voltage-voltage feedback

loading in voltage-voltage feedback

I tried to analyse this voltage-voltage feedback circuit. (file: voltage-voltage feedback) the book modified the circuit so that it is easier to find the product of the feedback factor and the open loop gain. (file: modified v-v feedback) from that diagram Vf = -Vt * (C1/(C1+C2))gm1 (ro2||ro4). Then, bAo =(C1/(C1+C2))gm1 (ro2||ro4), where b is the feedback factor(beta in the file: block diagram), and Ao is the open loop gain. So I tried to prove that -Vf/Vt = bAo from the block diagram(file: block digram) but I could not prove it. I placed Vout with Vf, Vin+ with ground, and Vin- with Vt along with that capcitor divider. Basically, I modified the block diagram so that it is equivalent to the circuit in the file: modified v-v feedback. Please give some suggestions, thanks in advance

From the block diagram to make it equivalent to your previous circuit you would need to give Vt at the input of the β block where you are breaking the loop so the output of the β block will be β Vt. Note here that Vt is in the polarity shown by Vout. Now Vin terminals are shorted (because Vin=0) so Ve = -βVt. So now Vf at the Vout terminals with Vout polarity will be -β Ao Vt = Vf thus -Vf/Vt = β Ao

First of all, thank you for replying. I understand how to prove the equation, now. However, there still one more question that I would like to ask.
You mentioned Vin is shorted since Vin =0. And Vin = 0, because it is connected to the ground. So everytime I have a terminal connected to the ground, I can assume that it is shorted. Is that right?

In your circuit Vin was the input voltage with respect to ground. Imagine it as a battery connected from the Vin terminal to the ground terminal. To make this Vin Battery 0 you need to make its voltage 0, to make its voltage 0 you need to short its terminals thus you need to short the Vin node to ground node. Had Vin been referenced to some other node rather than ground to make Vin 0 you would have had to short those 2 nodes. In your block diagram, the negative terminal is actually ground if you want to match it exactly to your circuit. So when we short it we are actually grounding the Vin terminal.

Thank you very much for another detailed explanation. I think I kinda understand it now. Before, I kept thinking that the positive terminal of Vin is connected to the gate of M1 and the negative terminal of Vin is connected to the gate of M2. Apparently, that is not correct. The negative terminal of Vin is connected to the ground. Since both terminals of Vin are conncted to the ground, Vin is shorted.

There is one question that came to my mind when I was thinking about all this: generally, how would you know where to put Vt and VF to find the loop gain, β*Ao?
To be honest, I dont really understand how was the modfied v-v feedback derived.

One thing I forgot to mention previously is that in your modified feedback circuit C1 and C2 are not labeled correctly. Since you broke C1's top terminal from Vout you would be giving Vt on C1is top terminal, so the capacitor connecting Vt should be C1.

The place where you break the loop has to be chosen considering the impedances loading the breakaway point. Break the loop at a point so you can include the loading effect of any impedances easily. Like in this case the loop path is through C1 and C2. So you need to break somewhere in that. So now there are 2 immediate options, either break top terminal of C1 like its done here or break the Gate connection of M2. I would have preferred to break the gate connection of M2 because it has a infinite resistance so even if you break the loop there C1 and C2 does not experience any loading change. Right now the loop was broken at top of C1 so in a way we have removed the C1 and C2 capacitor load on the amplifier. The proper analysis would have been to replicate the load on the amplifier by loading it with C1 and C2 and then find the loop gain, as shown in the attached figure.

Thank you very much for the explanation. I kinda it understand the general idea now. I guess I will need more practice to make myself to be good at analyzing feedback circuits. Anyway, thanks for helping to learn all these stuff. I appreciate it.