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In a photovoltaic cell is basically a pn-junction where free electrons are created by a photoelectric effect.

Now I am interested to understand the electric field inside the depletion layer for various situations:

a) no solar energy input; open circuit
b) no solar energy input; closed circuit
c) solar energy input; open circuit
d) solar energy input; closed circuit

For cases a) and b) I think we have just the same situation as in any pn-junction, as for example described in Chabay Sherwood:

enter image description here

enter image description here

So in this cases we have an (non uniform) electric field as described in the pictures above.

Since the depletion layer is only some hundred atoms thick, to produce a voltage in the magnitude of 1 V, the electric field needs to be pretty high. As calculated by Chabay Sherwood, in the order of 10^7 V/m.

How does the situation change in the case c) and d).

I guess that in c) the net electric field inside the depletion layer will be zero (the electric field from the space charge region and from the charge separation produced by this field an the photoelectric effect cancel). In d) the net field will be small, but in the other direction compared to the picture above (to overcome the internal resistance of the pn-junction).

Is this correct? If so, why? Can you give me some more details (in particular order of magnitudes of case d)) and references?

How does the net electric field depend on the intensity of solar input in the cases c) and d)?

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  • $\begingroup$ First, do NOT forget about holes, particularly in a semiconductor. They are equal partners in the behavior. Second, intensity of the incoming light matters. Making one electron-hole pair every second does nothing (intrinsic e-h generation rates are faster than that). Sending in enough photons to make the steady-state carrier concentrations in excess of the doping levels means there is no more junction. In between? Well, a good semiconductor textbook will have curves (I'm out of the office right now, sorry). $\endgroup$
    – Jon Custer
    Commented Mar 5 at 14:48

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Case c) open circuit: due to the built-in electric field, the light generated electrons flow to the n-side, the holes to the p-side producing a counteracting field which reduces the built-in voltage. This corresponds to an applied forward voltage. This forward voltage produces an internal forward diode current which exactly compensates the internal photoelectric electron-hole generation current, so that the total diode current becomes zero. The depletion layer field is reduced but generally not zero.

Case d) Short circuit: The built-in electric field in the depletion zone remains constant because the "applied voltage" is zero. At least as long as the light intensity isn't too high. The generated electron-hole pairs are separated in the build-in field and produce a photoelectric diode current in reverse direction.

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