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Below is an exercise with solution about feedback network. I am wondering how the feedback factor is calculated like that. Could anyone shed some light on this?

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emnha
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  • I don't accept their answers based on the fact that they appear to believe that there is an impedance from the input (base) to ground of 4k + 1k - what happened to Q4's rE affecting this? – Andy aka Jun 12 '17 at 10:55
  • Are you talking about breaking the loop to calculate open loop gain? – emnha Jun 12 '17 at 11:07
  • Yes, their way of doing it. – Andy aka Jun 12 '17 at 11:12
  • For calculate open loop gain (forward gain) should we insert input voltage right at the base after 500 ohms resistor because that point is the true input of forward amplifier? – emnha Jun 12 '17 at 11:19
  • I just look at it and see that the closed loop gain will be about 8. – Andy aka Jun 12 '17 at 11:30
  • how did you calculate it? – emnha Jun 12 '17 at 11:34
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    4000/500 = 8, just like an op-amp with negative feedback. This makes input Z = 500 and of course output Z is rE of the final transistor which is going to be somewhere between 10 ohm and 30 ohm. In other words, I'm past the point where I care to do precise computations on circuits like this LOL. – Andy aka Jun 12 '17 at 11:57
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    And I've just noticed in your question that Ic = 1mA hence rE will be 26 ohms (26mV/Ic). Va is infinity (in your question) hence the gain will be 8. I can make this an answer if you want. – Andy aka Jun 12 '17 at 12:51
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    Regarding the feedback factor: f=Vbase/Vout for Vin=0. – LvW Jun 12 '17 at 14:45
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    Regarding the feedback factor: f=Vb1/Vout for Vin=0. Assuming rbe1=h11=hie1=5k (Ic=1mA) we can calculate Ve4=ic*0.8k and Ve4/Vout=0.8k/4k. From this, f=Vb1/Vout=(4.45/8.45)0.8k/4k. So we arrive at a feedback factor of app. f=0.1. – LvW Jun 12 '17 at 14:57
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    More than that - I doubt if the shown method for open-loop gain calculation is correct. – LvW Jun 12 '17 at 15:35
  • @LvW: I think so too. I think that the open loop gain should be from the base to Vout not from Vin to Vout. Is that same as yours? – emnha Jun 12 '17 at 15:38
  • No - I think the calculation based on Vin is OK - however, in the right part of the 2nd figure I miss the influence of the finite base-emitter resistance (rbe=hie) for Q1.More than that, for the input resistace (second picture) where is the influence of the low impedance node of the emitter (Q4) ? – LvW Jun 12 '17 at 16:38
  • @LvW: I think these are just approximation. They assumed that bjt input resistance is zero. However, I am confused about loop gain, it is the gain of feedforward and feedback network. But from Vin to Vout is not feedforward. Should it be from Vbase to Vout? – emnha Jun 12 '17 at 18:31
  • Input resistance of ZERO??? Feedback factor is defined as percentage of output fed back to the input. – LvW Jun 12 '17 at 19:20
  • I don't see anyone with the correct answer with gain at the collector. = -80 – Tony Stewart EE75 Apr 17 '18 at 08:00
  • What book is this? – user367640 Jun 20 '18 at 15:34

2 Answers2

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Gain at Q4 collector is 40.

This can be seen as the the output 4k load is driven by a current source (Q4 collector). Current is found as input current Vin/.5k which flows through the 4k feedback R. Feedback drives emitter of Q4 to -Vin/.5k * 4k=-8*Vin causing the net emitter current to be (Vin)(1/0.5k+(4k/0.5k)(1/1k))=vin*10e-3. Multiply by 4k and divide by Vin to get 40.

Agustin Ochoa
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Each of the gain stages is 4K/(0.026 / 1mA) = 4K/26 = 150X or 42dB. Total low frequency gain is 126dB (2 Million), ignoring any loading by the next stage's DC rin (probably significant, but no beta is given, so ignore that gain-shunting).

Gain will be Rf/Rin = 8x (14dB). There is a rather effective Virtual Gnd activity, thus Rin sets the input resistance.

DC Rout will be 26 ohms/gain_margin = 26 / (2million/8) = 26/ 250,000 or 0.1 milliohm.

analogsystemsrf
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  • I think, loading of the following stages must not be neglected because the expected rin (rbe=hie) will be in the same order as a the collector resistors. – LvW Jun 12 '17 at 16:44
  • correct---but we have no beta data; if same (4K || 4K), the open loop gain remains at 8X, Rin still 500, and Rout increases from 100uOhm to 800uOhm. – analogsystemsrf Jun 12 '17 at 16:50
  • .... "if the output were on Q4 emitter" is missing at the beginning of your answer – carloc Nov 26 '18 at 17:04