Semiconductor Device Physics
Lecture 5
Dr.
Gaurav
Trivedi
,
EEE Department,
IIT Guwahati
Example: Energy

Band Diagram
For Silicon at 300 K, where is
E
F
if
n
= 10
17
cm
–
3
?
Silicon at 300 K,
n
i
= 10
10
cm
–
3
Consider a Si sample at 300 K doped with 10
16
/cm
3
Boron. What is its resistivity?
N
A
= 10
16
/cm
3
,
N
D
= 0
(
N
A
>>
N
D
p

type)
p
10
16
/cm
3
,
n
10
4
/cm
3
Consider a Si sample
doped with 10
17
cm
–
3
As.
How will its resistivity change when the
temperature is increased from
T
= 300 K to
T
= 400 K?
The temperature dependent factor in
(and therefore
) is
n
.
From the mobility vs. temperature curve for 10
17
cm
–
3
, we
find that
n
decreases from 770 at 300 K to 400 at 400 K.
As a result,
increases
by a factor of
: 770/400 =1.93
1.a.
(4.2)
Calculate the equilibrium hole concentration in silicon at
T
= 400 K if the Fermi
energy level is 0.27
eV
above the valence band energy.
1.b.
(E4.3)
Find the intrinsic carrier concentration in silicon at:
(i)
T
= 200 K and (ii)
T
= 400 K.
1.c.
(4.13)
Silicon at
T
= 300 K contains an acceptor impurity concentration of
N
A
= 10
16
cm
–
3
. Determine the concentration of donor impurity atoms that must
be added so that the silicon is
n

type and the Fermi energy level is 0.20 eV below
the conduction band edge.
What is the hole diffusion coefficient in a sample of silicon at 300 K with
p
= 410 cm
2
/ V
.
s ?
•
Remark:
kT/q
= 25.86 mV
at
room temperature
Consider a sample of Si doped with 10
16
cm
–
3
Boron, with recombination lifetime 1
μ
s
. It is
exposed continuously to light, such that electron

hole pairs are generated throughout the
sample at the rate of 10
20
per cm
3
per second,
i.e.
the
generation rate
G
L
= 10
20
/cm
3
/s
a)
What are
p
0
and
n
0
?
b)
What are
Δ
n
and
Δ
p
?
•
Hint
: In steady

state,
generation rate equals
recombination rate
Consider a sample of Si at 300 K doped with 10
16
cm
–
3
Boron, with recombination lifetime 1
μ
s
. It is exposed continuously to light, such that electron

hole pairs are generated
throughout the sample at the rate of 10
20
per cm
3
per second,
i.e.
the
generation rate
G
L
=
10
20
/cm
3
/s.
c)
What are
p
and
n
?
d)
What are
np
product?
•
Note
: The
np
product can be very
different from
n
i
2
in case of
perturbed/agitated semiconductor
Photoconductor
Photoconductivity
is an optical and electrical phenomenon in
which a material becomes more electrically conductive due to
the absorption of electro

magnetic radiation such as visible
light, ultraviolet light, infrared light, or gamma radiation.
When light is absorbed by a material like semiconductor, the
number of free electrons and holes changes and raises the
electrical conductivity of the semiconductor.
To cause excitation, the light that strikes the semiconductor
must have enough energy to raise electrons across the band
gap.
Photoconductor
Net Recombination Rate (General Case
)
Net
Recombination Rate (General Case)
•
E
T
: energy level of R
–
G cente
r
Chapter 3
For arbitrary injection levels and both carrier types in a non

degenerate semiconductor, the net rate of carrier
recombination is:
where
Net Recombination Rate (General Case)
Continuity Equation
J
N
(
x
)
J
N
(
x
+
dx
)
dx
Area
A
, volume
A.dx
Flow of current
Flow of electron
•
Taylor’s Series Expansion
The Continuity Equations
Continuity Equation
Minority Carrier Diffusion Equation
The minority carrier diffusion equations are derived from the
general continuity equations, and are applicable only for
minority carriers.
Simplifying assumptions:
The electric field is small, such that:
•
For
p

type material
•
For
n

type material
Equilibrium minority carrier concentration
n
0
and
p
0
are
independent of
x
(uniform doping).
Low

level injection conditions prevail.
Minority Carrier Diffusion Equation
Starting with the continuity equation for electrons:
Therefore
Similarly
Carrier Concentration Notation
The subscript “
n
” or “
p
” is now used to explicitly denote
n

type
or
p

type material
.
p
n
is the hole concentration in
n

type material
n
p
is the electron concentration in
p

type material
Thus, the minority carrier diffusion equations are:
Simplifications (Special Cases)
Steady state:
No diffusion current:
No thermal R
–
G:
No other processes:
Minority Carrier Diffusion Length
Similarly,
Consider the special case:
Constant minority

carrier (hole) injection at
x
= 0
Steady state, no light absorption for
x
> 0
The
hole diffusion length
L
P
is defined to be:
Minority Carrier Diffusion Length
The general solution to the equation is:
A
and
B
are constants determined by boundary conditions:
Therefore, the solution is:
•
Physically,
L
P
and
L
N
represent the average
distance that a minority carrier can diffuse
before it recombines with majority a
carrier.
Quasi

Fermi Levels
Whenever
Δ
n
=
Δ
p
≠
0
then
np
≠
n
i
2
and we are at non

equilibrium conditions.
In this situation, now we would like to preserve and use the
relations:
On the other hand, both equations imply
np
=
n
i
2
, which does
not apply anymore.
The solution is to introduce to quasi

Fermi levels
F
N
and
F
P
such that:
•
The quasi

Fermi levels is useful to describe the carrier
concentrations under non

equilibrium conditions
Example: Minority Carrier Diffusion
Length
Given
N
D
=10
16
cm
–
3
,
τ
p
= 10
–
6
s. Calculate
L
P
.
From the plot,
Example: Quasi

Fermi Levels
Consider a Si sample at 300 K with
N
D
= 10
17
cm
–
3
and
Δ
n
=
Δ
p
= 10
14
cm
–
3
.
•
The sample is an
n

type
a)
What are
p
and
n
?
b)
What is the
np
product?
Example: Quasi

Fermi Levels
Consider a Si sample at 300 K with
N
D
= 10
17
cm
–
3
and
Δ
n
=
Δ
p
= 10
14
cm
–
3
.
c)
Find
F
N
and
F
P
?
F
P
0.238 eV
F
N
0.417 eV
E
c
E
v
E
i
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