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can anyone explain to me why moving to configuration (b) overcame the problems I was having with the photodiode saturating?

Many thanks,

John

enter image description here

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Because Si junction has 0.6 V knee and is in parallel to the current source represented by photons reaching the PN junction. So if the voltage drop on the 50 ohm resistor caused by current generated by photons is greater than 0.6 V the parallel diode begins to conduct and seems that saturation is reached. When you place a 1 ohm resistor the maximum illumination can be 50 times greater before 0.6 V is reached.

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In photovoltaic mode, you are measuring the short circuit current produced by the diode. Ideally, that means you present it with 0 impedance (the short circuit). B works better than A since 1 Ω is closer to 0 than 50 Ω is.

The usual way to do this is with a active circuit that keeps the voltage across the diode at 0 while creating a voltage signal proportional to the diode current. Put another way, this is done with a transimpedance amplifier with 0 input voltage:

When the photodiode is illumnated, it will produce reverse current. The direction of that current is down in the schematic above. The opamp will drive the right side of R1 to whatever it takes to null out the voltage on its negative input, since the positive input is connected to 0 V. The result is that the opamp output voltage is proportional to the current thru the diode, with R1 being the proportionality constant. For example, if R1 were 100 kΩ, then the gain of this circuit is 100 kΩ, which could be expressed as 100 mV per microamp.

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Current flows across the photodiode from cathode to anode: potential is built up against the forward conduction direction of the diode, this is what actually gives you saturation. The answer from @krufra is wrong.

In this 'photovoltaic' mode charge builds up across the diode like a capacitor and is dissipated across your 50 Ω resistor (case A). The responsively drops as more light is incident. You can think of it as the electrons having to do more and more work charge is built up, the diode responds less as the voltage builds.

In case B you have simply lowered the resistor value making it easier for charge to flow away from the anode. With less of a potential build up across the diode then you would expect to see less of a non-linear drop off of the diodes responsively to light. I'd imaging that your second circuit might product a lot less signal.

The best way of getting linear (voltage proportional to amount of light) that doesn't saturate is by using a transimpedance amplifier. This can be relatively cheap with modern op amps. Olin's answer is mostly right but photovoltaic mode is when the diode is hooked up so that it develops a voltage across some resistive element where as photoconductive mode is when it is configured (with a transimpedance amplifier) to see 0 Ω resistance across its output.

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