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I drive an IR LED with a 50% duty cycle square wave and receive its signal with a photodiode. If I take the photodiode farther and farther from the LED, the signal's positive duty cycle is lower and lower. Why?

ocrdu
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DJDB999
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    I'm assuming that the IR LED you are using isn't as focused as a laser might be, the further you move the PD away from the LED it's likely you're losing power over distance and seeing less active time that the IR LED is actually able to correctly bias the PD leading to the duty cycle, although without knowing which LED and which PD it will be difficult to say for certain. – snowmanemperor Nov 30 '21 at 19:00
  • It's probably due to the photodiode signal having different rise and fall times depending on light intensity. Which photodiode are you using, what circuit are you using to detect the 'square wave', and what is its frequency? – Bruce Abbott Nov 30 '21 at 20:04

2 Answers2

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Because the photo-diode receives less light, because the solidangle (2d angle) of the LED causes the light that falls upon a unit area to deacrease with distance. If you were to draw a circle and measure the light next to the LED and then move it away, the same circle will receive less light. Same thing with the photo diode, it has a fixed area, the more you move it away from the LED, the less light it receives.

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Source: https://www.optiled.com/en-US/About/BasicLightingKnowledge

Voltage Spike
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Are you actually getting a 50% duty-cycle output from the receiving photodiode?

I had a similar project with a laser and a PIN photodiode. The laser saturated the photodiode circuit and the output took a while to fall after the laser was removed. The effect was that I saw a higher duty-cycle than that of the input.

And at higher frequencies I saw a 100% duty-cycle output for my 15% duty-cycle input, since the circuit never had a chance to decay.

bitsmack
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