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
E
F


E i
 kT ln
 
  n
0.56 8.62 10

5 
300 ln


10
17
10
10



eV

0.977 eV

 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 p

10 16 /cm 3 , n

10 4 /cm 3
(N
A
>> N
D
 p-type)
 


1 q
 n
  p p
1 q
 p p
 
19 16
(1.6 10 )(470)(10 )

1.330
 cm

1

 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
 n
.

(and therefore

) is
From the mobility vs. temperature curve for 10 find that

17 cm –3 , we 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.
Find the intrinsic carrier concentration in silicon at:
(i) T = 200 K and (ii) T = 400 K.
(E4.3)
 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 ?
D
P
 kT

 q 
 p

25.86 meV
 2 1 1
410 cm V s
1e
 cm
2
 2
10.603 cm /s
1 eV

1 V
1 e
1 eV
  
19
J
•
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 μs. 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
?
p
0 n
0

 n i
2

3 16
10 cm

 
2 p
0
10 10
10
16
 4
10 cm

3 b) What are Δn and Δp?
p n

G
L
  
10
20 
10

6  14
10 cm

3
• 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
μs. 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?
p
 p
0
  p n
 n
0
  n

10
16 
10
14

10
4 
10
14
 16 10 cm

3
 14
10 cm

3 d) What are np product?
np

10
16 
10
14
 30
10 cm

3  n i
2
• 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.

Chapter 3
 For arbitrary injection levels and both carrier types in a nondegenerate semiconductor, the net rate of carrier recombination is:
 p
 t
1 i
2
Net Recombination Rate (General Case) Net
 t
Recombination Rate (General Case) where n
1
 n e i
( E
T

E i
) kT p
1
 n e i
( E i

E
T
) kT
• E
T
: energy level of R –G center

Area A, volume A.dx
Flow of current
J
N
(x) J
N
(x+dx)
Flow of electron dx
Adx




 n



1 t q

J
N
( x
 dx )

J
N


J
N
( x
 dx )

J
N


J
N
 x dx
• Taylor’s Series Expansion

 n

1

J

N t q x
 The Continuity Equations

 n

1

J

N t q x

 p
 
1

J

P t q x

 n
 t thermal
R G

 n
 t other processes

 p
 t thermal
R G

 p
 t other processes

 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:
J
N
 q
 n n
E  qD
N

 n x
 qD
N

 n x
•
For p-type material
J
P
 q
 p p
E  qD
P
 p
 x
  qD
P
 p
 x
•
For n-type material
 Equilibrium minority carrier concentration n
0 independent of x (uniform doping).
 Low-level injection conditions prevail.
and p
0 are

 Starting with the continuity equation for electrons:

 t n

1 q

J
N
 x


 n n

G
L
 n
 t thermal
R G
 
 n
 n

( n
0
  n )
 t

1
 
 qD
N

( n
0
  n )
 x




 n

G
L n
 n
 t other processes

G
L
 Therefore
 Similarly
 n
 t

D
N
 
 x
2 n


 n n

G
L
 p
 t

D
P
 
 x
2 p


 p p

G
L

 The subscript “ n ” or “ p ” is now used to explicitly denote n -type or p -type material .
 p n
 n p is the hole concentration in n -type material is the electron concentration in p -type material
 Thus, the minority carrier diffusion equations are:
 n p
 t

D
N
 
 x
2 n p 

 n n p 
G
L
 p n
 t

D
P
 
 x
2 p n


 p p n

G
L

 Steady state:
 n p
 t

0,
 p n
 t

0
 No diffusion current:
 No thermal R –G:
D
N  x
2 n p 
0, D
P
 n p
 n

0,

 p n p

0
 x
2 p n

0
 No other processes: G
L

0

 Consider the special case:
 Constant minority-carrier (hole) injection at x = 0
 Steady state, no light absorption for x > 0
0

D
P
 
 x
 p n
(0)
2 p n

 

 p n0 p n p
G
L

0 for x

0
 x
2 p n

 p n
D

P p

 p n
L
P
2
 The hole diffusion length L
P is defined to be:
Similarly,
L
P

L
N

D

P p
D

N n


 p x n
( )

Ae
 x L
P

Be x L
P
 x p n
 p n
The general solution to the equation is:
L
P
 A and B are constants determined by boundary conditions:
 p n
( ) 0
 p n
(0)
  p n0

B

0

A
  p n0
 Therefore, the solution is:
 p x n
  p e n0
 x L
P
•
Physically, L
P and L
N represent the average distance that a minority carrier can diffuse before it recombines with majority a carrier.

 Whenever Δ n = Δ p ≠ 0 then np ≠ n i
2 equilibrium conditions.
and we are at non-
 In this situation, now we would like to preserve and use the relations: n
 n e i
( E
F

E i
) kT
, p
 n e i
( E i

E
F
) kT
 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: n
 n e i
( F
N

E i
) kT
F
N

E i
 kT ln
 
  p
 n e i
( E i

F
P
) kT
F
P

E i
 kT ln n i
• The quasi-Fermi levels is useful to describe the carrier concentrations under non-equilibrium conditions

 Given N
D
=10 16 cm –3 , τ p
= 10 –6 s. Calculate L
P
.
 From the plot,
  p
D
P
 kT q


 p
2 
 2 
L
P

D

P p

 
3.361 10 cm

 Consider a Si sample at 300 K with N
D
Δ n = Δ p = 10 14 cm
–3
.
= 10 17 cm
–3 and a) What are p and n ?
n
0 n


N n
0
D

  

3
•
The sample is an n-type p
0 n i
2
  n
0
3
10 cm
17 14  17
10 +10 10 cm

3

3 p
 p
0
   3 14  14
10 +10 10 cm

3 b) What is the np product?
np
 17  14 31
10 10 =10 cm

3

 Consider a Si sample at 300 K with N
D
Δ n = Δ p = 10 14 cm
–3
.
= 10 17 cm
–3 and
0.417 eV E c
F
N c) Find F
N and F
P
?
F
N

E i
 kT ln
 n n i

F
N

E i

8.62 10

5 

17 10 10

0.417 eV

E i
0.238 eV
F
E
P v
F
P

E i
 kT ln
 p n i

E i

F
P
  
5  

14
8.62 10 300 ln 10 10
10

0.238 eV
 np
 n e i

F
N

E i
 kT
0.417

10
10 e 0.02586
 n e i

E i

F
P
 kT
0.238

10
10 e 0.02586
  31
 
3