How is the input offset voltage a function of common mode voltage?

Question

I don't know about the internals of a opamps, but is it possible to explain how a change in common mode voltage changes input offset voltage Voff?

I'm asking because when the signal Vin is fed to the non-inverting input, the common mode voltage becomes equal to it Vcm = Vin. As far as I understand, if the Vcm varies the Voff varies and this would cause non linearity error at the output Vout. (Is that true?)

Is it possible to explain how the common mode voltage variation varies the input offset voltage Voff?

Answer 1

As far as I understand, if the Vcm varies the Voff varies and this would cause non linearity error at the output Vout.(Is that true?) Yes.

And is it possible to explain how the common mode voltage variation varies the input offset voltage Voff? Yes.

If you take a perfect constant voltage offset and add it to one of the VGS of the input differential pair and put that in a follower configuration, then look at the output response vs vin -- you will see the input referred offset is not a fixed constant, but varies with vin. This is because the fixed (applied) offset causes the device behavior and properties (e.g. vt), current and vds of the individual input devices to respond in an non-equivalent and non-linear manner across the input vcm range.

Here's another longer explanation from Toshiba.

Answer 2

Typical op-amps that exhibit this effect, use a complementary input stage with two differential pairs. One is trimmed for low offset, and is active over most of the Vicm range; the other is mediocre and takes over near VDD. Sometimes the other pair is trimmed after packaging (e.g. TI e-Trim(TM)), giving comparable performance over the full range.

There were also some early NMOS or CMOS op-amps (TLC324 et al?? I forget which ones were actually this awful), with the "long tail" under the diff pair made with a very poor structure (I don't even know what; metal-gate MOS?) so that the CMRR and Vicm range were just awful. Like, they even had plots of slew rate vs. Vicm and weird shit like that in the datasheet. Which, at least they're telling you about it... but man, something that rough almost shouldn't have even existed.

Anyway, modern CMOS amps are fine, they're largely well behaved, no phase reversal, Vios varies a few mV over Vicm, usually prioritizing "single supply" range (well behaved from just below VSS to a volt or two below VDD).

Answer 3

The correlation between Vcm and Voff depend on the opamp design, particularly the input stage.

The general cause is output impedance of the transistors in the input stage. In simple terms, as Vcm changes, VDS of only one of the devices in the input pair changes; thus it's VGS (required to run the bias current) changes (slightly) - this is a source of offset voltage.

It is usually not a strong effect.

Note that in many (mostly non-inverting) configurations, opamp GAIN (which is non-infinite) may appear to behave as an offset V. This is because as Vout changes, a (small) differential input V change is required to drive this.

Even if Voff depended (linearly) on Vcm, this would only cause an effective gain error in the opamp. Consider a follower - Vout = Vin + Voff. Vin is also = Vcm. If Voff changes, as Vin changes, it's equivalent to saying that the gain of the buffer isn't precisely 1.0000, but perhaps 0.9999 or 1.0001.

Answer 4

In my opinion, these are different errors that do not depend on each other.

Voff is a result of the difference between the characteristics of the two transistors building the differential pair.

The common-mode voltage error is a result of the imperfections of the current source in the emitters.

We can say that the offset voltage is an additive (constant) imperfection while the output voltage caused by the common-mode voltage is a multiplicative (proportional) error.

It is caused by the fact that the output of the differential stage is single-ended (only one of the two collector voltages is taken as the output). If the output was differential (both collector voltages were used as output), there would be no such error because they would change in the same manner and their difference would always be zero.

Strictly speaking, the non-inverting amplifier does not represent a "true" common mode since the second input voltage (applied to the inverting input) is not exactly equal to the first input voltage (applied to the non-inverting input)... and, as @jp314 noted, it is proportional to the op-amp output voltage.

You can observe a "true" common mode if, for example, you join the two inputs of a differential stage and apply a single input voltage to them.

If you want, I can illustrate my explanations with a circuit diagram.