1
\$\begingroup\$

I am having a hard time imagining the flow of current through the p-n junction. When a diode is forward biased, the holes on one side of the barrier and electrons from the other side try to cross the barrier. But as soon as they do, shouldn't the electrons fall into holes and fill them, making a neutral entity just like the barrier in the first place? How does the current flow then? How do the holes and electrons cross over without fusing into each other?

\$\endgroup\$
3
\$\begingroup\$

An electron "falling into a hole" generates two ions, an anion in the P-type material and a cation in the N-type material. The conduction occurs when the ion forces an electron out/draws an electron in with the assistance of an external electromotive force in in order to restore the ionically neutral rest state of the semiconductor material.

\$\endgroup\$
1
\$\begingroup\$

It is assumed that you have a voltage source that tries to keep the same voltage drop over the diode. In other words, the voltage source continues to shove electrons and holes into the diode.

Please note that electrons going in one direction is the equivalent of holes going in the other direction (a hole is created by sucking out an electron), so there is indeed a current flowing.

\$\endgroup\$
1
\$\begingroup\$

When you apply a bias, majority carriers cross the junction due to a reduction in the potential barrier. On reaching the other side, they become minority carriers.

At near the depletion region edges, the minority carriers are in excess. It is true that the minority carriers then begin to recombine as they travel along the semiconductor. This leads to a concentration gradient of the minority carriers between the depletion edge and the end-contact.

With this concentration gradient, the minority carriers begin to diffuse along the semiconductor towards the end contact, creating a diffusion current. I believe this is where your question comes in. This diffusion current is approximately 0 at the end contact.

What happens is, the current throughout the semiconductor must be constant and equal to the current flowing into it. The decrease in diffusion current is therefore made up for by a rise in majority carrier drift current.

So yes, they do "fuse into each other". But because the majority carrier concentration is so much greater than the minority carrier concentration, the decrease in minority carrier current near the end contacts is made up for by the majority carriers.

I hope this is sufficiently clear.

\$\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.