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I know that the offset voltage of an op amp can be multiplied by the voltage gain to reveal the error in an output voltage, but what happens when it's a transimpedance amplifier?

The gain of a transimpedance amplifier is \$ Vout/Iin \$.

The gain is dependent on the input current, not the input voltage. Maybe I'm thinking about this the wrong way?

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    \$\begingroup\$ Model your amplifier circuit with a DC source at the noninverting input for the input offset voltage, and then use basic analysis techniques to find the gain. The term with the input voltage offset in the gain tells you exactly how it affects your output. \$\endgroup\$
    – imthegman55
    Apr 11, 2022 at 23:09
  • \$\begingroup\$ it is no different than an Op Amp using Ib, Ios, Vos, Rfb, Rs for Vin+/-+/- \$\endgroup\$
    – Tony Stewart EE75
    Apr 12, 2022 at 4:17

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This is a typical TIA: -

enter image description here

Image from here

Because the input is a current source, the effective impedance of that current source is infinite hence, it can be modelled as having a very high value voltage as a source in series with a very high value resistance. Hence, the gain is minus unity (-1).

This means that any input offset will be seen as the same error voltage at the output.

Of course, if the current source is not ideal then it can be modelled with a shunt resistor in parallel with that source. Now, you can regard the input offset as being amplified by \$1 + \dfrac{R_f}{R_{shunt}}\$ and, that voltage will appear at the output.

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