Re: Photodiode Dark Current Measurement

On Sep 28, 1:35 pm, "Greg Neill" <gneill...@xxxxxxxxxxxxxxx> wrote:
"GraemeC" <graeme.cunning...@xxxxxxxxxxxxxx> wrote in message

news:1190977761.357722.184850@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx





I am trying to measure the dark current of a photodiode

I have Chathode of the PD connected to 1Meg ohm resistor which is
connected to +2V
The annode is connected to 0V

I obtain the following measurements
Voltage accross Resistor: 1.39V
Voltage accross Photodiode : 0.82V

I have assumed that the Dark Current (or indeed any current value) is
(2 - (1.39+0.82)) / 1Meg

=> I obtain a dark current of 210nA. This figrue however is less by
more than an order of magnitude than that I am expecting.

The photodiode and resistor are in series, so they
share the same current. Given that you measured a
1.39V drop across the 1M resistor, what's the current
in the resistor?- Hide quoted text -

- Show quoted text -

I don't understand that. Because the the value across the resistor
varies as you adjust the bias voltage. The bias voltage shouln't
effect the value of the photocurrent generated.

G

.

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