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I'm trying to build a transimpedance amplifier to convert the current from my photodiode to a voltage that can be read using an ESP32. The components I am using are:

  • OPA170AIDRLT Operational Amplifier (link)
  • QSB34CGR Onsemi Photodiode (link)
  • 97.6kΩ Resistor
  • 0.15uF Capacitor

I designed my circuit based on the Texas Instruments guide (link), and my circuit looks like the following schematic:

enter image description here

Like the title says, I am not measuring a voltage output from my circuit. More specifically, I am measuring 0V. I have checked the resistance over the photodiode to see whether it is in working order, and it is. Unfortunately, my circuit is soldered onto a PCB, so measuring the current output from the photodiode will not be very easy and I am leaving it as a last resort. Am I missing something obvious? What tests can I perform to find out what is wrong with my circuit? My lack of experience with this type of circuit is currently my downfall, so I would greatly appreciate any help or advice!

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  • \$\begingroup\$ it is better to use a TIA with the photodiode operating in the photoconductive mode \$\endgroup\$
    – Confused
    May 5, 2023 at 8:17

1 Answer 1

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Am I missing something obvious?

Yes, your photodiode is connected the wrong way round.

When light hits it, photocurrent will flow from 0 volts to the inverting input of the op-amp and, to maintain a virtual ground, the op-amp output would need to drop to a negative value on its output but, it can't do that because you have a single supply configuration.

Here's how it's normally done with a single rail op-amp and ground referenced photodiode: -

enter image description here

Image from Photodiode Amplifier Design using Photodiode Wizard - CN-0312 Spectroscopy Example supplied by Analog Devices (a very useful web page).

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  • \$\begingroup\$ If the photodiode absolutely needs a grounded cathode, then adding a DC negative supply voltage to pin 2 (Vee?) is an option. Detected light will cause a negative-going output voltage from the opamp. \$\endgroup\$
    – glen_geek
    May 4, 2023 at 12:46
  • \$\begingroup\$ I suspect that a 3.3 volt rail is used so that a positive output is desirable for feeding into an ADC input (that probably can't cope with negative voltages). \$\endgroup\$
    – Andy aka
    May 4, 2023 at 12:50

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