Input offset voltage transimpedance amplifier

I am currently creating a transimpedance amplifier for a small-area photodiode that creates around 0.15mA, 0.00015A, from it's associated laser diode at it's nominal operating point.

I am currently looking to use the transimpedance amplifier configuration with a feedback resistor for a gain of 10,000, by using a 10,000ohm resistor, therefore creating a voltage output of 1.5V. However I am looking at sources of error and one significant one seems to be the input offset voltage. Even with an extremely low input offset voltage, say 10microVolts, is this offset voltage not multiplied by the closed loop gain and therefore creates a theoretical offset of 6% of the voltage output, at 1.5V?

Furthermore even if the gain is reduced it still creates a 6% theoretical deviation as they are proportional to each other? Is this a problem that is required to be fixed with the use of the offset null pin on op-amps?

I have simulated this in LTspice with the use of an op-amp and a voltage source in series with the positive input terminal, however this only adds the 10microVolt different to the output. But this seems contrary to what I have read/understand the concepts to be?

Any help would be greatly appreciated!

Regards

• That's not a "gain" of 10,000, it's a transimpedance of 10,000 ohms (because V/A has units while V/V is dimensionless). And 10,000 is not a particularly high impedance unless you're working at high frequencies (MHz). I'll give you one more hint : what current might be generated by 10uV at the input? Jul 3 '16 at 11:13
• Hmmm is it the application of ohm's law? Taking the input resistance of the op-amp and dividing the voltage by this resistance? Jul 4 '16 at 9:03
• As a hint: If your application does not intend the diode to be biased you might need to take into account what the OPamp's input offset voltage causes when it develops across the photodiode. Jun 25 '18 at 13:27