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Bias voltages in terms of threshold voltage and overdrive voltage for folded cascode

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student_of_analog

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Hello,

I am trying to understand what the bias voltages Vb1, Vb2 etc of the attached opamp should be in terms of Vov and Vth.
Can anybody help me how to proceed with this ?

I could figure out that:
Vb2 = Vov2 + Vth2
Vb4 = Vov4 + Vth4.

But I dont know about the others.
 

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    opamp.png
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Vb2=Vg(Q7)=Vg(Q7)-Vs(Q7)+Vs(Q7)=Vgs(Q7)+VDD=Vov(Q7)+VDD+Vth7.

Vb4 is as written by you

In a similar way you can write an equation for Vb1 you asked for.
 
thanky you jimito, but if Vb2 = Vov(Q7) + VDD + Vth7, then Vb2 will be greater than supply voltage, but that is not possible.
I want to know how to write equations for Vb1, Vb3, Vb4, Vb5, Vb6
 

...but if Vb2 = Vov(Q7) + VDD + Vth7, then Vb2 will be greater than supply voltage, but that is not possible.

Well,things are not like that!Transistor Q7 is pmos so Vth7<0 and Vov(Q7)<0 as well,thus Vb2 = Vov(Q7) + VDD + Vth7 is always < Vdd.

Ok,as far as the other potentials i would suggest that you try to extract them by yourself,it is meaningless to serve you the final answer without you make any effort ;-)
Besides i saw you the way!
 
Thank you. I got confused because I have the habit of writng absolute values, I should have seen that it was a PMOS.
I have attached the bias block as well.
My confusion about the bias voltages was because the required Vb1, Vb2, Vb3, Vb4, Vb5, Vb6 in the opamp was not
matching the Vb1, Vb2, Vb3, Vb4, Vb5, Vb6 generated from the bias block.

For example, Vb3 (opamp) = Vov(Q1C) + Vth (Q1C) + Vov(Q5) ----- (1)
Vb3 (bias block) = Vov (Q25) + Vth (Q25) ----- (2)
I was confused that the Vb3 generated in bias block is one Vov short.
But now I understand that since the currents are different, both (1) and (2) work out to approxiamately same numerically. Is this analysis correct?

Then I want to understand on what criteria should the biasing scheme be chosen ? I can choose any biasing scheme as long as current mirroring happens
and absolute voltages required are approxiamately the same both in the bias block and the opamp ? Is this correct ?

Thank you for your time.bias.png
 

The expression you wrote about Vb3(opamp) is not correct.The valid one is : Vb3(opamp)=Vov(Q1C) + Vth (Q1C) + Vds(Q5).
It is apparently expected that the two equations of the biasing potential (opamp-bias circuitry) are different since the they are written for different circuits.In contrast,their numerical value is apparently the same.
 
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