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The differential output of fully differential amplifier is centered at Common mode voltage or Zero?

Ali263

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The signal at each differential output terminal is centered around Common mode voltage point with 180degree out of phase. But when we talk about differential signal is it centered around zero or common mode voltage?
 

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sutapanaki

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The signal at each differential output terminal is centered around Common mode voltage point with 180degree out of phase. But when we talk about differential signal is it centered around zero or common mode voltage?

If you are asking about Vout_p - Vout_m, then obviously it is around 0.
 

KlausST

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Hi,

EACH output is centered (referenced) at V_Com.
But the DIFFERENTIAL output voltage depends on input voltage. It does not depend on VCom.
V_OUT_DIFF = V_in * A

Let´s say
* V_in = the (differential) input voltage
* A = gain
* V_Com = common mode output voltage

then
* V_Out+ = V_Com + A * V_In / 2
* V_Out- = V_Com - A * V_In / 2

...at least for standard (symmetric) differential OPAMP circuits.

Klaus
 

FvM

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If you are asking about Vout_p - Vout_m, then obviously it is around 0.
Common mode amplifier is setting the output common mode voltage (Vout_p + Vout_m)/2 to Vocm reference. Differential voltage Vout_p - Vout_m only depends on the input signal.
 

MahmoudHassan

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For a fully differential OTA the output voltage is not well defined unless you provide another feedback loop knowing as common mode feedback loop to define the output voltage an settle to to a well defined voltage.
please check CMFB topic for more details
 

Ali263

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If you are asking about Vout_p - Vout_m, then obviously it is around 0.
I am asking about differential so yea it is " Vout_p - Vout_m" . It means if i want to see maximum allowable swing for output should i take the swing from 0 , right?
 

KlausST

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Hi,

From what "0"?
If Vout_p = Vout_m then Vout_diff = 0. Is this what you mean?

If not: please provide a drawing.

Klaus
 

Ali263

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Hi,

From what "0"?
If Vout_p = Vout_m then Vout_diff = 0. Is this what you mean?

If not: please provide a drawing.

Klaus
in figure you can see i get differential output of amplifier which is centered at 0. The supply voltage vdd is 1.2v so output saturates around 1 V. My simple question is that if i take differential output will it be centered around 0 volt or output common mode voltage set by CMFB stage
?
 

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KlausST

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Hi,

I think this already has been answered clearly. Where are your doubts?

If: differential input voltage = 0 then differential output voltage = 0.
If: differential input voltage differs from 0 then differential output voltage will differ from 0, too.


Klaus
 

FvM

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in figure you can see i get differential output of amplifier which is centered at 0. The supply voltage vdd is 1.2v so output saturates around 1 V. My simple question is that if i take differential output will it be centered around 0 volt or output common mode voltage set by CMFB stage
?
We can't see if the waveform shows differential output or anything else. But as already stated, the differential output isn't affected by CMFB circuit. It solely depends on the amplifier input signal. Unfortunately you don't show the signal source connected to the amplifier, so it's an almost useless discussion.
 

sutapanaki

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I am asking about differential so yea it is " Vout_p - Vout_m" . It means if i want to see maximum allowable swing for output should i take the swing from 0 , right?


This is a picture of what I said before. Each individual output is centered around Vcm, but when you take the output differentially, then of course it is around 0 and has double the peak amplitude.
 

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