# Applying a forward bias to a photodiode to shift output

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.

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

• As you already have a -ve supply, you can easily inject a known negative current into the virtual earth instead.
– user16324
Commented Sep 9, 2020 at 12:19
• 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? Commented Sep 9, 2020 at 13:27
• Or I could trim the other side of the divider with a smaller pot. But I'd still probably need the buffer Commented Sep 9, 2020 at 14:10
• No need for a buffer. Currents into a virtual earth sum directly.
– user16324
Commented Sep 9, 2020 at 14:25
• @ignoramusextraordinaire The standard way to do offset correction is on the op amp side, and it should not affect the response of the loop. See: electronics.stackexchange.com/questions/34071/… Commented Sep 9, 2020 at 18:46

You only need to inject/extract a small current into the summing junction, the inverting op-amp input. So, if your photodiode is producing 10μA into that junction, causing an offset at the output, then sink that same current to -9V:

simulate this circuit – Schematic created using CircuitLab

This works because the op-amp is actively maintaining a permanent virtual 0V at that junction, and KCL is the law. As a result, you don't change the diode's biasing (or, in this case non-biasing).

The only issue with using the supply potential in this way, is that supply noise will be directly injected into the signal. In practice you would use a precision current source, or a resistor and precision voltage source/reference:

simulate this circuit

If the supply isn't too noisy, you might get away with a voltage stabilised by a capacitor:

simulate this circuit

This has the disadvantage that there will be a short pulse at the output when power is first applied, while C2 initially charges, and you'll have to wait 0.1s or so for everything to settle at the DC operating point.

The -ve input of the non-inverting op amp is a current summing point. You can inject current here of either sign to trim the offset as you wish. Here is a typical trim circuit (photodiode not shown).

simulate this circuit – Schematic created using CircuitLab

You might not need R1 and R2, they just reduce the effective range of adjustment. Making R2 larger also reduces the amount of adjustment possible.

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.