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Engineering
Electrical Engineering
Understanding Photodiode Behavior in Reverse Bias Circuits
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[QUOTE="Paul Colby, post: 5467667, member: 584221"] Well, I don't understand what you mean by "needs 6V to work" in the above. For the most part resistors always "work" and there will be a a voltage drop given by Ohm's law, ##V=IR##. The voltage drop across the diode is that part of 15V that isn't dropped across the resistor. If current through the resistor is ##I=I_\text{leak}+I_\text{photo}## the voltage drop across ##R_A## is this current times ##R_A##. All this holds in the DC limit. So, if sufficient current is passing through the diode to make the voltage drop 6V across the resistor then the voltage drop across the diode will be ##9 = 15 - 6## volts. It might be best to look at the diode as a current source and not think in terms of voltages. For typical photodiodes ##R_A## would have to be very very large to drop 6V for the typical photocurrent. (By photodiode you don't mean solar cell do you? In that limit 6V might make some sense. You could help us out by sharing what you're trying to do) [/QUOTE]
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Forums
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Electrical Engineering
Understanding Photodiode Behavior in Reverse Bias Circuits
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