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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.)

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  • \$\begingroup\$ What speed do you need? \$\endgroup\$ – Leon Heller Nov 18 '11 at 12:49
  • \$\begingroup\$ ideally ns response times, ie GHz frequency discrimination. \$\endgroup\$ – Nic Nov 18 '11 at 17:14
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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.

enter image description here

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

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  • \$\begingroup\$ Yes, but a photodiode should be reverse biassed as per Konsalik's answer! \$\endgroup\$ – MikeJ-UK Nov 18 '11 at 13:08
  • \$\begingroup\$ @MikeJ-UK Ah - terminology brainfade - I'm well aware of the normal reverse biased mode - "conductive mode implied to me conductive whereas its just the opposite. A photodiode may be operated in either polarity - with different results in each case. Either may be preferred depening on requirement. What he wanted was in fact reverse biased mode, as you noted :-) \$\endgroup\$ – Russell McMahon Nov 18 '11 at 13:31
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Russel explains the calculation of the resistor in his post above. Here is a simple circuit:

enter image description here

Note that the diode is reverse biased for "photoconductive" mode as Mike explained in his comment on your question.

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  • \$\begingroup\$ As above :-). I'm well aware of the normal reverse biased mode - "conductive mode implied to me conductive whereas its just the opposite. What he wanted was in fact reverse biased mode, as you noted :-) \$\endgroup\$ – Russell McMahon Nov 18 '11 at 13:32

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