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# Explain me how does this circuit using CMFB work

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#### firsttimedesigning

##### Full Member level 1
recently i am studying CMFB. i have attached file that has a circuit uses CMFB.
my question is that when Id3 and Id4 increase, why does that ouptut CM get lowered?

I think it's because that there is a set amount of current flowing into R1 and R2 from PMOS transistors. There is also a set amount of current flowing out of R1 and R2, namely Id3 + Id4. Since Id3 + Id4 increase, there is more current flowing out of R1 and R2. Therefore, the output CM level decreases. I am not sure if I am right..

One thing that i should add is Ve is the difference between Vout,cm and Vref.
Vref is connected to the negative input of the amplifier.

I also would like to know why does the feedback circuit forces Vout,cm to be equal to Vref?

Re: CMFB

because it's negative feedback.

if it's not the case, then Ve will be very large or very small, then the circuit can not work correctly.

I think you should read Razavi's book more carefully, it is explained there.

Re: CMFB

first of all, thank you for replying.
and yes, i know it's negative feedback. as a matter of fact, i have read that paragraph many times and thought about this for at least half hour before i decided to put this question up here.
what i dont get is how does increasing Id3 and Id4 will cause Vout1 and Vout2 to drop.
i posted my reasoning up there. please read it and to see if there is any mistake.
thank you.

CMFB

no common mode current goes through the resistors, so if id3 and id4 increase together then the vout1==vout2 and no current will go through the resistors.
about y the vout goes down with the current increase , u can think about it as if the out node contain a cap. with a certain charge so if u increase the current to ground the vout will decrease (cap will discharge) while if u increase the current from vdd , vout will increase (cap. will charge)

Points: 2