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Adm in CMRR is a open loop or closed loop gain?

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BillQ

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CMRR is defined as Adm/Acm
where Adm is the differential gain of the amplifier

If the amplifier is used in a feedback configuration
At this time Adm refers to the open loop gain or the closed loop gain?

PSRR also has Adm term, so this should also apply to PSRR
 

If the amplifier is used in a feedback configuration
At this time Adm refers to the closed loop gain!

BillQ said:
PSRR also has Adm term, so this should also apply to PSRR
PSRR = Power Supply Rejection Ratio - simply is Vout(ac+noise)/Vsupply(ac+noise) expressed in dB (usually a negative value).
 
I also saw in some books or papers the equation of
CMRR=Aol/Acm, Aol seems like an open loop gain

But no book has specified the definition of CMRR for a closed loop amp

so I am really a little confused
Anyway, it sounds reasonable if this Adm refers to the closed loop gain
 

CMRR (and PSRR, too) should always be measured for/in the actually used configuration: for a comparator - which is normally used in open loop configuration - use the open loop gain. For opAmps, LDOs etc. - which are used in closed loop configuration - the closed loop gain is the proper value.
 
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    BillQ

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    donaldtsl

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thank you very much, it is so helpful for me

I had googled it for a couple of hours, with no result, because normally people just give a CMRR value but never explicit mention if it is closed loop gain or not
 

erikl said:
CMRR (and PSRR, too) should always be measured for/in the actually used configuration: for a comparator - which is normally used in open loop configuration - use the open loop gain. For opAmps, LDOs etc. - which are used in closed loop configuration - the closed loop gain is the proper value.

I disagree to the above.
Both parameters - CMRR and PSRR - are specified by opamp manufacturers in databooks (mostly as a function of frequency) for the devices without feedback.
Both values are important quality properties of the opamp.
Of course, if it is important for the user, both parameters can also be calculated/measured/simulated for any specific application - using the opamp data as an input.
 
LvW said:
erikl said:
CMRR (and PSRR, too) should always be measured for/in the actually used configuration: for a comparator - which is normally used in open loop configuration - use the open loop gain. For opAmps, LDOs etc. - which are used in closed loop configuration - the closed loop gain is the proper value.

I disagree to the above.
Both parameters - CMRR and PSRR - are specified by opamp manufacturers in databooks (mostly as a function of frequency) for the devices without feedback.
Both values are important quality properties of the opamp.
Of course. OpAmp manufacturers hardly have another choice, as they cannot know in which configuration their products will be used. But - AFAIR - I've also seen CMRR and PSRR values for certain gain configurations (20dB, 40dB), which of course differ from the corresponding open loop values (the former ones are better, of course).

LvW said:
Of course, if it is important for the user, both parameters can also be calculated/measured/simulated for any specific application - using the opamp data as an input.
That's what I meant. Of course I want to know these values for my individual configuration, because this is what counts. The open loop rejection ratio values are too bad for most cases.
 

    BillQ

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Hi!

i am learning Op amps. As you told i have took a op amp and simulated it for finding CMRR with closed Loop.

AS per the circuit,
the Output is Vout = Vp(1+R2/R1)-Vn(R2/R1);
So, generally gain is = A = (Vp+(R2/R1)(Vp-Vn))/(Vp-Vn)
For Acm = Vp/Vp = 1 always
for Adm = (Vp+R2/R1(Vp-Vn))/(Vp-Vn);
Even i have simulated with different input combinations , the results are following this relation.

But Acm = 1; ALways constant
Adm = Vp/(Vp-Vn)+R2/R1; depending on the external components + inputs
is it right.....

Strange, gain is depending on input voltage what we are applying......
 

malli_1729 said:
Hi!
i am learning Op amps. As you told i have took a op amp and simulated it for finding CMRR with closed Loop.

AS per the circuit,
the Output is Vout = Vp(1+R2/R1)-Vn(R2/R1);
So, generally gain is = A = (Vp+(R2/R1)(Vp-Vn))/(Vp-Vn)
For Acm = Vp/Vp = 1 always
for Adm = (Vp+R2/R1(Vp-Vn))/(Vp-Vn);
Even i have simulated with different input combinations , the results are following this relation.
.........
Strange, gain is depending on input voltage what we are applying......

Hi Malli !

I think, you have made an error:
The following equation is still correct:
Vout = Vp(1+R2/R1)-Vn(R2/R1)=Vp+R2/R1(Vp-Vn).

However, the voltage Vout does NOT linearly depend on the difference (Vp-Vn) because of the term Vp.
Therfore, you are NOT allowed to define a gain A as you did.

It is simple to see that for Vp=Vn you get Vout=Vp=Vn with Acm=1.
But you cannot define any Adm as a formula. Instead, you only can compute for a given set of actual values the output voltage.
 

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