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Semiconductor Device Physics
Lecture 5
Dr. Gaurav Trivedi,
EEE Department,
IIT Guwahati
Example: Energy-Band Diagram
 For Silicon at 300 K, where is EF if n = 1017 cm–3 ?
Silicon at 300 K, ni = 1010 cm–3
n
EF  Ei  kT ln  
 ni 
17


10
5
 0.56  8.62 10  300  ln  10  eV
 10 
 0.56  0.417 eV
 0.977 eV
 Consider a Si sample at 300 K doped with 1016/cm3 Boron. What is its resistivity?
(NA >> ND  p-type)
NA = 1016/cm3 , ND = 0
p  1016/cm3, n  104/cm3
1

qn n  qp p
1

qp p
 (1.6 10
)(470)(10 ) 
 1.330  cm
19
16
1
 Consider a Si sample doped with 1017cm–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 1017cm–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
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 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 cm2 / V.s ?
 kT 
DP  
 p
 q 
25.86 meV

 410 cm 2 V 1s 1
1e
cm 2
 25.86 mV  410
V s
 10.603 cm2 /s
1 eV
1 V
1e
1 eV  1.602 1019 J
• Remark: kT/q = 25.86 mV at
room temperature
 Consider a sample of Si doped with 1016 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 1020 per cm3 per second, i.e. the generation rate GL = 1020/cm3/s
a) What are p0 and n0?
p0  1016 cm3
10 2
2
10 

ni
4
3

10
cm

n0 
1016
p0
b) What are Δn and Δp?
p  n  GL   1020 106  1014 cm3
• Hint: In steady-state,
generation rate equals
recombination rate
 Consider a sample of Si at 300 K doped with 1016 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 1020 per cm3 per second, i.e. the generation rate GL =
1020/cm3/s.
c) What are p and n?
p  p0  p  1016  1014  1016 cm 3
n  n0  n  104  1014  1014 cm 3
d) What are np product?
30
3
np  1016 1014  10 cm  ni2 • Note: The np product can be very
different from ni2 in case of
perturbed/agitated semiconductor
Photoconductor
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.
Net Recombination Rate (General Case)
Chapter 3
 For arbitrary injection levels and both carrier types in a nondegenerate semiconductor, the net rate of carrier
recombination is:
ni2Rate
 np (General Case) Net
p Net Recombination
n


t R G
t RRecombination
 p (n  n1 )   n (Rate
p  p1(General
)
G
Case)
where n1  ni e( ET Ei ) kT
( Ei  ET ) kT
p1  ni e
• ET : energy level of R–G center
Continuity Equation
Area A, volume A.dx
Flow of current
JN(x)
JN(x+dx)
Flow of electron
dx
 n  1
Adx      J N ( x  dx)  J N ( x) A
 t  q
Continuity Equation
J N ( x)
J N ( x  dx)  J N ( x) 
dx
x
n 1 J N ( x)

t q x
• Taylor’s Series Expansion
 The Continuity Equations
n
1 J N ( x) n
n



t
q x
t thermal t other
R G
p
1 J P ( x) p


t
q x
t
processes
p

thermal
t other
R G
processes
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:
n
n
J N  q  n nE  qDN
 qDN
x
x
p
p
J P  qp pE  qDP
 qDP
x
x
• For p-type material
• For n-type material
 Equilibrium minority carrier concentration n0 and p0 are
independent of x (uniform doping).
 Low-level injection conditions prevail.
Minority Carrier Diffusion Equation
 Starting with the continuity equation for electrons:
n 1 J N ( x) n


 GL
t q x
n
(n0  n) 1  
(n0  n)  n

qDN

 GL


t
q x 
x
 n
 Therefore
n
 2 n n
 DN

 GL
2
t
x
n
 Similarly
p
 2 p p
 DP

 GL
2
t
x
p
n
t

thermal
R G
n
t other
processes
n
n
 GL
Carrier Concentration Notation
 The subscript “n” or “p” is now used to explicitly denote n-type
or p-type material.
 pn is the hole concentration in n-type material
 np is the electron concentration in p-type material
 Thus, the minority carrier diffusion equations are:
np
t
 DN
 2 np
x
2

np
n
 GL
pn
 2 pn pn
 DP

 GL
2
t
x
p
Simplifications (Special Cases)
 Steady state:
np
pn
 0,
0
t
t
 2 np
x 2
 2 pn
 0, DP
0
2
x
 0,
pn
 No diffusion current:
DN
 No thermal R–G:
np
 No other processes:
GL  0
n
p
0
Minority Carrier Diffusion Length
 Consider the special case:
 Constant minority-carrier (hole) injection at x = 0
 Steady state, no light absorption for x > 0
 2 pn pn
0  DP

2
x
p
pn (0)  pn0
GL  0 for x  0
 2 pn pn pn

 2
2
x
DP p LP
 The hole diffusion length LP is defined to be: LP  DP p
Similarly, LN  DN n
Minority Carrier Diffusion Length
 2 pn pn
 2 is:
 The general solution to the equation
2
x
LP
pn ( x)  Ae x LP  Bex LP
 A and B are constants determined by boundary conditions:
pn ()  0
 B0
pn (0)  pn0  A  pn0
 Therefore, the solution is:
pn ( x)  pn0e x LP
• Physically, LP and LN represent the average
distance that a minority carrier can diffuse
before it recombines with majority a
carrier.
Quasi-Fermi Levels
 Whenever Δn = Δp ≠ 0 then np ≠ ni2 and we are at nonequilibrium conditions.
 In this situation, now we would like to preserve and use the
relations:
n  ni e( EF Ei ) kT , p  nie( Ei EF ) kT
 On the other hand, both equations imply np = ni2, which does
not apply anymore.
 The solution is to introduce to quasi-Fermi levels FN and FP such that:
n  ni e( FN Ei ) kT
n
FN  Ei  kT ln  
 ni 
p  nie( Ei FP ) kT
 p
FP  Ei  kT ln  
 ni 
• The quasi-Fermi levels is useful to describe the carrier
concentrations under non-equilibrium conditions
Example: Minority Carrier Diffusion
Length
 Given ND=1016 cm–3, τp = 10–6 s. Calculate LP.
 From the plot,
p  437 cm2 V  s
kT
DP 
p
q
 25.86 mV  437 cm2 V  s
 11.3cm2 s
LP  DP p
 11.3cm 2 s 106 s
 3.361103 cm
= 33.61 m
Example: Quasi-Fermi Levels
 Consider a Si sample at 300 K with ND = 1017 cm–3 and
Δn = Δp = 1014 cm–3.
a) What are p and n? • The sample is an n-type
2
n
n0  ND  1017 cm3 , p0  i  103 cm3
n0
n  n0  n  1017 +1014  1017 cm3
p  p0  p  103 +1014  1014cm3
b) What is the np product?
np  1017 1014 =1031cm3
Example: Quasi-Fermi Levels
 Consider a Si sample at 300 K with ND = 1017 cm–3 and
Δn = Δp = 1014 cm–3.
0.417 eV
FN
c) Find FN and FP?
FN  Ei  kT ln  n ni 
Ei

FN  Ei  8.62 105  300  ln 1017 1010
 0.417 eV
FP  Ei  kT ln  p ni 
5

FP
Ev

0.238 eV
Ei  FP  8.62  10  300  ln 10 10
 0.238 eV
Ec
14
10

np  ni e
FN  Ei  kT
 1010 e
0.417
0.02586
 ni e Ei  FP  kT
1010 e
 1.000257 1031
 1031cm3
0.238
0.02586

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