PN junction shorted and barrier potential

Hi friends
I'm a second year B.Sc student. I've started studying PN junctions and I'm stuck somewhere. Here is my problem,
The two terminals of a PN junction are at different potentials so, if I connect a them with a wire i.e. short them, What would happen? Would the potential difference between the two ends becomes zero or what? As far I'm understanding, there would be a flow of electrons from the P region to the N region because of potential difference. But as soon as the PD becomes less, the barrier potential becomes low and therefore the diffusion process will start, in which electrons would move from N to P and so on....
But, this would constitute a current without any battery, etc.
And why can't we measure inbuilt potential difference of the PN junction in Lab using voltmeters.
I'd greatly appreciate any help.
Thank you

Hello!

First, why you say that you have the PN terminals a different potential? Are they connected to a circuit? If so, you should look to the state of the diode is directly or inversely polarised. And that tells you how is the electron flow. If not, then the potentials in the contacts are undefined, there is no reference, so you can not tell anything but that the system is in equilibrium, so the potential difference should be zero.

If you connect both terminals with a wire, the potential difference between contacts is zero. So no current will go through the diode.

The built-in potential is a different thing, it tells you the difference of the potential energies of the electrons in the P and N sides, and it appears due to the different concentration of "free" electrons on each side of the diode, which leads to a diffusion of electrons and the appearance of net charge in the depletion region. This will produce the creation and electric field which opposed the diffusion movement until equilibrium is achieved and charge net movement is zero. And the built-in potential is the potential associated to that electric field in the junction.

You can not measure the built-in potential because when you connect you voltmeter new built-in potentials will appear between the metal contacts and the p and n semiconductor that will make zero the potential difference in the loop. You can see in this book **broken link removed** in page 79. Moreover you should remember that what defines de voltage in your voltmeter is the difference between Fermi level in the extreme of each region.

Hope it helps,
E.

Thank you so much!! I think I've to revise my basic concepts.

ecomesana said:

You can see in this book **broken link removed** in page 79.

Click to expand...

I have also the same doubt as above. Please can you say which is this book. the link seems broken.

-Devanand T