# Feedback Amplifier

Discussion in 'General Electronics Chat' started by KCHARROIS, Mar 24, 2014.

1. ### KCHARROIS Thread Starter Member

Jun 29, 2012
292
1
Hello,

I've been looking at feedback amplifiers such as the one below and the overall gain of the amplifier is equal to R11/R3. Knowing this is great but I'm trying to understand the gain of each of the two stages. Figuring this out is easy without the feedback path, you calculate Zin, Zbase, gain of each stage, etc. But with feedback path its not so easy, so how should I go about calculating the gain of each stage?

Thanks

File size:
11.3 KB
Views:
70
2. ### kubeek AAC Fanatic!

Sep 20, 2005
4,688
806
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.

3. ### KCHARROIS Thread Starter Member

Jun 29, 2012
292
1
I can't figure what the gain of the first stage will be but for the second stage I think that:

R10||R9||(R11+R3)

Can someone confirm this for the second stage? Any ideas for the first stage?

Thanks

4. ### KCHARROIS Thread Starter Member

Jun 29, 2012
292
1
After playing around with the feedback resister R11 the gain of both stages is changing there for its safe to say that R11 has a direct effect on both amplifiers but still don't understand how. Not sure if this helps anyone?

Thanks

5. ### LvW Active Member

Jun 13, 2013
674
100
I am afraid, there is no other way than to apply Black´s classical feedback formula for the closed-loop gain Acl:

Acl=Aol/(1-T)

with Aol=gain without feedback and T=Loop gain (negative sign).

Loop gain is the gain around the complete feedback loop, which for the purpose of calculation must be opened at a suitable node. Suitable means: A node where a low output impedance is connected to a large input impedance. By doing this, loading errors are minimized. Otherwise, a kind of load mirror has to be connected.

EDIT 1: After looking into your circuit, I would propose to open the loop at the collector of Q1 and inject a test signal (for calculation of T) into the second amplifier circuit. However, dont forget to load the collector of Q1 with a good approximation for the input resistance of the second stage. This node provides the output voltage for loop gain calculation.
A a result, you will see that the overall closed-loop gain will not be equal to R11/R3.

EDIT 2: For calculation of Aol you must not forget to load Q2 with R11.

Last edited: Mar 25, 2014
6. ### LvW Active Member

Jun 13, 2013
674
100
With the aim to justify the above formula I have simulated the whole circuit (according to the mentioned procedure).
Results (for f>300Hz):
Acl=10.5
Aol=220
T=-20

Thus: Acl=Aol(1+20)=10.47 (good agreement with the measured value of 10.5).

Last edited: Mar 25, 2014
7. ### LvW Active Member

Jun 13, 2013
674
100
R11 is not only connected to R3. This node is also connected to the very low impedance emitter node. For this reason, you cannot simply isolate the voltage divider R11-R3.
More than that, I don`t think it makes sense to ask separately for the gain of the 1st stage because you have TWO signal inputs (base and emitter).
Therefore my former answer (suggestion) to use the loop gain concept.

8. ### KCHARROIS Thread Starter Member

Jun 29, 2012
292
1
I will try your loop method and see if I get the same results that you are getting.

Thanks

9. ### PRS Well-Known Member

Aug 24, 2008
989
35
The gain of the first stage is (Rin2//R5)/(R3+re1)
The gain of the second: (R10//R9)/(R8+re2)

These are the gains without feedback, of course. The overall gain without feedback is the gain of stage one times the gain of stage two.

10. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,990
1,115
To find open loop gain you need to use this small-signal diagram.

So if we assume Ic1 = Ic2 ≈ 1mA and Hfe = 150.
The first stage voltage gain is equal to

$Av1 = \frac{R5||R6||R7||(Hfe+1)*re2}{re1 +R3||R11} = \frac{918}{117} = 7.8V/V$

And Second stage voltage gain

$Av2 = \frac{R10||R9||(R3 + R11)}{re2} = \frac{777}{26} = 29.8V/V$

So the Aol = Av1 * Av2 = 232V/V

And the close loop gain (Acl)

Acl = Aol/(1 + Aol * B) =
232/(1 + 232*(0.1/1.1)) = 10.5V/V

where
Aol - open-loop voltage gain. The gain without the feedback.

B - feedback factor (feedback gain) in our example

B = Ve1/Vc2 = R3/(R3 + R11) = 100/(100Ω + 1KΩ) = 0.1/1.1 = 0.090

And is Aol>>1/B we can say that Acl ≈ 1 + R11/R3 ≈ 11V/V

• ###### 2.PNG
File size:
9.8 KB
Views:
82
Last edited: Mar 25, 2014
11. ### LvW Active Member

Jun 13, 2013
674
100
Hi Jony, good calculation.
Just one comment: I think at the output we should have R11 in series with R3||(1/gm1).
This would reduce the gain Aol to about 198. However, the gain Acl remains nearly the same.

12. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,300
335
Of course there's another way; analysis of the entire circuit. I explained how to do this in an older thread: http://forum.allaboutcircuits.com/showthread.php?t=26710

It makes sense if the question is asked precisely. V1 applies a signal to the base of Q1, which leads to signal voltages of various amplitudes at the several nodes.

One can ask, "What is the ratio of the signal at the collector of Q1 to the signal at the base of Q1?", and call this the base to collector gain of Q1.

Then ask "What is the ratio of the signal at the collector of Q1 to the signal at the emitter of Q1?", and call this the emitter to collector gain of Q1.

Using the method described in the older thread, we can set up the admittance matrix. I left R8 in the matrix even though it is bypassed, so that a non-zero emitter resistor for the Q2 stage could be added if desired later; it's set to a very low value to account for the effect of the bypass capacitor.

Node 1 is the base of Q1; node 2 is the emitter of Q1; node 3 is the collector of Q1 and the base of Q2; node 4 is the emitter of Q2; node 5 is the collector of Q2:

First let's analyze the circuit without feedback. We do this by setting R11 to ∞. Here's the overall voltage gain and the impedances at the nodes. Notice that without feedback, the impedance at the emitter of Q1 is 30.4852 ohms (assuming the source driving the base of Q1 has zero internal impedance).

Here are the individual stage gains as defined above (without feedback):

File size:
12.1 KB
Views:
74
File size:
14.4 KB
Views:
64
File size:
21 KB
Views:
66
13. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,300
335
Now repeat the analyses with R11 set to 1000 ohms. First the overall amplifier voltage gain, and the nodal impedances; the impedance at the emitter is now 1.719 ohms with feedback, a very low value:

And the individual stage gains as defined in the previous post:

File size:
14.4 KB
Views:
69
File size:
20.8 KB
Views:
60
14. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,990
1,115
I'm not so sure about that. Simulation show that R11 + R3 gives closer result for Acl.

Why you don't include R11 loading effect on the open loop gain?

15. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,300
335
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:

Edit: I added the effect of R11 to the emitter of Q1, and added 100 ohms in series with R11 at the collector of Q2 to make the circuit exactly the same as yours:

File size:
9.9 KB
Views:
61
• ###### Charrois7.png
File size:
11.3 KB
Views:
69
Last edited: Mar 25, 2014
16. ### KCHARROIS Thread Starter Member

Jun 29, 2012
292
1

Jony130, thanks this does make sense but here is what really gets to me. When I measure the signal voltage across R5 I measure 300uV and when I measure across R3 to ground I get ≈ 1mV there for the gain is about 1/3, how is that possible I can't see or calculate a gain of 1/3 for the first stage. (This is with feedback included)

Thanks to all other op's as well

17. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,300
335
Look at my analysis in post #13. I get a gain of about .32, emitter to collector.

18. ### KCHARROIS Thread Starter Member

Jun 29, 2012
292
1
Great seems like you got it but I can't figure out how you calculated that vale I also looked at post #12.

19. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,990
1,115
My analysis that I show at post 10 describes how to find Open Loop Gain.

But if you want the individual gains with the feedback loop closed you need to analysis the whole circuit at once.
See this small-signal representation, so all you need to do the small-signal analysis for the whole circuit.

File size:
19.9 KB
Views:
26
20. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,300
335
As Jony130 says, you have to analyze the whole circuit at once, with feedback included.

To fully understand what I'm doing in post #12 and #13, you really must read and understand the old thread I referenced: