Feedback factor of S/H

Hi,there
We know that the closed-loop gain is the reciprocal of feedback factor f, that is , Acl=1/f. Then from the architecture of the S/H shown in the pic1, we can conclude that the feedback factor is same as pic2. The S/H should have a closed-loop gain equal one, but at this time ,the reciprocal of feedback factor f is C\[{ S}_{ }\]+C\[{ F}_{ }\]+Copamp/C\[{ F}_{ }\]>1. I dont know why they conflict each other, any hint?




Regards
urian

urian said:

... why they conflict each other, any hint?

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There's no conflict: it's only important that both feedback factors are constant (over time).

erikl said:

There's no conflict: it's only important that both feedback factors are constant (over time).

Click to expand...

But, the closed-gain calculated from them is conflicted, why?

urian said:

Hi,there
We know that the closed-loop gain is the reciprocal of feedback factor f, that is , Acl=1/f. Then from the architecture of the S/H shown in the pic1, we can conclude that the feedback factor is same as pic2. The S/H should have a closed-loop gain equal one, but at this time ,the reciprocal of feedback factor f is C

+C

+Copamp/C

>1. I dont know why they conflict each other, any hint?
Regards
urian

Click to expand...

I suppose, there is a misunderstanding on your side.
The closed-loop gain of an opamp with feedback is the reciprocal of the feedback factor (in non-inverting operation) for time-continuous systems in the steady-state only.
However, in your case you have no time-continuous system but a sampled-data system that has to be analyzed with the tools and rules of DSP.

We know that the closed-loop gain is the reciprocal of feedback factor f, that is , Acl=1/f.

Click to expand...

That's only true for a non-inverting amplifier (and infinite closed loop gain, of course). In the general case, ideal Acl=a/f, where a is the forward transfer factor. a ≠ 1 in the present case.

The S/H should have a closed-loop gain equal one.

Click to expand...

Chose Cs = Cf to achieve it.

LvW said:

I suppose, there is a misunderstanding on your side.
The closed-loop gain of an opamp with feedback is the reciprocal of the feedback factor (in non-inverting operation) for time-continuous systems in the steady-state only.
However, in your case you have no time-continuous system but a sampled-data system that has to be analyzed with the tools and rules of DSP.

Click to expand...

So,the reciprocal is strictly restricted to time-continuous system and in non-inverting operation? It seems like the feedback factor we obtained in a sampled-data system is not the same as a time-continuous case. Then in a sampled-data system like a S/H circuit, why people use feedback factor to calculate amplifier's parameters such as GBW and so on? Does it mean that we can also use the feedback factor to derive the amp parameters except the closed-loop gain in a time-continuous case? It's valid for amp, not the system?

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That's only true for a non-inverting amplifier (and infinite closed loop gain, of course). In the general case, ideal Acl=a/f, where a is the forward transfer factor. a ≠ 1 in the present case.

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What is a and how to find it? I havent seen the equation before. In S/H case, what is a, and what is f?

Chose Cs = Cf to achieve it.

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Yeah, we choose the cs=cf to achieve the one time gain from a charge conservation viewpoint, and then the feedback factor is 1/2 in a charge redistritution SC circuit.It means that a =1/2?

urian, see here:

https://www.google.de/url?sa=t&rct=...2oHICA&usg=AFQjCNHdcDXhkCkdp84JWgly6muOQIv8wA

https://www.edaboard.com/threads/181687/ (in particular: post#6 and 7)

For a general description of feedback systems, I refer to this block diagram, kindly provided by LvW in a previous discussion, a pretty long thread about feedback factor definition: https://www.edaboard.com/threads/174717/



A1 is the amplifier open loop gain, ß is the feedback factor. The "forward factor" a is often assumed as 1 in text books when discussing feedback systems, but it's ≠ 1 in many real circuits, e.g. yours. I guess, you are referring to a simplified feedback block diagram with a = 1, which causes the confusion about closed loop gain.

I presume, that you can derive a yourself, it's calculated similar to f (or ß), finally you get Acl(ideal) = a/ß = Cs/Cf.

LvW said:

urian, see here:

https://www.google.de/url?sa=t&rct=...2oHICA&usg=AFQjCNHdcDXhkCkdp84JWgly6muOQIv8wA

https://www.edaboard.com/threads/181687/ (in particular: post#6 and 7)

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thx,LvW, the transfer function is sinx/x, so the magnitude of gain is one.

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FvM said:

For a general description of feedback systems, I refer to this block diagram, kindly provided by LvW in a previous discussion, a pretty long thread about feedback factor definition: https://www.edaboard.com/threads/174717/

A1 is the amplifier open loop gain, ß is the feedback factor. The "forward factor" a is often assumed as 1 in text books when discussing feedback systems, but it's ≠ 1 in many real circuits, e.g. yours. I guess, you are referring to a simplified feedback block diagram with a = 1, which causes the confusion about closed loop gain.

I presume, that you can derive a yourself, it's calculated similar to f (or ß), finally you get Acl(ideal) = a/ß = Cs/Cf.

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Yes, FvM, I always assume a=1 for there is no a in the feedback chapter in many textbooks. Ill try it.
BTW, your name has the same style with LvW

i think you should derive the transfer function according to charge conservation between these two phases.
you can refer to Razavi's book, Chap. 12 page 435~436