2
\$\begingroup\$

In a LED (light emitting diode,) a forward bias allows electrons to fill the vacancies in the p-type region, recombining with holes to emit a certain frequency of light.

After a set amount of recombination occurs, I visualize that all of the recombined electrons are at a lower energy, and other electrons that are getting pushed by the voltage difference flow directly towards the other end of the diode into a wire.

Do these electrons actually push out recombined electrons, and recombine with the created hole again? Isn't a lot of energy lost when achieving the activation energy of an electron (removing it from the valence band,) reducing the length of emitted wavelength?

Thank you for clearing up my confusion about electric current.

\$\endgroup\$
2
\$\begingroup\$

Before you drive a current through it, a semiconductor’s population of holes and electrons is (roughly) in balance between thermal excitation that supplies them, and recombination that removes them.

When you drive currents through the semiconductor, those processes are still working. Thermal excitation will continue to create holes, which the current can now fill to create light.

\$\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.