I have made a quick search but couldn't find anything specific around.
I need to design a rudimentary circuit for a photodiode in conductive mode. I am interested in high speed rather than sensitivity, a wavelength around 550nm.
I have a notional idea of a battery connected across the legs as a voltage bias, but how do calculate what resistor to use to avoid overloading the diode…
Sorry for being vague, I'm struggling to find out where to begin!
(nb. I have a background in physics so have a basic knowledge of electronics.)
Best Answer
I'll edit this answer severely as a general comment:
Photodiodes may be operated either forward or reverse biased.
Forward biased gives most output.
Reverse biased gives most speed. and is noisier.
In this mode Vsupply needs to be < Vreversebreakdown - hopefully in the data sheet.
Reverse biased mode is most usually used.
The Sharp BS120 is optimised for 560 nM operation. It specifies Vreverse_abs_max as 10 volts and has curves all the way up (or down) to -10V.
Wikipedia says
In "forward biased" mode the diode is usually operated with NO added bias and is used as a voltage source.
For a silicon diode with Vforwards = Vf = 0.6V . Say you want 5 mA current and are using a 5V supply then
I = V / R = = (Vsupply - Vdiode) / R = (5-0.6) / 0.005 = 880 ohms Say 1000 Ohms (which gives a notional 4.4 mA)
For 1 mA you'd use (5-0.6)/0.001 = 4400 Ohms