# LM324N transimpedance amp for photodiode: Too low output, smaller feedback resistor gives greater output, not lower

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

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.

Thanks!

• What sign is Vout? How far negative can Vout swing with 0V as the opamp's -ve supply rail? – user_1818839 Jul 8 '20 at 14:22
• 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. – Lukas K. Jul 8 '20 at 14:35
• I_in and Rf are both positive, so check your signs and try again. – user_1818839 Jul 8 '20 at 14:37
• Well, but my oscilloscope says V_out is positive... – Lukas K. Jul 8 '20 at 14:40
• Operating outside the linear range of an opamp's input, phase reversals can happen. eetimes.com/the-phase-reversal-story/# – user_1818839 Jul 8 '20 at 14:42

For Vout to successfully drag $$\V_{IN-}\$$ to equal $$\V_{IN+}\$$ (at 0 volts) it has to sink current from the photodiode (via $$\R_F\$$) towards a negative voltage. That can't happen because the lowest negative rail on the LM324 is 0 volts hence the output cannot force the circuit into equilibrium.
On the other hand, if you changed the pins on the photodiode, it should begin to work because then the op-amp has to source current into the photodiode via $$\R_F\$$ and all should be well. Now you will need the photodiode cathode connected to ground: -