One Question about FB in Razavi's Design of Analog CMOS

[andover]

Hi,
When I was reading razavi's book about feedback, I got a question about the attached circuit. (chapter 8 Fig8.3, 8.6).
From KCL, It's straight forward to get Vout/Vin=-C1/C2, and it's not hard to get Y/X=A/(1+BA)=1/B * (1-1/BA) as a general equation for feedback. However, when he applies loop break method (break at right side of C2)to calculate the open loop gain, he gets BA=C2/(C1+C2)gm1r01, where B feedback factor equals to C2/(C1+C2). My question is, when you apply this B back to Y/X, you will get (1+C1/C2) instead of (-C1/C2).
I am just wondering did I miss something in the deduction? I really appreciate if you can help to explain this.

Thanks

 

[dedalus]

Hi, andover

By comparing this circuit and block diagram in the Fig. 8.4. you can see that output of the feedback network and input of the amplifier don't match. Thus you cannot determine B considering just one voltage divider with input shorted, because in this case feedback signal (Vf=V(C1)) is a superposition of two signals resulting from input and output signals: Vf = B1*Vin + B2*Vout (B1 = Vf/Vin with Vout = 0, B2 = Vf/Vout with Vin = 0), Vf = B1*Vout/Gain + B2*Vout = (B1/Gain + B2)*Vout => B = Vf/Vout = (B1/Gain + B2). Gain Vout/Vin determined by feedback network is equal to 1/B => B = B1*B + B2 <=> B = B2/(1+B1). Applying these results to the circuit yields B1 = C2/(C1+C2), B2 = -C2/(C1+C2), B = -C2/C1.

There is another way to obtain this result by considering closed-loop system. Signal that is fed back by the feedback network C1, C2 is equal to the voltage across the capacitor C1 (B = V(C1)/Vout, V(C1) = Vin-Vx). In the case when product gm*ro approaches infinity, negative feedback will force voltage at the node X to be equal to 0. This means that the feedback network must deliver to the input exact copy of the input signal: V(C1) = Vin. B = V(C1)/Vout [gm*ro -> inf] = Vin/Vout [gm*ro -> inf] = 1/(Vout/Vin) [gm*ro -> inf] = 1/(-C1/C2) = -C2/C1.

I want to admit that block A (Fig. 8.4) does not represent initial amplifier without feedback (A != gm*ro). This is somewhat "unnatural" and there is another representation of the feedback system that eliminates this "drawback". You can read more on this here: (pay attention to my post, 14th April).

The reason why Razavi hasn't given expression for B is because we don't need it! To examine performance of feedback amplifier we must determine closed-loop gain (A/(1+A*B)) and loop gain (A*B). Both values can be calculated without separate calculation of A and B (this is what Razavi has done).

Hope this helped

 

[LvW]

.....................
From KCL, It's straight forward to get Vout/Vin=-C1/C2, and it's not hard to get Y/X=A/(1+BA)=1/B * (1-1/BA) as a general equation for feedback.
........
.........
I am just wondering did I miss something in the deduction? I really appreciate if you can help to explain this.
Thanks

Hi ANDOVER !

Yes, indeed, you have made an error.
But let me first say that I disagree with the (for my opinion relative involved) explanations from DEDALUS. I'ts much simpler.
DEDALUS is right by pointing your attention to another discussion in this forum dealing with the same subject. Unfortunately, up to now there was no agreement beween DEDALUS and me - but I am optimistic.
*Now, here is my answer to your problem:
What you have called a "general equation for feedback" can be applied only in case the input voltage is connected directly at the amplifiers input. In your circuit this is not the case because the input voltage has to pass a capacitive divider before arriving at the gate.
But this does not complicate the calculation at all. The same situation arises for each inverting opamp circuit:
The "general feedback equation" is supplemented by an additional forward factor F which in your case is simply F=-C1/(C1+C2). The rule to calculate F is to find the voltage at the gate caused by the input voltage - with zero output voltage. It is simply an application of the superposition theorem.
Thus: Y/X=-FA/(1+AB).
And for A infinite you get Y/X=-F/B=-C1/C2.
Does this answer your question?

Added somewhat later: To be specific - DEDALUS is (most probably) not wrong.
However, his explanation and his approach is based on an artificial block diagram representation which assumes that the factor F (as defined above) belongs to the amplifier A, which thereby is reduced. Of course, this approach has an impact on the actual feedback path. For my opinion, this method is (a) more complicated, (b) has lost the direct connection to the electrical circuit and (c) leads to a "feedback factor" which is not identical to the factor which gives the percentage which is fed back to the input (that means: wrong name?).
Therefore, my doubts: What is the advantage and the justification of this approach?.

 

[andover]

Thanks LvW,

For the forward factor F, by letting output voltage 0, how do you get the - sign? By the way, can you show me how to calculate the input impedance of this kind of circuit using FB technique?

Thanks again


The "general feedback equation" is supplemented by an additional forward factor F which in your case is simply F=-C1/(C1+C2). The rule to calculate F is to find the voltage at the gate caused by the input voltage - with zero output voltage.

 

[LvW]

Thanks LvW,
For the forward factor F, by letting output voltage 0, how do you get the - sign? By the way, can you show me how to calculate the input impedance of this kind of circuit using FB technique?

* That's a matter of definition. You can assign the minus sign to the gain A (perhaps this is the most logical way) or to the factor F. In this context you should remember the correct and general feedback formula which is valid for positive feedback (A>0) as well as for negative feedback (A<0)
Y/X=A/(1-AB). Watch the minus sign in the denumerator.
In the circuit under discussion : Y/X=-F*|A|/(1+|A|*B).

* The input impedance is calculated applying the classical KCL and KVL rules.
It is not recommended to use an equation like
Zin=Zin,o*K (Zin,o: Zin without feedback; K=correction factor due to feedback).
The reason is that the correction factor K not always is identical to
(1+|A|*B) or 1/(1+|A|*B). It depends on the specific circuitry.

 

[amriths04]

I doubt this circuit is of any use as it does not have a DC negative feedback and hence will not work unless this setup is within some bigger feedback loop which has a DC negative feedback around it. (Sorry for being off-topic, but wanted to mention that).