Feedback Amplifier

#12

Joined Nov 30, 2010
18,076
That's incidental to the mistakes of both conjectures, and it's not telling us much.
Pretending that one phrase or sentence is completely unrelated to the rest of the paragraph, pretending that you can't tell whether 3.6 is the open loop gain unless I say it is, and in spite of the fact that the last sentence in the paragraph says this is completely overridden my the closed loop components is intentionally disingenuous. I am interested in answering the question of the original poster (See post #1). "I'm trying to understand the gain of each of the two stages."

Your math shows exactly the gain of the two stages, and you want to argue against it???

You win.
 
Let me give a short description:

1.) Loop gain T
Open the circuit at the left side of C3 and inject a test signal.
The first part of the loop gain (starting at the base of Q2) measured at the collector of Q2 is

T1=-gm2*[R10||R9||(R11+R3)]

The second part between collector of Q2 and collector of Q1 is

T2=(R3||re1)/[R11+(R3||re1)]*gm1*(R5||rin2).

T=T1*T2.
If you break the loop at the left side of C3, then rin2 is no longer in parallel with R5, is it?

Maybe the loop should be broken just to the right of R6 and R7.
 
Pretending that one phrase or sentence is completely unrelated to the rest of the paragraph, pretending that you can't tell whether 3.6 is the open loop gain unless I say it is, and in spite of the fact that the last sentence in the paragraph says this is completely overridden my the closed loop components is intentionally disingenuous. I am interested in answering the question of the original poster (See post #1). "I'm trying to understand the gain of each of the two stages."

Your math shows exactly the gain of the two stages, and you want to argue against it???

You win.
It isn't my intention to be disingenuous or to "win". I saw some errors in Kubeek's and your conjectures. I said earlier "These conjectures are not borne out by analysis". I thought you felt that I was too vague, so I gave some details of the errors I saw, which I felt supported my assertion.

I'm sorry you were insulted by that.
 

LvW

Joined Jun 13, 2013
847
If you break the loop at the left side of C3, then rin2 is no longer in parallel with R5, is it?
Maybe the loop should be broken just to the right of R6 and R7.
When breaking a loop at at a low-resistive node (example: opamp output) it is not necessary - in many cases - to correct this loading error caused by disconnection of the feedback path.
However, in this case such a correction is necessary because we do not have such a low-resistive node in the loop.
By disconnecting the second stage from the 1st one we must retain the closed-loop loading conditions. That is a general rule for loop gain analyzes.
Thus, the load resistance for Q1 is R5 in parallel with rin,2.
 
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LvW

Joined Jun 13, 2013
847
How could R3 be in parallel with re1?
For loop gain calculation we need the voltage at the emitter of Q1.
This voltage is to be calculated based on a simple voltage divider rule.
This node is common to R3 and the dynamic input resistance of Q1 (re,1) at the emitter. Thus, both resistors are in parallel (seen from the emitter).
Because this resistance re,1 is very low it must not be neglected (re,1=1/gm1).
 

LvW

Joined Jun 13, 2013
847
And, in fact, in a circuit with multiple feedback loops, Black's formula becomes very difficult to apply.
Indeed, this is really an interesting point - worth to be discussed in a separate thread. If we have - let´s say - three feedback loops we have three different loop gain expressions. Question: Which loop mostly determines stability properties?

EDIT: In fact, we even have more than three alternatives to find the loop gains because opening loop 1 allows loops 2 and 3 either open or closed.
 
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