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

[brainbaby]

Hey guys please help me on this...

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?

 

[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°

 

[brainbaby]

The open loop gain is > 1 and the phase shift including the feedback is enough to cause feedback.

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,

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.

 

[meBigGuy]

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..

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

 

[Svein]

1. See https://en.wikipedia.org/wiki/Frequency_compensation
2. The purpose of compensation is to ensure that the open loop gain drops below unity before the phase shift gets too close to 180°

 

[brainbaby]

1. The purpose of compensation is to ensure that the open loop gain drops below unity before the phase shift gets too close to 180°

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…??

 

[meBigGuy]

http://www.ti.com/lit/ml/sloa077/sloa077.pdf

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.

 

[brainbaby]

http://www.ti.com/lit/ml/sloa077/sloa077.pdf

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.

Exactly bro...

 

[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.

 

[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.

 

[LvW]

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.
.

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).

 

[brainbaby]

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.

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...??

 

[brainbaby]

Loop gain: Gain of the complete loop (including feedback path) if it is assumed to be open.

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.

 

[Svein]

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.

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.

 

[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

 

[brainbaby]

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

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

 

[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)

 

[LvW]

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.

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 loading at the opening remains about the same , and
* 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.

 

[brainbaby]

Not quite going the right way.

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...

 

[LvW]

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...

* 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.

 

[Jony130]

Try carefully read and analyze this
http://educypedia.karadimov.info/library/acqt0131.pdf
http://e2e.ti.com/blogs_/archives/b...late-an-intuitive-look-at-two-frequent-causes
http://www.allaboutcircuits.com/technical-articles/negative-feedback-part-4-introduction-to-stability/

 

[brainbaby]

Try carefully read and analyze this
http://www.allaboutcircuits.com/technical-articles/negative-feedback-part-4-introduction-to-stability/

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 Aβ 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 = Aβ, 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..??

 

[LvW]

In case the loop gain Aβ 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-|Aβ|)
If Aβ=1 we have self-excitement (denominator approaches zero). Hence, the loop gain Aβ must be less than unity.

 

[brainbaby]

this reinforcement is not strong enough to cause self-excitement (oscillation).

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...

 

[LvW]

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...

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.

 

[brainbaby]

I hope, I was able to express myself clear enough.

Your last answer was quite impressive...hence I tried to illustrate it numerically...
My inference from my calculation..

For both cases when loop gain>1 and loop gain<1...Vout seems to be increasing...
In case first->
Vout raises from 2V to 4.2 V and in second case..
Vout raises from 2V to 3.6 V...hence in both the cases Vout is increasing that is against your inference which is for loop gain>1, Vout increases and for loop gain<1, Vout should decrease...

So please tell now where I am wrong...!

 

[LvW]

I only will comment case 2 (because the same method can be applied for case 1):
You have calculated the "first round trip" of the signals. As you can see, the output signal has increased from 2V to 3.6V.
This can be continued to further trips.
Finally you will arrive at a stable point, where no further increase will take place because we have an equilibrium between input and output.
This equilibrium is reached for:
Vin=0.2V
Vout=12V
Vfeedback=12/12=1v
V(amplifier,in)=0.2+1=1.2V
Vout=1.2*10=12V.
Result: 12/0.2=60 (the closed-loop gain is Acl=60 > A=10).

Formula: Acl=A/(1-A*k)=10/(1-0.8333)=60

 

[brainbaby]

I only will comment case 2 (because the same method can be applied for case 1):

But same thing would happen to case 1 also ..but at a more faster rate...so what difference would it make..??

 

[LvW]

But same thing would happen to case 1 also ..but at a more faster rate...so what difference would it make..??

Yes - at a faster rate, but you will not find such an equilibrium as in case 2 (input after addition x gain=output).
Try it.

 

[Jony130]

To see how signal decrease for Aβ < 1 apply a input signal only for one loop "revolution".
For example for Case 2 we have:
We have Vin = 0.2 and Vf = 0.16. But now Vin = 0 so we have Vf = 0 + 0.16 = 0.16 and Vf'1 = 0.13-->Vf'2--->0.11-->Vf'3-->0.09 and so on.
Because now you need more than 60 iterations to get to the stable point at Vin =1.2 and Vout = 20.
And for Case 1 the stable point do not exist.
http://wps.prenhall.com/chet_paynter_introduct_6/0,5779,426330-,00.html

 

[brainbaby]

Certainly I got the idea but my results aren't up-to the expected value..

Equilibrium means that the output stops increasing and the output is constant with constant input...ok
@LvW
According to you the equilibrium point is when Vin = 1.21 V but after that you can see in the table that Vout is constantly increasing ...however according to theory it should become constant making the input constant...So where is the equilibrium point ??
@JONY
From the table you can see that as Vin is decreasing.., so after 60 iterations the value of Vin will be very meagre ...but your stable point is at 1.2 V ..i.e at a higher Vin...
So what is this anomaly and where is the stable point??
and one more thing... at Vin=1.2V your Vout is 20V and mine is 12V how exactly??

 

[LvW]

Brainbaby - I am afraid, it is not easy to answer.
Let me try: You have startet this "step-by-step" approach in your post#26 - and I was following this approach (this way to explain what happens) in my reply.
However, let me say, that it is somewhat questionable if we are allowed to apply this "circling within a loop" procedure.
I know that there are general analyses for feedback systems which show that such a "deterministic" view is not allowed because - in reality - there is no signal which is circling n-times through such a loop.
Nevertheless, we have startet this (simplified) view - and the only answer I can now give is:
In your table, you have listed some discrete events (voltages) which are assumed to be existent at some discrete times.
But this is not the case.
Instead, we must assume a time-continuous process.
And anywhen during this process of increasing output voltage we arrive at a point, where we have an "equilibrium" (that means: Output voltage=Input voltage x gain).
In my post#27 I have given the corresponding final value: Vout=12V .
When this condition is reached the gain expression is "fulfilled" and there is no reason for a further increase of the output voltage.
That`s all I can say.
Again: It is a simplified view, which cannot explain all the effects to be observed.
But it can, perhaps, give a rough picture what is happening and why.

 

[Jony130]

From the table you can see that as Vin is decreasing.., so after 60 iterations the value of Vin will be very meagre ...but your stable point is at 1.2 V ..i.e at a higher Vin...
So what is this anomaly and where is the stable point??

You must have misunderstand my post or my last post was not clear for you. For the circuit with Aol = 10, β = 1/12 and Vin = 0.2V the steady state is reach as show by LvW at Vout = 12V and Vin + Vf = 1.2V (Vin is still equal to 0.2V). So the circuit with Aol*β<1 is stable.
To confirm this we can try different method. Let as apply Vin only for ONE iteration (a brief trigger signal) to see what will happens.
Vin'1 = 0.2V Vout'1 = 2V
Vin'2 = 0.16V Vout'2 = 1.6V
Vin'3 = 0.13V Vout'3 = 1.3V
Vin'4 = 0.11V Vout'4 = 1.1V
Vin'5 = 0.09V Vout'5 = 0.9V
Vin'6 = 0.08V Vout'6 = 0.8V
Vin'7 = 0.067 Vout'7 = 0.67V
Vin'8 = 0.055V Vout'8 = 0.55V
Vin'9 = 0.046V Vout'9 = 0.46V
Vin'10 = 0.038V Vout'10 = 0.38V
As you can see Vout will decay to zero. And this is why the amp is stable if Aol*β<1.

Know let as try the same think for case when Aol*β>1 and the the phase shift is 0°(positive feedback). So know we have Aol = 10, β = 1/9 and again as before let as apply a short trigger pulse to the input (first iteration).
Vin'1 = 0.2V Vout'1 = 2V
Vin'2 = 0.22V Vout'2 = 2.2V
Vin'3 = 0.24V Vout'3 = 2.4V
Vin'4 = 0.27V Vout'4 = 2.7V
Vin'5 = 0.3V Vout'5 = 3V
As you can see even without input signal output voltage keeps rising. And this is why our amp is unstable.

but after that you can see in the table that Vout is constantly increasing ...however according to theory it should become constant making the input constant

You must have a error in your iteration

 

[brainbaby]

Yeah exactly this I was expecting...thanks for it!
Yes now I understood the meaning of the word equilibrium which LvW mentioned about..it means that when loop gain<1 stability is said to be achieved because at a certain Vin the result is gain times Vout, which is equal to the closed loop gain which signifies that the maximum amplification that a system can attain when loop gain<1 is equal to the closed loop gain...however when loop gain>1 it means that Vout does not depends on neither A nor β...thats why the relation Acl=A/(1-A*k) doesn't hold true...

However what I am being wondering since the beginning of our discussion was that...as it seems very easy to accept that the product of A and β decides the stability of the system...which is though mathematically correct to infer...but it doesn't provide any intuitive insight to our discussion...like when the open loop gain (A) combines with β=1/12..the system got stabilised and when it combines with β=1/9 the system is unstable...so as we see that the difference between the two beta values is very trivial but it bring such a drastic stability change in the system..so what is this magic...as on superficial terms they look only numbers.

 

[LvW]

...it means that when loop gain<1 stability is said to be achieved because at a certain Vin the result is gain times Vout,.

I suppose you mean: ...Vout is the result of gain times Vin...

...however when loop gain>1 it means that Vout does not depends on neither A nor β...thats why the relation Acl=A/(1-A*k) doesn't hold true...

Not quite right... Vout still depends on A*k>1 - however, Vout is continuously rising (and the slope of this rising is prop. to A*k) until the voltage is clipped due to a finite supply voltage; hence, we cannot speak anymore of "gain" because the amplifier has left its linear region.

However what I am being wondering since the beginning of our discussion was that...as it seems very easy to accept that the product of A and β decides the stability of the system...which is though mathematically correct to infer...but it doesn't provide any intuitive insight to our discussion...

Perhaps the following helps:
Keep in mind that we have a so-called "oscillation condition" (given first by H. Barkhausen in the 1930th). This condition is A*k=1, and it constitutes a threshold between the two cases: stable and unstable (A*k<1 stable, A*k>1 unstable).
Another formulation is: A*k<0 neg. feedback; 0<A*k<1 stable pos. feedback; A*k>1 unstable pos. feedback.

...so as we see that the difference between the two beta values is very trivial but it bring such a drastic stability change in the system..so what is this magic...as on superficial terms they look only numbers.

Yes - correct. This effect is due to the above mentioned threshold A*k=1.

Visualation: A second-order closed-loop system has a pole pair in the complex s-plane.
* A*k<1: poles are in the left half of the s-plane (real part "sigma" negative) and the oscillation amplitude dies out continuously (system is stable)
* A*k>1: poles are in the right half of the s-plane (real part sigma" positive ) and the oscillation amplitude is rising continuously (system is unstable)

Finally: In the time domain, the real part of the pole ("sigma") appears as exp(sigma*t) which causes rising or falling amplitudes (depending on the sign).

 

[Jony130]

But this subject is intuitive.
Let us examine this case (points A and B are not connected)

At the beginning we apply input signal (1V) to the input of our inverting amplifier with gain equal to Aol = -29V/V.
The output voltage is of-course equal to Vout = Vin * -Aol = -29V, and we have 180° phase shift between Vin and Vout.

And now, if we add a feedback network which reduces the output signal the the level of the input signal (β = 1/29) and introduces another 180° voltage shift. We will have voltage at point B equal in amplitude and in phase to the input voltage. So now if we disconnect Vin and connect point A with point B we can hope that the circuit will start the oscillations at the frequency where these two conditions (amplitude and phase) are satisfied (Aol*β = 1 at Fo = √6/(2* pi *R*C)).
But if Aol*β < 1 but we meet the phase condition. The circuit will not oscillate because we don't have enough gain in our loop (voltage at point B is smaller than Vin).
And for Aol*β > 1 and phase are identical ( 0° phase shift between Vin and voltage at point B), the circuit will also start the osculation because now we have more gain then we need (Voltage at point B is larger than Vin.). Do you get it ?

 

[LvW]

And for Aol*β > 1 and phase are identical ( 0° phase shift between Vin and voltage at point B), the circuit will also start the osculation because now we have more gain then we need (Voltage at point B is larger than Vin.). Do you get it ?

Yes - I support this view.
Just a short comment: ...will start the oscillation - however, with continuous rising amplitudes until clipping occurs (hard limiting), unless there is another soft-limiting device within the loop (diodes or amplitude-controlled resistors,...).

 

[brainbaby]

we don't have enough gain in our loop (voltage at point B is smaller than Vin).

You have given me a new intuitive insight of the loop gain...I appreciate it...
Loop gain can be thought as Vf/Vin...and that solves all the problem...
So all deal is the loop gain ...if Vin is applied only for one iteration and then the system is realized we found that when loop gain<1 , Vout is attenuated by β and since now Vf is less than Vin, it can be thought of as for the next iteration Vin lowers and hence again this lowered Vin is further attenuated by β network and on and one the cycle perpetuates and the system got stabilised...vice versa when loop gain >1...

Last two questions...

From the beginning you have considered Vin for only one iteration...What would be your explanation when initial Vin is applied for multiple iterations...?
and
This question is somewhat not exactly of scientific temperament...but as we know that multiplication is also a form of addition...i.e its a repetitive addition ...for e.g
4*3=12...
if 4 is added 3 times it leads to 12 ...or if 3 is added 4 times it also leads to 12...
so what difference would if make if loop gain instead of being product of A and B would be the addition of A and B...??
I know it sounds so weird...but its the byproduct of my curiosity...

 

[Jony130]

From the beginning you have considered Vin for only one iteration...What would be your explanation when initial Vin is applied for multiple iterations...?
and

Because this case (only one iteration) is much more easier to understand and to analysis for our brain.

so what difference would if make if loop gain instead of being product of A and B would be the addition of A and B

10 * 1/12 = 0.83
10+1/12 = 10.083

 

[brainbaby]

Jony and LvW...

 

[jim hardy]

Once you have got that intuitive 'feel' for it

it lends new meaning to the old adage

closed loop transfer function = G/(1+GH)

even for the simplest closed loop, taking the case of Mother Nature's favorite shape sinewaves

out = in X 1/(1+1)
out/in = 0.5

Now add delay equal to 180 degrees phase shift at some frequency
180 degrees is just reversing the sign of a sinewave
so what was negative feedback became positive, that is GH became -1
so denominator of G/(1+GH) changes from sum to difference
transfer function becoming G/(1-GH)
out = in X 1/(1-1)
out/in = 1/0
AHA we can have output with zero input !

You saw that in grade school when the PA microphone got just the right distance from speaker to make that awful 'whistle'
that's the distance at which speed of sound over distance from speaker to mic gives delay equal to 180 degrees at the whistle frequency . The immediate fix is to turn down volume reducing loop gain GH below 1.
(New DSP PA systems detect oscillation and change their own settings. I looked for Behringer's specsheet but only found sales stuff from 3rd parties)

Any system in nature , including systems made by man, will find that frequency if one exists. It'll try to enter a state of sine wave oscillation but might shake itself apart first..
hope i wasnt overly redundant..
That little thought experiment locked in the concept for me. I need all the memory aids i can get

old jim

 

[brainbaby]

hope i wasnt overly redundant..

absolutely not Jim...

 

[brainbaby]

Vout still depends on A*k>1 - however, Vout is continuously rising (and the slope of this rising is prop. to A*k) until the voltage is clipped due to a finite supply voltage; hence, we cannot speak anymore of "gain" because the amplifier has left its linear region.

Can you provide me that curve where the slope of rising Vout is prop to loop gain..?? I couldn't google it..

 

[Jony130]

Can you provide me that curve where the slope of rising Vout is prop to loop gain..?? I couldn't google it..

Here you have

 

[LvW]

No - I don`t know if such a curve even does exist. However - why not trying by yourself?
Simulate a simple oscillator circuit (example: WIEN type) for different loop gains (1.1, 1.2,...) and observe the slope of the rising oscillation sgnal.

(Jony130 was quicker by some seconds, congratulations).)

 

[brainbaby]

(Jony130 was quicker by some seconds, congratulations).)

Jony might be celebrating the victory...

 

[LvW]

Can you provide me that curve where the slope of rising Vout is prop to loop gain..?? I couldn't google it..

Hi brainbaby - perhaps the following is interesting for you:
The loop gain is a function in the frequency domain and the slope of the rising oscillation amplitde can be observed in the time domain.
Hence, we need a "connection" of both domains.
This connection does exist in form of the closed-loop poles.
During the starting phase with rising amplitudes the poles of the closed-loop circuit are located in the right half of the s-plane (pos. real part σ).
Because these poles are the solutions of the characteristic equation they also appear as part of the solutions in the time domain.

That means: The oscillation is described by the expression:
Vout(t)=A*exp(s*t)=A*exp(σ+jω)t=A*exp(σt) * exp(jωt)
The first part describes the amplitude (rising for σ>0) and the second part is the sinewave.

As you can see, the pole location (better: The real part of the pole) determines the exponential function which describes the rising amplitudes.
It is the purpose of a nonlinear element in an oscillator to lower the loop gain for large amplitudes and to shift the poles back to the left half of the s-plane (σ<0) withthe consequence of decreasing amplitudes. As a result - the poles will swing between the left and right hallfs of the s-plane - and the amplitude will not be constant but periodically get a bit larger and smaller.

 

[brainbaby]

Hi brainbaby - perhaps the following is interesting for you:
The loop gain is a function in the frequency domain and the slope of the rising oscillation amplitde can be observed in the time domain.
Hence, we need a "connection" of both domains...
...This connection does exist in form of the closed-loop poles.

I'll certainly try to digest your text...but anyway for now your input is required in my following thread...

https://www.physicsforums.com/threa...n-dominant-pole-compensation-strategy.857590/

thanks...!