Electrical – photodiode circuit setup

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

According to this, the photodiode does indeed produce a current even when there is zero volts across it; it's the short circuit current. Note that the reference direction of \$I_S\$ in the question's diagram is opposite that of the \$I_{SC}\$ of the diode so the output voltage is:

\$V_{OUT} = - I_S \cdot R_F = I_{SC} \cdot R_F\$

I found the above here.

A reasonable question to ask is how can a current be produced with zero voltage?

Remember that there's an internal E field through the depletion region even when the diode terminals are shorted together. Briefly, light generated EHPs in the vicinity of the depletion region are separated by the E field resulting in charge accumulating in the P and N sides (that's how \$V_{OC}\$ is developed). A short circuit allows a current to restore charge balance.