# In this semi-ideal photodiode, what would be the physical interpretation of this current?

A real photodiode can be expressed as a circuit made of several ideal components, it looks something like this: simulate this circuit – Schematic created using CircuitLab

I copied it from here and I added a voltage source in reverse bias because it's a photodiode.

Now, lets lump the current source, the capacitance, and the shunt resistance into a semi-ideal photodiode, it would look something like this simulate this circuit

It was all well and good until I thought: "Wait a second... what is $$\I_D\$$?". It sounds simple until you consider that the series resistance is the resistance of the diode itself, of the crystal, the current flowing through that resistance is also flowing through the entire diode.

Perhaps this image will help: As you can see, the terminals of the diode are directly connected to the voltage source, just like in the circuit diagram, and yet the voltage drop across the series resistance is not $$\V_1\$$. But the most pressing issue here, is what is $$\I_D\$$ in physical terms?.

I imagine all the available electrons flowing from one side of the crystal to the other, and that's $$\I_O\$$, right?.

I'm considering that $$\I_D\$$ is not physical just as the Shunt Resistance or the Junction Capacitance are not physical, but rather arise as properties of the crystal, but I don't know.

I'd also like to understand why the Series Resistance is not in parallel with all the other elements of the real diode.

• The beginning is wrong: you don’t have to add a voltage source $V_1$. Lighting the photodiode leads to the photocurrent $I_1$. The voltage that you may observe is caused by the current that flows through the $D_1$ diode. – user2233709 Feb 26 '19 at 22:48 