In my design, a photo diode is connected in reverse bias with 12V. The LDO (low-dropout) voltage regulator which is used for biasing can deliver a maximum current of 500mA
My configuration:
simulate this circuit – Schematic created using CircuitLab
What I want to know is: If my photo diode gets saturated, how would it behave? Will it behave like a current source generating maximum rated current? or a short circuit allowing 500mA of supply current directly to the TIA (transimpedance amplifier)?
I'm asking because I want to put a protection diode before the TIA, which has maximum rated input current of 100mA.
I want to know the equivalent circuit of a saturated photo diode – is it a closed switch or a regulated current source?
Best Answer
OP: "...please explain how would a photo diode behave in saturation which would explain my doubt in total, is it like a close switch or a constant current source with current being peak current"
A quick Google gave me two useful results:
"Once the Saturation point is reached, the behavior of the photodiode becomes more and more non-linear until the Damage threshold point is reached. At this point the photodiode is no longer able to turn excess in incident optical power into a photocurrent output and all the extra energy received by the device would be absorbed, typically as heat. This heat inside the device can easily damage it irreversibly as it endangers the fragile wire bonds on the surface of the photodiode."
Source: http://www.osioptoelectronics.com/technology-corner/frequently-asked-questions/input-light-intensity.aspx
"As the photocurrent increases, first the non-linearity sets in, gradually increasing with increasing photocurrent, and finally at saturation level, the photocurrent remains constant with increasing incident light power."
Source: http://www.osioptoelectronics.com/application-notes/an-photodiode-parameters-characteristics.pdf
PS: Seems that the question is too old. I have no idea why it appeared in recent questions. Maybe a Community edit...