3
$\begingroup$

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?

$\endgroup$
1
  • $\begingroup$ 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? $\endgroup$
    – AHusain
    Jan 1 '17 at 13:47
3
$\begingroup$

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...)

$\endgroup$
1
  • $\begingroup$ 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! $\endgroup$
    – tau
    Jan 1 '17 at 4:04
2
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.