# Why inverting amp don't suffer from CM rejection ration?

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#### llbaobao

##### Member level 3
I found in the books that "
Inverting op amps do not suffer from common-mode rejection ration effects when
the amplifier inputs are permanently at ground for dual supply applications or half
the supply voltage in single supply applications. The non-inverting amplifier
configuration has a common-mode voltage equal to the input signal."

But I not know why? for in single supply application above, I think the common mode voltage is half of the supply.

llbaobao said:
"The non-inverting amplifier configuration has a common-mode voltage equal to the input signal."
Right.

llbaobao said:
... for single supply application above, I think the common mode voltage is half of the supply.
I think your understanding of the common mode voltage is not correct: It's not the absolute bias value of the inputs (e.g. "half of the supply"), but the common change of the inputs' voltage levels (away from the quiescent voltage level), caused by the input signal.
For a non-inverting amplifier, the inputs experience a common mode voltage change according to the input voltage change, whereas for an inverting opAmp configuration, the common voltage change at the inputs is much lower (by a factor (open_loop_gain/closed_loop_gain)).

### llbaobao

Points: 2
Thank you!

For I look in the book that
the common vlotage=(Vinp + Vinn)/2
then if give the inverting amp's inp half of the supply as reference,
then the inn is half supply, So I think the common vlot is half supply.

the common volt = ( (delta Vinn) + (delta Vinp))/2 is truely right?

llbaobao said:
the common volt = ( (delta Vinn) + (delta Vinp))/2 is truely right?
No. The common mode voltage range of an opAmp is the input voltage range, where (for Vinp ≈ Vinn) the amplifier works correctly.
The instantaneous common mode voltage is (max(Vinp,Vinn) - Vcm,quiescent), where Vcm,quiescent is the common mode voltage of the inputs without an external input signal (quite often half of the supply voltage).

### llbaobao

Points: 2

If an amp used in feedback, Vinn ≈ Vinp

then as you reply "The common mode voltage range of an opAmp is the input voltage range, where (for Vinp ≈ Vinn) the amplifier works correctly."

That is to say, the common mode voltage range = input voltage range?

llbaobao said:
That is to say, the common mode voltage range = input voltage range?
... where the OpAmp still works perfectly. Exact!
So-called "rail-to-rail" OpAmps dispose of both p- and n- differential input stages, hence their input common voltage range comprises the full operating voltage range, i.e. "rail-to-rail" - and possibly even a bit more.

Points: 2