So I've used the standard trans-impedance amplifier to sense the current of a photodiode. The photodiode is unbiased.


I went through the Spice simulations to calculate the stability and the value of the compensating capacitors I need and then I did the analog testing and then I made the pcb and it all works as designed.

However I encountered a possible failure mode that results in blowing the photodiode ( I did blow a test one) and I can't afford for that to happen ( there are 25 embeded photodidoes in a part that I can't replace easily and can't afford to).

I'm using a power supply (+-15V and ground) and if for whatever reason one of the power supply rails fails (simplest example is the +15 gets unplugged by mistake but originally I had a faulty power supply that resulted in this) the photodiode sees across it's terminals either + or - 15 V depending on which rail was unplugged (possibly 30 V ? if the ground cable is the disconnected ones?).

Either one of these ( 15V of forward or reverse bias) kills my photodiodes. As I said I have made my pcb so I don't have much space to implement some sort of current limiting. I considered fuses but I think they might be too slow or if I rate them low and the current peaks when the photodiode is illuminated by a particularly large source and it blows it would create the same failure mode as if I were to unplug one of the rails.

Anyone has any experience with this sort of problem ? The possible ideas I'd found so far are:

use a voltage regulators with a resistor to limit the current ( essentially becoming a current source) www.ti.com/lit/an/sbva011/sbva011.pdf

use some sort of current limiting circuit after the power supply. www.radio-electronics.com/info/circuits/transistor_current_limiter/transistor_current_limiter.php

Obviously I'd need two on both + and - side, but I'm not entirely sure how they would behave.

Any suggestions or advice is more than welcome.

P.s. links are not links since I don't have the rep to post links ...

  • \$\begingroup\$ Maybe an option is to place a couple of (non-photo) diodes in parallel with your sensitive photo diodes. I'm thinking 3 diodes in series so at around 1.8 V the diodes would start to conduct and protect the photo diodes. And another 3 diodes in parallel but for the other direction, like this: wiki.3rail.nl/images/d/d7/Dioden-antiparallel.jpg so the far left and right connections in parallel with your photo diodes. Indeed the solutions you propose are probably not fast enough. The diodes can be almost any type you have available. \$\endgroup\$ Commented Mar 30, 2016 at 12:35
  • 1
    \$\begingroup\$ I don't understand how that circuit, rigged up correctly, can blow a photodiode. How are you powering your op amp, and what is the allowable common mode input range for that power configuration with that op amp. I suggest using a positive and negative power supply, as the inputs are at the negative rail if you power with positive and ground. \$\endgroup\$ Commented Mar 30, 2016 at 12:37
  • \$\begingroup\$ @Blago You're thinking current limiting but my diode approach is different, it is voltage limiting but no excess current wil flow through your photo diodes thus they are protected. \$\endgroup\$ Commented Mar 30, 2016 at 12:37
  • \$\begingroup\$ FakeMoustache, I won't have the physical space for the solution of placing 6 component between the photodiodes and the op amps. \$\endgroup\$
    – Blago
    Commented Mar 30, 2016 at 13:14
  • \$\begingroup\$ Scott, I am using a positive and negative power supply. I had a faulty relatively cheap but portable power supply that somehow blew my test photodiode and this was the only failure mode I could reproduce. It is possible that someone trips or pulls on a cable or a connection gets loose \$\endgroup\$
    – Blago
    Commented Mar 30, 2016 at 13:16

1 Answer 1


The only possible high-current path that I see is the protection network on the op-amp. If you add a resistor of a few K in series with the (inverting) input it should limit the current and protect the diode. You can parallel the resistor with a small capacitor (100pF or something like that) if you need to speed it up (the slew rate should be limited enough by your supply bypass capacitors to limit the current through the capacitor).

Reverse breakdown is not damaging for diodes if the current is limited. The maximum allowable current would depend on the diode.

  • \$\begingroup\$ So that is equivalent to introducing a series resistance to the photodiode ? This would decrease the linearity of the photodiode and potentially introduce an error ? \$\endgroup\$
    – Blago
    Commented Mar 30, 2016 at 13:21
  • \$\begingroup\$ Could I use a a shunt voltage reference or a voltage regulator just after the power supply perhaps? That would get rid of some of the 50 hz noise as well ? I have not experience with either so not use it that would work. \$\endgroup\$
    – Blago
    Commented Mar 30, 2016 at 13:24
  • \$\begingroup\$ In series with the input, so not the same same as series with the PD. It would introduce a bit more noise, but probably not significant unless your feedback resistor is really low. \$\endgroup\$ Commented Mar 30, 2016 at 13:47
  • \$\begingroup\$ i am even having a similar problem of how to limit the current to TIA, what exactly happens is when a high/excess light is illuminated on photo diodes, they output a current of almost 10mA - 100mA which is directly fed to TIA would damage it, so how can i limit such current ? by the way my current is a pulse and of width only 10ns \$\endgroup\$
    – kakeh
    Commented May 19, 2016 at 8:03

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