SEMICONDUCTOR II: Extrinsic semiconductor statistics
(REF: Sze, McKelvey page 274 , and Kittel, W. Schockley: “Electrons and
holes”)
(1) INTRODUCTION
For extrinsic semiconductors, we can use impurity to control the carrier concentration. The
main purpose of this handout is to summarize how we calculate the free carrier concentrations and the position of Fermi level, and their temperature dependence at various useful
limits, for bulk semiconductors at thermal equilibrium. The properties are directly applicable in practical devices. (In a typical device course, e.g,, ENEE697, we often simplify the
complicated transport processes by assuming that we know the Fermi level everywhere. This
assumption is a big assumption, since devices under bias are not in thermal equilibrium. Fortunately, with extremely efficient electron-phonon and electron-electron scattering, it only
takes a few picoseconds to go from thermal equilibrium at room temperature to a thermal
equilibrium at a higher temperature. Heating up is very fast, but cooling down depends
more on heat sinking: how the excess heat is taken away by the sample mount.)
(2) CHARGE BALANCING EQUATION
For Si, Ge, and GaAs, here we rewrite the basic charge-balancing equation. Based on this
equation, we can numerically calculate the carrier concentrations. Particularly, we will pay
attention to the position of Fermi level as a function of temperature. Notice here that the
degeneracy for an acceptor state is now 4-fold: two spin state in the acceptor, and the
additional factor of two is coming from the degenerate heavy hole and light hole states at
k = 0. The degenerate hole valleys at Γ is a different situation from the degenerate electron
valleys (for Si, along the Σ direction, and for Ge, along the Λ direction), where the degenerate
states are at different k’s.
n + nd + Na = p + pa + Nd ,
(1)
where the definition of the terms are listed below:
1. n is free
electron concentration in the conduction band,
R up
n = Ec dEDe (E)f (E);
2. nd is (bound) electron concentration at the donor cites,
d
nd = 1+ 1 eENd −E
;
f /kT
2
3. Na is the concentration of acceptors, a known value when the sample is prepared;
4. p is free
hole concentration in the valence band,
R
dEDh (E)f (E);
p = Edown
v
5. pa is (bound) hole concentration at the acceptor cites; pa = Na − Na− =
1+ 14 e
Na
(Ed −Ef )/kT
6. Nd is the concentration of donors, a known value when the sample is prepared.
1
;
(3) CALCULATION FOR THE CASE OF DONORS ONLY (similar relations
hold for the case of acceptors only)
The band diagram (energy of electrons versus space) helps define the problem . Since Na = 0
and pa = 0, the charge balancing equation reduces to
n + nd = p + Nd .
(2)
That is, n = Nd+ + p. We can insert the full expression and numerically calculate the
concentration of electrons, holes, and the position of Fermi level. Remember that Ef = Ei
for the intrinsic case is close to the middle of the bandgap? Now the semiconductor has more
electrons, so the Ef will move closer to the conduction band. Based on np = n2i = constant
(well, actually a function of temperature and bandgap), if the electron concentration is
to increase, the hole concentration will be order of magnitudes smaller. For example, for
ni = 1010 /cm3 , with n = 1013 /cm3 , p would be 107 /cm3 , indeed much smaller than n.
(A concentration of 1013 /cm3 is barely measurable by today’s technology.) So, the charge
balancing equation for all practical purposes, can do without p, and we get:
n ∼ Nd+ = Nd −
1+
Nd
1 (Ed −Ef )/kT
e
2
=
1+
Nd
(E
2e f −Ed )/kT
.
(3)
In the non-degenerate regime (we will discuss degenerate case later, since the physics involves
impurity band and a new concept in density of states), we define:
n = Nc e(Ef −Ec )/kT , and
n1 = Nc e(Ed −Ec )/kT .
Using n and n1 to rewrite equation (3), we obtain
n=
2 2
Nd
)n + n − Nd = 0 .
n ⇒ (
1 + 2 n1
n1
n=
−1 ±
q
1 + 4 n21 Nd
4
n1
(4)
s
−n1 n1
8Nd
+
.
=
1+
4
4
n1
(5)
(3.1) Low temperature limit:
If the temperature is low enough, then, eventually,
n1
n1 ≪ Nd ⇒ n ∼
4
s
8Nd
1 q
= √ Nd Nc e(Ed −Ec )/2kT .
n1
2
(6)
At very low temperatures, electrons are not thermally excited from the bound donor to the
conduction band. We call this regime carrier freeze-out. Based on the definition that
n1 ≪ Nd , we can get the critical temperature to be:
T ≪
Ec − Ed
.
k(ln(Nc /Nd ) + 1)
(7)
2
d)
A log(n) plot versus 1/kT will show a slope of − (Ec −E
. This exponential decrease in
2
electron centration will make exponential increase in resistance of a semiconductor resistor.
The common carbon film resistor has such property: at low temperatures the resistance will
increase. Metal film resistors show the opposite temperature dependence, why?
(3.2) High temperature limit:
Since np = n2i , and ni has a strong temperature dependence of e−Eg /2kT , so at high temperature, n ∼ p ∼ ni ≫ Nd .
n ∼ ni =
q
Nc Nv e−Eg /2kT .
(8)
A log(n) plot versus 1/kT will show a slope of − E2g .
(3.3) Intermediate temperatures:
From equation (5), if n1 ≫ Nd , then n ∼
be 40 times of Nd , then when
T >
−n1
4
+
n1
(1
4
+
8Nd
)
2n1
∼ Nd . Suppose we define n1 to
Ec − Ed
,
k(ln(Nc /40 Nd ) + 1)
(9)
the free electron concentration will be totally determined by the doping concentration Nd .
In another words, all the donors are ionized and can contribute free electrons.
(3.4) Ef (T) when there is only donors
We go back to the charge balancing equation that n = Nd+ and plug in the full expressions:
1
Ef = Ed + kT ln[− +
4
s
1
Nd
Ec − Ed
+
exp(
)] .
16 2Nc
kT
(10)
At high temperatures, we are in the intrinsic limit, and Ef is pinned to somewhere near the
bandgap: Ef = Ei .
When temperature increases (to be greater than
Nc
).
kT ln( N
d
Ec −Ed
)
k(ln(Nc /40 Nd )+1)
and n ∼ Nd , Ef = Ec −
As temperature decreases, Ef gradually move from the room temperature value toward the
conduction band minimum.
Nd
At very low temperatures, where freeze-out occurs, Ef ∼ 12 (Ec + Ed ) + 12 kT ln( 2N
), with a
c
2
) 1 . Usually, this temperature is low, somewhere near 1K. Notice
peak at T = ( N4d )2/3 ( 2πh̄
m∗ k e
that at zero temperature, Ef = 21 (Ec + Ed ). And, Ef will peak as the temperature increases
from zero temperature.
(4) COMPENSATION: donors and acceptors both exist
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Any solid has some sort of impurities, defects, vacancies, etc. Such imperfections will always
introduce potential variations which are able to attract or repell electrons. It is not technically demonstrated that a crystal can be grown without impurities and defects. We now
look at the case where both shallow donors and shallow acceptors exist in the semiconductor. There are complications when multiple donor levels and multiple acceptors coexist, and
there is no reason why a donor state must be above an acceptor state.
(4.1) A typical case: Nd ≫ Na (Similar arguments holds for Nd ≪ Na )
Since the Fermi level lies above the middle of the bandgap, all the acceptor states must be
filled, i.e., p and pa are zero. So, n + nd + Na = Nd .
Thus, n + Na ∼ Nd − nd =
Nc e(Ed −Ec )/kT .
1+2e
Nd
(Ef −Ed )/kT
We therefore solve for n from n + Na =
1 n1
Na
n
=− (
+
)+
Nd
2 2Nd Nd
v
u n
u( 1
t 2Nd
Na 2
)
Nd
+
4
=
Nd
,
1+2n/n1
+
(1 −
Nd
,
1+2n/n1
where, again, n1 is defined by
and get:
Na n 1
)
N d Nd
2
.
(11)
(4.1.A) At high temperature, we will eventually go into the intrinsic regime.
(4.1.B) At intermediate temperatures, where n1 ≫ Nd , n ∼ Nd − Na . The free electron
concentration will be simply reduced. The effect of doping n-type impurity is compensated
by p-type impurities. The hole concentration can be calculated from p = n2i /n = n2i /(Nd −
Na ).
n1
(4.1.C) At low temperature where n1 ≪ Nd , n ∼ 2N
(Nd − Na ). We see that the n will
a
drop faster as temperature is lowered, since electrons can now fall into the acceptor levels as
an additional way of removing free electrons.
(4.1.D) At absolute zero temperature, all the acceptor states (below the donor level by
assumption) will be filled. Some donors are filled, and some are empty. Fermi level must be
pinned at the donor level at T = 0.
(4.1.E) Ef (T) = ? However, as temperature starts to rise from absolute zero, electrons at
donor level will be excited to the conduction band. The electrons at the acceptor states will
take too much energy to pump. The Fermi level must then be somewhere in between Ec
and Ed . At very low temperatures, n, p, and pa are negligibly small, so nd + Na = Nd
⇒ nd = Nd − Na .
For convenience, we define:
f (Ed∗ ) =
1
1+
1 (Ed∗ −Ef )/kT
e
2
=
1
(Ed∗ −Ef )/kT
1+e
So, nd = Nd − Na = f (Ed∗ )Nd ⇒ f (Ed∗ ) =
;
Nd −Na
.
Nd
4
(12)
Plugging this relation into equation (12),
we get
Ed∗ − Ef = kT ln(
Nd
Na
− 1) = kT ln(
)
Nd − Na
Nd − Na
⇒ Ef = Ed∗ + kT ln(
⇒ Ed + kT ln(
Nd − Na
Nd − Na
) = Ed + kT [ln(
) − ln(2)]
Na
Na
Nd − Na
)
2Na
(13)
(14)
(15)
1. If Nd = 3Na ⇒ When temperature is lowered, Ef will approach Ed from Ei without
showing a peak. That is, ∂Ef (T )/∂T = 0 at T = 0.
2. If Nd < 3Na ⇒ Ef will approach Ed from Ei , with ∂Ef (T )/∂T < 0 at T = 0.
3. If Nd > 3Na ⇒ When temperature is lowered, Ef will approach Ed from Ei with a
peak. That is, ∂Ef (T )/∂T > 0 at T = 0.
Homework: Use graphic method to work out the above results qualitatively. For example,
find the graphical solution of Ef at low temperatures for (1) n-type only; (2) p-type only;
and and (3) for compensated cases.
5