Transcript 投影片 1

EXAMPLE 3.1
OBJECTIVE
Calculate the probability that an energy state in the conduction band at E = Ec + kT is occupied by
an electron and calculate the thermal equilibrium electron concentration in silicon at T = 300 K.
Assume the Fermi energy is 0.20 eV below the conduction band energy Ec. The value of Nc for
silicon at T = 300 K is Nc = 2.8  1019 cm-3.
 Solution
The probability that an energy state at E = Ec + kT is occupied by an kelectron is given by
f F Ec  kT  
or

1  exp

1
  Ec  kT  E F  
 exp

Ec  kT  E F 
kT



kT

  2.20  0.0259
4
f F Ec  kT   exp

1
.
63

10

0.0259


The electron concentration is given by
or
  Ec  EF 
  0.20 
19
n0  N c exp

2
.
8

10
exp


kT
 0.0259



 Comment


n0 = 1.24  1016 cm-3
The probability of a state being occupied in the conduction band can be quite small, but the
thermal equilibrium value of electron concentration can be a reasonable value since the density of
states is large.
EXAMPLE 3.2
OBJECTIVE
Calculate the probability that an energy state in the valence band at E = Ev - kT is empty of an electron and calculate the thermal-equilibrium hole
concentration in silicon at T = 350 K.
Assume the Fermi energy is 0.25 eV above the valence-band energy. The value of Nv for silicon at T = 300 K is Nv =1.04  1019 cm-3.
 Solution
The parameter values at T = 350 K are found as

Nv  1.04 10
and
19

3/ 2
 350 


 300 
 1.311019 cm-3
 350
kT  0.0259
  0.0302eV
300


The probability that an energy state at E = Ev – kT is empty is given by
1  f F Ev  kT   1 
or

1  exp

1
  EF  Ev  kT  
 exp

Ev  kT  EF 
kT



kT

  0.25  0.0302
5
1  f F Ev  kT   exp

9
.
34

10

0.0302


The hole concentration is
  EF  Ev 
  0.25 
19
p0  N v exp

1
.
31

10
exp


kT
 0.0302




or
 Comment

p0 = 3.33  1015 cm-3
The parameter values at any temperature can easily be found using the 300 K values and the temperature dependence of the parameter.
EXAMPLE 3.3
OBJECTIVE
Calculate the intrinsic carrier concentration in silicon at T = 350 K and at T = 400 K.
The values of Nc and Nv vary as T3/2. As a first approximation, neglect any variation of bandgap energy with temperature.
Assume that the bandgap energy of silicon is 1.12 eV. The value of kT at 350 K is
 350
kT  0.0259
  0.0302eV
 300
And the value of kT at 400 K is
 400
kT  0.0259
  0.0345eV
 300 
Solution
Using Equation (3.23), we find for T = 350 K


ni2  2.8 1019 1.041019
so that
 350
  1.12 
22
exp



  3.6210
 300
 0.0302

3
ni (350 K) = 1.90  1011 cm-3
For T = 400 K, we find


ni2  2.8 1019 1.041019
so that
 Comment
 400
  1.12 
24

 exp
  5.5010
 300 
 0.0345

3
ni (400 K) = 2.34  1012 cm-3
We can note from this example that the intrinsic carrier concentration increases by approximately one order of magnitude for
each increase in temperature of 50C.
EXAMPLE 3.4
OBJECTIVE
Determine the position of the intrinsic Fermi level with respect to the center of the
bandgap in silicon at T = 300 K.
The density of states effective mass of the electron is mn* = 1.08m0 and that of the
hole is mp* = 0.56m0.
 Solution
The intrinsic Fermi level with respect to the center of the bandgap is
E Fi  Emidgap
or
*

m
3
p
 kT ln *
m
4
 n
 3
  0.0259 ln 0.56 
 4
 1.08 

EFi  Emidgap = -12.8 meV
 Comment
For silicon, the intrinsic Fermi level is 12.8 meV below the midgap energy. If we
compare 12.8 meV to 560 meV, which is one-half of the bandgap energy of silicon, we
can, in many applications, simply approximate the intrinsic Fermi level to be in the
center of the bandgap.
EXAMPLE 3.5
OBJECTIVE
Calculate the thermal equilibrium concentrations of electrons and holes for a given
Fermi energy.
Consider silicon at T = 300 K. Assume that the Fermi level is 0.25 eV above the
valence-band energy. If we assume the bandgap energy of silicon is 1.12 eV, then the
Fermi energy will be 0.87 below the conduction-band energy.
 Solution
Using Equation (3.19), we can write
  0.25 
14
-3
p0  1.041019 exp
  6.6810 cm
 0.0259


Using Equation (3.11),we can write

n0  2.8 10
19
  0.87 
4
-3
exp

7
.
23

10
cm

 0.0259

 Comment
The change in the Fermi level is a function of the donor and acceptor impurity
concentrations that are added to the semiconductor. This example shows that electron
and hole concentrations.
EXAMPLE 3.6
OBJECTIVE
Determine the hole concentration in silicon at T = 300 K given the electron
concentration.
Assume the electron concentration is n0 = 1  1016 cm-3.
 Solution
From Equation (3.43), we can write

or
ni2
1.5 1010
p0 

n0
11016

2
p0 = 2.25  104 cm-3
 Comment
As we have seen previously, the concentrations of electrons and holes can vary by
orders of magnitude. The charge carrier that has the greater concentration is referred to
as the majority carrier, and the charge carrier that has the lesser concentration is referred
to as the minority carrier. In this example, the electron is the majority carrier and the
hole is the minority carrier.
The fundamental semiconductor equation given by Equation (3.43) will prove to be
extremely useful throughout the remainder of the text.
EXAMPLE 3.7
OBJECTIVE
Determine the electron concentration using the Fermi-Dirac integral.
Assume that F = 3, which means that the Fermi energy is above the conductionband energy by approximately 77.7 meV at 300K.
 Solution
Equation (3.46) can be written as
n0 
2

N c F1/ 2  F 
From Figure 3.10, the Fermi-Dirac integral has a value of F1/2(3)  4. Then
n0 
2

2.8 10 4  1.2610
19
20
cm-3
 Comment
Note that if we had used Equation (3.11), the thermal-equilibrium value of the electron
concentration would be n0 = 5.62  1020 cm-3, which is a factor of approximately 4.5 too
large. When the Fermi level is in the conduction band, the Boltzmann approximation is
no longer valid so that Equation (3.11) is no longer valid.
EXAMPLE 3.8
OBJECTIVE
Determine the fraction of total electrons still in the donor states at T = 300 K.
Assume silicon is doped with phosphorus to a concentration of Nd = 5  1015 cm-3.
 Solution
Using Equation (3.55), we fine
nd

nd  n0
1
2.8 10
  0.045
1
exp

15
2 5 10
 0.0259 
19

 0.00203 0.203%

 Comment
This example shows that the vast majority of the donor electrons are in the conduction
band and, in this case, only approximately 0.2 percent of the donor electrons are still in
the donor states. For this reason, at room temperature, we can say that the donor states
are completely ionized.
EXAMPLE 3.9
OBJECTIVE
Determine the temperature at which 90 percent of acceptor atoms are ionized.
Consider p-type silicon doped with boron at a concentration of Na = 1016 cm-3.
 Solution
Find the ratio of holes in the acceptor state to the total number of holes in the valence
band plus acceptor state. Taking into account the Boltzmann approximation and
assuming the degeneracy factor is g = 4, we write
pa

p0  pa
For 90 percent ionization,
pa
 0.1 0 
p0  pa
1
Nv
  Ea  Ev  
1
exp

4Na
kT


1


 T 
1.0 4  1 0


3 0 0

1
4 1 016
19


3/ 2




 0.0 4 5

ex p
T 
 0.0 2 5 9



 3 0 0 


Using trial and error, we find that T = 193 K.
 Comment
This example shows that at approximately 100C below room temperature, we still
have 90 percent of the acceptor atoms ionized; in other words, 90 percent of the
acceptor atoms have “donated” a hole to the valence band.
EXAMPLE 3.10
OBJECTIVE
Determine the thermal-equilibrium electron and hole concentrations for a given doping
concentration.
Consider silicon at 300 K doped with phosphorus impurity atoms at a concentration
of Nd = 2  1016 cm-3. Assume Na = 0.
 Solution
From Equation (3.60), the majority-carrier electron concentration is
n0 
2 10
2
16

 2 10


2

16
2

10


1
.
5

10




2
 2 1016 cm3
The minority-carrier hole concentration is found as
p0 


10 2
n
1.5 10

n0
2 1016
2
i
 1.13104 cm-3
 Comment
In this example, Nd >> ni so that the thermal-equilibrium majority-carrier electron
concentration is essentially equal to the donor impurity concentration. This example
illustrates the fact that we can control the concentration of majority carriers and thus
the conductivity of the semiconductor by controlling the concentration of impurity
atoms added to the semiconductor material.
EXAMPLE 3.11
OBJECTIVE
Calculate the thermal-equilibrium electron and hole concentrations in a germanium
sample for a given doping density.
Consider a germanium sample at T = 300 K in which Nd = 5  1013 cm-3 and Na = 0.
Assume that ni = 2.4  1013 cm-3.
 Solution
Again, from Equation (3.60), the majority-carrier electron concentration is
 5 10
5 10
n0 
 
2
 2
13
13
2

  2.4 1013

The minority-carrier hole concentration is



2
 5.97 1013 cm3

13 2
n
2.4 10
12
-3
p0 


9
.
65

10
cm
n0 5.971013
2
i
 Comment
If the donor impurity concentration is not too different in magnitude from the intrinsic
carrier concentration, then the thermal-equilibrium majority carrier electron
concentration is influenced by the intrinsic concentration.
EXAMPLE 3.12
OBJECTIVE
Calculate the thermal-equilibrium electron and hole concentrations in a compensated ptype semiconductor.
Consider a silicon semiconductor at T = 300 K in which the impurity doping
concentrations Na = 2  1016 cm-3 and Nd = 5  1015 cm-3.
 Solution
Since Na > Nd, the compensated semiconductor is p type and the thermal-equilibrium
majority-carrier hole concentration is given by Equation (3.62), so that
p0 
2 10
16
 5 10
2
15

 2 1016  5 1015


2

2

10


1
.
5

10




2
Which yields
p0 = 1.5  1016 cm-3
The minority-carrier electron concentration is found to be

Comment

ni2
1.5 1010
n0 

p0
1.5 1016
2
 1.5 104 cm-3
If we assume complete ionization and if (Na  Nd) >> ni then .the majority-carrier hole
concentration is, to a very good approximation, just the difference between the acceptor
and donor impurity concentrations.
EXAMPLE 3.13
OBJECTIVE
Determine the required impurity doping concentration in a semiconductor material.
A silicon power device with n-type material is to be operated at T = 475 K. At this temperature, the intrinsic carrier concentration must
contribute no more than 3 percent of the total electron concentration. Determine the minimum doping concentration required to meet this
specification. (As a first approximation, neglect the variation of Eg with temperature.)
 Solution
At T = 475 K, the intrinsic carrier concentration is found from
  Eg 
ni2  N c N v exp
 kT 



or

 2.8  10
19
  1.12  300 
 475

 exp


 300 
 0.0259 475
1.0410 
19
ni2 = 1.59  1027
3
which yields
ni =3.99  1013 cm-3
For the intrinsic carrier concentration to contribute no more than 3 percent of the total electron concentration, we set n0 =
1.03Nd.
From Equation (3.60), we have
n0
or
1.0 3N d
which yields
 Comment
Nd


2
Nd


2
 Nd 


 2 
2
 Nd 


 2 
 ni2
2

 3.9 9  1 013

2
Nd = 2.27  1014 cm-3
If the temperature remains less than or equal to 475 K, or if the impurity doping concentration is greater than 2.27  1014 cm-3,
then the intrinsic carrier concentration will contribute less than 3 percent of the total electron concentration.
EXAMPLE 3.14
OBJECTIVE
Determine the required donor impurity concentration to obtain a specified Fermi energy.
Silicon at T = 300 K contains an acceptor impurity concentration of Na = 1016 cm-3.
Determine the concentration of donor impurity atoms that must be added so that the silicon is n
type and the Fermi energy is 0.20 eV below the conduction-band edge.
 Solution
From Equation (3.64), we have
which can be rewritten as

Nc
Ec  EF  kT ln
N N
a
 d




  Ec  EF 
N d  N a  N c exp

kT


Then
  0.20 
16
3
N d  N a  2.8 1019 exp

1
.
24

10
cm
 0.0259

or
 Comment
Nd = 1.24  1016 + Nd = 2.24  1016 cm-3
A compensated semiconductor can be fabricated to provide a specific Fermi energy level.
EXAMPLE 3.15
OBJECTIVE
Determine the Fermi-level position and the maximum doping at which the Boltzmann
approximation is still valid.
Consider p-type silicon, at T = 300 K, doped with boron. We can assume that the limit of the
Boltzmann approximation occurs when EF  Ev =3 kT (See Section 3.1.2.)
 Solution
From Table 3.3, we find the ionization energy is Ea  Ev = 0.045 eV for boron in silicon. If we
assume that EFi >> Emidgap, then from Equation (3.68), the position of the Fermi level at the
maximum doping is given by
EFi  EF 
Eg
2
 Ea  Ev   EF
 Na 
 Ea   kT ln
 n 

 i 
or
 Na
 ni
0.5 6  0.0 45 30.0 25 9  0.4 37  0.0 25 9 ln

We can then solve for the doping as
 Comment




 0.437 
17
3
N a  ni exp
  3.2 10 cm
 0.0259
If the acceptor (or donor) concentration in silicon is greater than approximately 3  1017 cm-3,
then the Boltzmann approximation of the distribution function becomes less valid and the
equations for the Fermi-level position are no longer quite as accurate.