 | While the picture may look complicated, it is actually just a means to keep track
of the number of charges in the semiconductor. Since there is an over-all charge neutrality,
we have |
| |
S
pos. charges = S neg. charges |
|
|  | This
equation can be used to calculate the exact position of the Fermi energy (as we will see).
Since almost everything else follows from the Fermi energy, let's look at this in some detail. |
 |
But wait – why are there positive
charges at all? Aren't we just discussing the occupation of electronic
levels/states? After all, electrons are negatively charged. And by which rule is it that some
states in the above figure count the one way (negative), but some the other way (positive)? |
|  | This
rule is simple: Everything counts relative to the individual behavior at zero kelvin
(*) where the conduction band is completely empty, the valence band is completely filled,
all donor states are occupied, and all acceptor states are empty. |
|  |
Of course, if both donor and acceptor states
were present together in a material, then at zero kelvin the electrons would not be distributed
like that. And that's why it is emphasized that one has to look at the individual behavior
of these states at zero kelvin, because charge neutrality is defined relative to the
latter (for which it obviously holds). |
|
 |
(* Yes: The units of measurement are spelled with small letters
– because it's only the natural persons' names that are spelled with capital letters.) |
 |
First, let's count the negative
charges which, following the above rule, means to look for those places where at zero
kelvin no electrons are to be found. |
|  | There
are, first, the electrons in the conduction band.
Their density, as spelled out before, was |
| | ne = Neffe
· exp | æ
ç è | – |
EC – EF
kT | ö ÷ ø | |
|
|  | More
generally and more correctly, however, we have |
|
| ne
= Neffe · f(EC, EF
,T) | | |
 | The Fermi energy EF is now included as a
variable in the Fermi function, because the density of electrons depends on its
precise value which we do not yet know. In this formulation the electron density comes out
always correct, no matter where the Fermi energy is positioned
in the band gap. |
|  |
Then there are, second, the negatively charged acceptor atoms
. Their density NA– is given by the density of
acceptor states (which is just the doping density N
A ) times the probability that the states are
occupied, and that is given by the value of the Fermi distribution at the energy of the acceptor
state, f(EA, EF,T). We have accordingly
|
| | |
|  | Note: The
Fermi distribution in this case should be slightly modified to be totally correct. The difference
to the straight-forward formulation chosen here is slight, however, so we will keep it in
this simple way. More about thisin an
advanced module. |
 | Now
let's count the positive charges which, following the
above rule, means to look for those places that at zero kelvin are completely
filled with electrons. |
|  | First , we have the
holes in the valence band. Their number is given by the
number of electrons that do not occupy states in the
valence band; in other words we have to multiply the effective density of states with the
probability that the state is not occupied. |
|  | The probability that a state
is not occupied is just 1
minus the probability that it is occupied, or simply 1 – f(E,
EF, T). This gives the density of holes to |
| |
nh = Neff
h · | æ
è | 1 – f(EV
, EF , T) |
ö ø | | |
|  | Second
, we have the positively charged donors, i.e.
the donor atoms that lost an electron. Their density ND+
is equal to the density of states at the donor level (which is again identical to the density
of the donors themselves) times the probability that the level is not
occupied; so, we have |
| | ND+ | =
| ND |
æ è | 1 – f(ED, E
F, T) | ö ø |
| |
 | Charge neutrality thus demands |
|
Neffe
· f(EC , EF, T) + NA·
f (EA, EF, T) = Neff
h · | æ è
| 1 – f(EV, EF,
T) | ö ø | + ND · | æ è | 1 – f(ED,
EF, T) | ö ø
| | |
|
 |
If we insert the expression for the Fermi distribution |
| | f( En,
EF, T) = | 1 | exp | æ è | En
– EF kT | ö ø
| + 1 | |
|
|  | where
En stands for EC,V,D,A , we have, for a
given material with a given doping, one equation for the
one unknown quantity EF ! |
 |
Solving this equation for any given semiconductor
and any density of (ideal) donors and acceptors will not only give us the exact value of the
Fermi energy EF for any temperature T, i.e. EF
(T), but all the carrier densities as specified
above. |
|  | Unfortunately, this is a messy transcendental equation – it has no
direct solution that we can write down. |
|
 |
Before the advent of cheap computers, this was a problem –
you had to do case studies and use approximations: High or low temperatures, only donors,
only acceptors and so on. This is still very useful, because it helps to understand the essentials. |
 |
However, here we could use a program (written by J. Carstensen) that solves the transcendental equation and
provides all functions and numbers required. |
|  | Unfortunately, the relevant JAVA applet
doesn't work anymore, so the link to this illustration module is deactivated. |
| Therefore, here we show only a screen shot
with the result for the typical case of Si with |
|
 |
1. An acceptor density
of 1015 cm–3 (red line). |
|  | 2.
A donor density of 1017 cm–3
(blue line). |
| The equations used for charge neutrality also make possible
to deduce an extremely important relation, the mass action law (for electrons and holes), as follows: |
|  | First
we consider the product |
| | ne · nh |
= | N effe ·
f(EC, EF, T) · Neffh ·
[1 – f(EV, E F,T)] | |
|
|  |
Then we insert the formula for the Fermi distribution and note that |
| |
1 – f(EV, T)
= 1 – | 1
| =
| 1 | exp
| æ
è | EV
– EF kT |
ö ø
| + 1 |
1 + exp |
æ è | EF
– EV kT
| ö
ø | | |
|  |
Using the Boltzmann approximation for the
Fermi distribution, we get the famous and very important mass action law (try
it yourself!): |
| |
ne
· nh = Neffe · Neffh
· exp | æ
ç è | – | EC – E V kT |
ö ÷ ø | =
ni 2 | | |
 |
We are now in a position to calculate the density
of majority and minority carriers with very good precision if we use the complete formula,
and with a sufficient precision for the appropriate temperature range where we can use the
following very simple relations: |
| |
nmaj |
= | Ndop | | | |
| | |
| | |
n min | =
| n i2 nmaj | = | ni2
Ndop | |
|
 | We know that the conductivity s
of the semiconductor is given by |
| | s = e ·
(n e · µe + nh·
µh) | | |
|  |
With µ = mobility of the carriers. |
| We now have the densities; obviously we now
have to consider the mobility of the carriers. |
| |
| Mobility |
| |
 |
Finding simple relations for the mobility of the
carriers is just not possible. Calculating mobilities
from basic material properties is a far-fetched task, much more complicated and involved than
the carrier density business. |
|  | However, the carrier densities (and their redistribution in contacts
and electrical fields) is far more important for a basic understanding of semiconductors and
devices than the carrier mobility. |
|  |
At this point we will therefore only give a cursory view of
the essentials relating to the mobility of carriers. |
 | The mobility µ
of a carrier in an operational sense is defined as the proportionality constant between the
average drift velocity vD of an ensemble
of carriers in the presence of an electrical field E: |
| | |
|  |
The average (absolute) velocity v of a carrier
and its drift velocity vD must not be confused; for a detailed discussion
consult the links to two basic modules: |
|
 |
The simple linear relationship between the drift velocity and
the electrical field as a driving force is pretty
universal – it is the requirement for ohmic behavior (look it up in the link) –
but not always obeyed. In particular, the drift velocity may saturate
at high field strengths, i.e. increasing the field strength does not increase vD
anymore. We come back to that later. |
 | Here we only want to get a feeling for the order of magnitudes of the mobilities
and the major factors determining these numbers. |
|
 |
As we (should) know, the prime factor influencing
mobility is the average time between scattering processes. In fact, the mobility µ
may be written as |
| |
|
|  | With
ts = mean scattering time. We thus
have to look at the major scattering processes in semiconductors. |
 |
There are three important mechanisms: |
| The first
(and least important one) is scattering at crystal defects
like dislocations or (unwanted) impurity atoms. |
|
 |
Since we consider only "perfect" semiconductors at
this point, and since most economically important semiconductors are pretty perfect in this
respect, we do not have to look into this mechanism here. |
|  |
However, we have to keep an open mind because semiconductors
with a high density of lattice defects are coming into their own (e.g. GaN or CuInSe2
) and we should be aware that the mobilities in these semiconductors might be impaired
by these defects. |
 | Second, we have the scattering
at wanted impurity atoms, in other word at the (charged) dopant
atoms. |
|  |
This is a major scattering process which leads to decreasing mobilities with increasing doping
density. The relation, however, is non-linear and the influence is most pronounced for larger
doping densities, say beyond 1017 cm–3 for Si. |
|  | Examples for the relation between doping
and mobilities can be found in the illustration. |
|
 |
As a rule of thumb for Si, increasing the doping level
by 3 orders of magnitude starting at about about 1015 cm–3
will decrease the mobility by one order of magnitude, so the change in conductivity will be
about only two orders of magnitude instead of three if only the carrier density would change. |
|  | The
scattering at charged dopant atoms decreases with increasing
temperature. (This results mainly from the scattering being Rutherford scattering via Coulomb
interaction, for which the scattering cross section reduces with increasing temperature.)
|
 |
Third, we have
scattering at phonons – the other important process.
Phonons are the (quantum-mechanical) "particles" corresponding
to the thermally stimulated lattice vibrations and thus strongly depend on temperature. |
|  | This
part scales with the density of phonons, i.e. it increases
with increasing temperature (with about T 3/2). It is thus not surprising
that it dominates at high temperatures (while scattering at dopant atoms may dominate at low
temperatures). |
 | Scattering
at phonons and dopant atoms together essentially dominate the mobilities. |
|  | The different
and opposing temperature dependencies almost cancel each other to a certain extent for medium
to high doping levels (see the illustration),
again a very beneficial feature for technical applications where one doesn't want strongly
temperature dependent device properties. |
|
| |
© H. Föll (Semiconductors - Script)