Feedback question from EE240's slides

[kidmanbasha]

feedback question

In this slide the "c" declares the ideal closed loop gain. What I really can't understand is how did the feedback factor "f" turn out to be that equation written down? Isn't that shunt-shunt FB (current in - voltage out)? So the feedback factor should be -Cf ?
I'm sure what's written in the slide isn't wrong because its also written in Stanford's slides. thanks

 

[sutapanaki]

Any feedback system has a transfer of Ao/(1+fAo)=1/f*[fAo/(1+fAo)] and we know that fAo=To - loop gain, while 1/f is the ideal closed loop gain if the amplifier were to have infinite gain. In the above circuit:

To=Ao*[Cf/(Cf+Ci+Cs)] - assuming it is an opamp not ota

and the ideal gain 1/f=-Cs/Cf

Acl=1/f*[fAo/(1+fAo)] =1/f*[To/(1+To)]=1/f*[1/(1+1/To)]=-(Cs/Cf)*[1/(1+1/FAo)]

Where F=Cf/(Cf+Ci+Cs) and To=FAo

Yes, the feedback is shunt-shunt, but don't mix F with f

 

[kidmanbasha]

You told me not to mix between F and f.

Acl=1/f*[fAo/(1+fAo)] =1/f*[To/(1+To)]=1/f*[1/(1+1/To)]=-(Cs/Cf)*[1/(1+1/FAo)]

and in the above equation you first assumed To=Ao*f then you assumed To=Ao*F

I understand that F is a capacitive divider. But how can we just replace f with F in the above?

 

[sutapanaki]

Sorry for the confusion. I, unwittingly, used the same notation for the open loop gain of the system and the gain of the opamp. To avoid this confusion again let's use slightly different notation this time.
Ao - open loop gain of the system
Avo - gain of the opamp
β - feedback factor of the system
F=Cf/(Cf+Ci+Cs)

1. Also, let's first establish a common ground in understanding. I assume we can agree that if the gain of the opamp is ∞, then the closed-loop gain of the system is

Ao∞=1/β=-Cs/Cf and consequently the feedback factor β=-Cf/Cs

2. As I said before, the loop-gain is:

To=Avo*[Cf/(Cf+Ci+Cs)]

This follows from circuit analysis if we break the loop at the output of the opamp.

Since by definition To=Ao*β and we know To and β, one can find if necessary Ao from here. I'm not going to do it.

3. Again by definition, for any feedback system

Acl=Ao/(1+fAo)=1/f*[fAo/(1+fAo)] =1/f*[To/(1+To)]=1/f*[1/(1+1/To)]

We already know that To=Avo*F but here Avo is not Ao and F is not β.

It follows that Acl=-(Cs/Cf)*[1/(1+1/FAo)]

4. While a non-inverting opamp configuration directly maps into the general feedback block diagram, the inverting one does not. If you want to map it you need to find the two main parameters, namely Ao and β. But those are not Avo and F. That's why I said not to mix F and β in this case. A shunt-shunt feedback configuration that is driven by a current input signal directly at the inverting input of the amplifier can map directly, but here we drive with a voltage through Ci and things change slightly.

I hope it's clear now.

 

[safwatonline]

Try using this Model for feedback (picture below) not the basic feedback model, here the feedback is series-shunt but the input to the summer is a scaled version of the input Vin=α*IN , where α=Cs/(Cf+Ci+Cs)

this should answer all ur questions