As I understand it, a photosensitive diode generates a small current with light. I ordered these off Amazon: https://www.amazon.com/gp/product/B01F3TU2XC/ref=ask_ql_qh_dp_hza

These change resistance but do not generate any current. I'm currently using a regular clear 5mm LED and my circuit is working. Of course, it does not work with the Amazon ones since they change resistance with light.

I am asking here since a google search produces two different descriptions for a photosensitive diode. Some say it changes resistance and others say it generates a current. Trying to understand...

  • \$\begingroup\$ Well, the item listing says they are photodiodes and they do look photodiodes, and they do not look like photoresistive sensor. How did you measure the resistance? How did you figure out they do not generate current? Maybe your method is not right. But you bought random Amazon item that comes with no data what it is, what are the specs and how to use it, and unfortunately this is not unpaid Amazon support. You should ask the seller about them, but most likely seller does not know either what those are. \$\endgroup\$
    – Justme
    Mar 4, 2021 at 0:22
  • \$\begingroup\$ Put your volt meter leads across the diode and then see if the voltage goes up in the sunlight and to zero in the dark. \$\endgroup\$ Mar 4, 2021 at 0:31
  • \$\begingroup\$ If it doesn't come with a manufacturer's part number, from a reputable source, then all you can count is that it's a thing that looks like an electronic component. With luck, it'll behave like the thing it's shaped like. Or it could be a factory reject and not work at all. \$\endgroup\$
    – TimWescott
    Mar 4, 2021 at 1:43
  • 2
    \$\begingroup\$ Black plastic photodiodes only respond to invisible IR light. They cannot see red LEDs, or green or blue or white. You must use an IR-emitting LED when using a black-colored photosensitive diode. Or, test it with incandescent light bulb, or with sunlight, since these contain much IR light. \$\endgroup\$
    – wbeaty
    Mar 4, 2021 at 1:54
  • 1
    \$\begingroup\$ @TimDuncklee Aha, maybe those are actually phototransistors? They would turn on in IR light. If so, then connect one in series with an LED, battery, and resistor (so if the phototransistor shorts out, the LED turns on.) Now shine some powerful IR on it, and if the LED lights up, then you've got an IR phototransistor, not a photodiode. Phototransistors often have only two wires, no base wire, and they're often inside an LED-type package. (If it didn't work, try reversing the black thing, since maybe the collector term wasn't the longer lead. It will still work backwards, but w/lower gain.) \$\endgroup\$
    – wbeaty
    Mar 5, 2021 at 2:20

2 Answers 2


These are Black daylight-blocking Infrared PD’s that typically generate 10~50uA with a remote control but that’s not what they use for receivers. You may test it reverse biased from 5V into a DMM in uA scale to gnd.

Like these https://www.mouser.com/datasheet/2/308/QSD2030-D-1814665.pdf

  • \$\begingroup\$ I bet it's a PT, not a PD. Phototransistor wouldn't give significant PV voltage, and with an ohmmeter, might be mistaken for a photoresistor. \$\endgroup\$
    – wbeaty
    Mar 5, 2021 at 2:25

There are two modes of operation of a PD. A PD that conducts a current when reverse-biased and illuminated will produce a voltage across the terminals when illuminated.

If you buy parts with proper datasheets (not random stuff without proper manufacturers and part numbers listed) they should characterize at least one mode of operation.

For example, this (probably similar) part

enter image description here

The photovoltaic mode shows an OC voltage of 400mV and a SC current of 35uA with the stated illumination wavelength and power density.

With the same illumination, and a -5V bias, the PD conducts 35uA, and when dark the current is typically 5nA.

From the fact that the parameters are only guaranteed for reverse bias operation, one might conclude that is the expected application.


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