add: $U_G = U_1 + U_2$ (Spannungsteiler, S.12); the Ah capacity stays that of one cell (outside the sheet).", + "revision": 5, + "explanation": "Six cells of $2\\,\\text{V}$ in series: the voltages add, $6\\cdot 2\\,\\text{V} = 12\\,\\text{V}$. A series connection does not change the charge capacity, so it stays at the single-cell $10\\,\\text{Ah}$. The pack is rated $12\\,\\text{V}/10\\,\\text{Ah}$. Hilfsmittel: series voltages add, $U_G = U_1 + U_2 + \\dots$ (Reihenschaltung, S.12).", "source": "https://50ohm.de/NEA_akku.html#AB209", - "confidence": 7 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AB210": { - "revision": 2, - "explanation": "The mAh value on an accumulator pack states how much charge it can nominally deliver, so it is the nominal capacity.", + "revision": 3, + "explanation": "The mAh figure is a charge: how much current the cell can supply over time — e.g. $2200\\,\\text{mA}$ for one hour, or $1100\\,\\text{mA}$ for two. This stored-charge rating is the nominal capacity (Nennkapazität). The 7.4 V is voltage and the 16.28 Wh is energy.", "source": "https://50ohm.de/NEA_akku.html#AB210", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AB211": { - "revision": 4, - "explanation": "Discharging only down to 10 percent leaves 90 percent usable: $0.9 \\cdot 60\\,\\text{Ah} = 54\\,\\text{Ah}$; $54\\,\\text{Ah} / 0.8\\,\\text{A} = 67.5\\,\\text{h}$.", + "revision": 5, + "explanation": "Discharge stops at $10\\%$, so only $90\\%$ of the charge is usable: $Q = 0.9\\cdot 60\\,\\text{Ah} = 54\\,\\text{Ah}$. At a draw of $0.8\\,\\text{A}$ the run time is $t = \\frac{Q}{I} = \\frac{54\\,\\text{Ah}}{0.8\\,\\text{A}} = 67.5\\,\\text{h}$, i.e. 67 h 30 min.", "source": "https://50ohm.de/NEA_akku.html#AB211", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AB212": { - "revision": 2, - "explanation": "A solar cell converts incident light or other radiation energy directly into electrical energy by freeing charge carriers.", + "revision": 3, + "explanation": "A solar cell converts the radiant energy of sunlight directly into electrical energy. The underlying mechanism is the photovoltaic effect.", "source": "https://50ohm.de/NEA_photovoltaik.html#AB212", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AB213": { - "revision": 4, - "explanation": "Input power is $12\\,\\text{V} \\cdot 2\\,\\text{A} = 24\\,\\text{W}$ and output power is $5\\,\\text{V} \\cdot 3\\,\\text{A} = 15\\,\\text{W}$; efficiency is $15/24 = 62.5\\%$.", + "revision": 5, + "explanation": "Efficiency is output power over input power. $P_\\mathrm{in} = 12\\,\\text{V}\\cdot 2\\,\\text{A} = 24\\,\\text{W}$, $P_\\mathrm{out} = 5\\,\\text{V}\\cdot 3\\,\\text{A} = 15\\,\\text{W}$, so $\\eta = \\frac{15\\,\\text{W}}{24\\,\\text{W}} = 0.625 = 62.5\\%$. Hilfsmittel: $P = U\\cdot I$ and $\\eta = P_\\mathrm{ab}/P_\\mathrm{zu}$ (Leistung / Wirkungsgrad, S.12).", "source": "https://50ohm.de/NEA_spannungswandler.html#AB213", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AB214": { - "revision": 4, - "explanation": "Input power is $5\\,\\text{V} \\cdot 3\\,\\text{A} = 15\\,\\text{W}$ and output power is $12\\,\\text{V} \\cdot 1\\,\\text{A} = 12\\,\\text{W}$; efficiency is $12/15 = 80\\%$.", + "revision": 5, + "explanation": "$P_\\mathrm{in} = 5\\,\\text{V}\\cdot 3\\,\\text{A} = 15\\,\\text{W}$ and $P_\\mathrm{out} = 12\\,\\text{V}\\cdot 1\\,\\text{A} = 12\\,\\text{W}$, so $\\eta = \\frac{12\\,\\text{W}}{15\\,\\text{W}} = 0.8 = 80\\%$. Hilfsmittel: $P = U\\cdot I$ and $\\eta = P_\\mathrm{ab}/P_\\mathrm{zu}$ (Leistung / Wirkungsgrad, S.12).", "source": "https://50ohm.de/NEA_spannungswandler.html#AB214", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AB301": { - "revision": 5, - "explanation": "For a sine current, $I_\\mathrm{eff} = I_\\mathrm{max} / \\sqrt{2}$; power is $I_\\mathrm{eff}^2 R = (0.5/\\sqrt{2})^2 \\cdot 20 = 2.5\\,\\text{W}$. Hilfsmittel: first $I_\\mathrm{eff} = \\hat{I}/\\sqrt{2}$ (from $\\hat{U} = U_\\mathrm{eff}\\cdot\\sqrt{2}$, Wechselspannung, S.12), then $P = I_\\mathrm{eff}^2\\cdot R$ (Leistung, S.12).", + "revision": 6, + "explanation": "AC power uses RMS values. From the peak current $\\hat{I} = 0.5\\,\\text{A}$, the RMS value is $I_\\mathrm{eff} = \\frac{\\hat{I}}{\\sqrt{2}}$, so $P = I_\\mathrm{eff}^2\\cdot R = \\left(\\frac{0.5\\,\\text{A}}{\\sqrt{2}}\\right)^2\\cdot 20\\,\\Omega = 0.125\\,\\text{A}^2\\cdot 20\\,\\Omega = 2.5\\,\\text{W}$. Hilfsmittel: $\\hat{U} = U_\\mathrm{eff}\\sqrt{2}$ (Wechselspannung, S.12) and $P = I^2 R$ (Leistung, S.12).", "source": "https://50ohm.de/NEA_wechselstrom_leistung.html#AB301", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AB302": { - "revision": 3, - "explanation": "Point X3 is three quarters of a cycle after zero, which is 270 degrees or $3 \\pi / 2$ radians.", + "revision": 4, + "explanation": "One full cycle $T$ spans the figure's four segments, so each segment is $\\frac{360^\\circ}{4} = 90^\\circ$. Point $X_3$ lies three segments in: $3\\cdot 90^\\circ = 270^\\circ$, which is $\\frac{3\\pi}{2}$ in radians.", "source": "https://50ohm.de/NEA_phase.html#AB302", - "confidence": 7 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AB303": { - "revision": 3, - "explanation": "The two sine waves are shifted by one eighth of a full cycle; $360^\\circ / 8 = 45^\\circ$.", + "revision": 4, + "explanation": "Signal $b$ completes one full period over eight divisions, so each division is $\\frac{360^\\circ}{8} = 45^\\circ$ (i.e. $\\frac{\\pi}{4}$). The two traces are offset by exactly one division, so the phase difference is $45^\\circ$.", "source": "https://50ohm.de/NEA_phase.html#AB303", - "confidence": 7 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AB401": { "revision": 3, @@ -300,16 +338,18 @@ "confidence": 8 }, "AB409": { - "revision": 3, - "explanation": "Noise power scales with bandwidth; changing from 2.5 kHz to 0.5 kHz is a factor of 5 reduction, and $10\\log_{10}(5)$ is about 7 dB.", + "revision": 4, + "explanation": "Thermal-noise power scales with bandwidth: $\\Delta p_R = 10\\cdot\\log_{10}\\left(\\frac{B_1}{B_2}\\right) = 10\\cdot\\log_{10}\\left(\\frac{0.5\\,\\text{kHz}}{2.5\\,\\text{kHz}}\\right) \\approx -7\\,\\text{dB}$. Narrowing the filter fivefold cuts the noise level by about 7 dB. Hilfsmittel: $\\Delta p_R = 10\\cdot\\log_{10}(B_1/B_2)$ (Thermisches Rauschen, S.16).", "source": "https://50ohm.de/NEA_rauschen.html#AB409", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AB501": { - "revision": 6, - "explanation": "Stored energy in watt-hours is voltage times ampere-hour capacity: $12\\,\\text{V} \\cdot 5\\,\\text{Ah} = 60\\,\\text{Wh}$. Hilfsmittel: $W = U\\cdot I\\cdot t$, i.e. $W = P\\cdot t$ with $P = U\\cdot I$ (Arbeit/Leistung, S.12); $U\\cdot\\text{Ah}$ gives Wh.", + "revision": 7, + "explanation": "Stored energy is voltage times charge: $W = U\\cdot Q = 12\\,\\text{V}\\cdot 5\\,\\text{Ah} = 60\\,\\text{Wh}$. (This follows from $W = P\\cdot t = U\\cdot I\\cdot t = U\\cdot Q$.) Hilfsmittel: $W = P\\cdot t$, $P = U\\cdot I$ (Arbeit / Leistung, S.12).", "source": "https://50ohm.de/NEA_akku.html#AB501", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AB502": { "revision": 6, @@ -330,112 +370,130 @@ "confidence": 7 }, "AC101": { - "revision": 2, - "explanation": "In an ideal capacitor the current is proportional to the rate of voltage change, so current reaches its extrema a quarter cycle before voltage: it leads by 90 degrees.", + "revision": 3, + "explanation": "In a lossless capacitor the current leads the voltage by $90^\\circ$. A handy German mnemonic: \"Beim Kondensat-ooo-r eilt der Strom v-ooo-r\" — at the capacitor, the current runs ahead.", "source": "https://50ohm.de/NEA_kondensator_2.html#AC101", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AC102": { - "revision": 3, - "explanation": "Capacitive reactance is negative in AC sign convention, and its magnitude is $1/(2 \\pi f C)$, so it depends on frequency and capacitance.", + "revision": 4, + "explanation": "From $|X_C| = \\frac{1}{\\omega C} = \\frac{1}{2\\pi f C}$, the reactance depends on frequency and capacitance, falling as either rises. It is written with a negative sign because in a capacitor the current leads the voltage by $90^\\circ$ (opposite to an inductor). Hilfsmittel: $X_C = 1/(\\omega C)$ (Kapazitiver Blindwiderstand, S.13).", "source": "https://50ohm.de/NEA_kondensator_2.html#AC102", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AC103": { - "revision": 2, - "explanation": "A pure reactance stores and returns energy instead of converting it to heat, so the ideal reactive resistance has no heat loss.", + "revision": 3, + "explanation": "An ideal reactance — capacitive or inductive — dissipates no heat. With voltage and current $90^\\circ$ apart, the instantaneous power is positive and negative in equal measure and averages to zero over a cycle, so no active (real) power is converted. Only the ohmic resistance turns power into heat.", "source": "https://50ohm.de/NEA_kondensator_2.html#AC103", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AC104": { - "revision": 3, - "explanation": "Use $X_C = 1/(2 \\pi f C)$; with 100 MHz and 10 pF this gives about 159 ohm.", + "revision": 4, + "explanation": "$X_C = \\frac{1}{\\omega C} = \\frac{1}{2\\pi f C} = \\frac{1}{2\\pi\\cdot 100\\,\\text{MHz}\\cdot 10\\,\\text{pF}} \\approx 159\\,\\Omega$. Mind the powers of ten: $f = 10^8\\,\\text{Hz}$ and $C = 10^{-11}\\,\\text{F}$, so $2\\pi f C \\approx 6.28\\cdot 10^{-3}$ and its reciprocal is about $159$. Hilfsmittel: $X_C = 1/(\\omega C)$ with $\\omega = 2\\pi f$ (Kapazitiver Blindwiderstand, S.13).", "source": "https://50ohm.de/NEA_kondensator_2.html#AC104", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AC105": { - "revision": 3, - "explanation": "Use $X_C = 1/(2 \\pi f C)$; with 145 MHz and 50 pF this is about 22 ohm.", + "revision": 4, + "explanation": "$X_C = \\frac{1}{\\omega C} = \\frac{1}{2\\pi f C} = \\frac{1}{2\\pi\\cdot 145\\,\\text{MHz}\\cdot 50\\,\\text{pF}} \\approx 22\\,\\Omega$. Hilfsmittel: $X_C = 1/(\\omega C)$ with $\\omega = 2\\pi f$ (Kapazitiver Blindwiderstand, S.13).", "source": "https://50ohm.de/NEA_kondensator_2.html#AC105", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AC106": { - "revision": 3, - "explanation": "Use $X_C = 1/(2 \\pi f C)$; with 100 MHz and 100 pF this gives about 15.9 ohm.", + "revision": 4, + "explanation": "$X_C = \\frac{1}{\\omega C} = \\frac{1}{2\\pi f C} = \\frac{1}{2\\pi\\cdot 100\\,\\text{MHz}\\cdot 100\\,\\text{pF}} \\approx 15.9\\,\\Omega$. Hilfsmittel: $X_C = 1/(\\omega C)$ with $\\omega = 2\\pi f$ (Kapazitiver Blindwiderstand, S.13).", "source": "https://50ohm.de/NEA_kondensator_2.html#AC106", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AC107": { - "revision": 3, - "explanation": "Use $X_C = 1/(2 \\pi f C)$; with 435 MHz and 100 pF this gives about 3.7 ohm.", + "revision": 4, + "explanation": "$X_C = \\frac{1}{\\omega C} = \\frac{1}{2\\pi f C} = \\frac{1}{2\\pi\\cdot 435\\,\\text{MHz}\\cdot 100\\,\\text{pF}} \\approx 3.7\\,\\Omega$. Hilfsmittel: $X_C = 1/(\\omega C)$ with $\\omega = 2\\pi f$ (Kapazitiver Blindwiderstand, S.13).", "source": "https://50ohm.de/NEA_kondensator_2.html#AC107", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AC108": { - "revision": 6, - "explanation": "First find reactance from $X_C = U/I = 16\\,\\text{V} / 0.032\\,\\text{A} = 500\\,\\Omega$; then $C = 1/(2 \\pi f X_C)$ gives about 6.37 microfarad. Hilfsmittel: first $X_C = U/I$ (Ohmsches Gesetz, S.11 for reactance), then $C = 1/(2\\pi\\cdot f\\cdot X_C)$ ($X_C = 1/(\\omega C)$, $\\omega = 2\\pi f$, S.13/12).", + "revision": 7, + "explanation": "First the reactance from Ohm's law: $X_C = \\frac{U}{I} = \\frac{16\\,\\text{V}}{32\\,\\text{mA}} = 500\\,\\Omega$. Then solve $X_C = \\frac{1}{2\\pi f C}$ for the capacitance: $C = \\frac{1}{2\\pi f X_C} = \\frac{1}{2\\pi\\cdot 50\\,\\text{Hz}\\cdot 500\\,\\Omega} \\approx 6.37\\,\\mu\\text{F}$. Hilfsmittel: $X_C = 1/(\\omega C)$ (Kapazitiver Blindwiderstand, S.13).", "source": "https://50ohm.de/NEA_kondensator_2.html#AC108", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AC109": { - "revision": 2, - "explanation": "Real capacitors are not ideal: dielectric loss and lead or ESR losses convert some energy into heat under AC operation.", + "revision": 3, + "explanation": "Yes. A real capacitor has loss resistance both in the dielectric (insulator leakage) and in the plates and leads (equivalent series resistance, ESR), so a small amount of power is turned into heat.", "source": "https://50ohm.de/NEA_kondensator_2.html#AC109", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AC110": { - "revision": 2, - "explanation": "Capacitor loss is commonly described by loss factor tan delta; high loss means low quality factor, with tan delta equal to the reciprocal of Q.", + "revision": 3, + "explanation": "Capacitor loss is usually quoted as the loss factor (loss tangent) $\\tan\\delta$, which is the reciprocal of the quality factor: $\\tan\\delta = 1/Q$. A smaller $\\tan\\delta$ (higher $Q$) means the capacitor is closer to ideal.", "source": "https://50ohm.de/NEA_kondensator_2.html#AC110", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AC111": { - "revision": 2, - "explanation": "An ideal capacitor draws reactive current but no real power in steady-state AC, so the real power is approximately zero.", + "revision": 3, + "explanation": "In steady state an ideal capacitor draws no active power: voltage and current are $90^\\circ$ apart, so energy only shuttles into and out of the electric field and averages to zero over a cycle. A real capacitor's small parasitic losses make the figure approximately, not exactly, $0\\,\\text{W}$.", "source": "https://50ohm.de/NEA_kondensator_2.html#AC111", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AC201": { - "revision": 2, - "explanation": "In an ideal inductor the magnetic field opposes current changes, so current lags the applied voltage by 90 degrees.", + "revision": 3, + "explanation": "In an ideal inductor the voltage leads the current by $90^\\circ$ — equivalently, the current lags the voltage by $90^\\circ$. Self-induction opposes every change of current and so delays it.", "source": "https://50ohm.de/NEA_spule_2.html#AC201", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AC202": { - "revision": 3, - "explanation": "Inductive reactance is positive in AC sign convention and has magnitude $X_L = 2 \\pi f L$, so it depends on frequency and inductance.", + "revision": 4, + "explanation": "$X_L = 2\\pi f L$ is positive and grows with both the frequency and the inductance. Hilfsmittel: $X_L = \\omega L$ with $\\omega = 2\\pi f$ (Induktiver Blindwiderstand, S.13).", "source": "https://50ohm.de/NEA_spule_2.html#AC202", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AC203": { - "revision": 2, - "explanation": "With DC only the winding resistance limits current; with AC the inductive reactance is added, so the total impedance is higher and current is smaller.", + "revision": 3, + "explanation": "With DC, after the switch-on transient only the coil's ohmic winding resistance limits the current. With AC the inductive reactance $X_L = 2\\pi f L$ adds to the opposition, so at the same voltage the AC current is smaller. Hilfsmittel: $X_L = \\omega L$ (Induktiver Blindwiderstand, S.13).", "source": "https://50ohm.de/NEA_spule_2.html#AC203", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AC204": { - "revision": 4, - "explanation": "Use $X_L = 2 \\pi f L$; $2 \\pi \\cdot 100\\,\\text{MHz} \\cdot 3\\,\\mu\\text{H}$ is about 1885 ohm.", + "revision": 5, + "explanation": "$X_L = \\omega L = 2\\pi f L = 2\\pi\\cdot 100\\,\\text{MHz}\\cdot 3\\,\\mu\\text{H} \\approx 1885\\,\\Omega$. The prefixes cancel ($10^6\\cdot 10^{-6} = 1$), leaving $2\\pi\\cdot 100\\cdot 3 \\approx 1885$. Hilfsmittel: $X_L = \\omega L$ with $\\omega = 2\\pi f$ (Induktiver Blindwiderstand, S.13).", "source": "https://50ohm.de/NEA_spule_2.html#AC204", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AC205": { - "revision": 5, - "explanation": "For a core with AL value, $L = N^2 \\cdot A_L$; $14^2 \\cdot 1.5\\,\\text{nH} = 294\\,\\text{nH} = 0.294\\,\\mu\\text{H}$. Hilfsmittel: apply $L = N^2\\cdot A_L$ (Ringkernspulen $L = N^2\\cdot A_L$, S.13).", + "revision": 6, + "explanation": "For a core given by its $A_L$ value, $L = N^2\\cdot A_L = 14^2\\cdot 1.5\\,\\text{nH} = 294\\,\\text{nH} = 0.294\\,\\mu\\text{H}$. Hilfsmittel: $L = N^2\\cdot A_L$ (Ringkernspulen, S.13).", "source": "https://50ohm.de/NEA_spule_2.html#AC205", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AC206": { - "revision": 5, - "explanation": "Use $L = N^2 \\cdot A_L$; $300^2 \\cdot 1250\\,\\text{nH} = 112500000\\,\\text{nH} = 112.5\\,\\text{mH}$. Hilfsmittel: apply $L = N^2\\cdot A_L$ (Ringkernspulen $L = N^2\\cdot A_L$, S.13).", + "revision": 6, + "explanation": "$L = N^2\\cdot A_L = 300^2\\cdot 1250\\,\\text{nH} = 1.125\\cdot 10^8\\,\\text{nH} = 112.5\\,\\text{mH}$. Hilfsmittel: $L = N^2\\cdot A_L$ (Ringkernspulen, S.13).", "source": "https://50ohm.de/NEA_spule_2.html#AC206", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AC207": { - "revision": 5, - "explanation": "Rearrange to $N = \\sqrt{L/A_L}$; $\\sqrt{2\\,\\text{mH} / 250\\,\\text{nH}} = \\sqrt{8000}$, about 89 turns. Hilfsmittel: rearrange $L = N^2\\cdot A_L \\to N = \\sqrt{L/A_L}$ (Ringkernspulen $L = N^2\\cdot A_L$, S.13).", + "revision": 6, + "explanation": "Solve $L = N^2 A_L$ for the turns: $N = \\sqrt{\\frac{L}{A_L}}$. Put both in the same unit — $2\\,\\text{mH} = 2\\cdot 10^6\\,\\text{nH}$ — so $N = \\sqrt{\\frac{2\\cdot 10^6\\,\\text{nH}}{250\\,\\text{nH}}} = \\sqrt{8000} \\approx 89$, i.e. about 90 turns. Hilfsmittel: $L = N^2\\cdot A_L$ (Ringkernspulen, S.13).", "source": "https://50ohm.de/NEA_spule_2.html#AC207", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AC208": { "revision": 6, @@ -444,118 +502,137 @@ "confidence": 8 }, "AC209": { - "revision": 2, - "explanation": "Coil losses are represented by an equivalent series resistance; the loss factor tan delta is used and equals the reciprocal of the quality factor Q.", + "revision": 3, + "explanation": "A real coil's losses fold into a series resistance $R_S$. Its quality is $Q = X_L/R_S$, and the loss factor is the reciprocal: $\\tan\\delta = 1/Q = R_S/X_L$. More loss means a larger $R_S$, a lower $Q$, and a larger $\\tan\\delta$.", "source": "https://50ohm.de/NEA_spule_2.html#AC209", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AC210": { - "revision": 2, - "explanation": "A conductive metal enclosure shields the electric field around the tuned-circuit coil and reduces unwanted radiation from it.", + "revision": 3, + "explanation": "To reduce radiation from (and pickup into) a tuned-circuit coil, house it in a conductive metal can; the conductive shield contains the electromagnetic field. A plastic enclosure provides no shielding.", "source": "https://50ohm.de/NEA_spule_2.html#AC210", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AC211": { - "revision": 2, - "explanation": "A choke core is normally ferrite because ferrite gives high magnetic permeability and high RF loss for unwanted common-mode currents.", + "revision": 3, + "explanation": "RF chokes use ferrite cores: ferrite combines high magnetic permeability with low eddy-current loss at high frequencies, so it raises the inductance without adding much loss. Steel would suffer heavy eddy-current losses; plastic or diamagnetic material barely affects the field.", "source": "https://50ohm.de/NEA_spule_2.html#AC211", - "confidence": 7 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AC301": { - "revision": 2, - "explanation": "Mutual induction needs a changing magnetic field, so a changing current in a magnetically coupled neighboring coil induces voltage in the other coil.", + "revision": 3, + "explanation": "Mutual induction produces a voltage whenever the magnetic field linking the coil changes — for instance when a changing current flows in a magnetically coupled neighbouring coil. A steady DC current creates no changing field and so induces nothing.", "source": "https://50ohm.de/NEA_uebertrager_2.html#AC301", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AC302": { - "revision": 6, - "explanation": "With losses neglected, primary and secondary power are equal: $6\\,\\text{V} \\cdot 1.15\\,\\text{A} = 6.9\\,\\text{W}$, and $6.9\\,\\text{W} / 230\\,\\text{V} = 0.030\\,\\text{A}$. Hilfsmittel: the transformer relation $U_P\\cdot I_P = U_S\\cdot I_S$ (from $\\ddot{u} = U_P/U_S = I_S/I_P$, S.13) gives the primary current; equivalently equal powers $P = U\\cdot I$ (S.12).", + "revision": 7, + "explanation": "Neglecting losses, input power equals output power, so $U_P I_P = U_S I_S$. Hence $I_P = \\frac{U_S I_S}{U_P} = \\frac{6\\,\\text{V}\\cdot 1.15\\,\\text{A}}{230\\,\\text{V}} \\approx 0.03\\,\\text{A} = 30\\,\\text{mA}$. Hilfsmittel: $U_P/U_S = I_S/I_P$ (Transformator-Übersetzungsverhältnis, S.13).", "source": "https://50ohm.de/NEA_uebertrager_2.html#AC302", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AC303": { - "revision": 4, - "explanation": "Impedance transforms with the square of the turns ratio; with 1:4, the input sees $16\\,\\text{k}\\Omega / 4^2 = 1\\,\\text{k}\\Omega$. Hilfsmittel: impedance transforms with the turns ratio squared (Übersetzungsverhältnis $\\ddot{u} = N_P/N_S = U_P/U_S = I_S/I_P = \\sqrt{Z_P/Z_S}$, S.13).", + "revision": 5, + "explanation": "Impedance transforms as the square of the turns ratio. With a $1:4$ winding, the secondary $16\\,\\text{k}\\Omega$ appears on the primary divided by $4^2 = 16$: $Z_P = \\frac{16\\,\\text{k}\\Omega}{16} = 1\\,\\text{k}\\Omega$. Hilfsmittel: $\\ddot{u} = \\sqrt{Z_P/Z_S}$ (Transformator-Übersetzungsverhältnis, S.13).", "source": "https://50ohm.de/NEA_uebertrager_2.html#AC303", - "confidence": 7 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AC304": { - "revision": 4, - "explanation": "The same 1:4 transformer reflects the secondary load by a factor of 16, so $6.4\\,\\text{k}\\Omega / 16 = 0.4\\,\\text{k}\\Omega$ at a-b. Hilfsmittel: impedance transforms with the turns ratio squared (Übersetzungsverhältnis $\\ddot{u} = N_P/N_S = U_P/U_S = I_S/I_P = \\sqrt{Z_P/Z_S}$, S.13).", + "revision": 5, + "explanation": "With the $1:4$ winding the impedance transforms by $4^2 = 16$: $Z_P = \\frac{Z_S}{16} = \\frac{6.4\\,\\text{k}\\Omega}{16} = 0.4\\,\\text{k}\\Omega$. Hilfsmittel: $\\ddot{u} = \\sqrt{Z_P/Z_S}$ (Transformator-Übersetzungsverhältnis, S.13).", "source": "https://50ohm.de/NEA_uebertrager_2.html#AC304", - "confidence": 7 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AC305": { - "revision": 3, - "explanation": "The impedance ratio is $450/50 = 9$; turns ratio is the square root of impedance ratio, so $\\sqrt{9} = 3$.", + "revision": 4, + "explanation": "Required impedance ratio: $\\frac{450\\,\\Omega}{50\\,\\Omega} = 9$. Because impedance goes as the square of the turns ratio, the turns ratio is $\\sqrt{9} = 3$, i.e. $3:1$. Hilfsmittel: $\\ddot{u} = \\sqrt{Z_P/Z_S}$ (Transformator-Übersetzungsverhältnis, S.13).", "source": "https://50ohm.de/NEA_uebertrager_2.html#AC305", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AC306": { - "revision": 2, - "explanation": "A 2.5 kOhm load against 50 ohm is about a 50:1 impedance ratio, close to 49:1, so the turns ratio is about 1:7.", + "revision": 3, + "explanation": "Required impedance ratio: $\\frac{2500\\,\\Omega}{50\\,\\Omega} = 50$. The turns ratio is the square root: $\\sqrt{50} \\approx 7$, i.e. about $1:7$. Hilfsmittel: $\\ddot{u} = \\sqrt{Z_P/Z_S}$ (Transformator-Übersetzungsverhältnis, S.13).", "source": "https://50ohm.de/NEA_uebertrager_2.html#AC306", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AC307": { - "revision": 7, - "explanation": "The wire area is $\\pi d^2/4 = \\pi \\cdot 0.5^2/4 = 0.196\\,\\text{mm}^2$; at $2.5\\,\\text{A/mm}^2$ the current is about $0.49\\,\\text{A}$. Hilfsmittel: first $A = d^2\\pi/4$ (S.11), then $I = S\\cdot A$ with $S \\approx 2{,}5\\,\\text{A/mm}^2$ (Belastbarkeit von Wicklungen, S.13).", + "revision": 8, + "explanation": "Cross-section first: $A_\\mathrm{Dr} = d^2\\cdot\\frac{\\pi}{4} = (0.5\\,\\text{mm})^2\\cdot\\frac{\\pi}{4} \\approx 0.196\\,\\text{mm}^2$. Then the permissible current from the current density: $I = S\\cdot A_\\mathrm{Dr} = 2.5\\,\\text{A/mm}^2\\cdot 0.196\\,\\text{mm}^2 \\approx 0.49\\,\\text{A}$. Hilfsmittel: $A_\\mathrm{Dr} = d^2\\pi/4$ (Widerstand von Drähten, S.11) and $I = S\\cdot A_\\mathrm{Dr}$ (Belastbarkeit von Wicklungen, S.13).", "source": "https://50ohm.de/NEA_uebertrager_2.html#AC307", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AC401": { - "revision": 2, - "explanation": "In forward bias, the depletion region is reduced and electrons can cross the PN junction from the N side to the P side.", + "revision": 3, + "explanation": "Forward bias shrinks the depletion layer so carriers can cross. Electrons flow from the n-side (electron surplus) into the p-side (electron deficit) — that is, from N to P.", "source": "https://50ohm.de/NEA_diode_2.html#AC401", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AC402": { - "revision": 2, - "explanation": "Electrons are the majority carriers in the N region, and in forward operation they move across the junction into the P region.", + "revision": 3, + "explanation": "In forward bias the barrier shrinks and electrons move from the n-region into the p-region (physical electron flow from − to +); the holes drift the opposite way.", "source": "https://50ohm.de/NEA_halbleiter_2.html#AC402", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AC403": { - "revision": 2, - "explanation": "As temperature rises, diode saturation current increases, so the forward voltage needed for a given current falls.", + "revision": 3, + "explanation": "A diode's forward voltage falls as temperature rises: warmer carriers need less energy to cross the junction. For a silicon diode it drops by roughly $2\\,\\text{mV}/^\\circ\\text{C}$.", "source": "https://50ohm.de/NEA_diode_2.html#AC403", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AC404": { - "revision": 2, - "explanation": "A varicap is reverse-biased; lower reverse voltage makes the depletion region narrower, which increases junction capacitance.", + "revision": 3, + "explanation": "A varicap is run reverse-biased, with the depletion layer acting as the capacitor gap. Lowering the reverse voltage narrows that layer and raises the capacitance (raising it widens the layer and lowers the capacitance). The plate-capacitor formula $C = \\varepsilon_0\\,\\varepsilon_r\\,\\frac{A}{d}$ makes it clear: $d$ is the depletion width, and smaller $d$ means larger $C$.", "source": "https://50ohm.de/NEA_diode_2.html#AC404", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AC405": { - "revision": 2, - "explanation": "Antiparallel silicon diodes clip the waveform when either polarity exceeds about 0.6 V, so the output is the sine wave limited at that threshold.", + "revision": 3, + "explanation": "Two silicon diodes sit antiparallel across the output, clipping it in both directions. While the voltage stays below about $0.6\\,\\text{V}$ both diodes block and the signal passes almost unchanged; once it exceeds a silicon diode's forward voltage, one diode conducts and holds the output near $\\pm 0.6\\,\\text{V}$. The output is therefore the input with its peaks flattened at $+0.6\\,\\text{V}$ and $-0.6\\,\\text{V}$.", "source": "https://50ohm.de/NEA_diode_2.html#AC405", - "confidence": 7 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AC406": { - "revision": 2, - "explanation": "Germanium diodes have a lower threshold, about 0.3 V, so the same limiter clips the waveform earlier and more strongly than silicon diodes.", + "revision": 3, + "explanation": "The same antiparallel clipper, but germanium diodes conduct already at about $0.3\\,\\text{V}$. The peaks are clipped earlier than with silicon, so the output is limited more tightly — to roughly $\\pm 0.3\\,\\text{V}$.", "source": "https://50ohm.de/NEA_diode_2.html#AC406", - "confidence": 7 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AC407": { - "revision": 2, - "explanation": "A photodiode generates electron-hole pairs when illuminated and can produce photocurrent from light.", + "revision": 3, + "explanation": "A photodiode uses the photoelectric effect: light hitting the PN junction frees charge carriers, so a current can flow — it generates current itself. A photoresistor merely changes its resistance and produces no current of its own.", "source": "https://50ohm.de/NEA_diode_2.html#AC407", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AC408": { - "revision": 2, - "explanation": "An optocoupler transfers a signal optically between an LED and a photosensitive device, giving galvanic isolation between the two circuits.", + "revision": 3, + "explanation": "An optocoupler pairs an LED with a light-sensitive semiconductor (e.g. a phototransistor) and passes the signal across as light. Input and output therefore share no electrical connection. This galvanic isolation shields circuits from interference, ground-loop currents and high voltages.", "source": "https://50ohm.de/NEA_diode_2.html#AC408", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AC501": { - "revision": 2, - "explanation": "In a bipolar transistor, a small base current controls a larger collector current, so it is current-controlled.", + "revision": 3, + "explanation": "A bipolar transistor needs a base current before the larger collector-emitter current can flow, so it is a current-controlled device. Field-effect transistors, by contrast, are voltage- (field-) controlled.", "source": "https://50ohm.de/NEA_transistor_2.html#AC501", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AC502": { "revision": 2, @@ -564,22 +641,25 @@ "confidence": 8 }, "AC503": { - "revision": 2, - "explanation": "An NPN transistor has an N emitter, P base, and N collector, so the p-doped region is the base.", + "revision": 3, + "explanation": "An NPN transistor is two n-doped regions with a thin p-doped region between them. That central p-region is the thin control layer and is connected to the base terminal.", "source": "https://50ohm.de/NEA_transistor_2.html#AC503", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AC504": { - "revision": 2, - "explanation": "A PNP transistor has a P emitter, N base, and P collector, so the n-doped region is the base.", + "revision": 3, + "explanation": "A PNP transistor is two p-doped regions surrounding a thin n-doped region. That central n-region is the control layer and is connected to the base terminal. (Memorise NPN; PNP is simply the reverse.)", "source": "https://50ohm.de/NEA_transistor_2.html#AC504", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AC505": { - "revision": 2, - "explanation": "For a bipolar transistor to conduct normally, the base-emitter junction is forward biased.", + "revision": 3, + "explanation": "For a bipolar transistor to conduct, its base-emitter junction must be forward-biased, just like an ordinary diode. Only then does a base current flow to switch the transistor on.", "source": "https://50ohm.de/NEA_transistor_2.html#AC505", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AC506": { "revision": 2, @@ -636,58 +716,67 @@ "confidence": 8 }, "AC515": { - "revision": 4, - "explanation": "Base current is $5\\,\\text{mA} / 298 = 16.8\\,\\mu\\text{A}$; with about 0.6 V base-emitter drop, $R_1 = (12 - 0.6)\\,\\text{V} / 16.8\\,\\mu\\text{A}$, about 680 kOhm. Hilfsmittel: first base current $I_B = I_C/B$ (from $B = I_C/I_B$, S.14), then $R = U/I$ (Ohmsches Gesetz, S.11); the 0,6 V base drop is added knowledge.", + "revision": 5, + "explanation": "Base current: $I_B = \\frac{I_C}{B} = \\frac{5\\,\\text{mA}}{298} \\approx 16.8\\,\\mu\\text{A}$. The base-emitter junction drops about 0.6 V, leaving $U_{R_1} = 12\\,\\text{V} - 0.6\\,\\text{V} = 11.4\\,\\text{V}$ across $R_1$, so $R_1 = \\frac{U_{R_1}}{I_B} = \\frac{11.4\\,\\text{V}}{16.8\\,\\mu\\text{A}} \\approx 680\\,\\text{k}\\Omega$. Hilfsmittel: $B = I_C/I_B$ (Transistor, S.14); the 0.6 V base-emitter drop is a memory item.", "source": "https://50ohm.de/NEA_transistor_2.html#AC515", - "confidence": 7 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AC516": { - "revision": 2, - "explanation": "Making the divider current much larger than base current keeps the base voltage mostly set by the divider, so transistor beta and temperature changes disturb the operating point less.", + "revision": 3, + "explanation": "If the divider current through $R_2$ is about ten times the base current, the base draws only a small fraction of it and the base voltage stays nearly constant. The operating point then barely moves when the current gain $B$ or the temperature changes, so the amplifier stays stable.", "source": "https://50ohm.de/NEA_transistor_2.html#AC516", - "confidence": 7 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AC517": { - "revision": 4, - "explanation": "Base current is $2\\,\\text{mA}/200 = 10\\,\\mu\\text{A}$; R2 carries ten times that, so R1 carries 110 microampere. With 1 V at the emitter, the base is about 1.6 V, giving $R_1 = 8.4\\,\\text{V}/110\\,\\mu\\text{A} = 76.4\\,\\text{k}\\Omega$. Hilfsmittel: first base current $I_B = I_C/B$ ($B = I_C/I_B$, S. 14); $R_2$ carries ten times that and $R_1$ the sum, then $R = U/I$ (Ohmsches Gesetz, S. 11). The 0,6 V base-emitter drop is added knowledge.", + "revision": 5, + "explanation": "Base current $I_B = \\frac{I_C}{B} = \\frac{2\\,\\text{mA}}{200} = 10\\,\\mu\\text{A}$; the divider carries $I_2 = 10\\,I_B = 100\\,\\mu\\text{A}$, so $R_1$ carries $I_1 = I_2 + I_B = 110\\,\\mu\\text{A}$. The base sits at the emitter-resistor drop plus the B-E drop, $U_B = 1\\,\\text{V} + 0.6\\,\\text{V} = 1.6\\,\\text{V}$, leaving $U_1 = 10\\,\\text{V} - 1.6\\,\\text{V} = 8.4\\,\\text{V}$ across $R_1$. Thus $R_1 = \\frac{8.4\\,\\text{V}}{110\\,\\mu\\text{A}} \\approx 76.4\\,\\text{k}\\Omega$. Hilfsmittel: $B = I_C/I_B$ (Transistor, S.14); 0.6 V B-E drop is a memory item.", "source": "https://50ohm.de/NEA_transistor_2.html#AC517", - "confidence": 7 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AC518": { - "revision": 4, - "explanation": "Base current is 10 microampere and R2 current is 100 microampere, so R1 current is 110 microampere; with the base near 0.6 V, $R_1 = 9.4\\,\\text{V}/110\\,\\mu\\text{A} = 85.5\\,\\text{k}\\Omega$. Hilfsmittel: first base current $I_B = I_C/B$ ($B = I_C/I_B$, S.14), then $R = U/I$ (Ohmsches Gesetz, S.11); the 0,6 V base drop is added knowledge.", + "revision": 5, + "explanation": "Base current $I_B = \\frac{I_C}{B} = \\frac{2\\,\\text{mA}}{200} = 10\\,\\mu\\text{A}$; the divider carries $I_2 = 10\\,I_B = 100\\,\\mu\\text{A}$, so $R_1$ carries $I_1 = I_2 + I_B = 110\\,\\mu\\text{A}$. With no emitter resistor here, only the B-E drop is subtracted: $U_1 = 10\\,\\text{V} - 0.6\\,\\text{V} = 9.4\\,\\text{V}$, so $R_1 = \\frac{9.4\\,\\text{V}}{110\\,\\mu\\text{A}} \\approx 85.5\\,\\text{k}\\Omega$. Hilfsmittel: $B = I_C/I_B$ (Transistor, S.14); 0.6 V B-E drop is a memory item.", "source": "https://50ohm.de/NEA_transistor_2.html#AC518", - "confidence": 7 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AC519": { - "revision": 2, - "explanation": "If R1 is open, the base receives no forward bias, the transistor switches off, and with no collector current the collector rises to the supply voltage.", + "revision": 3, + "explanation": "With $R_1$ broken, the base no longer receives positive bias from the divider; through $R_2$ it sits essentially at ground. The base-emitter voltage is then too small, no base current flows and the transistor blocks. With no collector current there is no drop across $R_C$, so the collector voltage rises to the full supply voltage.", "source": "https://50ohm.de/NEA_transistor_2.html#AC519", - "confidence": 7 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AC520": { - "revision": 2, - "explanation": "If R2 is open, the base is driven too strongly through R1, so the transistor saturates; collector current is then limited mainly by RC and collector voltage falls near saturation voltage.", + "revision": 3, + "explanation": "With $R_2$ broken, the lower half of the divider is gone and $R_1$ pulls the base nearly up to the supply. The base-emitter voltage rises sharply and the transistor saturates (fully on), so the collector current is now limited only by the collector resistor $R_C$ and the collector voltage drops to a very small value (about $0.1\\,\\text{V}$).", "source": "https://50ohm.de/NEA_transistor_2.html#AC520", - "confidence": 7 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AC521": { - "revision": 5, - "explanation": "The gate draws negligible current, so the divider gives $U_G = 44\\,\\text{V} \\cdot 1\\,\\text{k}\\Omega/(10\\,\\text{k}\\Omega + 1\\,\\text{k}\\Omega) = 4\\,\\text{V}$; with the source at reference, that is $U_\\mathrm{GS}$. Hilfsmittel: apply the divider $U_G = U\\cdot R/(R_1+R_2)$ (Spannungsteiler, S.12).", + "revision": 6, + "explanation": "A FET gate draws no current, so $R_1$ and $R_2$ form an unloaded divider: $U_\\mathrm{GS} = U_B\\cdot\\frac{R_2}{R_1 + R_2} = 44\\,\\text{V}\\cdot\\frac{1\\,\\text{k}\\Omega}{10\\,\\text{k}\\Omega + 1\\,\\text{k}\\Omega} = 44\\,\\text{V}\\cdot\\frac{1}{11} = 4\\,\\text{V}$. Hilfsmittel: unloaded divider $U_2/U_G = R_2/(R_1 + R_2)$ (Spannungsteiler, S.12).", "source": "https://50ohm.de/NEA_transistor_2.html#AC521", - "confidence": 7 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AC522": { - "revision": 5, - "explanation": "For a divider, $R_2 = R_1 U_G/(U_B - U_G)$; $10\\,\\text{k}\\Omega \\cdot 2.8/(44 - 2.8)$ gives about 680 ohm. Hilfsmittel: rearrange the divider for $R_2$ (Spannungsteiler, S.12).", + "revision": 6, + "explanation": "The gate divider is unloaded, so $U_\\mathrm{GS} = U_B\\cdot\\frac{R_2}{R_1 + R_2}$. Solving for $R_2$: $2.8\\,\\text{V} = 44\\,\\text{V}\\cdot\\frac{R_2}{10\\,000\\,\\Omega + R_2}$, hence $2.8\\cdot(10\\,000\\,\\Omega + R_2) = 44\\,R_2$, so $28\\,000\\,\\Omega = 41.2\\,R_2$ and $R_2 \\approx 680\\,\\Omega$. Hilfsmittel: unloaded divider $U_2/U_G = R_2/(R_1 + R_2)$ (Spannungsteiler, S.12).", "source": "https://50ohm.de/NEA_transistor_2.html#AC522", - "confidence": 7 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AC523": { - "revision": 5, - "explanation": "Conduction loss is $P = I^2 R$; $25^2 \\cdot 0.004\\,\\Omega = 2.5\\,\\text{W}$. Hilfsmittel: apply $P = I^2\\cdot R$ (Leistung, S.12).", + "revision": 6, + "explanation": "Conduction loss in the on-resistance: $P = I^2\\cdot R = (25\\,\\text{A})^2\\cdot 4\\,\\text{m}\\Omega = 625\\cdot 0.004\\,\\Omega = 2.5\\,\\text{W}$. Hilfsmittel: $P = I^2 R$ (Leistung, S.12).", "source": "https://50ohm.de/NEA_transistor_2.html#AC523", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AC524": { "revision": 3, @@ -720,112 +809,130 @@ "confidence": 8 }, "AD101": { - "revision": 4, - "explanation": "Series capacitors add by reciprocals: $1/C = 1/100\\,\\text{pF} + 1/47\\,\\text{pF} + 1/22\\,\\text{pF}$, giving about 13.0 pF. Hilfsmittel: series caps via $1/C_G = \\sum 1/C_i$ (Kapazität, S.13).", + "revision": 5, + "explanation": "Capacitors in series combine like resistors in parallel: $\\frac{1}{C_\\mathrm{ges}} = \\frac{1}{C_1} + \\frac{1}{C_2} + \\frac{1}{C_3}$. With $C_1 = 0.10\\,\\text{nF} = 100\\,\\text{pF}$: $\\frac{1}{C_\\mathrm{ges}} = \\left(\\frac{1}{100} + \\frac{1}{47} + \\frac{1}{22}\\right)\\frac{1}{\\text{pF}} \\approx 0.077\\,\\frac{1}{\\text{pF}}$, so $C_\\mathrm{ges} \\approx 13.0\\,\\text{pF}$. Hilfsmittel: $1/C_G = 1/C_1 + \\dots$ (C Reihe, S.13).", "source": "https://50ohm.de/NEA_reihe_parallel_gemischt.html#AD101", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AD102": { - "revision": 5, - "explanation": "Series inductances add directly: $2200\\,\\text{nH}$ is $2.2\\,\\mu\\text{H}$, $0.033\\,\\text{mH}$ is $33\\,\\mu\\text{H}$, so the sum is $2.2 + 33 + 150 = 185.2\\,\\mu\\text{H}$. Hilfsmittel: inductances in series add: $L_G = \\sum L_i$ (Induktivität, S.13).", + "revision": 6, + "explanation": "Inductances in series add. Convert to a common unit — $2200\\,\\text{nH} = 2.2\\,\\mu\\text{H}$ and $0.033\\,\\text{mH} = 33\\,\\mu\\text{H}$ — then $L_\\mathrm{ges} = 2.2\\,\\mu\\text{H} + 33\\,\\mu\\text{H} + 150\\,\\mu\\text{H} = 185.2\\,\\mu\\text{H}$. Hilfsmittel: $L_G = L_1 + L_2 + \\dots$ (L Reihe, S.13).", "source": "https://50ohm.de/NEA_reihenschaltung_spule.html#AD102", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AD103": { - "revision": 4, - "explanation": "The shown capacitances are effectively parallel, so they add: $100\\,\\text{pF} + 1500\\,\\text{pF} + 220\\,\\text{pF} + 1\\,\\text{pF} = 1821\\,\\text{pF}$. Hilfsmittel: parallel caps add: $C_G = \\sum C_i$ (Kapazität, S.13).", + "revision": 5, + "explanation": "In a parallel connection capacitances add, and the coil's self-capacitance lies in parallel too. Converting everything to pF: $C_\\mathrm{ges} = 100 + 1500 + 220 + 1 = 1821\\,\\text{pF}$. Hilfsmittel: $C_G = C_1 + C_2 + \\dots$ (C Parallel, S.13).", "source": "https://50ohm.de/NEA_reihe_parallel_gemischt.html#AD103", - "confidence": 7 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AD104": { - "revision": 4, - "explanation": "At 1 MHz and 1 nF, $X_C$ is about 159 ohm; the series impedance magnitude is $\\sqrt{100^2 + 159^2}$, about 188 ohm. Hilfsmittel: first $X_C = 1/(\\omega C)$ ($X_C = 1/(\\omega C)$ bzw. $X_L = \\omega L$, S.13), then $|Z| = \\sqrt{R^2+X^2}$ ($Z = \\sqrt{R^2+X^2}$, S.12).", + "revision": 5, + "explanation": "First the capacitive reactance: $X_C = \\frac{1}{2\\pi f C} = \\frac{1}{2\\pi\\cdot 1\\,\\text{MHz}\\cdot 1\\,\\text{nF}} \\approx 159\\,\\Omega$. In a series R-C, resistance and reactance add by Pythagoras: $|Z| = \\sqrt{R^2 + X_C^2} = \\sqrt{(100\\,\\Omega)^2 + (159\\,\\Omega)^2} \\approx 188\\,\\Omega$. Hilfsmittel: $X_C = 1/(\\omega C)$ (S.13) and $Z = \\sqrt{R^2 + X^2}$ (Wechselspannung, S.12).", "source": "https://50ohm.de/NEA_reihe_parallel_gemischt.html#AD104", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AD105": { - "revision": 4, - "explanation": "At 1 MHz and 100 microhenry, $X_L$ is about 628 ohm; $|Z| = \\sqrt{100^2 + 628^2}$, about 636 ohm. Hilfsmittel: first $X_L = \\omega L$ ($X_C = 1/(\\omega C)$ bzw. $X_L = \\omega L$, S.13), then $|Z| = \\sqrt{R^2+X^2}$ ($Z = \\sqrt{R^2+X^2}$, S.12).", + "revision": 5, + "explanation": "First the inductive reactance: $X_L = 2\\pi f L = 2\\pi\\cdot 1\\,\\text{MHz}\\cdot 100\\,\\mu\\text{H} \\approx 628\\,\\Omega$. In a series R-L, resistance and reactance add by Pythagoras: $|Z| = \\sqrt{R^2 + X_L^2} = \\sqrt{(100\\,\\Omega)^2 + (628\\,\\Omega)^2} \\approx 636\\,\\Omega$. Hilfsmittel: $X_L = \\omega L$ (S.13) and $Z = \\sqrt{R^2 + X^2}$ (Wechselspannung, S.12).", "source": "https://50ohm.de/NEA_reihe_parallel_gemischt.html#AD105", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AD106": { - "revision": 3, - "explanation": "If 1 mA flows through R3, the parallel section has 10 V across it; R2 draws another 1 mA, so 2 mA through R1 drops 20 V, making the total 30 V. Hilfsmittel: resistors add in series / combine in parallel (Reihen-/Parallelschaltung, S.12); voltages via $U = R\\cdot I$ (Ohmsches Gesetz, S.11).", + "revision": 4, + "explanation": "All three resistors are $10\\,\\text{k}\\Omega$. Since $R_2 \\parallel R_3$ are equal, the $1\\,\\text{mA}$ through $R_3$ is matched by $1\\,\\text{mA}$ through $R_2$, so $I_{R_1} = 2\\,\\text{mA}$. The total resistance is $R_1 + (R_2 \\parallel R_3) = 10\\,\\text{k}\\Omega + 5\\,\\text{k}\\Omega = 15\\,\\text{k}\\Omega$, giving $U = I\\cdot R = 2\\,\\text{mA}\\cdot 15\\,\\text{k}\\Omega = 30\\,\\text{V}$. Hilfsmittel: series $R_G = R_1 + R_2$ and parallel $R_G = R_1 R_2/(R_1+R_2)$ (Widerstandsnetzwerke, S.12).", "source": "https://50ohm.de/NEA_reihe_parallel_widerstandsnetz_2.html#AD106", - "confidence": 7 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AD107": { - "revision": 4, - "explanation": "R2 and R3 in parallel give $5\\,\\text{k}\\Omega$, in series with R1 gives $15\\,\\text{k}\\Omega$; $15\\,\\text{V} / 15\\,\\text{k}\\Omega = 1\\,\\text{mA}$ total, split equally so R3 has $0.5\\,\\text{mA}$. Hilfsmittel: parallel then series (Reihen-/Parallelschaltung, S.12); current via $I = U/R$ (Ohmsches Gesetz, S.11).", + "revision": 5, + "explanation": "The total resistance is $R_1 + (R_2 \\parallel R_3) = 10\\,\\text{k}\\Omega + 5\\,\\text{k}\\Omega = 15\\,\\text{k}\\Omega$, so the total current is $I_\\mathrm{ges} = \\frac{U}{R_\\mathrm{ges}} = \\frac{15\\,\\text{V}}{15\\,\\text{k}\\Omega} = 1\\,\\text{mA}$. It then splits equally between the two equal resistors $R_2$ and $R_3$, so $I_3 = \\frac{1\\,\\text{mA}}{2} = 0.5\\,\\text{mA}$. Hilfsmittel: series/parallel combination and the current divider (Widerstandsnetzwerke, S.12).", "source": "https://50ohm.de/NEA_reihe_parallel_widerstandsnetz_2.html#AD107", - "confidence": 7 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AD108": { - "revision": 4, - "explanation": "The total current is 1 mA, so the parallel section has 5 V across it; R2 power is $5^2/10000 = 0.0025\\,\\text{W} = 2.5\\,\\text{mW}$. Hilfsmittel: reduce the network (Reihen-/Parallelschaltung, S.12), then $P = U^2/R$ (Leistung, S.12).", + "revision": 5, + "explanation": "The total current is $I_\\mathrm{ges} = \\frac{15\\,\\text{V}}{15\\,\\text{k}\\Omega} = 1\\,\\text{mA}$, which splits equally, so $I_2 = 0.5\\,\\text{mA}$. The power in $R_2$ is then $P_2 = I_2^2\\cdot R_2 = (0.5\\,\\text{mA})^2\\cdot 10\\,\\text{k}\\Omega = 2.5\\,\\text{mW}$. Hilfsmittel: $P = I^2 R$ (Leistung, S.12).", "source": "https://50ohm.de/NEA_reihe_parallel_widerstandsnetz_2.html#AD108", - "confidence": 7 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AD109": { - "revision": 4, - "explanation": "The input is $200\\,\\Omega$ plus $100\\,\\Omega$ in parallel with $200\\,\\Omega + R$; at $R = 0$ this is about $267\\,\\Omega$, and at $R = 1\\,\\text{k}\\Omega$ it is about $292\\,\\Omega$. Hilfsmittel: series adds, parallel $= R_1 R_2/(R_1+R_2)$ (Reihen-/Parallelschaltung, S.12).", + "revision": 5, + "explanation": "A fixed $200\\,\\Omega$ is in series with a parallel pair: $100\\,\\Omega$ alongside $(200\\,\\Omega + R)$. With $R = 0$: $100\\,\\Omega \\parallel 200\\,\\Omega \\approx 67\\,\\Omega$, so $R_\\mathrm{min} = 200\\,\\Omega + 67\\,\\Omega = 267\\,\\Omega$. With $R = 1\\,\\text{k}\\Omega$: $100\\,\\Omega \\parallel 1200\\,\\Omega \\approx 92\\,\\Omega$, so $R_\\mathrm{max} = 200\\,\\Omega + 92\\,\\Omega = 292\\,\\Omega$. The input resistance spans $267\\text{–}292\\,\\Omega$. Hilfsmittel: series/parallel combination $R_G = R_1 R_2/(R_1+R_2)$ (Widerstandsnetzwerke, S.12).", "source": "https://50ohm.de/NEA_reihe_parallel_widerstandsnetz_2.html#AD109", - "confidence": 7 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AD110": { - "revision": 3, - "explanation": "Each side branch is $2.2\\,\\text{k}\\Omega + 220\\,\\Omega = 2420\\,\\Omega$, and two equal branches in parallel give half that value, $1210\\,\\Omega$. Hilfsmittel: series adds, then two equal branches in parallel halve (Reihen-/Parallelschaltung, S.12).", + "revision": 4, + "explanation": "The centre link makes the two midpoints one node, so $R_1$ and $R_3$ are in parallel (between a and the midpoint) and $R_2$ and $R_4$ are in parallel (midpoint to b), the two pairs then in series. Equal pairs halve: $R_{13} = \\frac{2.2\\,\\text{k}\\Omega}{2} = 1.1\\,\\text{k}\\Omega$ and $R_{24} = \\frac{220\\,\\Omega}{2} = 110\\,\\Omega$. In series: $R_\\mathrm{ges} = 1100\\,\\Omega + 110\\,\\Omega = 1210\\,\\Omega$. Hilfsmittel: series and parallel combination (Widerstandsnetzwerke, S.12).", "source": "https://50ohm.de/NEA_reihe_parallel_widerstandsnetz_2.html#AD110", - "confidence": 7 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AD111": { - "revision": 4, - "explanation": "A bridge has zero branch voltage when the two divider ratios are equal, which gives $R_1/R_2 = R_3/R_4$. Hilfsmittel: the bridge balances when the two divider ratios match: $U_1/U_2 = R_1/R_2$ (Spannungsteiler, S.12).", + "revision": 5, + "explanation": "The bridge is balanced — no voltage across the meter — when its two unloaded dividers have equal ratios: $\\frac{R_1}{R_2} = \\frac{R_3}{R_4}$. Then points A and B sit at the same potential. Hilfsmittel: unloaded voltage divider $U_2/U_G = R_2/(R_1+R_2)$ (Spannungsteiler, S.12).", "source": "https://50ohm.de/NEA_brueckenschaltung.html#AD111", - "confidence": 7 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AD112": { - "revision": 3, - "explanation": "All four resistors are equal, so the two divider midpoints sit at the same potential; the bridge voltage from A to B is 0 V. Hilfsmittel: equal divider ratios give equal midpoints (Spannungsteiler, S.12); the bridge voltage is 0.", + "revision": 4, + "explanation": "With all four resistors equal, the left and right dividers are identical: each splits the $10\\,\\text{V}$ evenly, so $U_1 = U_2 = 5\\,\\text{V}$. A and B sit at the same potential, hence $U_\\mathrm{AB} = 0\\,\\text{V}$. Hilfsmittel: unloaded voltage divider (Spannungsteiler, S.12).", "source": "https://50ohm.de/NEA_brueckenschaltung.html#AD112", - "confidence": 7 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AD113": { - "revision": 2, - "explanation": "The left divider gives point A at 10 V and the right divider gives point B at 1 V, so measured from A to B the bridge voltage is +9 V. Hilfsmittel: each side is a voltage divider (Spannungsteiler, S.12); subtract the two midpoint voltages.", + "revision": 3, + "explanation": "Treat each side as an unloaded divider of the $11\\,\\text{V}$. Left ($R_1 = 1\\,\\text{k}\\Omega$ over $R_2 = 10\\,\\text{k}\\Omega$): A sits at $10\\,\\text{V}$. Right ($R_3 = 10\\,\\text{k}\\Omega$ over $R_4 = 1\\,\\text{k}\\Omega$): B sits at $1\\,\\text{V}$. So $U_\\mathrm{AB} = 10\\,\\text{V} - 1\\,\\text{V} = 9\\,\\text{V}$ (A more positive). Hilfsmittel: unloaded voltage divider (Spannungsteiler, S.12).", "source": "https://50ohm.de/NEA_brueckenschaltung.html#AD113", - "confidence": 7 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AD114": { - "revision": 6, - "explanation": "The load is parallel to R2: $2.2\\,\\text{k}\\Omega \\parallel 8.2\\,\\text{k}\\Omega$ is about 1.73 kOhm. The divider output is $12\\,\\text{V} \\cdot 1.73/(10 + 1.73)$, about 1.8 V. Hilfsmittel: first the load in parallel with $R_2$ (Reihen-/Parallelschaltung, S.12), then the divider (Spannungsteiler, S.12).", + "revision": 7, + "explanation": "The load $R_L$ sits in parallel with $R_2$: $R_\\mathrm{2L} = \\frac{R_2 R_L}{R_2 + R_L} = \\frac{2.2\\,\\text{k}\\Omega\\cdot 8.2\\,\\text{k}\\Omega}{2.2\\,\\text{k}\\Omega + 8.2\\,\\text{k}\\Omega} \\approx 1.74\\,\\text{k}\\Omega$. The output is then the divider of $R_1$ and $R_\\mathrm{2L}$: $U_2 = U_B\\cdot\\frac{R_\\mathrm{2L}}{R_1 + R_\\mathrm{2L}} = 12\\,\\text{V}\\cdot\\frac{1.74\\,\\text{k}\\Omega}{10\\,\\text{k}\\Omega + 1.74\\,\\text{k}\\Omega} \\approx 1.8\\,\\text{V}$. Hilfsmittel: parallel combination and voltage divider (Widerstandsnetzwerke / Spannungsteiler, S.12).", "source": "https://50ohm.de/NEA_spannungsteiler_2.html#AD114", - "confidence": 7 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AD115": { - "revision": 2, - "explanation": "Adding the load lowers the effective lower resistance of the divider, increasing the supply current through R1; with higher current, R1 dissipates more heat.", + "revision": 3, + "explanation": "Loading the divider puts $R_L$ in parallel with $R_2$, lowering the lower arm and hence the total resistance. The total current $I_1$ therefore rises, and since it flows through $R_1$, more power is dissipated as heat in $R_1$.", "source": "https://50ohm.de/NEA_spannungsteiler_2.html#AD115", - "confidence": 7 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AD201": { - "revision": 5, - "explanation": "An RC high-pass cutoff is $f_g = 1/(2 \\pi R C)$; with 4.7 kOhm and 2.2 nF this is about 15.4 kHz. Hilfsmittel: apply $f_g = 1/(2\\pi\\cdot R\\cdot C)$ (RC-Hochpass, S.14).", + "revision": 6, + "explanation": "For an RC high-pass the corner frequency is $f_g = \\frac{1}{2\\pi R C} = \\frac{1}{2\\pi\\cdot 4.7\\,\\text{k}\\Omega\\cdot 2.2\\,\\text{nF}} \\approx 15.4\\,\\text{kHz}$. Hilfsmittel: $f_g = 1/(2\\pi R C)$ (RC TP/HP, S.14).", "source": "https://50ohm.de/NEA_schwingkreis_2.html#AD201", - "confidence": 7 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AD202": { - "revision": 5, - "explanation": "An RC low-pass has the same cutoff formula, $f_g = 1/(2 \\pi R C)$; with 10 kOhm and 47 nF this is about 339 Hz. Hilfsmittel: apply $f_g = 1/(2\\pi\\cdot R\\cdot C)$ (RC-Tiefpass, S.14).", + "revision": 6, + "explanation": "For an RC low-pass the corner frequency is $f_g = \\frac{1}{2\\pi R C} = \\frac{1}{2\\pi\\cdot 10\\,\\text{k}\\Omega\\cdot 47\\,\\text{nF}} \\approx 339\\,\\text{Hz}$. Hilfsmittel: $f_g = 1/(2\\pi R C)$ (RC TP/HP, S.14).", "source": "https://50ohm.de/NEA_schwingkreis_2.html#AD202", - "confidence": 7 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AD203": { - "revision": 5, - "explanation": "The relevant low-pass is R1 with C1; C2 is supply decoupling and the amplifier input is very high impedance. $1/(2 \\pi \\cdot 4.7\\,\\text{k}\\Omega \\cdot 6.8\\,\\text{nF})$ is about 5 kHz. Hilfsmittel: apply $f_g = 1/(2\\pi\\cdot R\\cdot C)$ (RC-Tiefpass, S.14).", + "revision": 6, + "explanation": "Only $R_1$ and $C_1$ form the audio low-pass; $C_2$ is a supply-decoupling capacitor and the amplifier's 1 MHz corner and 1 MΩ input impedance are negligible here. So $f_g = \\frac{1}{2\\pi R_1 C_1} = \\frac{1}{2\\pi\\cdot 4.7\\,\\text{k}\\Omega\\cdot 6.8\\,\\text{nF}} \\approx 5\\,\\text{kHz}$. Hilfsmittel: $f_g = 1/(2\\pi R C)$ (RC TP/HP, S.14).", "source": "https://50ohm.de/NEA_schwingkreis_2.html#AD203", - "confidence": 7 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AD204": { "revision": 2, @@ -852,34 +959,39 @@ "confidence": 7 }, "AD208": { - "revision": 4, - "explanation": "Use Thomson's formula $f = 1/(2 \\pi \\sqrt{L C})$; with 1.2 microhenry and 6.8 pF the result is about 55.7 MHz. Hilfsmittel: apply $f_0 = 1/(2\\pi\\sqrt{L\\cdot C})$, S.14.", + "revision": 5, + "explanation": "Resonance follows Thomson's formula (the series resistance does not affect it): $f = \\frac{1}{2\\pi\\sqrt{L C}} = \\frac{1}{2\\pi\\sqrt{1.2\\,\\mu\\text{H}\\cdot 6.8\\,\\text{pF}}} \\approx 55.7\\,\\text{MHz}$. Hilfsmittel: $f_0 = 1/(2\\pi\\sqrt{L C})$ (Schwingkreis, S.14).", "source": "https://50ohm.de/NEA_schwingkreis_2.html#AD208", - "confidence": 7 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AD209": { - "revision": 5, - "explanation": "The resistor does not set the ideal resonant frequency; $1/(2 \\pi \\sqrt{10\\,\\mu\\text{H} \\cdot 1\\,\\text{nF}})$ is about 1.592 MHz. Hilfsmittel: apply $f_0 = 1/(2\\pi\\sqrt{L\\cdot C})$, S.14 (R does not set $f_0$).", + "revision": 6, + "explanation": "Thomson's formula (the resistance is irrelevant for resonance): $f = \\frac{1}{2\\pi\\sqrt{L C}} = \\frac{1}{2\\pi\\sqrt{10\\,\\mu\\text{H}\\cdot 1\\,\\text{nF}}} \\approx 1.592\\,\\text{MHz}$. Hilfsmittel: $f_0 = 1/(2\\pi\\sqrt{L C})$ (Schwingkreis, S.14).", "source": "https://50ohm.de/NEA_schwingkreis_2.html#AD209", - "confidence": 7 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AD210": { - "revision": 4, - "explanation": "Using $f = 1/(2 \\pi \\sqrt{L C})$ with 100 microhenry and 0.01 microfarad gives about 159 kHz. Hilfsmittel: apply $f_0 = 1/(2\\pi\\sqrt{L\\cdot C})$, S.14.", + "revision": 5, + "explanation": "Thomson's formula: $f = \\frac{1}{2\\pi\\sqrt{L C}} = \\frac{1}{2\\pi\\sqrt{100\\,\\mu\\text{H}\\cdot 0.01\\,\\mu\\text{F}}} \\approx 159\\,\\text{kHz}$. Hilfsmittel: $f_0 = 1/(2\\pi\\sqrt{L C})$ (Schwingkreis, S.14).", "source": "https://50ohm.de/NEA_schwingkreis_2.html#AD210", - "confidence": 7 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AD211": { - "revision": 5, - "explanation": "For the parallel resonant circuit, $f = 1/(2 \\pi \\sqrt{2.2\\,\\mu\\text{H} \\cdot 56\\,\\text{pF}})$, giving about 14.34 MHz. Hilfsmittel: apply $f_0 = 1/(2\\pi\\sqrt{L\\cdot C})$, S.14.", + "revision": 6, + "explanation": "Thomson's formula (resistance irrelevant for resonance): $f = \\frac{1}{2\\pi\\sqrt{L C}} = \\frac{1}{2\\pi\\sqrt{2.2\\,\\mu\\text{H}\\cdot 56\\,\\text{pF}}} \\approx 14.34\\,\\text{MHz}$. Hilfsmittel: $f_0 = 1/(2\\pi\\sqrt{L C})$ (Schwingkreis, S.14).", "source": "https://50ohm.de/NEA_schwingkreis_2.html#AD211", - "confidence": 7 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AD212": { - "revision": 4, - "explanation": "The parallel capacitances add to about 1.82 nF; with 1.2 mH, $1/(2 \\pi \\sqrt{L C})$ gives about 107.7 kHz. Hilfsmittel: first add the parallel caps (S.13), then $f_0 = 1/(2\\pi\\sqrt{L\\cdot C})$, S.14.", + "revision": 5, + "explanation": "The three capacitors are in parallel, so they add: $C_\\mathrm{ges} = 0.1\\,\\text{nF} + 1.5\\,\\text{nF} + 220\\,\\text{pF} = 1.82\\,\\text{nF}$. Then $f = \\frac{1}{2\\pi\\sqrt{L C}} = \\frac{1}{2\\pi\\sqrt{1.2\\,\\text{mH}\\cdot 1.82\\,\\text{nF}}} \\approx 107.7\\,\\text{kHz}$. Hilfsmittel: $C_G = C_1 + C_2 + \\dots$ (C Parallel, S.13) and $f_0 = 1/(2\\pi\\sqrt{L C})$ (Schwingkreis, S.14).", "source": "https://50ohm.de/NEA_schwingkreis_2.html#AD212", - "confidence": 7 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AD213": { "revision": 3, @@ -948,16 +1060,18 @@ "confidence": 8 }, "AD224": { - "revision": 4, - "explanation": "For the parallel case, $B = 1/(2 \\pi R C)$; with 1 kOhm and 56 pF this is about 2.84 MHz. Hilfsmittel: apply $B = 1/(2\\pi R C)$ (Parallelschwingkreis, S.14).", + "revision": 5, + "explanation": "For this parallel circuit: $f_0 = \\frac{1}{2\\pi\\sqrt{L C}} \\approx 14.34\\,\\text{MHz}$ (from $L = 2.2\\,\\mu\\text{H}$, $C = 56\\,\\text{pF}$). Then $X_L = 2\\pi f_0 L \\approx 198\\,\\Omega$, the quality is $Q = \\frac{R_p}{X_L} = \\frac{1\\,\\text{k}\\Omega}{198\\,\\Omega} \\approx 5.05$, and the bandwidth is $B = \\frac{f_0}{Q} \\approx 2.84\\,\\text{MHz}$. Hilfsmittel: $f_0 = 1/(2\\pi\\sqrt{L C})$, $X_L = \\omega L$, parallel-circuit $Q = R_p/X_L$ and $B = f_0/Q$ (Schwingkreis, S.14).", "source": "https://50ohm.de/NEA_schwingkreis_2.html#AD224", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AD225": { - "revision": 4, - "explanation": "For the series circuit, Q is resonant frequency divided by bandwidth; about $159\\,\\text{kHz} / 15.9\\,\\text{kHz} = 10$. Hilfsmittel: apply $Q = f_0/B$ (Schwingkreis, S.14).", + "revision": 5, + "explanation": "The quality formulas hold only at resonance, so first $f_0 = \\frac{1}{2\\pi\\sqrt{L C}} \\approx 159.2\\,\\text{kHz}$ (from $L = 100\\,\\mu\\text{H}$, $C = 0.01\\,\\mu\\text{F}$). Then $X_L = 2\\pi f_0 L \\approx 100\\,\\Omega$, and for the series circuit $Q = \\frac{X_L}{R_s} = \\frac{100\\,\\Omega}{10\\,\\Omega} = 10$. Hilfsmittel: $f_0 = 1/(2\\pi\\sqrt{L C})$, $X_L = \\omega L$, series-circuit $Q = X_L/R_s$ (Schwingkreis, S.14).", "source": "https://50ohm.de/NEA_schwingkreis_2.html#AD225", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AD226": { "revision": 4, @@ -984,64 +1098,74 @@ "confidence": 8 }, "AD301": { - "revision": 6, - "explanation": "In each series string the cell voltages add, giving $30 \\cdot 0.6\\,\\text{V} = 18\\,\\text{V}$; four identical strings in parallel add their short-circuit currents to 4 A. Hilfsmittel: series voltages add ($U_G = U_1+U_2$) and parallel currents add ($I_G = I_1+I_2$), both S.12; the 0,6 V cell voltage is given.", + "revision": 7, + "explanation": "Each row is 30 cells in series, so the voltages add: $30\\cdot 0.6\\,\\text{V} = 18\\,\\text{V}$, while a series string's short-circuit current stays $1\\,\\text{A}$. The four rows are in parallel, so their currents add: $4\\cdot 1\\,\\text{A} = 4\\,\\text{A}$. The module gives $18\\,\\text{V}$ open-circuit and $4\\,\\text{A}$ short-circuit.", "source": "https://50ohm.de/NEA_photovoltaik.html#AD301", - "confidence": 7 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AD302": { - "revision": 5, - "explanation": "The unloaded smoothing capacitor charges close to the peak of the secondary AC voltage; about 15 V RMS times $\\sqrt{2}$ gives roughly 21 V. Hilfsmittel: the cap charges to the peak: $\\hat{U} = U_\\mathrm{eff}\\cdot\\sqrt{2}$ (Wechselspannung, S.12).", + "revision": 6, + "explanation": "The capacitor charges through the diode to the peak of the AC voltage: $\\hat{U} = U_\\mathrm{eff}\\cdot\\sqrt{2} = 15\\,\\text{V}\\cdot\\sqrt{2} \\approx 21\\,\\text{V}$ (the diode's forward drop is neglected). So $U_\\mathrm{AB} \\approx 21\\,\\text{V}$. Hilfsmittel: $\\hat{U} = U_\\mathrm{eff}\\sqrt{2}$ (Wechselspannung, S.12).", "source": "https://50ohm.de/NEA_gleichrichter_2.html#AD302", - "confidence": 7 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AD303": { - "revision": 5, - "explanation": "A 20:1 transformer gives $230\\,\\text{V} / 20 = 11.5\\,\\text{V}$ RMS; the peak is about $16.3\\,\\text{V}$, and adding 50 percent safety gives about $24.4\\,\\text{V}$, so choose at least $25\\,\\text{V}$. Hilfsmittel: first $U_S = U_P\\cdot N_S/N_P$ (Übersetzungsverhältnis $\\ddot{u} = N_P/N_S = U_P/U_S = I_S/I_P = \\sqrt{Z_P/Z_S}$, S.13), then $\\hat{U} = U_\\mathrm{eff}\\cdot\\sqrt{2}$ (Wechselspannung, S.12).", + "revision": 6, + "explanation": "The secondary RMS voltage is $\\frac{230\\,\\text{V}}{20} = 11.5\\,\\text{V}$. The capacitor charges to the peak $\\hat{U} = 11.5\\,\\text{V}\\cdot\\sqrt{2} \\approx 16.3\\,\\text{V}$; with the $50\\%$ safety margin, $16.3\\,\\text{V}\\cdot 1.5 \\approx 24.5\\,\\text{V}$, so a $25\\,\\text{V}$ rating is the minimum. Hilfsmittel: $\\hat{U} = U_\\mathrm{eff}\\sqrt{2}$ (Wechselspannung, S.12).", "source": "https://50ohm.de/NEA_gleichrichter_2.html#AD303", - "confidence": 7 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AD304": { - "revision": 2, - "explanation": "A 5:1 transformer gives 46 V RMS, or about 65 V peak; the diode must withstand about twice that peak plus 20 percent, giving about 156 V.", + "revision": 3, + "explanation": "The secondary RMS voltage is $\\frac{230\\,\\text{V}}{5} = 46\\,\\text{V}$. In this rectifier the diode must block twice the peak: $U_\\mathrm{Sperr} = 2\\hat{U} = 2\\cdot 46\\,\\text{V}\\cdot\\sqrt{2} \\approx 130\\,\\text{V}$; with the $20\\%$ margin, $130\\,\\text{V}\\cdot 1.2 \\approx 156\\,\\text{V}$. Hilfsmittel: $\\hat{U} = U_\\mathrm{eff}\\sqrt{2}$ (Wechselspannung, S.12).", "source": "https://50ohm.de/NEA_gleichrichter_2.html#AD304", - "confidence": 7 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AD305": { - "revision": 2, - "explanation": "In a bridge rectifier, two diodes conduct on each half-cycle so current through the load always has the same polarity; the correct diagram has all four diodes oriented for that path.", + "revision": 3, + "explanation": "In a bridge rectifier the four diodes are arranged so the output keeps the same polarity for either half-cycle of the AC. The two cathodes that meet form the positive terminal; the two anodes that meet form the negative terminal; the AC feeds the other two corners.", "source": "https://50ohm.de/NEA_brueckengleichrichter.html#AD305", - "confidence": 7 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AD306": { - "revision": 5, - "explanation": "The secondary peak is the mains peak divided by 8: $230\\,\\text{V} \\cdot 1.414 / 8$ is about 40.6 V, which is the unloaded capacitor voltage. Hilfsmittel: first scale by the turns ratio (Übersetzungsverhältnis $\\ddot{u} = N_P/N_S = U_P/U_S = I_S/I_P = \\sqrt{Z_P/Z_S}$, S.13), then $\\hat{U} = U_\\mathrm{eff}\\cdot\\sqrt{2}$ (Wechselspannung, S.12).", + "revision": 6, + "explanation": "The smoothing capacitor charges to the peak of the secondary voltage. With the 8:1 turns ratio the secondary RMS is $\\frac{230\\,\\text{V}}{8} = 28.75\\,\\text{V}$, whose peak is $\\hat{U} = 28.75\\,\\text{V}\\cdot\\sqrt{2} \\approx 40\\,\\text{V}$. Hilfsmittel: $\\hat{U} = U_\\mathrm{eff}\\sqrt{2}$ (Wechselspannung, S.12) and the turns-ratio relation $U_P/U_S$ (Transformator, S.13).", "source": "https://50ohm.de/NEA_brueckengleichrichter.html#AD306", - "confidence": 7 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AD307": { - "revision": 2, - "explanation": "A full-wave rectifier uses both half-cycles and routes them through the load with the indicated same output polarity.", + "revision": 3, + "explanation": "For the indicated output polarity, the two diode cathodes must join at the positive output terminal (and the anodes at the negative). That is the arrangement giving full-wave rectification with the shown polarity.", "source": "https://50ohm.de/NEA_vollweggleichrichter.html#AD307", - "confidence": 7 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AD308": { - "revision": 2, - "explanation": "The rectifier output is pulsating DC: the negative half-cycles are folded to the same polarity rather than appearing as negative voltage.", + "revision": 3, + "explanation": "With the two anodes joined to the output terminal, the rectified voltage is negative relative to the transformer's centre tap. Both half-waves therefore appear below the zero line as pulsating DC.", "source": "https://50ohm.de/NEA_vollweggleichrichter.html#AD308", - "confidence": 7 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AD309": { - "revision": 3, - "explanation": "The ripple span is the difference between the high and low points, 3 V, and a full-wave rectifier on 50 Hz mains produces ripple at 100 Hz.", + "revision": 4, + "explanation": "The ripple period is $10\\,\\text{ms}$, so the hum frequency is $f = \\frac{1}{T} = \\frac{1}{10\\,\\text{ms}} = 100\\,\\text{Hz}$ (full-wave-rectified 50 Hz mains). The ripple amplitude is one division at $3\\,\\text{V}$ per division, i.e. $3\\,\\text{V}$.", "source": "https://50ohm.de/NEA_restwelligkeit.html#AD309", - "confidence": 7 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AD310": { - "revision": 3, - "explanation": "A full-wave rectifier produces one output pulse for each half-cycle, so 50 Hz mains becomes 100 Hz ripple frequency.", + "revision": 4, + "explanation": "A full-wave rectifier uses both half-cycles, producing two output pulses per mains period. The fundamental therefore doubles: $f_\\mathrm{out} = 2\\cdot f_\\mathrm{in} = 2\\cdot 50\\,\\text{Hz} = 100\\,\\text{Hz}$.", "source": "https://50ohm.de/NEA_restwelligkeit.html#AD310", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AD311": { "revision": 2, @@ -1068,46 +1192,53 @@ "confidence": 7 }, "AD315": { - "revision": 3, - "explanation": "The Z-diode regulator clamps the output near the Zener voltage, so the output between A and B is approximately 5 V despite the varying input.", + "revision": 4, + "explanation": "This is a series regulator: the Zener diode pins the transistor's base at about $5.6\\,\\text{V}$, and the emitter (output A) follows one base-emitter drop lower. So $U_\\mathrm{A} \\approx U_\\mathrm{Z} - U_\\mathrm{BE} \\approx 5.6\\,\\text{V} - 0.6\\,\\text{V} \\approx 5\\,\\text{V}$.", "source": "https://50ohm.de/NEA_spannungsstabilisierung.html#AD315", - "confidence": 7 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AD316": { - "revision": 2, - "explanation": "A linear regulator needs headroom: its input voltage must be higher than the regulated output voltage so the pass element can control the drop.", + "revision": 3, + "explanation": "A linear regulator can only stabilise its output if the input voltage stays above the desired output. It needs some headroom (dropout) across itself to keep regulating and to absorb input variations.", "source": "https://50ohm.de/NEA_spannungsstabilisierung.html#AD316", - "confidence": 7 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AD317": { - "revision": 2, - "explanation": "A fixed 12 V regulator absorbs the allowed input variation as internal voltage drop, so the output variation is nearly zero while it remains in regulation.", + "revision": 3, + "explanation": "A 12 V fixed regulator holds its output essentially constant as long as the input stays sufficiently above it. The input swings between $15\\,\\text{V}$ and $18\\,\\text{V}$ — always above 12 V plus dropout — so the output stays at $12\\,\\text{V}$ and the output variation is nearly zero.", "source": "https://50ohm.de/NEA_spannungsstabilisierung.html#AD317", - "confidence": 7 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AD318": { - "revision": 6, - "explanation": "The load current is $5\\,\\text{V} / 10\\,\\Omega = 0.5\\,\\text{A}$ and the regulator drops $13.8\\,\\text{V} - 5\\,\\text{V} = 8.8\\,\\text{V}$; loss is $8.8\\,\\text{V} \\cdot 0.5\\,\\text{A} = 4.4\\,\\text{W}$. Hilfsmittel: first $I = U/R$ (Ohmsches Gesetz, S.11), then (voltage drop) $\\times\\,I$ via $P = U\\cdot I$ (Leistung, S.12).", + "revision": 7, + "explanation": "Load current $I = \\frac{U}{R} = \\frac{5\\,\\text{V}}{10\\,\\Omega} = 0.5\\,\\text{A}$. The regulator drops $U_\\mathrm{V} = 13.8\\,\\text{V} - 5\\,\\text{V} = 8.8\\,\\text{V}$ at that current, so its dissipation is $P_\\mathrm{V} = U_\\mathrm{V}\\cdot I = 8.8\\,\\text{V}\\cdot 0.5\\,\\text{A} = 4.4\\,\\text{W}$. Hilfsmittel: $P = U\\cdot I$ (Leistung, S.12).", "source": "https://50ohm.de/NEA_spannungsstabilisierung.html#AD318", - "confidence": 7 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AD319": { - "revision": 6, - "explanation": "A linear regulator dissipates the voltage drop times current: $(13.8\\,\\text{V} - 9\\,\\text{V}) \\cdot 0.9\\,\\text{A} = 4.32\\,\\text{W}$. Hilfsmittel: apply $P = (U_\\mathrm{in} - U_\\mathrm{out})\\cdot I$, i.e. $P = U\\cdot I$ (Leistung, S.12).", + "revision": 7, + "explanation": "A linear regulator burns the excess voltage as heat at essentially the same current in and out: $P_\\mathrm{V} = (U_\\mathrm{in} - U_\\mathrm{out})\\cdot I = (13.8\\,\\text{V} - 9\\,\\text{V})\\cdot 0.9\\,\\text{A} = 4.32\\,\\text{W}$. Hilfsmittel: $P = U\\cdot I$ (Leistung, S.12).", "source": "https://50ohm.de/NEA_spannungsstabilisierung.html#AD319", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AD320": { - "revision": 4, - "explanation": "Efficiency is output power over input power: $5\\,\\text{V} \\cdot 0.450\\,\\text{A}$ divided by $13.8\\,\\text{V} \\cdot 0.455\\,\\text{A}$ is about 0.36.", + "revision": 5, + "explanation": "Efficiency is output power over input power: $\\eta = \\frac{U_\\mathrm{out} I_\\mathrm{out}}{U_\\mathrm{in} I_\\mathrm{in}} = \\frac{5\\,\\text{V}\\cdot 450\\,\\text{mA}}{13.8\\,\\text{V}\\cdot 455\\,\\text{mA}} = \\frac{2.25\\,\\text{W}}{6.28\\,\\text{W}} \\approx 0.36$. Hilfsmittel: $P = U\\cdot I$ and $\\eta = P_\\mathrm{ab}/P_\\mathrm{zu}$ (Leistung / Wirkungsgrad, S.12).", "source": "https://50ohm.de/NEA_spannungsstabilisierung.html#AD320", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AD321": { - "revision": 5, - "explanation": "The load power is $4.7\\,\\text{V} \\cdot 10\\,\\text{mA} = 47\\,\\text{mW}$; input power is $13.8\\,\\text{V} \\cdot (10 + 15) \\text{mA} = 345\\,\\text{mW}$, so efficiency is about 0.14. Hilfsmittel: apply $\\eta = P_\\mathrm{ab}/P_\\mathrm{zu}$ (Wirkungsgrad, S.12) with $P = U\\cdot I$ for each side.", + "revision": 6, + "explanation": "Output side: $U_\\mathrm{L} = R_\\mathrm{L} I_\\mathrm{L} = 470\\,\\Omega\\cdot 10\\,\\text{mA} = 4.7\\,\\text{V}$, so $P_\\mathrm{L} = U_\\mathrm{L} I_\\mathrm{L} = 47\\,\\text{mW}$. The input current is load plus Zener current, $I_\\mathrm{IN} = 10\\,\\text{mA} + 15\\,\\text{mA} = 25\\,\\text{mA}$, so $P_\\mathrm{IN} = 13.8\\,\\text{V}\\cdot 25\\,\\text{mA} = 345\\,\\text{mW}$. Hence $\\eta = \\frac{47\\,\\text{mW}}{345\\,\\text{mW}} \\approx 0.14$ — the Zener wastes most of the power. Hilfsmittel: $P = U\\cdot I$, $\\eta = P_\\mathrm{ab}/P_\\mathrm{zu}$ (Leistung / Wirkungsgrad, S.12).", "source": "https://50ohm.de/NEA_spannungsstabilisierung.html#AD321", - "confidence": 7 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AD322": { "revision": 2, @@ -1344,10 +1475,11 @@ "confidence": 7 }, "AD503": { - "revision": 2, - "explanation": "In an envelope detector the extra output from the rectified envelope can be filtered into a DC control voltage, which is used as an AGC/regulating voltage.", + "revision": 3, + "explanation": "This is an envelope demodulator. The IF signal enters on the left, so X is not the IF output. Point X sits after an RC low-pass built with a high-capacitance electrolytic capacitor; the large $C$ pushes $f_g = \\frac{1}{2\\pi R C}$ very low, so neither the audio nor the oscillator signal gets through — only a slowly varying, near-DC voltage. That low-frequency voltage at X is used as a control (AGC) voltage. Hilfsmittel: $f_g = 1/(2\\pi R C)$ (RC-Tiefpass, S.14).", "source": "https://50ohm.de/NEA_modulatoren.html#AD503", - "confidence": 7 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AD504": { "revision": 3, @@ -1722,10 +1854,11 @@ "confidence": 8 }, "AE310": { - "revision": 4, - "explanation": "Narrowband FM in a 12.5 kHz channel uses a typical peak deviation, German Hub/Frequenzhub, of about 2.5 kHz. That leaves room for the modulation sidebands inside the channel spacing.", + "revision": 6, + "explanation": "Carson's rule bounds the deviation: $B \\approx 2\\cdot(\\Delta f_T + f_\\mathrm{mod,max})$, so for a $12.5\\,\\text{kHz}$ channel $\\Delta f_T + f_\\mathrm{mod,max} \\approx 6.25\\,\\text{kHz}$ — the peak deviation must be well under that, which already rules out the larger options. The exact figure is a convention rather than a calculation: narrow-band FM in the $12.5\\,\\text{kHz}$ raster uses a typical peak deviation of $2.5\\,\\text{kHz}$ (with audio up to ~3 kHz, Carson then gives $B \\approx 11\\,\\text{kHz}$, fitting the channel). Hilfsmittel: Carson bandwidth $B \\approx 2(\\Delta f_T + f_\\mathrm{mod,max})$ (FM, S.16).", "source": "https://50ohm.de/NEA_fm_3.html#AE310", - "confidence": 8 + "confidence": 7, + "provenance": "50ohm-loesungsweg" }, "AE311": { "revision": 7, @@ -2412,22 +2545,25 @@ "confidence": 7 }, "AF425": { - "revision": 4, - "explanation": "The resistor must drop $13.5\\,\\text{V} - 4\\,\\text{V} = 9.5\\,\\text{V}$ at 10 mA, so $R = 9.5\\,\\text{V} / 0.010\\,\\text{A} = 950\\,\\Omega$. Hilfsmittel: apply $R = U/I$ (Ohmsches Gesetz, S.11) to the resistor's drop and current.", + "revision": 5, + "explanation": "The bias resistor must drop the rest of the supply: $U_R = U_\\mathrm{CC} - U_D = 13.5\\,\\text{V} - 4\\,\\text{V} = 9.5\\,\\text{V}$, at the operating current $I_D = 10\\,\\text{mA}$. So $R_\\mathrm{BIAS} = \\frac{9.5\\,\\text{V}}{10\\,\\text{mA}} = 950\\,\\Omega$. Hilfsmittel: Ohm's law $R = U/I$ (S.11).", "source": "https://50ohm.de/NEA_integrierte_schaltkreise.html#AF425", - "confidence": 7 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AF426": { - "revision": 4, - "explanation": "The bias resistor drops $13.8\\,\\text{V} - 4\\,\\text{V} = 9.8\\,\\text{V}$ at 15 mA; $9.8\\,\\text{V} / 0.015\\,\\text{A} = 653\\,\\Omega$, so the nearest listed standard value is 680 ohm. Hilfsmittel: apply $R = U/I$ (Ohmsches Gesetz, S.11); then pick the nearest E-series value (S.12).", + "revision": 5, + "explanation": "The bias resistor drops $U_R = U_\\mathrm{CC} - U_D = 13.8\\,\\text{V} - 4\\,\\text{V} = 9.8\\,\\text{V}$ at $I_D = 15\\,\\text{mA}$: $R_\\mathrm{BIAS} = \\frac{9.8\\,\\text{V}}{15\\,\\text{mA}} \\approx 653\\,\\Omega$. The nearest standard value is $680\\,\\Omega$. Hilfsmittel: Ohm's law $R = U/I$ (S.11); E-series (Vorzugsreihen, S.12).", "source": "https://50ohm.de/NEA_integrierte_schaltkreise.html#AF426", - "confidence": 7 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AF427": { - "revision": 5, - "explanation": "With a 4 V MMIC drop, the resistor current is $(9\\,\\text{V} - 4\\,\\text{V}) / 470\\,\\Omega = 10.6\\,\\text{mA}$; MMIC heat is $4\\,\\text{V} \\cdot 10.6\\,\\text{mA} = 42.6\\,\\text{mW}$, about 43 mW. Hilfsmittel: first $I = U/R$ (Ohmsches Gesetz, S.11), then heat $P = U\\cdot I$ (Leistung, S.12).", + "revision": 6, + "explanation": "The bias resistor drops $U = U_\\mathrm{CC} - U_D = 9\\,\\text{V} - 4\\,\\text{V} = 5\\,\\text{V}$, setting the current $I_D = \\frac{5\\,\\text{V}}{470\\,\\Omega} \\approx 10.6\\,\\text{mA}$. The MMIC then dissipates $P_D = U_D\\cdot I_D = 4\\,\\text{V}\\cdot 10.6\\,\\text{mA} \\approx 43\\,\\text{mW}$. Hilfsmittel: $R = U/I$ (S.11) and $P = U\\cdot I$ (S.12).", "source": "https://50ohm.de/NEA_integrierte_schaltkreise.html#AF427", - "confidence": 7 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AF428": { "revision": 4, @@ -2556,10 +2692,11 @@ "confidence": 8 }, "AF619": { - "revision": 2, - "explanation": "A 4 kHz speech bandwidth needs just over 8 ksample/s by Nyquist; 9600 samples/s is the smallest listed rate above that limit.", + "revision": 3, + "explanation": "The Nyquist condition requires $f_s > 2\\cdot f_\\mathrm{max} = 2\\cdot 4\\,\\text{kHz} = 8\\,\\text{kHz}$. The smallest offered rate above $8000\\,\\text{Hz}$ is $9600\\,\\text{Sps}$. Hilfsmittel: sampling theorem $f_\\mathrm{abtast} > 2\\cdot f_\\mathrm{max}$ (Abtasttheorem, S.18).", "source": "https://50ohm.de/NEA_abtasttheorem.html#AF619", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AF620": { "revision": 2, @@ -3186,10 +3323,11 @@ "confidence": 8 }, "AG401": { - "revision": 2, - "explanation": "Maximum power transfer occurs when load impedance equals the source impedance; a 50 ohm source therefore wants a 50 ohm load.", + "revision": 3, + "explanation": "Maximum power transfer needs the load impedance to equal the source's output impedance: $Z_\\mathrm{L} = Z_\\mathrm{i} = 50\\,\\Omega$. (Matching condition — a memory item.)", "source": "https://50ohm.de/NEA_innenwiderstand.html#AG401", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AG402": { "revision": 3, @@ -3654,10 +3792,11 @@ "confidence": 8 }, "AI105": { - "revision": 2, - "explanation": "A thermal power sensor converts RF heating into a measurement, so it reads true effective power over a wide frequency range into the GHz region.", + "revision": 4, + "explanation": "A measuring instrument with a thermal converter (Thermoumformer) senses the heat a current produces in a conductor. That heating follows $P = I_\\mathrm{eff}^2\\cdot R$ — it depends on the mean-square current — so the instrument is calibrated to read the true RMS value, and hence true power in a known load, independent of waveform within its specified frequency range. Because that range reaches well into the GHz region, it suits accurate effective-power measurement of modulated signals up there, where rectifier-based meters misread non-sinusoidal waveforms.", "source": "https://50ohm.de/NEA_wechselstrom_leistung.html#AI105", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AI201": { "revision": 2, @@ -3732,16 +3871,18 @@ "confidence": 8 }, "AI305": { - "revision": 6, - "explanation": "From the trace the peak voltage is 100 V; $P_\\mathrm{PEP} = (100/\\sqrt{2})^2 / 50$ gives 100 W. Hilfsmittel: first $U_\\mathrm{eff} = \\hat{U}/\\sqrt{2}$ (from $\\hat{U} = U_\\mathrm{eff}\\cdot\\sqrt{2}$, Wechselspannung, S.12), then $P = U_\\mathrm{eff}^2/R$ (Leistung, S.12). Both formulas are on S.12; there is no ready-made PEP formula.", + "revision": 7, + "explanation": "The scope shows $50\\,\\text{V}$ per division and the envelope peaks at about two divisions, so (1:1 probe) $\\hat{U} = 2\\cdot 50\\,\\text{V} = 100\\,\\text{V}$. The RMS value is $U_\\mathrm{eff} = \\frac{\\hat{U}}{\\sqrt{2}} \\approx 70.7\\,\\text{V}$, and into the $50\\,\\Omega$ load $P_\\mathrm{PEP} = \\frac{U_\\mathrm{eff}^2}{R} = \\frac{(70.7\\,\\text{V})^2}{50\\,\\Omega} \\approx 100\\,\\text{W}$. Hilfsmittel: $\\hat{U} = U_\\mathrm{eff}\\sqrt{2}$ and $P = U^2/R$ (Wechselspannung / Leistung, S.12).", "source": "https://50ohm.de/NEA_ssb_3.html#AI305", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AI306": { - "revision": 6, - "explanation": "A 10:1 probe means the real peak voltage is ten times the displayed value; $P_\\mathrm{PEP} = (60/\\sqrt{2})^2 / 50$ gives 36 W. Hilfsmittel: correct the 10:1 probe (×10), then $U_\\mathrm{eff} = \\hat{U}/\\sqrt{2}$ (Wechselspannung, S.12) and $P = U_\\mathrm{eff}^2/R$ (Leistung, S.12).", + "revision": 7, + "explanation": "The scope shows $3\\,\\text{V}$ per division, peaking at two divisions, so the measured peak is $6\\,\\text{V}$. The 10:1 probe means the true peak is ten times larger: $\\hat{U} = 60\\,\\text{V}$. Then $U_\\mathrm{eff} = \\frac{60\\,\\text{V}}{\\sqrt{2}} \\approx 42.4\\,\\text{V}$, and into $50\\,\\Omega$: $P_\\mathrm{PEP} = \\frac{(42.4\\,\\text{V})^2}{50\\,\\Omega} \\approx 36\\,\\text{W}$. Hilfsmittel: $\\hat{U} = U_\\mathrm{eff}\\sqrt{2}$ and $P = U^2/R$ (Wechselspannung / Leistung, S.12).", "source": "https://50ohm.de/NEA_ssb_3.html#AI306", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AI401": { "revision": 2, @@ -3792,34 +3933,39 @@ "confidence": 8 }, "AI506": { - "revision": 4, - "explanation": "0.01% is $10^{-4}$; $29\\,\\text{MHz} \\cdot 10^{-4} = 2900\\,\\text{Hz}$.", + "revision": 5, + "explanation": "The deviation is the frequency times the relative accuracy: $\\Delta f = 29\\,\\text{MHz}\\cdot 0.01\\% = 29\\cdot 10^6\\,\\text{Hz}\\cdot 10^{-4} = 2900\\,\\text{Hz}$ (0.01% is $10^{-4}$).", "source": "https://50ohm.de/NEA_frequenzgenauigkeit.html#AI506", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AI507": { - "revision": 3, - "explanation": "0.00001% is $10^{-7}$; $14100\\,\\text{kHz} \\cdot 10^{-7}$ gives a maximum error of 1.410 Hz.", + "revision": 4, + "explanation": "The accuracy $0.00001\\%$ is $10^{-7}$. With $14100\\,\\text{kHz} = 14.1\\cdot 10^6\\,\\text{Hz}$, the maximum deviation is $\\Delta f = 14.1\\cdot 10^6\\,\\text{Hz}\\cdot 10^{-7} = 1.41\\,\\text{Hz}$.", "source": "https://50ohm.de/NEA_frequenzgenauigkeit.html#AI507", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AI508": { - "revision": 2, - "explanation": "One ppm is one part in a million; at 100 MHz that is 100 Hz.", + "revision": 3, + "explanation": "1 ppm is one part in $10^6$. At the top of the range, $\\Delta f = 100\\,\\text{MHz}\\cdot\\frac{1}{10^6} = 100\\,\\text{Hz}$.", "source": "https://50ohm.de/NEA_frequenzgenauigkeit.html#AI508", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AI509": { - "revision": 3, - "explanation": "10 ppm of 145 MHz is 1450 Hz, so the counter may read 145 MHz plus or minus 0.00145 MHz.", + "revision": 4, + "explanation": "10 ppm of the reading is $\\Delta f = 145\\,\\text{MHz}\\cdot\\frac{10}{10^6} = 1450\\,\\text{Hz}$. The displayed value therefore lies within $\\pm 1450\\,\\text{Hz}$: from $144.99855\\,\\text{MHz}$ to $145.00145\\,\\text{MHz}$.", "source": "https://50ohm.de/NEA_frequenzgenauigkeit.html#AI509", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AI510": { - "revision": 3, - "explanation": "Protecting the 144.400 MHz beacon edge needs room for 2.7 kHz USB audio plus 1 ppm frequency error of 144 Hz, totaling 2.844 kHz below the edge.", + "revision": 4, + "explanation": "The suppressed carrier must clear the beacon edge by the highest audio frequency plus the frequency tolerance. The 1 ppm tolerance is $\\Delta f = 144.4\\,\\text{MHz}\\cdot 10^{-6} \\approx 144\\,\\text{Hz} = 0.144\\,\\text{kHz}$. Adding the top audio frequency: $2.7\\,\\text{kHz} + 0.144\\,\\text{kHz} \\approx 2.844\\,\\text{kHz}$.", "source": "https://50ohm.de/NEA_frequenzgenauigkeit.html#AI510", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AI511": { "revision": 2, @@ -4182,118 +4328,137 @@ "confidence": 8 }, "AK101": { - "revision": 2, - "explanation": "In the near field, electric and magnetic fields are not locked to the fixed far-field phase and impedance relationship, so E and H cannot be simply converted.", + "revision": 3, + "explanation": "In the near field the electric and magnetic field components have no fixed phase or amplitude relationship, so one cannot be derived from the other (unlike the far field, where $E/H = Z$). In the reactive near field ($d < \\frac{\\lambda}{2\\pi}$) the fields are strongly position-dependent, and even in the radiating near field ($\\frac{\\lambda}{2\\pi} \\le d \\lesssim 4\\lambda$) the field structure still depends on distance and direction. So no simple conversion — and no simplified safety-distance calculation — is possible there.", "source": "https://50ohm.de/NEA_nahfeld.html#AK101", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AK102": { - "revision": 2, - "explanation": "In the far field, E and H are tied by the wave impedance of the medium, about 377 ohms in free space.", + "revision": 3, + "explanation": "In the far field the magnitudes of $E$ and $H$ are tied together by the wave impedance of the medium: $\\frac{E}{H} = Z$. For free space $Z_0 = \\sqrt{\\frac{\\mu_0}{\\varepsilon_0}} \\approx 120\\pi\\,\\Omega \\approx 377\\,\\Omega$ (air is practically identical). One field component therefore fixes the other — which is not possible in the near field.", "source": "https://50ohm.de/NEA_personenschutzabstand_3.html#AK102", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AK103": { - "revision": 5, - "explanation": "The simple distance formula assumes far-field behaviour; below roughly $\\lambda/(2 \\pi)$ or for electrically small antennas, measurement or near-field modelling is needed. Hilfsmittel: the simple field/power relation only holds in the far field, $d > \\lambda/(2\\pi)$ ($E = \\sqrt{30\\,\\Omega\\cdot P_\\mathrm{EIRP}}/d$, Feldstärke im Fernfeld, S.15).", + "revision": 7, + "explanation": "The far-field formula $d = \\frac{\\sqrt{30\\,\\Omega\\cdot P_\\mathrm{EIRP}}}{E}$ is valid for $d > \\frac{\\lambda}{2\\pi}$ for most antenna types (e.g. dipoles). In the radiating near field it generally gives a conservative over-estimate of the field, so it stays on the safe side — which is why it is allowed down to $\\frac{\\lambda}{2\\pi}$; only in the reactive near field ($d < \\frac{\\lambda}{2\\pi}$) is it not permitted. For electrically small or magnetically dominated antennas, though, the real near-field strength can exceed the predicted value. Where validity is not assured (such antennas, or shorter distances), determine the safety distance instead by field-strength measurement or by a near-field calculation (simulation), as the BNetzA BEMFV guidance prescribes.", "source": "https://50ohm.de/NEA_nahfeld.html#AK103", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AK104": { - "revision": 3, - "explanation": "Feed-line loss reduces transmitter power before it reaches the antenna, so antenna input power is transmitter power multiplied by the loss factor. Hilfsmittel: feed-line loss in dB subtracts from transmitter power before the antenna; $a = 10\\cdot\\log_{10}(P_\\mathrm{in}/P_\\mathrm{out})$ (Pegel, S.15).", + "revision": 4, + "explanation": "The antenna-input power is the transmitter output reduced by the feedline loss. Write the loss as a factor $D = \\frac{P_\\mathrm{Ant}}{P_\\mathrm{Sender}}$ (with $0 < D < 1$), then $P_\\mathrm{Ant} = D\\cdot P_\\mathrm{Sender}$. The greater the cable loss, the smaller $D$ and the less power reaches the antenna.", "source": "https://50ohm.de/NEA_personenschutzabstand_3.html#AK104", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AK105": { - "revision": 4, - "explanation": "Safety distance scales as $d \\propto \\sqrt{P_\\mathrm{EIRP}}/E$. A 6 dB pattern attenuation cuts EIRP in that direction by a factor of 4, so the distance shrinks by $\\sqrt{4}=2$: 20 m becomes 10 m.", + "revision": 5, + "explanation": "A $6\\,\\text{dB}$ drop is a power factor of $0.25$, so off-axis $P = 0.25\\,P_\\mathrm{EIRP}$. Because the safety distance $d = \\frac{\\sqrt{30\\,\\Omega\\cdot P}}{E}$ scales with $\\sqrt{P}$, it shrinks by $\\sqrt{0.25} = 0.5$: from $20\\,\\text{m}$ to $10\\,\\text{m}$. Hilfsmittel: $d = \\sqrt{30\\,\\Omega\\cdot P_\\mathrm{EIRP}}/E$ (Feldstärke Fernfeld, S.15); $-6\\,\\text{dB} \\to \\times 0.25$ from the dB↔ratio table (S.15).", "source": "https://50ohm.de/NEA_personenschutzabstand_richtantennen.html#AK105", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AK106": { - "revision": 2, - "explanation": "Use the person-protection distance formula with dipole gain and 100 W; solving against the 28 V/m limit gives about 2.50 m.", + "revision": 4, + "explanation": "A half-wave dipole has $g_i = 2.15\\,\\text{dBi}$, so the EIRP from the $100\\,\\text{W}$ transmitter power is $P_\\mathrm{EIRP} = 100\\,\\text{W}\\cdot 10^{2.15/10} \\approx 164\\,\\text{W}$. Then $d = \\frac{\\sqrt{30\\,\\Omega\\cdot 164\\,\\text{W}}}{28\\,\\text{V/m}} \\approx \\frac{70.2\\,\\text{V}}{28\\,\\text{V/m}} \\approx 2.50\\,\\text{m}$. The far-field condition $d > \\frac{\\lambda}{2\\pi} \\approx 1.59\\,\\text{m}$ holds. Hilfsmittel: $d = \\sqrt{30\\,\\Omega\\cdot P_\\mathrm{EIRP}}/E$ and $P_\\mathrm{EIRP} = P_S\\cdot 10^{(g_i - a)/(10\\,\\text{dB})}$, where $P_S$ is the transmitter power (Pegel / Antennen, S.15).", "source": "https://50ohm.de/NEA_naeherungsformel_2.html#AK106", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AK107": { - "revision": 6, - "explanation": "Rearrange $E = \\sqrt{30\\,\\Omega \\cdot P_\\mathrm{EIRP}}/d$ for the radiated power: $P_\\mathrm{EIRP} = (E \\cdot d)^2/(30\\,\\Omega) = (28 \\cdot 5)^2/(30\\,\\Omega) = 140^2/(30\\,\\Omega) \\approx 653\\,\\text{W}$ EIRP. Convert the antenna gain to an isotropic factor: $6\\,\\text{dBd} = 8.15\\,\\text{dBi} \\to G = 10^{8.15/10} \\approx 6.53$. The transmitter output is then $P = P_\\mathrm{EIRP}/G = 653/6.53 \\approx 100\\,\\text{W}$. Hilfsmittel: rearrange $E = \\sqrt{30\\,\\Omega\\cdot P_\\mathrm{EIRP}}/d$ (Feldstärke im Fernfeld, S.15) for power, using the gain factor.", + "revision": 8, + "explanation": "Rearranging the distance formula gives $P_\\mathrm{EIRP} = \\frac{(d\\cdot E)^2}{30\\,\\Omega} = \\frac{(5\\,\\text{m}\\cdot 28\\,\\text{V/m})^2}{30\\,\\Omega} \\approx 653\\,\\text{W}$. The question asks for the transmitter output power; with antenna gain $g_d = 6\\,\\text{dBd}$ (so $g_i = g_d + 2.15 = 8.15\\,\\text{dBi}$): $P_S = \\frac{P_\\mathrm{EIRP}}{10^{(g_d + 2.15)/10}} = \\frac{653\\,\\text{W}}{6.53} \\approx 100\\,\\text{W}$ (which corresponds to about $398\\,\\text{W}$ ERP). Hilfsmittel: $d = \\sqrt{30\\,\\Omega\\cdot P_\\mathrm{EIRP}}/E$ and $P_\\mathrm{EIRP} = P_S\\cdot 10^{(g_d + 2.15 - a)/(10\\,\\text{dB})}$, with $P_S$ the transmitter power (Pegel / Antennen, S.15).", "source": "https://50ohm.de/NEA_personenschutzabstand_3.html#AK107", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AK108": { - "revision": 2, - "explanation": "After 0.5 dB cable loss, the dipole ERP/EIRP term in the safety-distance formula gives about 4.10 m for the 28 V/m limit.", + "revision": 4, + "explanation": "For a dipole $g_i = 2.15\\,\\text{dBi}$ with feedline loss $a = 0.5\\,\\text{dB}$, the EIRP from the $300\\,\\text{W}$ transmitter power is $P_\\mathrm{EIRP} = 300\\,\\text{W}\\cdot 10^{(2.15 - 0.5)/10} \\approx 439\\,\\text{W}$. Then $d = \\frac{\\sqrt{30\\,\\Omega\\cdot 439\\,\\text{W}}}{28\\,\\text{V/m}} \\approx \\frac{114.7\\,\\text{V}}{28\\,\\text{V/m}} \\approx 4.10\\,\\text{m}$. Hilfsmittel: $d = \\sqrt{30\\,\\Omega\\cdot P_\\mathrm{EIRP}}/E$ and $P_\\mathrm{EIRP} = P_S\\cdot 10^{(g_i - a)/(10\\,\\text{dB})}$, with $P_S$ the transmitter power (Pegel / Antennen, S.15).", "source": "https://50ohm.de/NEA_naeherungsformel_2.html#AK108", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AK109": { - "revision": 2, - "explanation": "The same distance formula scales with the square root of power; raising input power from 300 W to 700 W increases the distance to about 6.26 m.", + "revision": 4, + "explanation": "For a dipole $g_i = 2.15\\,\\text{dBi}$ with feedline loss $a = 0.5\\,\\text{dB}$, the EIRP from the $700\\,\\text{W}$ transmitter power is $P_\\mathrm{EIRP} = 700\\,\\text{W}\\cdot 10^{(2.15 - 0.5)/10} \\approx 1024\\,\\text{W}$. Then $d = \\frac{\\sqrt{30\\,\\Omega\\cdot 1024\\,\\text{W}}}{28\\,\\text{V/m}} \\approx \\frac{175.2\\,\\text{V}}{28\\,\\text{V/m}} \\approx 6.26\\,\\text{m}$. Hilfsmittel: $d = \\sqrt{30\\,\\Omega\\cdot P_\\mathrm{EIRP}}/E$ and $P_\\mathrm{EIRP} = P_S\\cdot 10^{(g_i - a)/(10\\,\\text{dB})}$, with $P_S$ the transmitter power (Pegel / Antennen, S.15).", "source": "https://50ohm.de/NEA_naeherungsformel_2.html#AK109", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AK110": { - "revision": 6, - "explanation": "Net isotropic gain is $11.5\\,\\text{dBd} + 2.15 - 1.5\\,\\text{dB} = 12.15\\,\\text{dBi}$, a factor $G = 10^{12.15/10} \\approx 16.4$. So $P_\\mathrm{EIRP} = 75\\,\\text{W} \\cdot 16.4 \\approx 1231\\,\\text{W}$. Then $d = \\sqrt{30\\,\\Omega \\cdot P_\\mathrm{EIRP}}/E = \\sqrt{30\\,\\Omega \\cdot 1231}/28 \\approx 192/28 \\approx 6.86\\,\\text{m}$. Hilfsmittel: rearrange $E = \\sqrt{30\\,\\Omega\\cdot P_\\mathrm{EIRP}}/d$ (Feldstärke im Fernfeld, S.15) for distance: $d = \\sqrt{30\\,\\Omega\\cdot P_\\mathrm{EIRP}}/E$.", + "revision": 8, + "explanation": "With $g_d = 11.5\\,\\text{dBd}$ and $a = 1.5\\,\\text{dB}$, the EIRP from the $75\\,\\text{W}$ transmitter power is $P_\\mathrm{EIRP} = 75\\,\\text{W}\\cdot 10^{(11.5 + 2.15 - 1.5)/10} \\approx 1230\\,\\text{W}$. Then $d = \\frac{\\sqrt{30\\,\\Omega\\cdot 1230\\,\\text{W}}}{28\\,\\text{V/m}} \\approx \\frac{192.1\\,\\text{V}}{28\\,\\text{V/m}} \\approx 6.86\\,\\text{m}$. Hilfsmittel: $d = \\sqrt{30\\,\\Omega\\cdot P_\\mathrm{EIRP}}/E$ and $P_\\mathrm{EIRP} = P_S\\cdot 10^{(g_d + 2.15 - a)/(10\\,\\text{dB})}$, with $P_S$ the transmitter power (Pegel / Antennen, S.15).", "source": "https://50ohm.de/NEA_naeherungsformel_2.html#AK110", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AK111": { - "revision": 2, - "explanation": "With 100 W, 10.5 dBd gain and 1.5 dB cable loss, the far-field safety-distance formula against 28 V/m gives about 7.1 m.", + "revision": 4, + "explanation": "With $g_d = 10.5\\,\\text{dBd}$ and $a = 1.5\\,\\text{dB}$, the EIRP from the $100\\,\\text{W}$ transmitter power is $P_\\mathrm{EIRP} = 100\\,\\text{W}\\cdot 10^{(10.5 + 2.15 - 1.5)/10} \\approx 1303\\,\\text{W}$. Then $d = \\frac{\\sqrt{30\\,\\Omega\\cdot 1303\\,\\text{W}}}{28\\,\\text{V/m}} \\approx \\frac{197.7\\,\\text{V}}{28\\,\\text{V/m}} \\approx 7.1\\,\\text{m}$. Hilfsmittel: $d = \\sqrt{30\\,\\Omega\\cdot P_\\mathrm{EIRP}}/E$ and $P_\\mathrm{EIRP} = P_S\\cdot 10^{(g_d + 2.15 - a)/(10\\,\\text{dB})}$, with $P_S$ the transmitter power (Pegel / Antennen, S.15).", "source": "https://50ohm.de/NEA_naeherungsformel_2.html#AK111", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AK112": { - "revision": 2, - "explanation": "The 18 dBd dish gain minus 2 dB feed loss gives the EIRP term; using the 61 V/m limit yields about 4.6 m in the main beam.", + "revision": 4, + "explanation": "With $g_d = 18\\,\\text{dBd}$, $a = 2\\,\\text{dB}$ and the higher limit $E = 61\\,\\text{V/m}$, the EIRP from the $40\\,\\text{W}$ transmitter power is $P_\\mathrm{EIRP} = 40\\,\\text{W}\\cdot 10^{(18 + 2.15 - 2)/10} \\approx 2613\\,\\text{W}$. Then $d = \\frac{\\sqrt{30\\,\\Omega\\cdot 2613\\,\\text{W}}}{61\\,\\text{V/m}} \\approx \\frac{280\\,\\text{V}}{61\\,\\text{V/m}} \\approx 4.6\\,\\text{m}$. Hilfsmittel: $d = \\sqrt{30\\,\\Omega\\cdot P_\\mathrm{EIRP}}/E$ and $P_\\mathrm{EIRP} = P_S\\cdot 10^{(g_d + 2.15 - a)/(10\\,\\text{dB})}$, with $P_S$ the transmitter power (Pegel / Antennen, S.15).", "source": "https://50ohm.de/NEA_naeherungsformel_2.html#AK112", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AK113": { - "revision": 8, - "explanation": "12.15 dBi is a linear gain of about 16.4; $\\sqrt{30\\,\\Omega \\cdot 250\\,\\text{W} \\cdot 16.4}/30\\,\\text{m}$ is about 11.7 V/m. Hilfsmittel: apply $E = \\sqrt{30\\,\\Omega\\cdot P_\\mathrm{EIRP}}/d$ (Feldstärke im Fernfeld, S.15) after converting 12,15 dBi to a factor ($G = 10^{g/(10\\,\\text{dB})}$, S.15).", + "revision": 9, + "explanation": "Solve the far-field formula for $E$. With $g_i = 12.15\\,\\text{dBi}$ and direct feed ($a = 0$): $P_\\mathrm{EIRP} = 250\\,\\text{W}\\cdot 10^{12.15/10} \\approx 4101\\,\\text{W}$. Then $E = \\frac{\\sqrt{30\\,\\Omega\\cdot 4101\\,\\text{W}}}{30\\,\\text{m}} \\approx 11.7\\,\\text{V/m}$. Hilfsmittel: $E = \\sqrt{30\\,\\Omega\\cdot P_\\mathrm{EIRP}}/d$ and $P_\\mathrm{EIRP} = P_S\\cdot 10^{(g_i - a)/(10\\,\\text{dB})}$ (Pegel / Antennen, S.15).", "source": "https://50ohm.de/NEA_personenschutzabstand_3.html#AK113", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AK114": { - "revision": 4, - "explanation": "A vertical dipole has about 2.15 dBi gain; applying the free-space field formula at 10 W and 10 m gives roughly 2.2 V/m. Hilfsmittel: apply $E = \\sqrt{30\\,\\Omega\\cdot P_\\mathrm{EIRP}}/d$ (Feldstärke im Fernfeld, S.15) with the dipole gain ($g_i = 2{,}15$ dBi, S.15).", + "revision": 5, + "explanation": "Use the field-strength formula with the dipole's linear gain $G_i = 1.64$: $E = \\frac{\\sqrt{30\\,\\Omega\\cdot P_A\\cdot G_i}}{d} = \\frac{\\sqrt{30\\,\\Omega\\cdot 10\\,\\text{W}\\cdot 1.64}}{10\\,\\text{m}} \\approx 2.2\\,\\text{V/m}$. Hilfsmittel: $E = \\sqrt{30\\,\\Omega\\cdot P_A\\cdot G_i}/d$ and dipole $G_i = 1.64$ (Feldstärke Fernfeld / Gewinn, S.15).", "source": "https://50ohm.de/NEA_personenschutzabstand_3.html#AK114", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AK115": { - "revision": 7, - "explanation": "Convert 100 W ERP to about 164 W EIRP, then $\\sqrt{30\\,\\Omega \\cdot 164\\,\\text{W}}/100\\,\\text{m}$ gives about 0.7 V/m. Hilfsmittel: first ERP→EIRP (×1,64), then apply $E = \\sqrt{30\\,\\Omega\\cdot P_\\mathrm{EIRP}}/d$ (Feldstärke im Fernfeld, S.15).", + "revision": 8, + "explanation": "ERP relates to EIRP by $P_\\mathrm{EIRP} = P_\\mathrm{ERP}\\cdot 1.64 = 100\\,\\text{W}\\cdot 1.64 = 164\\,\\text{W}$. Then $E = \\frac{\\sqrt{30\\,\\Omega\\cdot 164\\,\\text{W}}}{100\\,\\text{m}} \\approx 0.7\\,\\text{V/m}$. Hilfsmittel: $E = \\sqrt{30\\,\\Omega\\cdot P_\\mathrm{EIRP}}/d$ and $P_\\mathrm{EIRP} = P_\\mathrm{ERP}\\cdot 1.64$ (Pegel / Antennen, S.15).", "source": "https://50ohm.de/NEA_personenschutzabstand_3.html#AK115", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AK201": { - "revision": 2, - "explanation": "Charged power-supply capacitors can remain dangerous after unplugging; a suitably rated high-value resistor discharges them without a violent short-circuit current.", + "revision": 3, + "explanation": "Power-supply capacitors hold a dangerous charge after switch-off, so discharge them in a controlled way through a high-value resistor of adequate wattage. A high resistance limits the initial discharge current $I_0 = \\frac{U_0}{R}$, preventing arcing and damage that a short or low resistance would cause. The resistor must handle enough power: the stored energy $E = \\frac{1}{2}C U_0^2$ all becomes heat in it, with an initial dissipation $P_0 = \\frac{U_0^2}{R}$ (e.g. $\\frac{(400\\,\\text{V})^2}{100\\,\\text{k}\\Omega} = 1.6\\,\\text{W}$).", "source": "https://50ohm.de/NEA_elektrische_geaete_oeffnen_2.html#AK201", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AK202": { - "revision": 2, - "explanation": "Low-resistance bonding keeps exposed conductive parts at nearly the same potential, reducing touch-voltage risk to people.", + "revision": 3, + "explanation": "A low-resistance equipotential bond brings all simultaneously touchable conductive parts to nearly the same potential. That suppresses dangerous touch voltages between cases and metal parts, so when a person touches two parts at once no hazardous body current flows — it is a personal-safety measure.", "source": "https://50ohm.de/NEA_schutzerdung_2.html#AK202", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AK203": { - "revision": 2, - "explanation": "A separate earth lead near a quarter wavelength can resonate as an RF conductor, creating a voltage maximum at the equipment end.", + "revision": 3, + "explanation": "If the earth lead is electrically about $\\lambda/4$ long (or an odd multiple) at the operating frequency, it acts as a quarter-wave transformer: the low-impedance ground appears as a high impedance at the transmitter end. Standing waves form on the lead and a voltage maximum can build up at the case, putting noticeable RF voltage on the chassis (a \"hot\" case) even though it is correctly earthed for DC and mains frequency.", "source": "https://50ohm.de/NEA_schutzerdung_2.html#AK203", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "AK204": { - "revision": 2, - "explanation": "A transmitting antenna can have high RF voltage at current nodes or voltage maxima even with only a few watts, so touching it is unsafe.", + "revision": 3, + "explanation": "Even at low transmit powers of only a few watts, high RF voltages can appear at antennas and feedlines. Through $P = \\frac{U^2}{R}$, a high or mismatched local impedance (standing waves) can produce hundreds of volts from just 5–10 W. RF voltage can cause skin burns and deep heating, often without the familiar shock sensation, so a real hazard exists already at small powers.", "source": "https://50ohm.de/NEA_antennen_beruehrung_2.html#AK204", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "BA101": { "revision": 3, @@ -7176,16 +7341,18 @@ "confidence": 8 }, "EG220": { - "revision": 3, - "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.", + "revision": 4, + "explanation": "The \"i\" in $\\text{dBi}$ gives it away: the reference is the isotropic radiator — an ideal point source radiating equally in all directions.", "source": "https://50ohm.de/NEA_antennengewinn.html#EG220", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "EG221": { - "revision": 5, - "explanation": "dBd is referenced to a half-wave dipole, and a dipole already has $2.15$ dBi of gain over isotropic. So convert by adding that offset: $5\\ \\text{dBd} + 2.15\\ \\text{dB} = 7.15$ dBi. Whenever you see dBi vs dBd, the gap is always $2.15$ dB. Hilfsmittel: add the 2,15 dB dipole offset: $g_i = g_d + 2{,}15$ dB (Antennen, Pegel, S.15).", + "revision": 6, + "explanation": "$\\text{dBd}$ is referenced to a half-wave dipole, $\\text{dBi}$ to an isotropic radiator, and a dipole has $2.15\\,\\text{dB}$ gain over isotropic. So just add it: $5\\,\\text{dBd} + 2.15\\,\\text{dB} = 7.15\\,\\text{dBi}$. Hilfsmittel: $g_i = g_d + 2.15\\,\\text{dB}$ (Gewinn, S.15).", "source": "https://50ohm.de/NEA_antennengewinn.html#EG221", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "EG222": { "revision": 3, @@ -7236,16 +7403,18 @@ "confidence": 8 }, "EG307": { - "revision": 5, - "explanation": "Attenuations in dB add along the signal path. Treat each cable or inline loss as a positive dB loss and sum them; the losses in the shown station layout total 5 dB before antenna gain is considered. Hilfsmittel: losses in dB add along the path; each is $a = 10\\cdot\\log_{10}(P_\\mathrm{in}/P_\\mathrm{out})$ (Pegel, S.15).", + "revision": 6, + "explanation": "Attenuations in dB simply add along the chain, so the two cable sections give $2\\,\\text{dB} + 3\\,\\text{dB} = 5\\,\\text{dB}$ total loss.", "source": "https://50ohm.de/NEA_kabeldaempfung_1.html#EG307", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "EG308": { - "revision": 5, - "explanation": "With SWR $= 1$ nothing is reflected, so the missing power is pure cable loss. Going from $100$ W to $50$ W is a factor of $2$, and $10\\log_{10}(2) \\approx 3$ dB of attenuation. (Attenuation is a positive loss figure, and dBm would be an absolute power, not a loss — so those options are wrong.) Hilfsmittel: apply $a = 10\\cdot\\log_{10}(P_\\mathrm{in}/P_\\mathrm{out})$ (Pegel, S.15); ×2 power = 3 dB (table, S.15).", + "revision": 6, + "explanation": "Halving the power, $100\\,\\text{W} \\to 50\\,\\text{W}$, is a factor of 2, which corresponds to $3\\,\\text{dB}$. As this is a loss, the line attenuation is $3\\,\\text{dB}$. Hilfsmittel: the dB↔ratio table — power factor 2 is 3 dB (S.15).", "source": "https://50ohm.de/NEA_kabeldaempfung_1.html#EG308", - "confidence": 8 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "EG309": { "revision": 5, @@ -8346,10 +8515,11 @@ "confidence": 8 }, "NB505": { - "revision": 2, - "explanation": "Resistance is found from $R = U/I$; applying the voltage and current shown in the figure gives 40 ohm.", + "revision": 3, + "explanation": "By Ohm's law $R = \\frac{U}{I} = \\frac{12\\,\\text{V}}{0.3\\,\\text{A}} = 40\\,\\Omega$. Hilfsmittel: $R = U/I$ (Ohmsches Gesetz, S.11).", "source": "https://50ohm.de/NEA_ohmsches_gesetz.html#NB505", - "confidence": 7 + "confidence": 8, + "provenance": "50ohm-loesungsweg" }, "NB601": { "revision": 5, diff --git a/test_amateurfunk_anki.py b/test_amateurfunk_anki.py index adecff7..5f7ccbd 100644 --- a/test_amateurfunk_anki.py +++ b/test_amateurfunk_anki.py @@ -420,6 +420,50 @@ class TestAnkiBuild(unittest.TestCase): self.assertNotIn("af-explanation-low-confidence", back_high) self.assertNotIn("low confidence", back_high) + def test_cc_by_credit_gated_on_provenance_not_source_domain(self): + """The CC BY derivative-work credit must follow explicit + provenance, not the citation domain: an explanation merely + citing a 50ohm.de page is NOT a derivative work and must not + carry the credit, while one marked as derived from a worked + solution must.""" + derived = { + "revision": 2, + "explanation": "Worked-solution port.", + "source": "https://50ohm.de/NEA_x.html#AD106", + "confidence": 8, + "provenance": aa.PROVENANCE_50OHM_SOLUTION, + } + cited_only = { + "revision": 1, + "explanation": "Independently written, just cites a study page.", + "source": "https://50ohm.de/NEA_x.html#XX101", + "confidence": 8, + } + derived_html = aa.render_explanation(derived) + cited_html = aa.render_explanation(cited_only) + self.assertIn("af-explanation-credit", derived_html) + self.assertIn("CC BY 4.0", derived_html) + self.assertIn("50ohm.de-Autorenteam", derived_html) + self.assertNotIn("af-explanation-credit", cited_html) + self.assertNotIn("translated", cited_html) + + def test_root_deck_description_carries_attribution(self): + self._build_all() + apkg = self.out_dir / "amateurfunk-betriebliche-kenntnisse.apkg" + db_path, _media, _names = extract_collection(apkg, self.root / "b") + conn = sqlite3.connect(db_path) + try: + decks = json.loads(conn.execute("select decks from col").fetchone()[0]) + finally: + conn.close() + descs = [d["desc"] for d in decks.values() if d["desc"]] + self.assertEqual(len(descs), 1) # only the root deck carries it + desc = descs[0] + self.assertIn("CC BY 4.0", desc) + self.assertIn("govdata.de/dl-de/by-2-0", desc) + self.assertIn("50ohm.de-Autorenteam", desc) + self.assertIn("Bundesnetzagentur", desc) + def test_missing_explanation_leaves_card_unchanged(self): self._build_all() apkg = self.out_dir / "amateurfunk-technische-kenntnisse-e.apkg" @@ -501,6 +545,48 @@ class TestAnkiBuild(unittest.TestCase): self._build_all() self.assertIn("author", str(ctx.exception)) + def test_schema_accepts_valid_provenance(self): + path = self.root / "exp_provenance.json" + path.write_text( + json.dumps({"NA101": { + "revision": 1, "explanation": "ok", "source": "ok", + "confidence": 5, "provenance": aa.PROVENANCE_50OHM_SOLUTION, + }}), + encoding="utf-8", + ) + loaded = aa.load_explanations(path) + self.assertEqual( + loaded["NA101"]["provenance"], aa.PROVENANCE_50OHM_SOLUTION, + ) + + def test_schema_rejects_unknown_provenance_value(self): + path = self.root / "exp_badprov.json" + path.write_text( + json.dumps({"NA101": { + "revision": 1, "explanation": "ok", "source": "ok", + "confidence": 5, "provenance": "made-up", + }}), + encoding="utf-8", + ) + with self.assertRaises(aa.AnkiBuildError): + aa.load_explanations(path) + + def test_schema_rejects_unhashable_provenance(self): + # A malformed but valid-JSON provenance value (list/object) is + # unhashable; the validator must surface AnkiBuildError, never a + # raw TypeError from the set-membership test. + for bad in ([], {}): + path = self.root / "exp_unhashable.json" + path.write_text( + json.dumps({"NA101": { + "revision": 1, "explanation": "ok", "source": "ok", + "confidence": 5, "provenance": bad, + }}), + encoding="utf-8", + ) + with self.assertRaises(aa.AnkiBuildError): + aa.load_explanations(path) + def test_invalid_explanation_schema_raises_build_error(self): self.explanations_path.write_text( json.dumps({ @@ -617,6 +703,34 @@ class TestRealExplanationsFile(unittest.TestCase): ) self.assertEqual([], offenders) + def test_safety_distance_block_uses_transmitter_power_not_erp(self): + """AK106-AK112 must feed the EIRP formula with transmitter power. + + The relation is `P_EIRP = P_S · 10^((g_d+2.15-a)/10)` with `P_S` + the transmitter (Sender) power; writing `P_EIRP = P_ERP · 10^…` + double-counts the antenna gain (ERP already includes it). The + upstream 50ohm solutions carry exactly this mislabel, so guard + the whole block against it regressing back in. + """ + repo = Path(__file__).resolve().parent + explanations_path = repo / "explanations.json" + if not explanations_path.exists(): + self.skipTest("explanations.json not available") + explanations = aa.load_explanations(explanations_path) + block = [f"AK1{n:02d}" for n in range(6, 13)] # AK106..AK112 + for key in block: + if key not in explanations: + continue + body = explanations[key]["explanation"] + self.assertIn( + "P_S", body, + f"{key}: EIRP should be computed from transmitter power P_S", + ) + self.assertNotIn( + r"P_\mathrm{EIRP} = P_\mathrm{ERP}", body, + f"{key}: EIRP must not be derived by scaling ERP by the gain", + ) + if __name__ == "__main__": unittest.main()