Transcript Document
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
qn n qp p
1
qp 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 1019 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 cm3
10 2
2
10
ni
4
3
10
cm
n0
1016
p0
b) What are Δn and Δp?
p n GL 1020 106 1014 cm3
• 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 qp 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
B0
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 106 s
3.361103 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 cm3 , p0 i 103 cm3
n0
n n0 n 1017 +1014 1017 cm3
p p0 p 103 +1014 1014cm3
b) What is the np product?
np 1017 1014 =1031cm3
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 8.62 105 300 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
1031cm3
0.238
0.02586