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OTA and OPAMP output impedance

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parkpika

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I know that OTA and Opamp differs in that Opamp has a output buffer with low output impedance while OTA has high output impedance (ro).

Why is it that OTA can't drive a resistor? I get that the gain drops from gm*ro to gm*R, if R is small but isn't this the same for an Opamp??
 

I know that OTA and Opamp differs in that Opamp has a output buffer with low output impedance while OTA has high output impedance (ro).
Why is it that OTA can't drive a resistor? I get that the gain drops from gm*ro to gm*R, if R is small but isn't this the same for an Opamp??

I suppose, with ro you mean the internal OTA output resistance, right?
Normally, the "gain" of the OTA is described by the transconductance gm.
Of course, one can calculate a fictive output voltage gm*ro, but this makes no sense because each loading (pure load or feedback) will change this value.
Thus, it is straight-forward and logical to treat the OTA as a (non-ideal) voltage-controlled current source and calculate the output voltage using the gain expression G=gm*(ro||Rload) .
In contrast, the opamp has a low output resistor and - as long as the load resistor Rload is not too small - the output voltage is independent on Rload.
 

No. A standard op amp's open loop gain (which is Ao, not gm) is little affected by the output load (within it's output load capability), since it has a low output impedance.
 

Thanks for your reply. But I'm still confused of the opamp's output. Isn't opamp's output also Vo = (gm* (ro||Rl)) * Vi??
 

Thanks for your reply. But I'm still confused of the opamp's output. Isn't opamp's output also Vo = (gm* (ro||Rl)) * Vi??

In your expression for the opamp`s output, what is the meaning of gm ?
The term gm is a transconductance desribing the ouput current-to-input voltage ratio of the device.
However, the opamp provides a VOLTAGE at the output (across a rather low resistance) and not a current (like the OTA).
Thus, the opamp`s output voltage is Vout=A*Vin (independent on the connected load).
The gain A is either the closed-loop gain Acl (with feedback, normal case) or Aol (open-loop gain, just theoretically, no practical relevance).
 

Opamp with diff pair with single ended output with Rl as the load would give you Aol = gm*(ro||Rl). no?? or is this OTA?

I know that opamp just has a buffer at the output so what would be the VOLTAGE of the opamp with the above opamp with buffer?

Thanks!
 

Opamp with diff pair with single ended output with Rl as the load would give you Aol = gm*(ro||Rl). no?? or is this OTA?
Thanks!

This is simply a diff. pair - neither opamp nor OTA.
Opamps and OTAs are specialized devives either with a very large (OTA) or a very small output resistance (opamp).
 

Opamps DO NOT HAVE a single-ended collector output in class-A. Their output is produced by very low output impedance complementary emitter-followers in class-AB.
 

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