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tresca

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Hey
So I was organizing some components I had lying around, and I found this black square type component. It has two long leads and at the tip, is a square black thing.

I thought maybe it had something to do with temperature or light because the other components around it where heat/light sensors. So i measured its resistance and it was about 4.14Megaohms. I used a lighter (thinking it was a temperature sensor) and the resistance shot up to like 16Mega. It reacts to light more than temperature. I moved the flame a good distance away, enough to not feel the heat, and it still shot up.

It seems like its an LDR, but instead of resistance having an inverse relation to light, it seems to be proportional. When I completely covered it up with my hand, it dropped to about 500K.

Do they make LDR that increase resistance ? Is it meant to detect IR light, since its coated in some, what appears to be glossy plastic casing ?

Anyways, I threw it in with my other LDRs until a better explanation could be found.
 

ericgibbs

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Most Helpful Member
Is the square black block at the end rectangular about 3 or 4mm and the does the black block have a corner cut off.???
 

ericgibbs

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Most Helpful Member
Sweeet ! you kno what it is.....

what is it ?
It sounds like an IR detector.

The change in 'resistance' due to the light, could be caused by the voltage generated by the detector, if the DVM is set to ohms

Try the same test ie: light and dark but set your DVM to dc volts.. try the DVM leads both ways to the two wires... do it one way and then the other, tell us what you see.:)
 
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tresca

Member
AAAH !! That makes sense. I switched my DVM to dc, and in complete or near darkness it hovers around 0. And with a flame relatively close to it, its about 0.3V. Ambient light puts it anywhere from 0.1 - 0.15V.

I guess I should have known that it would generated a dc voltage since one lead is shorter than the other. Polarity wouldnt make much sense for anything else.

Thanks man !
 

ericgibbs

Well-Known Member
Most Helpful Member
AAAH !! That makes sense. I switched my DVM to dc, and in complete or near darkness it hovers around 0. And with a flame relatively close to it, its about 0.3V. Ambient light puts it anywhere from 0.1 - 0.15V.
This is the photo voltaic mode
I guess I should have known that it would generated a dc voltage since one lead is shorter than the other. Polarity wouldnt make much sense for anything else.

Thanks man !
hi,
It can also be used with a resistor as a load in leakage current mode.

Connect one end of a 10K resistor to the anode and the other end of the resistor to 0V, connect the cathode to say +9V.

Measure with a DVM on volts at the junction of the anode/resistor, do light/dark test.
 
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