If a photodiode with responsivity R is operated in photo-conductive mode with reverse bias voltage Vb and load resistance Rld what relation gives the output current from the photo-diode flowing trough Rld?
I know that generated current from the incoming optical power P is Io = R*P, but I am not sure how is Io affected by Vb and Rl and when could reverse saturation current be neglected.
I am using this photo-diode BPW 34 FS with incident power levels of about 10 - 1000 w/m^2.
As a first-order approximation, the reverse bias and loading resistor don't affect the photocurrent at all, so you have the correct formula:
$$I_p = P_o {\mathcal R}$$
In reality, these will have a small effect, but to account for it, you have to know the diode's effective shunt conductance. Then you can model the photodiode as a current source in parallel with the shunt conductance (and some capacitance if you want a dynamic model).
You may be able to estimate the shunt conductance from the photo-current and open-circuit voltage curves given in your device's datasheet. At 100 uW/cm2 (1 W/m2), I'm estimating about 32 uS conductance, or 30 kohms resistance. The conductance increases as optical intensity increases.
