Instead of moving up and down the V=0 line, you'll be operating about some negative voltage, but there won't be much difference in the output behavior (aside from a larger dark current and faster response). If you attach the anode to a negative voltage instead of ground, it's now reverse biased, and will be faster. I suppose you can most accurately call your circuit "photovoltaic", but that's still a transconductance amp. The "bias" of a photodiode refers to the voltage across it, not the direction of current through it (though the two are certainly related). The output of your circuit will be the product of the photocurrent and the feedback resistor. Note that photocurrent is a leakage current, from Cathode to Anode, always. In the feedback circuit you show, the voltage drop across the diode is zero, thus you are moving up and down the V=0 line in the figure above (from the same document) as light intensity changes. The advantage of reverse biasind a photodiode is higher speed and dynamic range, the disadvantage is higher dark current. The Cathode is at a higher potential than the Anode. In the figure above, from a superb application note from Sharp (which I can no longer find at Sharp, but it is available at ) the diode is reverse biased. In fact, both the cathode and anode are at ground, so there is zero bias. No, your photodiode is not reverse biased in the circuit shown. Simulate this circuit – Schematic created using CircuitLab The PD has high capacitance when the voltage across it is near zero so the output will tend to be noisy and the op-amp may even oscillate without a cap across Rf.Įdit: To try to make the word soup a bit more clear, below is what I am trying to describe in the comments. This kind of circuit is very simple, however it is not very suitable for some applications. You can also test the PD with the diode test function on your multimeter. It won't hurt anything in this circuit if you get it wrong so if you for some reason don't have a datasheet, just try it. Most (perhaps not all) are probably similar to this. You can see that the longer lead is the anode (pin 2), so you would connect the longer lead to ground. Your PD datasheet should indicate which lead is which (eg. The output voltage is (ignoring op-amp imperfections) the PD current multiplied by Rf. The PD (as shown in your schematic) effectively tries to pull the inverting input below ground, and the op-amp supplies current through Rf to try to maintain the inverting input at 0V (it is a virtual ground). If not, then it will rail near (or fairly near) ground, depending on the type of op-amp. If you flip the PD in the shown circuit you'll get a negative output when light falls on the PD (assuming the op-amp has bipolar supplies such as +/-5V). ![]() When light falls on the photodiode, currents flows in such a direction to forward bias the PD. Is that true that any photodiode with voltage across it not equal to 0 is in photoconductive mode?Īnother question is that: if the photodiode is in photoconductive mode, if I change the polarity (example: anode initially connected to ground, then cathode to ground), does the mode (photoconductive / photovoltaic) change, too? ![]() ![]() What I'm confused about is the concept about the photoconductive mode. ![]() So, is the reversed bias here photoconductive mode of photodiode?Ī problem that came to my mind is that, in the figure above, do I connect the longer lead of photodiode to ground or to the negative node of the op amp?Īnd does the current direction through photodiode flow opposite to its indication in any mode (photoconductive or voltanic), for example like in picture above (photodiode "points up" while current goes from negative node to ground)? Photodiode is in reversed biased while LED is in forward biased I found a topic on the Internet which indicated that I tried to search for them in Google and got some definitions, but it still confuses me a lot. After reading about some of the photodiode characteristics on the Internet I got really confused with the photodiode working principle.
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