# Numericals on semiconductors - EngineeringDuniya.com

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1 Νοε 2013 (πριν από 4 χρόνια και 6 μήνες)

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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

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
;

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

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

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