1
$\begingroup$

I had questions about holes in semiconductor, when i checked the net it is said holes are said to be equivalent positive charge and they say because the hole moves from one place to another when it is occupied by electron the electron leaves holes and etc... But i couldn't find the clear answer in " P type " semi conductor it is written the holes in the valence band increase, my problem is i see that lags and reduces the current not increases it because the holes somehow will attract electrons and get them from conduction band to valence band and since the conduction band is the band that does current , isn't that will reduce current not increase it ?

$\endgroup$
4
  • $\begingroup$ Conduction band electrons can contribute to the electric current but valance band holes can too. In intrinsic material, both contribute equally. In P type material, due to the doping, it is primarily holes in the valance band that contribute. $\endgroup$ Feb 28, 2015 at 13:08
  • $\begingroup$ I pointed out that i know that but i can't understand it since i feel that holes attract electrons to valence band from conduction band hence i feel it reduces it, that is the question. $\endgroup$ Feb 28, 2015 at 13:12
  • 2
    $\begingroup$ Mohamed, as I wrote, the conduction band electrons are not the only contribution to electric current. Reducing the number of conduction band electrons is irrelevant as long as there are far more valence band holes to participate in a current. $\endgroup$ Feb 28, 2015 at 13:36
  • 1
    $\begingroup$ Potential answerers please note: this question was cross-posted on EE.SE. $\endgroup$
    – The Photon
    Mar 1, 2015 at 2:37

2 Answers 2

3
$\begingroup$

In a semiconductor, there are contributions to the total electric current from both electrons in the conduction band and holes in the valence band.

For N-type material, there are far more electrons in the conduction band than there are holes in the valence band. Thus, almost all of the electric current is due to drift of conduction band electrons.

However, in P-type material, there are far more holes in the valence band than there are conduction band electrons and, thus, almost all of the electric current is due to drift of valence band holes.

So, as long as there are plenty of holes to participate in an electric current, the reduction in conduction band electrons is irrelevant.

In summary, there are two components to the electric current through a semiconductor; one due to drift of conduction band electrons and one due to drift of valence band holes. So, it isn't correct to say

the conduction band is the band that does current

$\endgroup$
1
$\begingroup$

because the holes somehow will attract electrons and get them from conduction band to valence band

The reason that a p-type semiconductor is p-type is that it contains acceptor impurities. These are atoms that tend to capture electrons in localized states around their nucleus. For example, group III boron is a typical acceptor impurity in silicon.

Because the captured electrons are in localized states, they aren't free to contribute to conduction.

But, consider if we start with intrinsic material and start to increase the density of acceptor impurities. In intrinsic material, the conduction band has very low occupancy, so the electrons can't be captured from there. Instead, they're captured from the valence band, leaving holes behind.

And indeed, these holes do attract electrons from the conduction band, but to make p-type material you typically add many more (orders of magnitude more) impurities than the intrinsic carrier density, so there simply aren't enough conduction band electrons to fill the acceptor states or to fill the holes resulting from the acceptors attracting valence electrons.

Another way to look at this is to look at Fermi level. As the acceptors capture electrons and create holes, then we know the occupancy of the valence states has decreased. Since the valence state occupancy is reduced we realize the Fermi level must be closer to the valence band edge than in the intrinsic material. And since the Fermi level is closer to the valence band edge, it must be farther from the conduction band edge, resulting in even fewer conduction band electrons than in intrinsic material. In (quasi-)equilibrium, we find a Fermi level that gives a statistical balance between density of valence band holes, occupancy of acceptor states, and density of conduction band electrons. And this gives tells us in what ratio holes and electrons are available to carry current.

$\endgroup$

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.