Timeline for Sziklai pair feedback loop
Current License: CC BY-SA 4.0
22 events
| when toggle format | what | by | license | comment | |
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| Jan 7, 2020 at 9:59 | vote | accept | Lucky-Luka | ||
| Jan 7, 2020 at 9:59 | |||||
| Jan 4, 2020 at 16:31 | vote | accept | Lucky-Luka | ||
| Jan 4, 2020 at 16:31 | |||||
| Jan 2, 2020 at 9:57 | comment | added | LvW | @peufeu...I am sure we can agree to the following definition. A current.feedback amplifier (CFA) has the following properties: (1) The "open-loop-gain" is a transimpedance (V/A); (2) the feedback factor has the unit (1/ohm); (3) the effective input signal is not a voltage difference (up-un=ud) as for voltage opamps but a current difference (ip-in=id) which is mirrored internally and then transferred into a voltage (across the transimpedance). If your circuit is in accordance with these properties it can be called CFA. | |
| Jan 1, 2020 at 21:47 | comment | added | bobflux | @jonk just ordered it. It's wonderful to preserve the knowledge from the golden days of analog! btw if you don't mind, I have a question opened about quite a similar topic... ;) | |
| Jan 1, 2020 at 21:23 | comment | added | jonk | @peufeu Did you buy a copy and enjoy it?? (Nice fix!) | |
| Jan 1, 2020 at 21:11 | comment | added | bobflux | @jonk thanks, fixed, that was a dumb mistake. Thanks for the book recommendation on modelling bipolar transistors you made in another question btw | |
| Jan 1, 2020 at 21:09 | history | edited | bobflux | CC BY-SA 4.0 |
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| Jan 1, 2020 at 20:47 | comment | added | jonk | @peufeu This question has lots of answers and is done and over, so all I was doing was skimming. But by background process for detecting violations of dimensional analysis is almost perfect and it immediately triggered on this: \$\frac{v_{\text{b}_1}-v_{\text{out}_1}}{g_m}\$. You assert that's a current. But there is no possible way that's true. The dimensions don't work out. It's more like a "Volt-Ohm." Just FYI. | |
| Jan 1, 2020 at 17:55 | comment | added | G36 | I suspect that this circuit is called a CFA because of the fact that the "error" current at the input stage is "mirrored" and further processed by the circuit. But if we look at the circuit from classical defined a feedback theory point of view. This topology of a circuit is a voltage-series feedback amplifier (series-shunt). | |
| Jan 1, 2020 at 17:44 | comment | added | bobflux | @LvW If I follow your logic, then I can grab any current-feedback opamp on the market and say it is voltage feedback because what matters is the Vbe of its input transistors... I'm not saying you're incorrect but there have been lots of arguments about the actual definition of "current feedback". I've added more explanations to the answer. | |
| Jan 1, 2020 at 17:37 | comment | added | Lucky-Luka | Since i prefer reading things on paper I've printed this answer 3 times up to now. Who knows if a 4th print will come? :) | |
| Jan 1, 2020 at 17:35 | comment | added | bobflux | @Circuitfantasist I found your website, gonna read it. Thanks! | |
| Jan 1, 2020 at 17:32 | history | edited | bobflux | CC BY-SA 4.0 |
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| Jan 1, 2020 at 16:54 | comment | added | LvW | @peufeu....It is a well-known fact that an emitter resistance causes current-controlled VOLTAGE feedback (even at a low-impedance input). This is true for a simple npn transistor as well as a compound transistor (Darlington, Sziklay) which can be treated as a single transistor. This is also in accordance with the drawings as given in the answer from Circuit fantasist. | |
| Jan 1, 2020 at 16:52 | comment | added | Circuit fantasist | @peufeu, There is something mystic in the idea to separate the "emitter" of the compound Sziklai "transistor" from the true emitter of the input transistor... since this solves the otherwise unsolvable task to take a part of the output voltage of the emitter follower and subtract it from the input voltage... i.e., to transform it to a non-inverting amplifier... | |
| Jan 1, 2020 at 16:23 | comment | added | LvW | @peufeu....I must admit that I cannot follow your reasoning - as far as the "current-feedback" concept is concerned. Your "explanation" would hold also for a simple npn BJT- correct? But it cannot be true. Of course, the emitter represents a low-impedance input - however, this does not automatically mean that it is a current input. In your answer you say: " If v(out1) is too high, Q1 collector current decreases". This is correct!. But you do not say WHY the collector current decreases! Answer: v(out1) is the emitter voltage which makes Vbe smaller - and THIS effect causes Ic to decrease. | |
| Jan 1, 2020 at 16:07 | comment | added | Circuit fantasist | Everything I have understood and explained to my students and web readers has been uploaded to the web. I am used to using these resources in the discussions I have on the web by linking to them. But, as I explained below, this was not allowed here. And at the beginning of the month, my account was suspended for a week. So now I am careful and avoid putting such links... Regarding the CFA, it really deserves a special attention... | |
| Jan 1, 2020 at 16:03 | history | edited | bobflux | CC BY-SA 4.0 |
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| Jan 1, 2020 at 15:51 | history | edited | bobflux | CC BY-SA 4.0 |
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| Jan 1, 2020 at 15:43 | comment | added | bobflux | Thanks! Sure, send a link, that'll be interesting! | |
| Jan 1, 2020 at 15:41 | comment | added | Circuit fantasist | I fully support your reasoning and admire the way you put it. I myself have the same idea about the so-called "current-feedback amplifier" (CFA) and have repeatedly defended it in other forums on the web... but have not understood. I would be happy to link to these resources... but this would be interpreted as a self-promotion with negative consequences for my contribution here. It's great that there are people like you with such a creative thinking... | |
| Jan 1, 2020 at 14:03 | history | answered | bobflux | CC BY-SA 4.0 |