A common photodiode amplifier circuit (Horowitz and Hill 2nd edition, pg. 253 figure J) looks like this:


simulate this circuit – Schematic created using CircuitLab

I am currently working on a project that requires detection of power in an infrared (1550nm) laser. The photodiode/phototransistor is from an older project in the lab I work in and I cannot find its part anywhere online.

My question is this: There are 3 leads. One of them is presumably for grounding, and the other two have the signal across them. Assume I operate the photodiode in photovoltaic mode. Since the two leads that have the signal also have the characteristic diode drop of .7 Volts, this amplifier will have a DC offset which will cause it to rail. How can this possibly be a functioning circuit?

I suspect that my misunderstanding stems from the fact that this is a transimpedance amplifier, so the DC offset is somehow irrelevant, but I would like some clarification and insight.

Edited my question to include a question!

  • 2
    \$\begingroup\$ What is your question? You say "My question is this" and then never actually ask a question. \$\endgroup\$
    – kpnz
    Sep 29, 2016 at 0:16
  • \$\begingroup\$ As JohnD mentions, a 3-lead diode is uncommon, it may be a photo-transistor, or contain two diodes with a common lead. The amplifier biases the diode at zero volts, so its not running in photovoltaic mode. Photo current is scaled to output volts by 4.7x10^6. R1 compensates for any bias current, but adds noise. Keep D1 close to OA1 to reduce stray capacitance. \$\endgroup\$
    – glen_geek
    Sep 29, 2016 at 1:20

3 Answers 3


In this circuit the junction is generating a photocurrent, and in the above circuit the op amp will generate just enough current through the feedback resistor to cancel the photocurrent such that the voltage across the junction is zero volts (in the ideal case.) That's because any difference in voltage between the + and - inputs gets amplified by the very large voltage gain of the op-amp and the feedback serves to servo the voltage back to zero. So the only offset you will have is the input offset voltage of the op-amp.

Note that this is not the fastest response circuit that you can get. The junction capacitance is large with no bias, so if you need response speed you may want to explore a TIA with reverse bias of the photodiode.

Also, it's relatively uncommon for a photodiode to have 3 leads. Are you sure it's a photodiode?

  • \$\begingroup\$ It's not all that uncommon in lab-grade photodiodes. As the OP suggests, the third lead is the case, and is used for grounding/shielding. See, for instance, the Thorlabs FGA and FDS lines of photodiodes thorlabs.com/newgrouppage9.cfm?objectgroup_id=285 \$\endgroup\$ Sep 29, 2016 at 1:30
  • \$\begingroup\$ @WhatRoughBeast true, good point- I was wondering though if he actually had a phototransistor with the base lead pinned out instead of a photodiode since he calls it a photodiode/phototransistor at one point. \$\endgroup\$
    – John D
    Sep 29, 2016 at 1:44
  • \$\begingroup\$ I found a data sheet for a similar photodiode, and as WhatRoughBeast suggests, the third lead is probably for grounding. I'm operating under that assumption. \$\endgroup\$ Sep 30, 2016 at 22:08

Maybe it could be better to buy a new photo-diode. Normally, you should also be able to have the data-sheet, and in the data-sheet you should found some information to use the photodiode correctly.

  • \$\begingroup\$ Quality photodiodes for 1550 nm are pretty expensive in the searches I have made (> 50 dollars) \$\endgroup\$ Sep 30, 2016 at 22:08
  • \$\begingroup\$ An alternative could be to read the data-sheet of photo-diode that could be similar. For this one (example): thorlabs.com/drawings/…, the information output should be in the leakage current, that give a voltage on RL resistance. And a DC voltage should be apply on the diode to work correctly. \$\endgroup\$
    – ABU
    Oct 2, 2016 at 18:04

A multimeter applies a small current to the diode and measures the drop to be 0.7 volts. However, this current is much larger than the current generated by the light incident on the diode. Therefore, when my photodiode is operating in photovoltaic mode, it doesn't have this offset. It has a much smaller voltage signal that is proportional to the number of photons incident on the diode.

I didn't understand how the multimeter measured the diode drop, so I thought that the 0.7V voltage drop across the photodiode was always present (like a battery). This is not the case, so the amplifier works just fine.


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