Consider the potential distribution across a PN junction. The difference in the Fermi levels between the differently doped semiconductors leads to a 'built-in' voltage across the semiconductor junction.

What happens to the potential distribution if we short-circuit the device i.e. connect the two contacts together?

Given there is a difference in the potential values does a current flow until the potentials at the contacts match up? Then it feels like we're getting current from nothing.

Or does the external circuitry instead not see the potential difference? After all if no current is flowing across the device, how can it suddenly start flowing if we short circuit.

Thanks in advance for any help.

  • $\begingroup$ Since the Fermi levels have been brought into equilibrium (through the built-in potential), there is no potential difference. $\endgroup$ – Jon Custer Nov 4 '20 at 16:57
  • $\begingroup$ I think I see where your coming from. So when I see diagrams showing a potential gradient through a semiconductor junction, I'm to treat that as only affecting the charge carriers in the semiconductor (with the corresponding drift current being balanced out by the diffusion current in equilibrium), whereas the external circuitry won't see any potential difference across the device? $\endgroup$ – Chris Nov 4 '20 at 17:04

Externally short circuiting the pn junction will not force the electric potentials on the p and n side to be equal. The pn junction will remain in equilibrium, with zero current flowing through and the equilibrium built-in potential difference across it.

So how can the potentials on the two sides be different? The answer is in the contacts. If you short the pn junction with a metal wire, the potential difference between the metal and the semiconductor is higher across the p-side metal-semiconductor junction than the n-side junction. This way, there is still no net electric field or potential drop across the wire.


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