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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?

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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.

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Here's another longer explanation from Toshiba.

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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).

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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.

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    \$\begingroup\$ See this e2e.ti.com/blogs_/b/analogwire/posts/… It says " The common-mode input voltage affects the bias point of the input differential pair. Because of the inherent mismatches in the input circuitry, changing the bias point changes the input offset voltage (VOS), which, in turn, changes the output voltage. In other words, as you change your common-mode voltage you will see a change in input offset voltage." \$\endgroup\$
    – floppy380
    Jun 26 at 21:17
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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.

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    \$\begingroup\$ See this e2e.ti.com/blogs_/b/analogwire/posts/… It says " The common-mode input voltage affects the bias point of the input differential pair. Because of the inherent mismatches in the input circuitry, changing the bias point changes the input offset voltage (VOS), which, in turn, changes the output voltage. In other words, as you change your common-mode voltage you will see a change in input offset voltage." \$\endgroup\$
    – floppy380
    Jun 26 at 21:17
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    \$\begingroup\$ @floppy380, I cannot agree that "the common-mode input voltage affects the bias point of the input differential pair". If there is a perfect current source in the emitters, the "bias point" will not change when both input voltage simultaneously vary (common mode). The common emitter voltage will vary but the current source will keep the emitter current constant. So, the collector currents will not change and the bias point (output voltage) will not change as well. This assertion would be valid for the simple differential pair with an emitter resistor instead of a current source. \$\endgroup\$ Jun 26 at 21:33
  • \$\begingroup\$ I dont know anything about the opamp internals its just I came across the article. If you say so maybe the article is wrong then. Thanks for the input. \$\endgroup\$
    – floppy380
    Jun 26 at 21:44
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    \$\begingroup\$ @Circuitfantasist When is a perfect current source not a perfect current source? When it's made from real components. Typical op-amps that exhibit this effect use a complementary input stage with two diff pairs, one of which is trimmed for low offset and is active over most of the range, the other which is mediocre (sometimes trimmed after packaging (TI e-Trim(TM)) and takes over near VCC. \$\endgroup\$ Jun 26 at 21:58
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    \$\begingroup\$ That's a good way to use the terms; but this effect is neither -- it's a nonlinear function of Vicm. I would say it's easiest to model as, say, a smooth step function of Vicm, added to Vios. Taking the average over Vicm range, or locally in the smooth (transition) region, it does look multiplicative, but it's a bit of both overall. \$\endgroup\$ Jun 27 at 15:49

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