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# how to understand the output current of an OTA?

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#### bhl777

##### Full Member level 6
Hi All, if we have an OTA, which can generate a current proportional to (Vin-Vout), I was confused from the detailed transistor level implementation.

(1) If we have a capacitance load at the output, does that mean Iout=gm (Vin-Vout) only valid in AC domain? In another word, Iout will never has a DC component, is that true? Then a more correct decription is Iout(s)=gm [Vin(s)-Vout(s)] ?
(2) Is there any way we can mirror this Iout? For example, how can we have a current source/sink, with a fixed DC level, plus the AC component from the Iout?
Thank you!

Hi All, if we have an OTA, which can generate a current proportional to (Vin-Vout), I was confused from the detailed transistor level implementation.
View attachment 122012
(1) If we have a capacitance load at the output, does that mean Iout=gm (Vin-Vout) only valid in AC domain? In another word, Iout will never has a DC component, is that true? Then a more correct decription is Iout(s)=gm [Vin(s)-Vout(s)] ?

At first, the output current of an (idealized) OTA is Iout=gm(V+ - V-) and Vout is the voltage produced by Iout across a load impedance.
This equation holds for dc as well as ac.
Real OTAs have a finite Rout which has to be taken into account in parallel to the load.

bhl777

Points: 2