# Why don't solar panels eventually reach electrical charge equilibrium?

Here is my understanding: Solar panels work based on electrons from an N-type semiconductor losing electrons that travel from a metal grid, through whatever you're powering, to the back grid where they fill in the holes in a P-type semiconductor.

What prevents the electrons in the N-type semiconductor from just filling in the holes of the P-type directly instead of travelling through the front grid?

Won't this process eventually stop when all of the holes in the P-type semiconductor are full of electrons from the N-type?

• You are thinking about going all the way to the ground state. So notice the temperature dependence in Mihai's answer. What happens as you cool down? Jan 1 '17 at 13:47

Your understanding is not wrong - except for the bit where the electrons end up with a greater potential. This means that when you close the circuit, they will want to travel around the circuit - and at that point they are back in the N-doped part of the junction. This is what drives the current. When a solar panel is not part of a circuit, what you describe would happen (barring some diffusion back in the other direction...)

• why dont they stop at the p-type? why do they have to travel through the circuit first instead of going directly to the p-type? thanks!
– tau
Jan 1 '17 at 4:04

Electrons migrate from $N$ side to $P$ side until a equilibrium is reached.

If we consider the width of each side like $-w_p$ as the width of $P$ side, $w_n$ as the width of $N$ side, with $0$ being the coordinate of the contact between $P$ and $N$ sides.
The length of the $PN$ diode will be $[-w_p, w_n ]$

If we use the charge depletion approximation, a diode will have two types of regions:

(1) the space charge region - SCR which forms in between the P and N sides; the length of this region is measured as $[-x_p, x_n]$

(2) two quasi neutral regions - QNR; one in between the ohmic contact of the P side and the

The concentration of excess populations will be : $$n'(-x_p) = \left( \frac{n_i^2}{N_{Ap}} \right) (e ^ {\frac{qv_{AB}}{k_B T}} - 1)$$

$$p'(x_n) = \left( \frac{n_i^2}{N_{Dn}} \right) (e ^ {\frac{qv_{AB}}{k_B T}} - 1)$$

In a photovoltaic cell electrons from SCR are freed by photons and separated immediately by the electric field present in the SCR, holes to $P$ and electrons to $N$ side. A potential will appear between the P and the N side : $v_{AB}$. The cell is engineered such that for the direct current $I_d(v_{AB})$ for the $v_{AB}$ is as little as possible (while balancing other factors too, etc).

What prevents the electrons in the N-type semiconductor from just filling in the holes of the P-type directly instead of travelling through the front grid?

The field created near SCR is opposite to the field created by the free electrons in the $N$ side forcing them to stay in the $N$ side, in a so called equilibrium (and also the field in SCR is forcing holes to stay in the $P$ region). You can compare this situation with a virtual capacitor.
Only the excess carriers will flow throw the SCR from $N$ side to $P$ side.
SCR length will diminish with the increase of $v_{AB}$ under the action of light. This means that fewer and fewer electrons/holes can be freed from the SCR with the increase of $v_{AB}$ due to photoelectric effect.

Some graphics could help too and so I recommend you to read the following OpenCourse lecture from MIT 6-012 Microelectronic Devices and Circuits and see lecture 6

Regarding $I_d$ see Losses occurring in solar cells