0
\$\begingroup\$

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.

transimpedance amplifier

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

\$\endgroup\$
  • \$\begingroup\$ 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? \$\endgroup\$ – Brian Drummond Jul 3 '16 at 11:13
  • 1
    \$\begingroup\$ Hmmm is it the application of ohm's law? Taking the input resistance of the op-amp and dividing the voltage by this resistance? \$\endgroup\$ – ConfusedCheese Jul 4 '16 at 9:03
  • \$\begingroup\$ 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. \$\endgroup\$ – stowoda Jun 25 '18 at 13:27
2
\$\begingroup\$

10,000 ohms is the transimpedance gain of the circuit. It applies when a current is input and you want to know the voltage response at the output.

The voltage gain of the circuit is not the same value.

Rather than do your work for you, I'll leave it as an exercise to calculate the voltage gain of the circuit.

\$\endgroup\$
  • \$\begingroup\$ Ahh okay, was thinking about it incorrectly, thank you. So the current will still induce x volts across the resistor dependent on it's resistance but the voltage offset is not multiplied by this gain and just subtracts from this x volts? \$\endgroup\$ – ConfusedCheese Jul 2 '16 at 16:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.