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I'm trying to build an optical power meter using a photodiode in photoconductive mode and a LM324N OpAmp:

schematic

simulate this circuit – Schematic created using CircuitLab

However, the output voltage signal is very low. Exchanging R_f for a smaller resistor or even shorting it gives a higher output. According to https://en.wikipedia.org/wiki/Transimpedance_amplifier, V_out / I_in = -R_f, so I would expect a smaller R_f to decrease the gain, not increase it.

Please help me understand what I'm doing wrong.

Thanks!

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    \$\begingroup\$ What sign is Vout? How far negative can Vout swing with 0V as the opamp's -ve supply rail? \$\endgroup\$
    – user16324
    Commented Jul 8, 2020 at 14:22
  • \$\begingroup\$ Vout is positive and increasing with illumination, up to ~1 V when illuminated with a bright LED close to the photodiode. I don't think the output could swing lower than 0V for this device and circuit. \$\endgroup\$
    – LimaKilo
    Commented Jul 8, 2020 at 14:35
  • \$\begingroup\$ I_in and Rf are both positive, so check your signs and try again. \$\endgroup\$
    – user16324
    Commented Jul 8, 2020 at 14:37
  • \$\begingroup\$ Well, but my oscilloscope says V_out is positive... \$\endgroup\$
    – LimaKilo
    Commented Jul 8, 2020 at 14:40
  • \$\begingroup\$ Operating outside the linear range of an opamp's input, phase reversals can happen. eetimes.com/the-phase-reversal-story/# \$\endgroup\$
    – user16324
    Commented Jul 8, 2020 at 14:42

1 Answer 1

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For Vout to successfully drag \$V_{IN-}\$ to equal \$V_{IN+}\$ (at 0 volts) it has to sink current from the photodiode (via \$R_F\$) towards a negative voltage. That can't happen because the lowest negative rail on the LM324 is 0 volts hence the output cannot force the circuit into equilibrium.

enter image description here

The red box in the drawing show the problem - the LM324 needs to have a negative rail when the photodiode is connected as shown.

On the other hand, if you changed the pins on the photodiode, it should begin to work because then the op-amp has to source current into the photodiode via \$R_F\$ and all should be well. Now you will need the photodiode cathode connected to ground: -

enter image description here

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  • \$\begingroup\$ This'll work, but the OP wants photoconductive mode, while this is photovoltaic. \$\endgroup\$
    – user16324
    Commented Jul 8, 2020 at 14:36
  • \$\begingroup\$ Correct. I could add a second supply for a negative bias to the diode, but first I would like to know if it is feasible with a single side supply. \$\endgroup\$
    – LimaKilo
    Commented Jul 8, 2020 at 14:38
  • \$\begingroup\$ @BrianDrummond the OP is building an optical power meter so I'm assuming that having a reverse bias on the PD is irrelevant for speed. If speed was important, then an LM324 would be out of the question. Plus, dark current would be a lot greater (I believe) and this is counter to the requirements for optical power measurement. \$\endgroup\$
    – Andy aka
    Commented Jul 8, 2020 at 14:40
  • \$\begingroup\$ @Andyaka I'm actually building this for speed, which is why I chose reverse bias. I already found out that the LM324 is not good for higher speeds, but I wanted to at least get the circuit working before looking for alternative amps. Dark current is not a problem, as I'm actually dealing with quite high intensities. \$\endgroup\$
    – LimaKilo
    Commented Jul 8, 2020 at 14:51
  • \$\begingroup\$ @LukasK. what PD are you using - please link to the data sheet. Also, what speed are you actually talking about here? How quickly do you need to compute the power or, over what time frame are you averaging the power for a single measurement? \$\endgroup\$
    – Andy aka
    Commented Jul 8, 2020 at 15:17

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