how to determine feedback factor

[jimito13]

I have seen in various papers that describe application specific opamps,the term "feedback factor β".How can i calculate this factor for the very simple circuit of a miller integrator?

feedback factor is always<1?

Thanks in advance.Any helpful answer woulb be appreciated.

 

[LvW]

Normally, the feedback factor is ≤1 .
In case of a Miller integrator it is R/((1/jωC)+R).
Feedback factor is calculated using the rule of impedance divider as seen from the opamp output.

 

[dedalus]

How can i calculate this factor for the very simple circuit of a miller integrator?

In this simple case you can use Acl = 1/β, (A -> inf), where Acl - closed-loop gain, A - gain of the basic amplifier. For Miller integrator Acl = -1/sC, thus β = -sC.

Refer to Baker - CMOS Circuit Design, Layout, and Simulation. In chapter 23 - Feedback Amplifiers general methodology for calculating all parameters for various feedback topologies is presented.

feedback factor is always<1?

Not always, but this is the most common case, e.g. feedback factor of Miller integrator is greater than 1 for frequencies greater than 1/C.

 

[FvM]

dedalus, you're shaking the foundations of network theory. Please reconsider.

Regarding the specific question, |β| > 1 is possible for transformer or resonant LC feedback circuits, normally it's ≤1, as said.

 

[dedalus]

dedalus, you're shaking the foundations of network theory. Please reconsider.

Please, specify your remark.

 

[FvM]

Simply, none of the expressions given in your post is correct. Starting with the (ideal) inverting integrator gain, which is
actually -1/sRC. The missing R is possibly a typo, but then you're concluding a wrong β expression, because
you apply A = 1/β, which is incorrect for the integrator, and so on. Finally, you conclude that |β| can be > 1 for large frequencies.
That's too much.

Simply consider, that LvW has given the correct β expression, you can rewrite it as sRC/(1+sRC), if you like.

 

[dedalus]

1) I assumed that under simple Miller integrator jimeece13 had meant circuit in the attached figure, so β = -sC isn't a typo. Sorry for don't specifying this in my first post.

2) Why Acl = 1/β is incorrect for the integrator? This is general expression for all feedback amplifiers with ideal basic amplifier, doesn't it?

3) Let's consider integrator with voltage input (with resistor, second attached figure). Vs - source voltage, Vf - voltage fed back to input by feedback network, Vo - output voltage. Feedback factor is defined as Vf/Vo. Clearly Vf - voltage across resistor: Vf = I1*R. Assuming ideal op amp I1 = I2. I2 = -Vo*s*C, thus Vf = -Vo*s*C*R <=> Vf/Vo = β = -s*C*R. I want to emphasize that Acl = -1/(s*C*R) = 1/β.

4) LvM shorted the input to ground, this is incorrect for series-shunt feedback (please refer to Baker). Correct derivation of β using impedance divider: Vf = (Vs-Vo)*R/(R+1/(sC)) = (Vs-Vf+Vf-Vo)*R/(R+1/(sC)) = (Vo/A + Vf - Vo)*R/(R+1/sC). Assuming ideal amplifier A -> inf, Vo/A -> 0. Vf = (Vf-Vo)*R/(R+1/(sC)) <=> Vf/Vo = β = -s*C*R. And again Acl = -1/(s*C*R) = 1/β.

 

[FvM]

I admit, that the term miller integrator isn't clear. But I won't say, that it's mainly designating an ideal transimpedance amplifier
integrator as you assumed.

If you refer to a transimpedance amplifier (your figure 1), then β is no longer a dimensionless factor but has the dimension of
an admittance. In so far I don't agree with β = -sC.

Regarding the other points, I'll restrict myself to 4). You claimed, that the β expression derived by LvW is wrong. I don't
understand, what's the role of A in your considerations. β is simply a voltage ratio, it doesn't depend on A. Any attempt to disprove
the said expression can't but fail.

I wondered for a moment, if your literature possibly involves a different definition of any usual term as β, Acl or Aol? I don't
know Baker, but I assume, that the representation isn't actually different from my literature, e.g. Gray/Meyer, Razavi. I think
the misunderstanding is caused by the divider for the input signal, that must be considered for the integrator gain as well.
Unfortunately, a similar factor (we can call it "alpha") isn't present in the popular literature's feedback schemes.

 

[dedalus]

5) About |β| > 1.

Let's consider integrator integrator using conceptual feedback amplifier, shown in the atached figure. So = ∫Si(t)dt.

For example Si(t) = sin(2*pi*f*t) => So(t) = -cos(2*pi*f*t)/(2*pi*f) = sin(2*pi*f*t-90)/(2*pi*f). In ideal case a (open-loop gain) -> inf and Se(t) = 0 <=> Si(t) = Sf(t). So feedback factor is equal to Si(t)/So(t) = 2*pi*f (and also feedback must shift back phase to 0 degrees) = s. Thus |β| of ideal integrator will be greater than 1 for all frequencies greater than 1 Hz.

 

[dedalus]

Please answer the question, why Acl = 1/β is incorrect for integrator?

If you refer to a transimpedance amplifier (your figure 1), then β is no longer a dimensionless factor but has the dimension of
an admittance. In so far I don't agree with β = -sC.

So what is wrong? β = -sC has dimension of admittance, doesn't it?

what's the role of A in your considerations

A is the open-loop gain of basic amplifier (op amp in this particular case).

I wondered for a moment, if your literature possibly involves a different definition of any usual term as β, Acl or Aol? I don't
know Baker, but I assume, that the representation isn't actually different from my literature, e.g. Gray/Meyer, Razavi. I think
the misunderstanding is caused by the divider for the input signal, that must be considered for the integrator gain as well.
Unfortunately, a similar factor (we can call it "alpha") isn't present in the popular literature's feedback schemes.

I use the same definitions of β, Acl or Aol as in Gray/Meyer.

 

[LvW]

...............
...............
3) Let's consider integrator with voltage input (with resistor, second attached figure). Vs - source voltage, Vf - voltage fed back to input by feedback network, Vo - output voltage. Feedback factor is defined as Vf/Vo. Clearly Vf - voltage across resistor: Vf = I1*R. Assuming ideal op amp I1 = I2. I2 = -Vo*s*C, thus Vf = -Vo*s*C*R <=> Vf/Vo = β = -s*C*R. I want to emphasize that Acl = -1/(s*C*R) = 1/β.

4) LvM shorted the input to ground, this is incorrect for series-shunt feedback (please refer to Baker). Correct derivation of β using impedance divider: Vf = (Vs-Vo)*R/(R+1/(sC)) = (Vs-Vf+Vf-Vo)*R/(R+1/(sC)) = (Vo/A + Vf - Vo)*R/(R+1/sC). Assuming ideal amplifier A -> inf, Vo/A -> 0. Vf = (Vf-Vo)*R/(R+1/(sC)) <=> Vf/Vo = β = -s*C*R. And again Acl = -1/(s*C*R) = 1/β.

Hi DEDALUS,

Just a comment to points 3) and 4) as cited above:

I think there is no doubt about it, that the feedback factor is defined as a portion of the output signal fed back to the input of an amplifier. This implies, of course, that the input signal is zero!
*Sergio Franco: β is the "gain of the feedback network"
* Jerry Graeme: "This defines β as simply the voltage divider ratio of the feedback network".

I am involved since more than 40 years in control systems and feedback theory, but - up to now - I never have heard or read, that "shorting the input to ground" during calculation, measurement or simulation of the feedback factor should be "incorrect".
As an example take the calculation/simulation of loop gain for a system with feedback - and the feedback factor is a fundamental part of the loop gain expression. There are various methods to find a loop gain expression (exact or approximate), however, each of these procedures require to set the input signal to zero!
May be you misunderstood Baker? Or Baker is wrong?
What do you think of my arguments?
Regards
LvW

 

[jimito13]

Ok,thanks for your answers but i need a more clear prove of the feedback factor equation.Let's consider the miller integrator as depicted below.The symbols i will use are on the schematic below.

[As i read in various literature (Baker,Gray-Meyer etc...) everybody agrees that feedback factor β=Vfeedback/Vout , so there is not another kind of definition as suspected in the above posts].

Let's suppose a finite Gain,GBW opamp.

Uo=A(U+ - U_) => Uo=-AU_ since U+=0, where A=(A0*ω-3dB)/(s+ω-3dB) first order model for the opamp.

I1=I2<=>...<=>U_=(sRCUo+Uin)/(sRC+1).

Which exactly is the feedback voltage on this circuit in order to take the fraction β=Vfeedback/Vo and derive β as said in the first post??

I notice that β=sRC/sRC+1 as said above apears in the equation i wrote.

Thanks in advance.

 

[jimito13]

Ok LvW it's clear know.Setting Uin=0 proves that β=sRC/(sRC+1) as mentioned above.Consequently Ufeedback is U_,am i right??

Why do we set Uin=0 for β calculation?Maybe,it is a silly question but i am not expert in feedback theory...

And a final question,when in papers they mention β=1/2 for example,they mean the |β|,right??

 

[LvW]

Hi jimeece13,

as long as the opamp input impedance is high enough to be neglected, the transfer function of the opamp (ideal or real) does not influence the feedback factor at all.
It is defined simply as a portion of the output signal appearing at the input.
LvW

 

[jimito13]

And what about my final question,when in papers they mention β=1/2 for example,they mean the |β|,right??

Sorry if i bother you...I am grateful for your answers.

 

[FvM]

Yes, β=R/((1/jωC)+R) and β=sRC/(1+sRC) are synonymous.

Regarding β sign, I understand the usual feed back schemes in the way, that negative feedback is represented by a positive
β value, because of the additional substraction operator. But it's a matter of definition and would work in the other way as well.

For the other questions, I leave the final say to LvW, who's also teaching the subject, while I'm only a practical man.

 

[LvW]

Ok LvW it's clear know.Setting Uin=0 proves that β=sRC/(sRC+1) as mentioned above.Consequently Ufeedback is U_,am i right??

Why do we set Uin=0 for β calculation?Maybe,it is a silly question but i am not expert in feedback theory...

And a final question,when in papers they mention β=1/2 for example,they mean the |β|,right??

OK, there is a good reason for your question. I'll try an answer:
* I suppose you have heard about the superposition theorem of circuit theory (more than one signal source within a circuit). This is the background. To calculate the voltage at the inverting terminal you make use of this rule and calculate two portions: (a) portion resulting from the input with output shorted, and (b) portion (often called β) as a result from the output with input shorted. That's all!
* Yes, normally we use only negative feedback (there are few exceptions), and as I have learned, some people define the factor as negative and some as positive.
This leads to two different kinds of block diagrams showing either a subtracting block or a summing block to combime the input with the feedback signal.

 

[jimito13]

Ok thanks a lot for your answers,absolutely helpful!

Added after 42 minutes:

LvW,maybe you misunderstood my second question.We proved that β=R/((1/jωC)+R).When it is mentioned that β=1/2 they mean |β|=|R/((1/jωC)+R)|=1/2??
I ask again because generally β is a function of jω.Thanks again.

 

[dedalus]

...............
...............
3) Let's consider integrator with voltage input (with resistor, second attached figure). Vs - source voltage, Vf - voltage fed back to input by feedback network, Vo - output voltage. Feedback factor is defined as Vf/Vo. Clearly Vf - voltage across resistor: Vf = I1*R. Assuming ideal op amp I1 = I2. I2 = -Vo*s*C, thus Vf = -Vo*s*C*R <=> Vf/Vo = β = -s*C*R. I want to emphasize that Acl = -1/(s*C*R) = 1/β.

4) LvM shorted the input to ground, this is incorrect for series-shunt feedback (please refer to Baker). Correct derivation of β using impedance divider: Vf = (Vs-Vo)*R/(R+1/(sC)) = (Vs-Vf+Vf-Vo)*R/(R+1/(sC)) = (Vo/A + Vf - Vo)*R/(R+1/sC). Assuming ideal amplifier A -> inf, Vo/A -> 0. Vf = (Vf-Vo)*R/(R+1/(sC)) <=> Vf/Vo = β = -s*C*R. And again Acl = -1/(s*C*R) = 1/β.

Hi DEDALUS,

Just a comment to points 3) and 4) as cited above:

I think there is no doubt about it, that the feedback factor is defined as a portion of the output signal fed back to the input of an amplifier. This implies, of course, that the input signal is zero!
*Sergio Franco: β is the "gain of the feedback network"
* Jerry Graeme: "This defines β as simply the voltage divider ratio of the feedback network".

I am involved since more than 40 years in control systems and feedback theory, but - up to now - I never have heard or read, that "shorting the input to ground" during calculation, measurement or simulation of the feedback factor should be "incorrect".
As an example take the calculation/simulation of loop gain for a system with feedback - and the feedback factor is a fundamental part of the loop gain expression. There are various methods to find a loop gain expression (exact or approximate), however, each of these procedures require to set the input signal to zero!
May be you misunderstood Baker? Or Baker is wrong?
What do you think of my arguments?
Regards
LvW

Ok. Suppose I'm wrong.

I think there is no doubt that closed loop gain of voltage integrator is Acl = -1/(s*R*C). According to your derivation β=R/((1/jωC)+R). So why 1/β=1/(R/((1/jωC)+R)) = 1+1/(s*C*R) isn't equal to -1/(s*R*C)?

FvM wrote that Acl = 1/β cannot be applied for integrator. Why???

 

[LvW]

LvW,maybe you misunderstood my second question.We proved that β=R/((1/jωC)+R).When it is mentioned that β=1/2 they mean |β|=|R/((1/jωC)+R)|=1/2??
I ask again because generally β is a function of jω.Thanks again.

Yes, sorry. Indeed, there was a misunderstanding.
When you read β=1/2 (independent on sign) this can only mean that either it is a fixed portion (caused by resistors, independent on frequency) or it is a frequency dependent portion given at a specific frequency.