Look at this data sheet, in the second page we have a graph for the spectral response.

Now, I have learned that photodiodes must be reverse-biased in order to get a decent signal from them, but then someone pointed out to me that all of the measurements in the data sheet are made at 0 volts, not at the reverse bias voltage, and so it would stand to reason that the spectral response was measured with 0 volts too.

Now, there must be a difference in the signal of a biased and an unbiased photodiode, right? A biased one produces a larger signal to the same amount of light intensity, right? That would mean that the spectral responses of biased and unbiased diodes are different.

Then the obvious question is, what is that difference? And for that matter, I would like to know all the properties of my photodiode at the voltage I'm going to use it at. How can I calculate those values?

  • \$\begingroup\$ Photodiodes don't need to be reverse biased to yield a decent signal. See my comment on your previous question. \$\endgroup\$
    – Andy aka
    Feb 27 '19 at 15:00
  • \$\begingroup\$ One of the main reasons for reverse bias is to reduce the space-charge capacitance of the junction, so higher speed and less noise peaking with transimpedance amplifiers. \$\endgroup\$
    – John D
    Feb 27 '19 at 15:23
  • \$\begingroup\$ What do you mean by "spectral response" exactly? Do you mean the response to the optical wavelength of the incident light, or the speed of the diode, i.e. how fast it reacts to variation of the level of incident light? \$\endgroup\$ Feb 27 '19 at 16:01
  • \$\begingroup\$ Lorenzo, by spectral response I refer to the first thing you said, that's how it's labeled in the datasheet I linked to \$\endgroup\$ Feb 27 '19 at 16:28
  • \$\begingroup\$ The photodiode response vs wavelength can be changed in many ways. This include anything that changes the thickness of the deletion region, like reverse bias voltage. Most only look at the reduced capacitance, but wavelength is also affected. The mechanism is pretty easy to gather. Charges are generated at different depths vs wavelength. If they are in the deletion region, they are more likely collected. Thinner depletion regions respond more poorly to longer wavelength, since more of their charges are generated outside that region. \$\endgroup\$
    – jonk
    Feb 28 '19 at 7:15

The answer is hardly at all.

The overall quantum efficiency might be slightly higher (a few per cent?) due to the carriers being swept out of the depletion region quicker and having less chance to recombine before contributing to terminal current.

The cut-off wavelength may be shifted very slightly so that if your optical wavelength is just on the edge of cut-off you could see a stronger effect.

But neither of these effects is really strong enough to be characterized on a datasheet. Your uncertainty about the optical coupling efficiency from whatever beam or fiber is delivering light to the detector will be greater than the change in quantum efficiency. And the variation in cut-off wavelength due to temperature changes will probably be greater than (or at least comparable to) the effect due to bias voltage.

a biased one produces a larger signal to the same amount of light intensity, right?

Not really. One carrier pair per absorbed photon is one carrier pair per absorbed photon, regardless of bias voltage.

The main trade-off is modulation frequency response vs noise. A biased photodiode will have slightly faster response than an ubiased one, but will produce more dark current (and thus more noise).


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