Photodiode circuit

[aiq25]

Hello. I'm having a hard time understanding the attached photodiode circuit.

I can understand the circuit without R2: Vout = R1•Ipd and the photodiode is reverted bias. But I'm not sure what Vout would be with R2. It's creating a voltage divider but what's the output equation?

Any feedback would be appreciated. Thanks.

[Zero999]

Hint: the output impedance of the voltage divider is equal to both the resistors in the divider connected in parallel.

[aiq25]

Hint: the output impedance of the voltage divider is equal to both the resistors in the divider connected in parallel.

The equation should be: R2/(R1 + R2). But is voltage just the output impedance•Ipd? I guess I'm confused as to how the photodiode current generated and the current from VCC affects the output voltage.

Would photodiode current would generate a voltage across just across R2? Since the photodiode is in reverse bias, would R1 have an effect on about?

[Andy Watson]

Look up Thevenin (and Norton).

[Zero999]

Hint: the output impedance of the voltage divider is equal to both the resistors in the divider connected in parallel.

The equation should be: R2/(R1 + R2). But is voltage just the output impedance•Ipd? I guess I'm confused as to how the photodiode current generated and the current from VCC affects the output voltage.

Would photodiode current would generate a voltage across just across R2? Since the photodiode is in reverse bias, would R1 have an effect on about?

Model the potential divider as an ideal voltage source, V = VCC*R2/(R1 + R2), with an output impedance of R1|R2 in series with the photodiode.