Given the following circuit, enter image description here

I need to redraw it as a Series Shunt feedback amplifier, i.e. with the following topology: enter image description here

Identifying the AC equivalent circuit seems to be easy: 12V and -12V become GND and current generator turns into an open circuit. However I have troubles identifying the amplifier net and the feedback net and the series shunt topology. Could you help me redrawing the circuit?


I have troubles identifying the amplifier net and the feedback net.

"P" or \$v_O\$ is the feedback net. The error amplifier is formed by Q1-Q4 and this also provides all the error voltage amplification. Q5 is a voltage follower and provides current gain (should it be deeded to drive the load impedance R.

  • \$\begingroup\$ I'm sorry, I don't understand exactly where the series shunt topology is... \$\endgroup\$ – Surfer on the fall Feb 11 '18 at 19:45
  • \$\begingroup\$ You said you were having difficulty figuring the amplifier and the feedback net so my answer is advising you of these. I’m not going to do your full homework for you but I thought I might help a bit. \$\endgroup\$ – Andy aka Feb 11 '18 at 20:27
  • \$\begingroup\$ sorry, I don't want the homework.. I'd like to understand because the textbook doesn't cover the details of how to recognize networks! I don't get "P is the feedback net".. do you mean that R is the feedback net? \$\endgroup\$ – Surfer on the fall Feb 11 '18 at 21:16
  • \$\begingroup\$ R is the load resistor. P I assume is the name of the net. \$\endgroup\$ – Andy aka Feb 11 '18 at 22:35

I do not like the notation "series-shunt"-feedback because it does not clearly say what`s really going on. Therefore, I prefer to describe the feedback scheme as shown in the circuit as "voltage-controlled voltage feedback".

(1) For finding the open-loop gain of the amplifier we must open the feedback path (node "P") - however, we must nor change the loading at this point. Therefore, gain calculation must take into account the fact that the dynamic input resistance at the base of Q2 is in parallel to the emitter resistor R (but this influence is not very strong, because the gain of the emitter follower is close to unity). hence, the open-loop gain of the whole amplifier is primarily determined by the differential amplifier (nothing new, formulas available).

(2) For the feedback network we have to answer the following question: Which percentage of the output voltage vo is fed back to the input of the ampifier? I think, the answer is clear and obvious!

So - with these information, it should be not a problem to write the corresponding expressions into the two boxes.

  • \$\begingroup\$ I think that no percentage of vo is fed back.. isn't it? \$\endgroup\$ – Surfer on the fall Feb 12 '18 at 10:15
  • \$\begingroup\$ What means "no percentage"? Do you think, that there is no feedback at all? \$\endgroup\$ – LvW Feb 12 '18 at 11:21

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