# Why is this op-amp outputting such low voltage?

Why am I not getting the expected gain out of my op-amp?

I m using a 16-pin TI OPA4322SAIPW op-amp with a Advanced Photonix 019-141-411-R photodiode and this circuit:

where Rin is 1.8 kΩ and Rf is 27 kΩ. My Vin measured across the photodiode is ~0.14 V in ambient lighting conditions. I have also connected my op-amp to 3.3 V from a Pi Pico and to ground. I would expect to get Vout = -(27/1.8)·0.14 V = -2.1 V. However, when I read the output using a Raspberry Pi Pico ADC pin, I get 0.34 V. Is there something off with my op-amp here? I also replaced the op-amp with a fresh one and got the same behavior.

My goal is to get around 3.3 V out, but no matter what I do, I can't seem to get above 1 V out of my op-amp. Am I missing something?

Update: I've got everything working now, The issue was that my photodiode was a reverse bias diode. I got it working using the circuit from this answer although Spehros answer to this would have worked for any forward bias diode.

• Please show all connections and pin numbers. Aug 3 at 1:23
• I would first realize that a photodiode is best thought of as a light dependent current source, not a voltage source. I would look at transimpedance amplifier design to start. Aug 3 at 1:23
• Are you sure the polarity of the photodiode is correct? Voltage to correct pins of op-amp? Is the inhibit pin connected properly? Is the input of the Raspberry Pi loading down the signal? Aug 3 at 1:28
• @PStechPaul Ive tested my PiPico ADC input by feeding 3.3v directly from a power supply, and it reads out 3.3v correctly so I dont think its loading down the signal, Ive tried the polarity of the photodiode both ways to make sure ive got the anode and cathode correct Aug 3 at 1:48
• Show us your actual circuit, including the photodiode & power supply, not the standard example opamp configurations. Aug 3 at 2:08

You need enough power supply voltage to accommodate your input and output voltages. You have not specified what you are using, but if you are using a single supply such as 3.3V/ground you can only get positive voltages output. The open-circuit voltage you measured probably has a short circuit current in the 0.1-1uA range under normal ambient lighting. You can get a rough measure of it with a multimeter.

If you use a transimpedance amplifier configuration you'd want to do something like this (plus a bypass capacitor on the power):

simulate this circuit – Schematic created using CircuitLab

Try something in the M$$\\Omega\$$ for Rf.

Here D1's negative current in photovoltaic mode works into a virtual ground with the current supplied via Rf. C1 helps with stability. The output is 1V/uA of PD current with a 1M$$\\Omega\$$ resistor and proportionally more for higher values of Rf.

• Thank you for posting this! I got the best response so far, but there is still something wrong. I used a 1Mohm resistor and 220pf capacitor, and got 0.3 v output in the dark and 0v when I turn a light on. Would it be helpful if I posted a picture of my circuit? Im 99% sure Im using the right pins on the op amp but you never know Aug 3 at 1:42
• Try flipping the PD. Aug 3 at 2:01
• Flipping the pd (cathode to ground, anode to -input/resistor/capacitor) gives me the original values I had, 0.1v in the dark and .34v with a flashlight on it. arg! Aug 3 at 2:13
• Maybe let's see the circuit. You could also try 10M. Aug 3 at 2:18
• Try replacing the PD with a 1.5V battery in series with a resistor the same value as Rf (eg. 1M). You should get 1.5V in one direction and 0V in the other. Aug 3 at 3:19

It looks like a photodiode needs to be reverse biased with about 1-5 VDC for it to operate, and then it will produce up tp 400 uA when exposed to light. Here is an article that explains a bit: