Wolfram Alpha:

```Intrinsic and Extrinsic Semiconductors
--------------------------------------

Electron and Hole Densities
---------------------------

Since the E-k curve is parabolic near the band edges we can
use the free electron theory to calculate the density of
electrons near the bottom of the conduction band and holes
near the top of the valence band.

The density of states function is given by:

g(E) = (1/2π2h3)(2m)3/2E1/2

The number of electrons/m3 with energies between E and E + dE
at the bottom of the conduction band is:

n = ∫[1/2π2h3)(2me*)3/2E1/2][1/(1 + exp(Eg/kT)]dE

Where me* is effective mass of the electron.

Set the zeo energy point at Ec.  Replace E by E - Ec

E
^
| ////////// CB
---------- Ec
^
.|........ Ef (Fermi energy)
| Eg
v
---------- Ev
/////////  VB

n = ∫[1/2π2h3)(2me*)3/2(E - Ec)1/2][1/(1 + exp(E - Ef))/kT]dE

Eg for Silicon ~ 0.6eV.  If we evaluate this integral under
the assumption that Ef is at mid gap we get such that
exp(E - Ef)/kT >> 1 we get:

ne = 2(me*KT/2h2π)3/2exp(-(Ec - Ef)/KT)

= Ncexp(-(Ec - Ef)/KT)

Where Nc is the effective density of states in the conduction
band.

W can regard a hole as the absence of an electron.  Therefore,
the Fermi statistic becomes:

F(E)h = 1 - F(E)e

By performing a similar procedure to the above we arrive at:

nh = 2(mh*KT/2h2π)3/2exp(-(Ef - Ev)/KT)

= Nvexp(-(Ec - Ef)/KT)

Where Nv is the effective density of states in the valence
band.

Intrinsic versus Extrinsic
--------------------------

In an intrinsic semiconductor ne = nh.  It is possible to
increase the conductivity of Germanium or Silicon by adding
small amounts of certain elements.  P, As or Sb have 5 valence
electrons (Group V). The impurity atom substitutes for a Ge or
Si atom in the lattice. 4 of the electrons satisfy the covalent
bonding requirements while the 5th is localized but easily
removed.  The energy of this electron is just below the CB edge.
The net result is that these electrons in the CB dominate the
conduction process.  The impurities are refered to as DONORS
and the semiconductor is said to be n-type with electrons as
the majority carriers.

Conversely, there are certain elements that behave in the opposite
way.  B, Ga and In have 3 valence electrons (Group III).  These
atoms trap electrons from the VB in order to fullfill the covalent
bonding requirements.  In the process they leave behind holes.
The energy of these holes is just above the VB.  The net result
is that these holes in the VB dominate the conduction process.
The impurities are refered to as ACCEPTORS and the semiconductor
is said to be p-type with holes as the majority carriers.

////////// CB
---------- Ec
---------- Ed

---------- Ea
---------- Ev
/////////  VB

We can use the formulas from above to calculate the number of
electrons/holes in at Ed and Ea respectively.

nd  = Ndexp(-(Ed - Ef)/KT)

and

na  = Naexp(-(Ef - Ea)/KT)

Where Nd and Na are the density of states at Ed and Ea.

The condition for charge neutrality is:

p + (Nd - nd) = n + na

# of holes + # of ionized donors
= # of electrons + # of ionized acceptors

Fermi Level
-----------

In an intrinsic semiconductor the Fermi level is at midgap.
In n type semiconductors the Fermi level is shifted towards
the CB reflecting the increased probability of finding electrons
in the CB.  In p type semiconductors the Fermi level is shifted
towards the VB reflecting the increased probability of finding
holes in the VB.

Hall Effect
-----------

The Hall effect can be used to measure the majority carrier
concentration.  A current is passed through a semiconductor
and a magnetic field, B, is applied in a perpendicular fashion.
The magnetic field deflects the carriers and causes a build
up of charge on one side.

............ VH ..........
:                          :
:            ^ I           :
:      ------|-------      :
:   l/    A  |       /|    :
:    ---------------  |    :
:...|        ^      | .....:
|        |v     | |
|               | |
B      d| <- Fm   Fe -> | |      B
(out of page) |               | |   (out of page)
|       w       |/
--------------
^
| I

Let n = number of charges/unit volume, A = area of a conductor
and d = length.  The mobile charge, Q, in d is:

Now t = d/v where v is the drift velocity of the charge.
Therefore,

I = Q/t = neAd/(d/v) = neAv or v = I/neA

Now Fm = e(v x B) and FE = eEH = VH/w

At equilibrium Fm = FE.  Therefore,

VH/w = IB/neA

VH = IBw/neA

But A = lw.  Therefore,

VH = IB/nel

We can now experimentally measure VH, B, I and the sample
dimensions to arrive at a value for n.  The sign of VH is
used to establish whether the sample is p or n type.```