Electron Hole Pair Production (EHPs)

[CoolDude420]

Homework Statement

I'm taking a module in Solid State Electronics and in the first chapter we went through energy band diagrams, conduction band, valence band, fermi energy level, forbidden gap, etc. Now in the notes it starts to derive some formulae for electron and hole current densities

Jn = q.n.vn
Jn = electron current density
q = electron charge
n = average number of CB electrons per unit volume, basically the electron concentration
vn = speed with each electron in CB moves in response an applied electric field.

Jp = q.p.vp
Jn = hole current density
q = electron charge, same as hole charge
p= average number of VB holes per unit volume, basically the hole concentration
vp = speed with each hole in CB moves in response an applied electric field.

We were explained EHPs. Basically if a substance gets heated enough so that an electron in the VB has enough energy to cross into the CB and contribute to current flow. Now there is an empty state in the VB, another electron in the VB fills this and contributes to current flow. Now my notes say that "holes also contribute to current flow." I don't understand that. Isn't it the electron that was promoted that is going to contribute to current flow? The hole isn't a particle, its just an empty state. If this is true then shouldn't vp = vn, since electron are the carriers of charge in both cases?

Homework Equations

The Attempt at a Solution

 

[Simon Bridge]

The hole is a site of net positive charge... if you've ever seen a row of lights set up to switch on and off so they "chase", then the electrons are the lit bulbs and the holes are the ones that are not lit.

A whole bunch of electrons moving in one direction is the same as a few holes traveling the other way. You'll see this in chasing lights when most of the bulbs are lit at once: it looks like a dark spot moving around the setup.
It is often more convenient to treat the holes as the charge carriers in the semiconductor... since that's what it looks like.

There's a lot of this stuff in solid state - designed to simplify the maths.
eg. the band edges are not smooth in fact - the real potential is bumpy, but it turns out we can pretend the potential is smoothly curves and put the effect of the bumpyness into the "effective mass" of the particle. So the "electron" of solid state band theory is not the same object as in atomic theory.

You can tell if it is the electron or the hole that is carrying charge by exploiting the hall effect.

 

[NLB]

Homework Statement

I'm taking a module in Solid State Electronics and in the first chapter we went through energy band diagrams, conduction band, valence band, fermi energy level, forbidden gap, etc. Now in the notes it starts to derive some formulae for electron and hole current densities

Jn = q.n.vn
Jn = electron current density
q = electron charge
n = average number of CB electrons per unit volume, basically the electron concentration
vn = speed with each electron in CB moves in response an applied electric field.

Jp = q.p.vp
Jn = hole current density
q = electron charge, same as hole charge
p= average number of VB holes per unit volume, basically the hole concentration
vp = speed with each hole in CB moves in response an applied electric field.

We were explained EHPs. Basically if a substance gets heated enough so that an electron in the VB has enough energy to cross into the CB and contribute to current flow. Now there is an empty state in the VB, another electron in the VB fills this and contributes to current flow. Now my notes say that "holes also contribute to current flow." I don't understand that. Isn't it the electron that was promoted that is going to contribute to current flow? The hole isn't a particle, its just an empty state. If this is true then shouldn't vp = vn, since electron are the carriers of charge in both cases?

Homework Equations

The Attempt at a Solution

To understand this, you need to understand the difference between p-type and n-type semiconductors. In an n-type semiconductor, there are lots of free electrons in the conduction band, that are free to flow if an electric force pulls on them. These free electrons are caused by the doping of the semiconductor. For n-type silicon, specifically, the doping is phosphorus, which has five valence band electrons. The silicon crystal only wants four of those electrons, so the fifth electron of that phosphorus atom is pretty much free to wander around with the slightest electric force applied to it. So if an electric field is applied to the bulk of that n-type semiconductor, these free electrons in the conduction band can move, and as a result, there is an electric current.
On the other hand, in a p-type semiconductor, there are lots of "holes". Please understand that these holes are not protons, these hole are not positrons, they are simply holes where an electron should normally be within the silicon crystal lattice. They are like bubbles in the water. These holes are due to the doping of the semiconductor, for p-type specifically it is boron. Boron has only three valance electrons instead of four. Thus there is an electron missing in the crystal lattice, where an electron should be. You can think of this as a hole. If I apply an electric field to the bulk of of that p-type semiconductor, this hole can move around, and as a result, there is an electric current.
As you might expect, the mobility of the hole is not as great as the mobility of the electron, so p-type silicon has different properties than n-type silicon. A bipolar transistor and an MOS transistor are made from a combination of both n-type and p-type doped silicon.