I have (in the example circuit) a photo diode with the following data sheet: http://www.ibsg-st-petersburg.com/datasheet/PD/PD24-01-HS.pdf.
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Now I was wondering (after there is no data about it in the data sheet): Why is it important how high the negative voltage bias is (except for determining the dark current)? Where does the negative voltage drop come from, and how do I calculate it?

In the circuit I have the resistor R1 with the resistance 1 kOhm, the diode D1 with a voltage drop of approximately 0.7 V, the diode D2 with resistance of approximately 70 kOhm (according to the data sheet) and R2 with "low resistance". Then I get a voltage drop of approximately 2.9 V over the diode. Is the calculation correct, or did I forget something?


Reverse bias puts a photodiode into photoconductive mode, as opposed to photovoltaic mode. This results in increased sensitivity, decreased capacitance, and increased speed. This is at the expense of increased dark current.

Good review at https://www.thorlabs.com/tutorials.cfm?tabID=31760

The best review I know of is the Sharp Application Note on Optoelectronics. I'm having trouble finding it on the Shape site, but one link is http://denethor.wlu.ca/pc300/projects/sensors/photdiod.pdf

In particular, that Sharp AN shows the voltage/current relationships as a function of the reverse bias.

FWIW, I'd recommend an active amplifier circuit for your photodiode just on general principles, but I have no idea what your application is.

  • \$\begingroup\$ Shouldn't it be "increased dark current", when increasing the bias? Furthermore, how can I calculate the bias in the circuit shown above? \$\endgroup\$ – arc_lupus Aug 3 '16 at 12:48
  • \$\begingroup\$ @arc_lupus - you do your calculations by reading the data sheet. See the "dark current vs reverse voltage" curve. \$\endgroup\$ – WhatRoughBeast Aug 3 '16 at 14:42

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