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Feedback amplifier circuit redrawing

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bznm

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Given the following circuit,

bcM56.jpg

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

Y4niB.png

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?
 

The output is vo.
The feedback is node P.

You should be able to figure out the rest from that
 
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    bznm

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crutschow said:
The output is vo.
The feedback is node P.

You should be able to figure out the rest from that
Click to expand...
Thanks for the answer!
However I am not able to understand it... Feedback net should be a two port network.. and I can't see what are these ports.
Unfortunately my textbook (and I didn't have any luck searching on other books) doesn't cover the details about identifying the topology.

Thanks again for the help
 

bznm said:
.......... Feedback net should be a two port network.. and I can't see what are these ports.
Unfortunately my textbook (and I didn't have any luck searching on other books) doesn't cover the details about identifying the topology.
Click to expand...
In the general case, a feedback net can have two ports but it doesn't have to be (and most feedback networks don't).
It this case the second port is the common (ground) connection.
 

crutschow said:
In the general case, a feedback net can have two ports but it doesn't have to be (and most feedback networks don't).
It this case the second port is the common (ground) connection.
Click to expand...
I suppose that the feedback box contains Q5, while Q1,Q2,Q3,Q4 are the amplifier box, because the signal is input in Q1, isn't it?
 

No - Q5 is part of the main amplifier, because the output vo is located at the emitter of Q5.
 

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