# Role of gain in amplifier's frequency compensation...

1. Feb 2, 2016

### brainbaby

The text says that we have to alter the open loop gain in order to compensate an amplifier. The open loop gain should be reduced below unity before sufficient phase shift develop to cause oscillation in the amplifier at higher frequencies..

My question is why they tell to alter the open loop gain...rather it should be closed loop gain..because
i) Open loop gain(gain of op-amp without feedback) is constant..can’t be change.

ii) Though parasitic capacitance and inductance may be present in the internal circuitry of op-amp (which causes frequency dependent phase shift)….but they are also present outside as well like a capacitive load and RC filter(delay) formed by feedback resistance and input capacitance.

So we should alter the closed loop gain....shouldn't we?

2. Feb 2, 2016

### meBigGuy

The open loop gain is > 1 and the phase shift including the feedback is enough to cause feedback. No matter what amplitude losses are in the external feedback loop, it will then oscillate.

from https://en.wikipedia.org/wiki/Frequency_compensation:

A more precise statement of this is the following: An operational amplifier will oscillate at the frequency at which its open loop gain equals its closed loop gain if, at that frequency,

1. The open loop gain of the amplifier is ≥ 1 and
2. The difference between the phase of the open loop signal and phase response of the network creating the closed loop output = −180°. Mathematically,
ΦOL – ΦCLnet = −180°

3. Feb 2, 2016

### brainbaby

So according to your explanation its more reasonable and correct to say that "it should be closed loop gain which is to be reduced lower than unity not the open loop gain"...because open loop gain is a constant value during the time of manufacture of op-amp and this value is very high...so in order to balance that value, negative feedback is applied.....
and oscillation happens when that negative feedback changes to a positive feedback......so compensation is applied in the presence of a feedback...in other words preventing feedback to change from negative to positive.

4. Feb 2, 2016

### meBigGuy

You can quibble about these words. Changing the compensation is technically altering the open loop gain.

5. Feb 3, 2016

6. Feb 3, 2016

### brainbaby

In the picture you are talking about dominant pole compensation method...I'll come to it later on but first I need to understand this.....
Technically keeping my query simple ...I meant to say that why they tell to roll off the open loop gain…rather than it should be closed loop gain because oscillation develops when negative feedback is turning to positive at higher frequencies.. so feedback anyhow is necessary to cause oscillation..and when we talk about feedback we mean about close loop gain…then from where open loop gains comes into picture…??

7. Feb 3, 2016

### meBigGuy

http://www.ti.com/lit/ml/sloa077/sloa077.pdf [Broken]

BTW, You are correct, that we need to talk loop gain, not open loop gain. I kept trying to convince myself that somehow it was about open loop gain, but obviously it isn't.

Last edited by a moderator: May 7, 2017
8. Feb 3, 2016

### brainbaby

Exactly bro....

Last edited by a moderator: May 7, 2017
9. Feb 3, 2016

### LvW

It is our goal to ensure stability of the closed-loop circuit for all feedback factors.
The most critical feedback factor is k=1 for a closed-loop gain of Acl=1 (0dB)
In principle, there are two basic methods to ensure stability:
(1) internal or (2) external frequency compensation.

Some opamps are available which are NOT internally compensated.
These units must be (a) externally compensated or (b) they are used for higher gain applications only (small feedback factor) where the phase shift does not touch the range of critical values.

However, most opamp units are internally compensated.
They have an open-loop response with a dominating low-frequency pole so that the gain function Aol=f(jw) approaches a first-order response within the active gain range (slope of -20dB/dec up to Aol=0dB).
In this case, the phase shift of the loop gain - caused by the open-loop response (assuming resistive feedback) - will not come too close to the critical value of -180deg.

Example: If the second pole of Aol is directly on the frequency axis (at Aol=0dB) the loop gain phase shift for 100% feedback (closed-loop gain Acl=1) will be approximately assume -(90+45)=-135 deg. Thus, we have a phase margin of PM=45deg.

Remark:
Aol:gain of the opamp with open loop (no feedback);
Loop gain: Gain of the complete loop (including feedback path) if it is assumed to be open.

10. Feb 3, 2016

### meBigGuy

I guess I've been pretty sloppy in my thinking/definitions. When one says "open loop gain" one thinks of the amplifier's open loop gain, separate from the feedback path. But, that is not correct. You need to think of the SYSTEM's open loop gain, which is referred to as the loop gain.

You need to think differently about where the open loop gain is measured. If you break the loop at the end of the feedback network (where you would connect back to the input), that is the point where the SYSTEM gain must be less than 1 at 360 degrees. As LvW said, the loop gain is the amplifier gain times the feedback path gain. It is not the closed loop gain we are concerned with, but rather the SYSTEM open loop gain (called loop gain) including the amplifier and the feedback path.

I expect if you go back to the textbook you will see either that they are talking about a unity gain closed loop system, or they are measuring loop gain as amplifier gain times feedback gain.

11. Feb 3, 2016

### LvW

Yes - and in this context we remember the "Barkhausen Criterion" for oscillation (instability):
When a circuit with feedback oscillates the loop gain is LG(jw)=1 (0dB).

12. Feb 3, 2016

### brainbaby

The text further says that ….
“The way to prevent oscillation is to ensure that the loop gain( open loop gain minus the feedback factor) falls below unity before phase shift reaches 180 degrees..........”
Feedback factor..??
It is the feedback network gain...
What I think.....
Open loop gain is the "uncontrolled gain" without feedback
and feedback gain is the "attenuation" that is provided to the open loop gain.......ok
so if we take difference of these two we are left with a controlled gain i.e gain of the amplifier after feedback is applied.....which is the loop gain.
Is this controlled gain is the SYSTEM gain which you are talking about....??

Last edited: Feb 3, 2016
13. Feb 3, 2016

### brainbaby

You seems to confuse me...
I agree to half of your statement i.e Loop gain is the gain of the complete loop (including feedback path)...
but why you have assumed it to be open....?
If we disconnect something that means that it has been eliminated from the system, similarly if we assume feedback path to be open then it means that the feedback network don't have any effect on the system.

14. Feb 4, 2016

### Svein

There is always capacitive feedback due to package capacitance, wiring capacitance etc. These are in effect even if you do not have any deliberate feedback.

15. Feb 4, 2016

### meBigGuy

There is no absolute need for the feedback path to be open other than it make measurement possible. It is the same gain (amplifier X feedback) whether the path is open or closed. If you look at the TI link I posted earlier, in figure 5.7 they show how the feedback loop is broken to calculate the loop gain.
http://www.ti.com/lit/ml/sloa077/sloa077.pdf [Broken]

Last edited by a moderator: May 7, 2017
16. Feb 4, 2016

### brainbaby

what do you say about post 12...
Am I going the right way.....

Last edited by a moderator: May 7, 2017
17. Feb 4, 2016

### meBigGuy

Not quite going the right way. Try reading the TI note. The gain equations are shown in the beginning.

Fig 5-4 shows the amplfier gain (A) and the feedback gain (B). The closed loop gain is A/(1+AB) (equation 5-5) which reduces to 1/B when AB >> 1 (for example, an opamp has very large gain A midband). A*B is the loop gain.

Try to apply your concept to a feedback gain of 1 which yields a unity gain amplifier in an opamp circuit.

In a feedback controlled opamp circuit the overall closed loop gain is totally controlled by the feedback gain (1/B), independent of the opamp component's gain (A). That's why we use feedback, to make the overall system gain independent of the opamp internal gain. (independent of compensation issues)

18. Feb 4, 2016

### LvW

You are asking why we open the loop?
Simple answer: For injecting a test signal in order to be able to calculate, measure or simulate the loop gain.
For defining a gain I need an input and an output. Hence, the loop must be opened.
This is the classical procedure described in many articles because the opening can cause some problems:
We must ensure that
* the DC operating point is restored.
For this opening of the loop different procedures are described (the most universal was invented by Middlebrook).
If needed, I can give you several references.

Last edited: Feb 4, 2016
19. Feb 4, 2016

### brainbaby

It seems that we are getting little deviated from the topic.....
The main source of my query was two conflicting definition of compensation written in the textbook..please refer the attachment.....
First definition.
"The compensation of the amplifier is the tailoring of the open loop gain.............."
Second definition.
“The way to prevent this (oscillation) is to ensure that the loop gain( open loop gain minus the feedback factor) falls below unity before phase shift reaches 180 degrees"

How come at the same time compensation is about open as well as closed loop gain..??
This is the source of my confusion...

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20. Feb 4, 2016

### LvW

* At first, what is the meaning of your last sentence (...."closed loop gain"...) ?
I cannot find any mentioning of the "closed loop gain" in both definitions.
* Secondly, who told you that the loop gain would be "open loop gain minus the feedback factor" ?
The loop gain L(s) is nothing else than the product "open-loop gain x feedback factor ß": (L(s)=Aol(s)*ß)
* Therefore, both "definitions" are telling you the same: Tailoring the open-loop gain Aol(s) will at the same time tailor the loop gain L(s).

EDIT: Only now I have seen the enclosure (excerpt from a book).
There is a typing error.
As I wrote, the loop gain is L(s)=Aol(s)*ß=Aol/(1/ß) >> in dB: Aol(s)db-(1/ß)db.
Therefore (in words): The loop gain L(s) is "open-loop gain (in dB) minus inverse feedback factor (in dB)".
I think, this can clarify your confusion.

Last edited: Feb 4, 2016
21. Feb 4, 2016

### Jony130

22. Feb 8, 2016

### brainbaby

After many days of research on the topic I found that I was confusing my self with loop gain and closed loop gain...."so it was the loop gain which should be take into consideration".......
Thanks guys especially Jony130 for a very useful article (above) which you have provided me.....But there is some text in the article to which I partially agree... The text is as follows:-
"If is less than unity, the signals will be gradually attenuated into insignificance despite the fact that they are reinforcing each other at the input"

If the open loop gain (A) depends upon the difference of the signal at the input, loop gain depends upon A since loop gain = , thus its clear that loop gain also depends upon the difference of the signal at the input..so how can we say that signals will be attenuated despite they reinforce each other at input..??

23. Feb 8, 2016

### LvW

In case the loop gain is already less than unity at the critical frequency (where the additional phase shift reaches a value of -180deg) the closed-loop will be stable.
That means: The input signal and the feedback signal "reinforce" each other - however, this reinforcement is not strong enough to cause self-excitement (oscillation).
Remember the closed-loop gain Acl=A/(1+Aβ) >>>> for 180deg phase shift: Acl=A/(1-||)
If Aβ=1 we have self-excitement (denominator approaches zero). Hence, the loop gain Aβ must be less than unity.

Last edited: Feb 8, 2016
24. Feb 9, 2016

### brainbaby

what do you mean by that the reinforcement is not strong...do you mean that the difference at the input is decreasing with every cycle...or something like that....

but reinforcement itself means that the input difference is becoming larger....so weak reinforcement is not making sense to me right now...

25. Feb 10, 2016

### LvW

When we speak about stabiliy it is the LOOP GAIN which matters only. That is the total gain around the loop.
Let`s assume we have a gain block with A=10 and a feedback network with an attenuation (feedback factor) of k=1/12.
In this case, no self-excitement is possible because there is no amplification factor larger than unity within the loop (loop gain=10/12<1).
However, for k=1/9 we have a loop gain of 10/9>1 which means: Any increase of the amplifiers output voltage will be attenuated on his way back to the input only by a factor of 1/9 and the amplified again by a gain factor of 10. Therefore, the output signal grows larger and larger (self-excitement).
In contrary, for k=1/12 the gain of 10 cannot compensdate the loss in the feeedback path and, therefore, the signal cannot further rise. In contrary - it will die out.
I hope, I was able to express myself clear enough.

Comment: As far as the term "weak reinforcement" is concerned the meaning is as follows:
Slight positive feedback (in the above example: k=1/12) will increase the total gain to a larger value (example: Acl=10/(1-10/12)=60), but the circuit will remain stable.

Last edited: Feb 10, 2016