# Diode contribution in photo-diode's model

A common model used to represent a photo-diode is shown below: It is then said that the current through Rsp (series resistance) is ip+id, where ip is the current proportional to the incident luminous flux and id the dark current (constant at a given temperature).

My question is: In the model it is always shown a diode Dp, that accounts for the rectifying effect of the junction. But it can't be an ideal diode, otherwise ip+idwould flow through it and not through Rps. What is the role of this diode in the model and how does it affect the current through Rsp?

• Ip+Id WILL flow through the diode ... to the extent that the voltage across the diode permits. There is a well known equation relating the current through a diode and the voltage across it; this diode should follow that equation. Any current left over is available via Rsp at the output. – Brian Drummond Aug 23 '17 at 16:14

Your assumption about all the current going through the diode is simply not right. Do this thought experiment. If the voltage across the diode, $V_D$, is fixed, as it often is in photodiode circuits, the current through it must also be fixed. Therefore, a change in $i_p$ must propogate down the circuit.
More generally, there is a fixed relationship between the current going through the diode and the voltage across the diode. Changing $i_p$ does not change that relationship. If the voltage across the diode does change, then if the change in $i_p$ does not correspond exactly to the change in diode current that is dictated, part of the change in $i_p$ needs to go somewhere other than through the diode.