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I want to use a photodiode to measure light intensity, but I am not sure if the photodiode should be used in photoconductive or photovoltaic mode.

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From my understanding the photovoltaic configuration will have a leakage current proportional to light intensity and the photoconductive configuration will produce a current proportional to the light intensity. Both seem like the would work for my application.

Will either configuration work? and is there a benefit to using one over the other?

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Right, except for photovoltaic it's not leakage current... just photocurrent. The biased version will give a little more dark current. The big advantage of biasing the photodiode is that you reduce it's capacitance. (typical numbers might be by a factor of 4 or 5. at 10V) And this makes the detector faster. Which is almost always a good thing. Oh one final disadvantage to the bias case is that any noise on the bias voltage appears on the output. (so filter the bias supply.)

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Photovoltaic will give you near zero dark current and very high linearity over a wide current range. Photoconductive can't do that.

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Use the photovoltaic topology when you care about dark current (the current through the photodiode in complete darkness) being as close as possible to zero. In darkness the voltage vs. current curve of the photodiode should pass through the origin, which is the operating point with this topology (\$V=0\$).

The junction capacitance decreases when reverse voltage is applied to the photodiode. This makes the photoconductive topology more suitable when you want a better frequency response out of the detector.

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