# Opamp input offset voltage...common mode?

Looking at the datasheet for an OPA197 opamp, I see this: What is the circled expression trying to tell me?

The common mode voltage is equal to the voltage at the non-inverting input minus 1.5 V which is equal to plus or minus 10 µV? I can't follow that.

Where does the 1.5 V come from? What is the point of this section in the datasheet?

The circled expression is under the "Test Conditions" heading so it is describing the conditions under which the input offset voltage was measured.

V+ is the name of the positive power supply pin: The non-inverting and inverting inputs are called +IN A/B and -IN A/B, respectively.

The expression is therefore telling you that the input offset voltage was measured with a common mode voltage set to 1.5V below the V+ voltage. This information is necessary because the top of the table (which you can see in your screenshot) says that

$$\V_{\text{CM}} = V_{\text{OUT}} = V_{\text{S}} / 2\$$ ... (unless otherwise noted)

Without this note you would assume the common mode voltage was $$\V_{\text{S}} / 2\$$ when the input offset voltage was measured. That measurement is given in the first three lines of the input offset voltage specification (where each of those lines is for different temperatures).

• This answers my question very nicely, thank you. I will wait a bit before accepting to give others a chance. Jul 1, 2019 at 15:55
• @evildemonic Happy to help.
– Null
Jul 1, 2019 at 15:59

To add a little continuation to @Null 's answer. It also talks about the voltage range in section 7.3.6 of this datasheet. and also it tells you the modes of operation with these transistors on the next page... • +1 Note to OP, the PFET diff pair is trimed in final test, the NFET diff pair is not. This is visible in Fig 11 in the datasheet Jul 1, 2019 at 16:40

If you operate the opamp in non-inverting gain circuit, that 50 microvolt bump may be a bad thing for your signal processing.