Feedback Amplifier

LvW

Joined Jun 13, 2013
875
But if you want the individual gains with the feedback loop closed you need to analysis the whole circuit at once.
I think, The Electrician did a really good job - and he has used the superposition principle for analyzing the first stage for closed-loop operation..
However, I still believe that - for hand calculations - it is easier to apply the loop gain concept. In this case, we only need to find the loop gain T(s) and the gain Aol without feedback. I did it with pencil and paper and it wasn´t too involved.
 

Thread Starter

KCHARROIS

Joined Jun 29, 2012
302
All I can do is come up with the gain of the second stage which is equal to:

(R10||(R11+R3)||R9) / re2

which = 32.87

When I calculate the FIRST stage I do:

((re2 * 100)||R7||R6||R5) / (re1 + (R3||(R11 + R9)

which = 6.37

I'm not sure what I'm doing wrong?
 

LvW

Joined Jun 13, 2013
875
I could, but I used the phrase "no feedback" to mean that R11 is not there.

When preparing to analyze the way you and LvW did, the presence of R11 as a load on the output is necessary to preserve the impedances at various nodes when the loop is open to be the same as when the loop is closed.

It's easy enough to analyze the circuit with R11 present as a load, but not connected as feedback:
I think, it is an interesting question if - for calculating the gain Aol without feedback - the feedback elements are to be considered or not (without feedback injection at the first stage).
The Electrician has demonstrated both methods.

For my opinion, these elements must be considered as a load to the 2nd transistor because:

1.) Undoubtly, the closed loop gain is Acl=10.5 (approximately).
2.) As can be shown by hand calculations as well as simulation the loop gain is approximately T=-20.
3.) From this it follows directly that the open-loop gain is (approximately)
Aol=Acl*(1+T)=211.
That is the value all of us (The Technician, Jony, LvW) have found in case the 2nd stage is loaded with feedback elements.
 
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LvW

Joined Jun 13, 2013
875
All I can do is come up with the gain of the second stage which is equal to:

(R10||(R11+R3)||R9) / re2

which = 32.87

When I calculate the FIRST stage I do:

((re2 * 100)||R7||R6||R5) / (re1 + (R3||(R11 + R9)

which = 6.37

I'm not sure what I'm doing wrong?
What value did you use for re1 and re2?
 
I'm not sure what I'm doing wrong?
You have one end of R11 connected to the emitter of Q1 and the other end grounded, but in the closed loop circuit the other end is not connected to ground; it's connected to the collector of Q2, and there is a large signal there. That modifies the apparent value of R11; think Miller effect.
 

#12

Joined Nov 30, 2010
18,159
From the local idiot's point of view, the first transistor has a gain of 3.6 and the second one has a gain of Beta (minus some loss in R6 and R7), but that is all over-ridden by R11 and R3 (give or take some Xc).

The R11/R3 circuit is going to reduce the gain of Q1 until it is just enough to drive Q2.
Q2 will continue to run at the gain of the transistor, which is quite a lot, so the gain of Q1 will be turned down quite a lot by the feedback loop. (What Kubeek said in post #2)
 
When I calculate the FIRST stage I do:

((re2 * 100)||R7||R6||R5) / (re1 + (R3||(R11 + R9)

which = 6.37

I'm not sure what I'm doing wrong?
I calculate the effective value of R11 as seen from the Q1 emitter (with the loop closed) as -104.5605 ohms; that's right, a negative value.

So forget R9 in the formula because that's already in this effective value of R11. Evaluate the expression:

((re2 * 100)||R7||R6||R5) / (re1 + (R3||R11))

with R11 set to -104.5605 and see what you get.
 
However, I still believe that - for hand calculations - it is easier to apply the loop gain concept. In this case, we only need to find the loop gain T(s) and the gain Aol without feedback. I did it with pencil and paper and it wasn´t too involved.
I suspect that I can form the admittance matrix about as quickly as you can do the Black style analysis. With practice, the matrix can be quickly written by inspection.

The further advantage to forming the admittance matrix is that once you have it, everything you could want to know about the circuit can be calculated--the impedance at every node, the signal amplitude at every node and from that, the gain between every pair of nodes, the impedance between every pair of nodes, the current gains as well as the voltage gains, etc., etc.

One of the problems with Black style analysis is maintaining the proper impedances when opening the loop. For example, in Jony130's circuit, he has a 1k resistor (R11) from emitter of Q1 to ground. But is the resistance seen at the emitter end of R11 just 1K when the loop is closed? R11 is not connected to ground at the end away from the emitter of Q1; it's connected to the collector of Q2, and the impedance there is not zero ohms. And the same problem occurs at the collector of Q2 with the 1k (representing R11) connected to ground; when the loop is closed, the other end is connected to the emitter of Q1 and the impedance there is not zero.

These are relatively small errors, and this is a relatively simple circuit and the result obtained is a good approximation to the exact results, but I think they are errors nonetheless. The admittance matrix method doesn't make these errors.
 
I think that Q2 will have the same gain as without feedback, and Q1 will have the rest of what is needed to maintain the overall gain set by R11/R3.
From the local idiot's point of view, the first transistor has a gain of 3.6 and the second one has a gain of Beta (minus some loss in R6 and R7), but that is all over-ridden by R11 and R3 (give or take some Xc).

The R11/R3 circuit is going to reduce the gain of Q1 until it is just enough to drive Q2.
Q2 will continue to run at the gain of the transistor, which is quite a lot, so the gain of Q1 will be turned down quite a lot by the feedback loop. (What Kubeek said in post #2)
These conjectures are not borne out by analysis.
 
Ok so the gain of the first stage is approx .3 but how did u get the negative resistance of R11???
Full analysis with the loop closed, and β1=β2=100, re1=re2=26 ohms, shows that with an input voltage of 1 volt, the signal amplitude at the collector of Q2 = 10.4454 volts, and at the emitter of Q1 the signal is .988788 volts.

This means that the voltage across R11 is 9.456612 volts. Since R11 = 1000 ohms, the current in R11 is .009456612 amps.

At the emitter of Q1, a current of .009456612 amps is being driven into that node even though the voltage there is a positive .988788 volts; one would expect that a positive voltage connected to a resistor would drive a current out of the node, rather than in to it. That fact means that the R11 resistor appears to be a negative resistor.

If the resistor R11 were going to ground at that point, what value of resistance would supply .009456612 amps with .988788 volts across it? .988788 volts/-.009456612 amps = -104.5605 ohms.

But you'll notice that there is no way to know this ahead of time. You have to do the full analysis to figure out what you needed to do to make your formula work.

Feedback circuits, and this is a really simple one, can be tricky, but a full analysis never fails.
 

LvW

Joined Jun 13, 2013
875
LvW could you please show your work?
Thanks
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.

Note that transconductances gm=1/re and rin,2=R6||R7||rbe,2

Numerically: For Ic1=Ic2=1mA and gm1=gm2=0.04 A/V (re=25 ohms)
and rin,2=1.1 kohms the loop gain is calculated to be T=-19.1.

2.) Open-loop gain Aol

First stage: A1=-(R5||rin,2)/[re1+R3]
Second stage: A2=-gm2*[R10||R9||(R11+R3)]

Aol=A1*A2=198

3.) Closed-loop gain Acl:

Acl=Aol/(1-T)=198/20.1=9.85

4.) Comment
Note that the above approximate results are based on some rough assumptions for gm and rin,2. Nevertheless, these results are rather close to the gain values obtained with simulation (see my earlier posts).
Perhaps it helps to understand how to analyze such a circuit by pencil and paper (without symbolic simulation programs).
 

Thread Starter

KCHARROIS

Joined Jun 29, 2012
302
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.

Note that transconductances gm=1/re and rin,2=R6||R7||rbe,2

Numerically: For Ic1=Ic2=1mA and gm1=gm2=0.04 A/V (re=25 ohms)
and rin,2=1.1 kohms the loop gain is calculated to be T=-19.1.

2.) Open-loop gain Aol

First stage: A1=-(R5||rin,2)/[re1+R3]
Second stage: A2=-gm2*[R10||R9||(R11+R3)]

Aol=A1*A2=198

3.) Closed-loop gain Acl:

Acl=Aol/(1-T)=198/20.1=9.85

4.) Comment
Note that the above approximate results are based on some rough assumptions for gm and rin,2. Nevertheless, these results are rather close to the gain values obtained with simulation (see my earlier posts).
Perhaps it helps to understand how to analyze such a circuit by pencil and paper (without symbolic simulation programs).
How could R3 be in parallel with re1?
 

#12

Joined Nov 30, 2010
18,159
These conjectures are not borne out by analysis.
Post #13 by The Electrician states that the base to collector gain of Q1 is -.321085 and the base to collector gain of Q2 is -32.6338

I believe this bears out the conjectures that measurements will show a small voltage gain from the base to the collector of Q1 and a larger voltage gain from the base to the collector of Q2.
 
Post #13 by The Electrician states that the base to collector gain of Q1 is -.321085 and the base to collector gain of Q2 is -32.6338

I believe this bears out the conjectures that measurements will show a small voltage gain from the base to the collector of Q1 and a larger voltage gain from the base to the collector of Q2.
That's incidental to the mistakes of both conjectures, and it's not telling us much. We could well suppose that Q1's gain will be less than that of Q2 given that Q1 has some unbypassed emitter resistance, and Q2 doesn't, . The really relevant parts of both conjectures are so far off the mark that it's reasonable to say that the conjectures are incorrect.


Kubeek said "I think that Q2 will have the same gain as without feedback, and Q1 will have the rest of what is needed to maintain the overall gain set by R11/R3." It's true that part of his statement, namely "Q1 will have the rest of what is needed to maintain the overall gain set by R11/R3" is true, but it's a tautology; it goes without saying. The important part is what he said about the gain of Q2, and it's incorrect.

You said "From the local idiot's point of view, the first transistor has a gain of 3.6..."

It isn't clear whether you meant without or with feedback (I think you meant without), but in neither case is the first transistor's gain 3.6, so this part of your conjecture is incorrect.

You also said "the second one has a gain of Beta (minus some loss in R6 and R7)", but this not exactly true as a general proposition. The Q2 gain with a β of 100 is -110.86, which is coincidentally close to the β of 100. But saying the gain is Beta is just a guess relying on a coincidence, because if β is 20, the Q2 gain is -106.637, and if β is 500 the Q2 gain is -111.745. The Q2 gain without feedback is quite independent of β.

You further said (referring to the case with feedback, I believe) "Q2 will continue to run at the gain of the transistor" which is also incorrect, the same mistake Kubeek made.

But thanks for playing, anyway. :)
 
Perhaps it helps to understand how to analyze such a circuit by pencil and paper (without symbolic simulation programs).
Black's method of analysis is useful. It can help one's intuition about what's going on in a feedback circuit. Signal flow graphs similarly are good to use.

I simply wanted to point that feedback circuits can be solved in another way. And, in fact, in a circuit with multiple feedback loops, Black's formula becomes very difficult to apply.
 

PRS

Joined Aug 24, 2008
989
Dog gone it, Kcharrios, think about it. If you implement this circuit on a bread board or on some other board, unless you use 1 percent resistors and known values of Betas for your transistors, you are not needful of fine equations. You only need the ones I gave you. I'll give you equations for Rin and re upon request.

In a real world situation you plug in 5 percent resistors and any old transistors you have and then measure the results. The equations I gave you are simple and close enough.

Incidentally, notice we have a negative feedback in this circuit. If the feedback went to the base of Q1 then we would have positive feedback and an unstable overall amplifier. The negative feedback stabilizes this amplifier.
 
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Dog gone it, Kcharrios, think about it. If you implement this circuit on a bread board or on some other board, unless you use 1 percent resistors and known values of Betas for your transistors, you are not needful of fine equations. You only need the ones I gave you. I'll give you equations for Rin and re upon request.
The equations you gave him are for the no feedback case. He said in the very first post that he is able to get those equations himself. Where he's having trouble is the case with feedback.

I also noticed that the equation you gave for the second stage gain:

(R10//R9)/(R8+re2)

is incorrect, because R8 is bypassed with C4.
 

Thread Starter

KCHARROIS

Joined Jun 29, 2012
302
Thank you backing me up, I will try the admittance matrix method because I see this method being very good with other methods. Ill post back sometime to show my results if I need any help.

Thanks everyone
 
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