I have been working on a circuit which involves a photosensitive element soldered on a PCB. I am confused whether it is a photodiode or a phototransistor. Is there any means by which I can figure out whether its a photodiode or phototransistor without desoldering it?
its a two terminal device.
I am attaching its picture.
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\$\begingroup\$ Shine light on it and see if current comes out? \$\endgroup\$– PlasmaHHJun 17, 2015 at 10:35
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1\$\begingroup\$ If you are working on this circuit then please give part numbers of the device. \$\endgroup\$– Andy akaJun 17, 2015 at 10:47
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\$\begingroup\$ What is the device? An safety device for a punch press? \$\endgroup\$– Spehro PefhanyJun 17, 2015 at 14:06
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2 Answers
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There are a few ways you can do this:
- Connect a voltmeter across the leads and shine a bright light or a laser pointer on it. If the voltmeter reads a voltage (not just noise) then it is probably a photodiode.
- Connect a diode tester across the leads. While this isn't ideal as they are in-circuit, it may still work. If your diode tester doesn't read anything, swap the leads around. If you still don't see a reading, then it's probably a phototransistor.
- If the board has some sort of amplifier (an op-amp, most likely) connected close to these devices, they're probably photodiodes. Otherwise they could be phototransistors, which are just switches and wouldn't need amplification.
- Do the PCB designators show a "D" or a "Q"? "D" is for diode, "Q" is for transistor
In order to get a more accurate reading (particularly with the diode test) it would be best to desolder one leg of the device (to take it out of the circuit) before measuring. However, it looks like there is some sort of gunk holding it to the board so you may need to remove that first.
Obviously a better view of the (entire) board and part numbers would be very helpful in this case.
answered Jun 17, 2015 at 12:10
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\$\begingroup\$ thanks a lot for the help. I have figured out that its a photodiode. \$\endgroup\$ Jun 18, 2015 at 17:15
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\$\begingroup\$ Now I am facing another problem.I believe that the connecting circuit, consisting a BJT, must work as an amplifier circuit. I expect it to be something like !this. However, on reverse engineering, the actual circuit turns out to be something like !this.I am confused whether I am thinking in the right direction or not. Please help @derstrom8 \$\endgroup\$ Jun 18, 2015 at 18:11
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\$\begingroup\$ That schematic doesn't seem to make much sense. What type of transistor is it (do you have a part number)? Seeing as the emitter is connected to Vcc, I am wondering if it is a PNP? \$\endgroup\$ Jun 18, 2015 at 23:04
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\$\begingroup\$ Also, I don't think you ever answered, what is this circuit used for? It might help us figure out what it is doing \$\endgroup\$ Jun 18, 2015 at 23:04
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\$\begingroup\$ Yes, it is a PNP transistor, its part no. is BC857. I expect it to be a common emitter config. of an amplifier but since the emitter is connected to Vcc, i dont think it would be a common emitter config. what do you say? \$\endgroup\$ Jun 19, 2015 at 7:08
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Just from the looks of the active area, it certainly looks like a photodiode.
However, since it's a 2-lead device, it doesn't matter which it is - it will be used the same way in either case. The difference between the two is generally that photodiodes are faster but less sensitive, while phototransistors are more sensitive but slower. Phototransistors also have greater dark current. In any case, each is used in a circuit which converts photocurrent to voltage. In the case of a phototransistor, this is usually a simple resistor, while photodiodes (with their lower currents) ordinarily use some sort of amplifier.
As I say, the large active area which is visible says they are probably photodiodes.
answered Jun 17, 2015 at 15:57
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1\$\begingroup\$ That would be my guess, I've never seen a photo transistor with such a big active area. (I wouldn't say that a phototransistor is more sensitive than a photo diode... one photon makes one e-h pair, the photo transistor just has built in gain. but maybe that is only semantics.) \$\endgroup\$ Jun 17, 2015 at 16:18