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| Photodiodes saturating in ambient light |
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| ejeffrey:
Increase your bias voltage or decrease the shunt resistor. You aren't saturating the photodiode you are saturating the battery+resistor. Drop the resistor to 1k and add an extra gain stage and should be fine. |
| ejeffrey:
Also, always draw a photodiode as a diode not just a circle. |
| ejeffrey:
Wait: is your sensor a photodiode reverse biased (ie in photoconductive mode) or a photoconductive sensor or LDR? |
| TimFox:
Your circuit will "saturate" when the DC photocurrent through the 13k resistor produces a voltage that overcomes your bias battery voltage and the parasitic diode of the photodiode forward biases, shorting out the photocurrent and preventing it from reaching your output meter. |
| tggzzz:
--- Quote from: aussie_laser_dude on May 10, 2020, 02:28:29 pm ---Thanks guys, I've attached a circuit diagram for a single diode (photodiode is BPW34S high speed silicon PIN photodiode). Thanks Tim and David, I've tried ac coupling via a simple 1nf capacitor, it's working well to remove the slow changing offset caused by ambient light. The fast changing signal is all that's wanted here. Unfortunately the signal amplitude drops to near zero with ambient sunlight :( a dc restoration loop? I'll investigate that. Thanks tggzzz, i'm trying to solve in terms of current, doing the math is proving challenging. Intuitively i expect either a) the photodiode actually saturates from too much sunlight and the current is a flat dc curve even with the extra pulsed light. OR b) it doesn't actually saturate, there's a big dc current with tiny periodic spikes from the pulsed light. The capacitor in my circuit should remove the dc offset and keep the spikes... I don't see the ac spikes when measuring with ambient sunlight so i guess that means it's case a, the photodiode can't create anymore free electrons? ie. It's optically saturated!? --- End quote --- That's half a circuit diagram: it doesn't indicate which way round the photodiode is connected. Yes, it matters! Assuming it is reverse biassed, then the arithmetic is simple. If you have x watts of light falling on the PD then it will have a current of i=x/2 amps. The voltage drop across the resistor will then be 13000*i. If we assume that we want >1V reverse bias across the PD, then the max voltage across the resistor can be 2.3V, and from that you can work out the maximum allowable light before saturation. You should also do research on opamp transimpedance amplifiers. That will help you improve the sensitivity and frequency response. You would also benefit from removing the 1nF cap in series with the resistor (why aren't you using AC coupling?). Then monitor the voltage while the 10us light pulses are there. Then gradually increase the ambient light (torch?) and observe the voltage. That should help you understand what is happening. |
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