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I am using this TIA circuit and photodiode in photovoltaic mode. I'd like to modify the circuit such that at zero light a positive voltage is produced at Vout. And as more light falls on the photodiode, the circuit would produce a greater negative voltage.

enter image description here

I can add a bias voltage to the positive input to the op amp to achieve the desired effect. But I am not sure about the downside of applying a positive voltage to the photodiode. I have added C4 here to reduce the resistor noise of the bias circuit.

The photodiode is expected to generate 100uA at the maximum light level

enter image description here

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    \$\begingroup\$ As you already have a -ve supply, you can easily inject a known negative current into the virtual earth instead. \$\endgroup\$ – user16324 Sep 9 '20 at 12:19
  • \$\begingroup\$ Thank you for the response. I have -9V available to in the design. To trim out 0.5uA of current I'd need a 18MEG resistor. I could use a voltage voltage divider but then I'd need a buffer to avoid loading the photodiode. The amount I need to trim out is never the same so I'd need a large value trimmer or a resistor in series with a trimmer. Am I missing something with your approach or is this an unintended consequence of this approach? \$\endgroup\$ – ignoramusextraordinaire Sep 9 '20 at 13:27
  • \$\begingroup\$ Or I could trim the other side of the divider with a smaller pot. But I'd still probably need the buffer \$\endgroup\$ – ignoramusextraordinaire Sep 9 '20 at 14:10
  • \$\begingroup\$ No need for a buffer. Currents into a virtual earth sum directly. \$\endgroup\$ – user16324 Sep 9 '20 at 14:25
  • \$\begingroup\$ Your second diagram is exactly what I have used, but with Vin replaced by a cheap I2C DAC. Then I just set the offset as required and program it into the DAC. Makes the resistor precision irrelevant. \$\endgroup\$ – user1850479 Sep 9 '20 at 15:36
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enter image description here

Image from https://tallerelectronica.com/fotodiodo/

We can start with characteristic of a photodiode, in the first circuit the voltage accross the photodiode is 0V, so the working points in the graph are on the y axis (current), increasing the light will increase the amount of current almost linearly.

If you give a positive voltage on the cathode of the photodiode you are moving to the left on the graph, so it should not have to be a big problem until you reach the reverse voltage of the diode.

If you give a positive voltage on the anode you will turn on the diode (as a regular diode) but I guess this is not the case.

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