Numericals on semiconductors
1.
Calculate
the
total
number
of
energy
states
per
unit
volume,
in
silicon,
between
the
lowest
level
in
the
conduction
band
and
a
level
kT
above
this
level,
at
T
=
300
K
.
The
effective
mass
of
the
electron
in
the
conduction
band
is
1
.
08
times
that
of
a
free

electron
.
Given (E

Ec) = kT =
1.38
10

23
J/K
300
The number of available states between E
c
and (E
c
+ kT) is given by
Numericals on semiconductors
2
Calculate
the
probability
that
an
energy
level
(a)
kT
(b)
3
kT
(c)
10
kT
above
the
fermi

level
is
occupied
by
an
electron
.
Probability that an energy level E is occupied is given by f(E) =
For (E

E
F
) = kT , f(E) =
For (E

E
F
) = 3kT, f(E) =
For (E

E
F
) = 10kT, f(E) =
Numericals on semiconductors
3
The fermi

level in a semiconductor is 0.35 eV above the valence band.
What is the probability of non

occupation of an energy state
at the top
of
the valence band, at (i) 300 K (ii) 400 K ?
The probability that an energy state in the valence band is not occupied is
1

f(E) =
=
1

f(E)
Alternate method: for E
F

E
V
> kT
(i) T=300K
(ii) T=400K
1

f(E)
Numericals on semiconductors
4 The fermi

level in a semiconductor is 0.35 eV above the valence band.
What is the probability of non

occupation of an energy state at a level kT
below
the
top of the valence band, at (i) 300 K (ii) 400 K?
The probability that an energy state in the valence band is not occupied is
1

f(E) =
for E
F

E
> kT
(i) T=300K
(ii) T=400K
1

f(E)
Note (E

E
F
) is

ve
Numericals on semiconductors
5
For
copper
at
1000
K
(a)
find
the
energy
at
which
the
probability
P(E)
that
a
conduction
electron
state
will
be
occupied
is
90
%
.
(b)
For
this
energy,
what
is
the
n(E),
the
distribution
in
energy
of
the
available
state?
(c)
for
the
same
energy
what
is
n
0
(
E)
the
distribution
in
energy
of
the
occupied
sates?
The
Fermi
energy
is
7
.
06
eV
.
The fermi factor f(E) =
= 0.90
Numericals on semiconductors
5
Density of available state n (E)
contd
Numericals on semiconductors
5
contd
i.e n
o
(E) = n (E) f(E)
The density of occupied states is =
(The density of states at an energy E )
( probability of
occupation of the state E)
Numericals on semiconductors
6
An
intrinsic
semiconductor
has
energy
gap
of
(a)
0
.
7
eV
(b)
0
.
4
eV
.
Calculate
the
probability
of
occupation
of
the
lowest
level
in
the
conduction
band
at
(i)
0
C
(ii)
50
C
(iii)
100
C
.
f(E) =
a) (i)
(ii)
(iii)
Numericals on semiconductors
6
An
intrinsic
semiconductor
has
energy
gap
of
(a)
0
.
7
eV
(b)
0
.
4
eV
.
Calculate
the
probability
of
occupation
of
the
lowest
level
in
the
conduction
band
at
(i)
0
C
(ii)
50
C
(iii)
100
C
.
f(E) =
b) (i)
(ii)
(iii)
Numericals on semiconductors
7
The
effective
mass
of
hole
and
electron
in
GaAs
are
respectively
0
.
48
and
0
.
067
times
the
free
electron
mass
.
The
band
gap
energy
is
1
.
43
eV
.
How
much
above
is
its
fermi

level
from
the
top
of
the
valence
band
at
300
K?
Fermi energy in an Intrinsic
semiconductor is
The fermi level is 0.75eV above the top of the VB
Write
Numericals on semiconductors
8
Pure
silicon
at
300
K
has
electron
and
hole
density
each
equal
to
1
.
5
10
16
m

3
.
One
of
every
1
.
0
10
7
atoms
is
replaced
by
a
phosphorous
atom
.
(a)
What
charge
carrier
density
will
the
phosphorous
add?
Assume
that
all
the
donor
electrons
are
in
the
conduction
band
.
(b)
Find
the
ratio
of
the
charge
carrier
density
in
the
doped
silicon
to
that
for
the
pure
silicon
.
Given
:
density
of
silicon
=
2330
kg
m

3
;
Molar
mass
of
silicon
=
28
.
1
g/mol
;
Avogadro
constant
N
A
=
6
.
02
10
23
mol

3
.
No of Si atoms per unit vol =
Carriers density added by P =
Ratio of carrier density in
doped Si to pure Si =
Numericals on semiconductors
9
The
effective
mass
of
the
conduction
electron
in
Si
is
0
.
31
times
the
free
electron
mass
.
Find
the
conduction
electron
density
at
300
K,
assuming
that
the
Fermi
level
lies
exactly
at
the
centre
of
the
energy
band
gap
(=
1
.
11
eV)
.
Electron
concentration
in
CB
is
=
Numericals on semiconductors
10
In
intrinsic
GaAs,
the
electron
and
hole
mobilities
are
0
.
85
and
0
.
04
m
2
V

1
s

1
respectively
and
the
effective
masses
of
electron
and
hole
respectively
are
0
.
068
and
0
.
50
times
the
electron
mass
.
The
energy
band
gap
is
1
.
43
eV
.
Determine
the
carrier
density
and
conductivity
at
300
K
.
Intrinsic carrier concentration is given by
n
i
Numericals on semiconductors
10
In
intrinsic
GaAs,
the
electron
and
hole
mobilities
are
0
.
85
and
0
.
04
m
2
V

1
s

1
respectively
and
the
effective
masses
of
electron
and
hole
respectively
are
0
.
068
and
0
.
50
times
the
electron
mass
.
The
energy
band
gap
is
1
.
43
eV
.
Determine
the
carrier
density
and
conductivity
at
300
K
.
Conductivity
of
a
semiconductor
is
given
by
mho / m
Numericals on semiconductors
11
A
sample
of
silicon
at
room
temperature
has
an
intrinsic
resistivity
of
2
.
5
x
10
3
Ω
m
.
The
sample
is
doped
with
4
x
10
16
donor
atoms/m
3
and
10
16
acceptor
atoms/m
3
.
Find
the
total
current
density
if
an
electric
field
of
400
V/m
is
applied
across
the
sample
.
Electron
mobility
is
0
.
125
m
2
/V
s
.
Hole
mobility
is
0
.
0475
m
2
/V
.
s
.
Effective doped concentration is
Numericals on semiconductors
From
charge
neutrality
equation
From law of mass action
Solving for p and choosing the right value for p as minority carrier concentration
Numericals on semiconductors
Since the minority carrier concentration p < n
i
Conductivity is given by
From Ohm’s law
Numericals on semiconductors
12
A
sample
of
pure
Ge
has
an
intrinsic
charge
carrier
density
of
2
.
5
x
10
19
/m
3
at
300
K
.
It
is
doped
with
donor
impurity
of
1
in
every
10
6
Ge
atoms
.
(a)
What
is
the
resistivity
of
the
doped

Ge?
Electron
mobility
and
hole
mobilities
are
0
.
38
m
2
/V
.
s
and
0
.
18
m
2
/V
.
s
.
Ge

atom
density
is
4
.
2
x
10
28
/m
3
.
(b)
If
this
Ge

bar
is
5
.
0
mm
long
and
25
x
10
–
12
m
2
in
cross

sectional
area,
what
is
its
resistance?
What
is
the
voltage
drop
across
the
Ge

bar
for
a
current
of
1
A?
No
of
doped
carriers
=
Since all the atoms are ionized, total electron density in Ge
N
d
+
= 4.2 x 10
2 2
/m
3
Numericals on semiconductors
•
From law of mass action
Electrical conductivity =
Numericals on semiconductors
12 Contd
Resistance
Of the Ge bar R =
Voltage drop across the Ge bar =
Numericals on semicodnuctors
13
A
rectangular
plate
of
a
semiconductor
has
dimensions
2
.
0
cm
along
y
direction,
1
.
0
mm
along
z

direction
.
Hall
probes
are
attached
on
its
two
surfaces
parallel
to
x
z
plane
and
a
magnetic
field
of
1
.
0
tesla
is
applied
along
z

direction
.
A
current
of
3
.
0
mA
is
set
up
along
the
x
direction
.
Calculate
the
hall
voltage
measured
by
the
probes,
if
the
hall
coefficient
of
the
material
is
3
.
66
10
–
4
m
3
/C
.
Also,
calculate
the
charge
carrier
concentration
.
Hall
voltage
is
given
by
Charge carrier density =
Z
(B)
Y (

V
H
)
X
(I)
t
w
L
Numericals on semiconductors
14
A
flat
copper
ribbon
0
.
330
mm
thick
carries
a
steady
current
50
.
0
A
and
is
located
in
a
uniform
1
.
30

T
magnetic
field
directed
perpendicular
to
the
plane
of
the
ribbon
.
If
a
Hall
voltage
of
9
.
60
V
is
measured
across
the
ribbon
.
What
is
the
charge
density
of
the
free
electrons?
Charge
carrier
density
n
is
given
by
n =
Numericals on semiconductors
15 The conductivity of intrinsic silicon is 4
.
17 x 10
–
5
/Ω m and 4
.
00 x 10
–
4
/
Ω m, at 0
C and 27
C respectively. Determine the band gap energy of
silicon.
Intrinsic conductivity
=
Numericals on semiconductors
•
15 Contd
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