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A common model used to represent a photo-diode is shown below: enter image description here

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?

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  • \$\begingroup\$ 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. \$\endgroup\$ – Brian Drummond Aug 23 '17 at 16:14
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The diode in the model provides the regular diode current contribution, i.e. the exponential current flow that you expect if you forward bias any normal diode. Let's call the current through that ideal diode I_Dp. Then the current through the load (including the parasitics Cp, Rpp, Rsp) would be the sum contribution of ip+id+I_Dp, where ip is photocurrent, id is dark current, and I_Dp is the exponential current Is(exp(qV_A/kT)-1). In other words, I_Dp != ip+id.

I think the confusion might come from the somewhat recursive-looking nature of the model. That or the notation of id for dark current.

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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.

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