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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??
 

parkpika said:
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??
Click to expand...

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??
 

parkpika said:
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??
Click to expand...

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!
 

parkpika said:
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!
Click to expand...

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.
 

OTA...Iout=(Vin+ - Vin-)gm...current source
OPA...Vout=(Vin+ - Vin-)Ao...voltage source
 

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