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# How to get an expression to the output resistance of the super cascode opamp shown?

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How can i get an expression to the output resistance of the super cascode opamp shown? any help.
Thanks

Re: Output impedance

You have made error at your picture, because you have forgotten MOS source current. Vg=-A*Vx, Vs=Vx=> ids =-gm(A+1). So I1 is not the current trough ro.
(I1=Iro+Ids)
This circuit has feedback, and output impedance also could be easily eavaluated by Blackman formula.
Loook at page 310 of Razavi book Design mpof analog integrated circuits.

Re: Output impedance

One can get the Rout expression if doing small signal analysis. Intitutively we can derive on the same result. The op-amp is increasing the transconductance of the NMOS with the factor of (A+1) so the new transconductance is gm(A+1). The remaining structure (taking out the op-amp) is cascode with ro1 as source degeneration resistance. The Rout of cascode structure is given as gm1ro1ro2 but here the same expression will be gm(A+1)ro1ro2.
Other way of looking the same is Vx is pretty much constant DC and Vb will keep it the same voltage. Any variation at Vx node will not change the current value in ro1 or the Rout is high impedance value.
thanks
-Bharat

Re: Output impedance

you can refer the book than analyze it in entail from two aspects.

Re: Output impedance

I think
Rout ≈ ro (with very good estimation)
because op-amp gain is very high and point of X is a voltage source with very
low impedance its impedance ≈ rot/A. on the other hand we can consider transistor M act as current
source.
equivalent resistance of transistor can be parallel with resistor ro

Regards
Davood.

Re: Output impedance

Davood Amerion said:
I think
Rout ≈ ro (with very good estimation)
because op-amp gain is very high and point of X is a voltage source with very
low impedance. on the other hand we can consider transistor M act as current
source.
equivalent resistance of transistor can be parallel with resistor ro

Regards
Davood.

I am sorry but your think is false, I just mentioned that to clarify.
Output resistance doesn't equal ro but like i mentioned above in picture.
anyway, i could prove the above relation (in picture) it can be simply got if you think in the above MOSFET as another one with higher gm i.e. Gm new =(1+A)gm of the MOSFET. hence Rout = Gm new x (....) like mentioned above.

And thank you Davoood for reply,

Output impedance

excuse me ! but i think my thought is not false!

Re: Output impedance

How can i get an expression to the output resistance of the super cascode opamp shown? any help.
Thanks

In the figure the terminals shown of the op-amp sould be interchanged. The positive terminal of the op-amp should be connected to the node X to ensure the negative feedback as the the CS config. will have inverted output of the given input.

thanks
-Bharat

Re: Output impedance

If the +ive and -ive terminals are interchanged the op-amp wont work as an amplifier.

So the pupose of the circuit cant be acheived. Rather it might start working as a kind of schmitt trigger circuit.

Prakash.

Re: Output impedance

If the +ive and -ive terminals are interchanged the op-amp wont work as an amplifier.

So the pupose of the circuit cant be acheived. Rather it might start working as a kind of schmitt trigger circuit.

This is not true. There is no inversion from the gate to source of the transistor, so the opamp is providing negative feedback (as desired) and boosting the output impedance of the circuit.

Re: Output impedance

Davood Amerion wrote:

Davood Amerion said:
I think
Rout ≈ ro (with very good estimation)
because op-amp gain is very high and point of X is a voltage source with very
low impedance its impedance ≈ rot/A. on the other hand we can consider transistor M act as current
source.
equivalent resistance of transistor can be parallel with resistor ro

poit of X can't be consided as fake ground.because point of X is the cross of two branch(one is OPamp,the other is the M),so the Voltage of X can be changed.
so the Rout can't be estimated as ro.

liccAMS

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