I always wondered why the most common pinout for laser diodes with a monitoring photodiode was so strange. It wasn't sure how it was intended to be used until I ran into this circuit here:

enter image description here https://www.youtube.com/watch?v=MUdro-6u2Zg

In my mind, it seems a bit strange that the current source is shared by both the laser diode and the photodiode since it seems that this coupling would cause the laser diode and photodiode to distort each other's output.

The only reason I can think for this is to reduce component count by using the laser diode's own forward voltage voltage drop to reverse bias the phtoodiode to eliminate the need for a bias supply, and that the tiny photodiode current is considered negligible relative to the full laser diode current.

Is my reasoning correct? Or is there a useful purpose to this coupling?


1 Answer 1


That isn't the usual way to use this laser diode configuration.

A more typical application would look something like this:


simulate this circuit – Schematic created using CircuitLab

The important difference is, the common node between the laser and photodiode is a low-impedance node, not high impedance.

  • \$\begingroup\$ Well, that just makes sense. lol. Not sure why it never occurred to me that you could use a voltage source there and not a current source. But does that mean there's no good way to use this pinout in photovoltaic mode without a negative supply? Or is photovoltaic mode simply not used with laser monitoring because bandwidth tends to be more important than low noise? \$\endgroup\$
    – DKNguyen
    Aug 2, 2019 at 17:37
  • \$\begingroup\$ @DKNguyen, you could use a virtual ground on the op-amp's non-inverting input, equal to the +V voltage. Might want to use a second differential or in-amp stage to be sure +V variations don't cause current sense errors. \$\endgroup\$
    – The Photon
    Aug 2, 2019 at 17:54
  • \$\begingroup\$ That was my original approach but it felt gross. \$\endgroup\$
    – DKNguyen
    Aug 2, 2019 at 18:02

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