03.05.2012 - Erwin Sitompul

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Semiconductor Device Physics
Lecture 4
Dr.-Ing. Erwin Sitompul
President University
http://zitompul.wordpress.com
President University
Erwin Sitompul
SDP 4/1
Chapter 3
Carrier Action
Direct and Indirect Semiconductors
E-k Diagrams
Ec
Ec
Phonon
Photon
GaAs, GaN
Photon
Ev
Si, Ge
(direct semiconductors)
(indirect semiconductors)
• Little change in momentum is
required for recombination
• Momentum is conserved by
photon (light) emission
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Ev
• Large change in momentum is
required for recombination
• Momentum is conserved by
mainly phonon (vibration)
emission + photon emission
Erwin Sitompul
SDP 4/2
Chapter 3
Carrier Action
Excess Carrier Concentrations
Deviation from
equilibrium values
Values under
arbitrary conditions
Equilibrium values
n  n  n0
p  p  p0
n ,  p  0
 Positive deviation corresponds to a carrier excess, while
negative deviation corresponds to a carrier deficit.
n ,  p  0
 Charge neutrality condition:
n  p
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Erwin Sitompul
SDP 4/3
Chapter 3
Carrier Action
“Low-Level Injection”
 Often, the disturbance from equilibrium is small, such that the
majority carrier concentration is not affected significantly:
 For an n-type material p  n0 ,
n
n0
p  p0
n  p0 ,
p
p0
n  n0
 For a p-type material
• Low-level injection condition
 However, the minority carrier concentration can be significantly
affected.
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Erwin Sitompul
SDP 4/4
Chapter 3
Carrier Action
Indirect Recombination Rate
 Suppose excess carriers are introduced into an n-type Si
sample by shining light onto it. At time t = 0, the light is turned
off. How does p vary with time t > 0?
 Consider the rate of hole recombination:
p
 cp NT p
t R
NT : number of R–G centers/cm3
Cp : hole capture coefficient
 In the midst of relaxing back to the equilibrium condition, the
hole generation rate is small and is taken to be approximately
equal to its equilibrium value:
p
p
p
 cp N T p0


t G t G-equilibrium
t R-equilibrium
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Erwin Sitompul
SDP 4/5
Chapter 3
Carrier Action
Indirect Recombination Rate
 The net rate of change in p is therefore:
p
p
p
 cp N T p  cp N T p0  cp NT  p  p0 


t R G t R t G
p
p
 cp N T p  
t R G
p
1
where  p 
cp NT
• For holes in
n-type material
 Similarly,
n
n
 cn NT n  
t R G
n
where  n 
1
cn N T
• For electrons
in p-type material
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Erwin Sitompul
SDP 4/6
Chapter 3
Carrier Action
Minority Carrier Lifetime
1
p 
cp NT
1
n 
cn NT
 The minority carrier lifetime τ is the average time for excess
minority carriers to “survive” in a sea of majority carriers.
 The value of τ ranges from 1 ns to 1 ms in Si and depends on
the density of metallic impurities and the density of
crystalline defects.
 The deep traps originated from impurity and defects capture
electrons or holes to facilitate recombination and are called
recombination-generation centers.
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Erwin Sitompul
SDP 4/7
Chapter 3
Carrier Action
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.
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Erwin Sitompul
SDP 4/8
Chapter 3
Carrier Action
Example: Photoconductor
 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.
a) What are p0 and n0?
p0  1016 cm 3
10 2
2
10 

ni
4
3

10
cm
n0 

p0
1016
b) What are Δn and Δp?
p  n  GL   1020 106  1014 cm 3
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Erwin Sitompul
• Hint: In steady-state,
generation rate equals
recombination rate
SDP 4/9
Chapter 3
Carrier Action
Example: Photoconductor
 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
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Erwin Sitompul
SDP 4/10
Chapter 3
Carrier Action
Net Recombination Rate (General Case)
 For arbitrary injection levels and both carrier types in a nondegenerate semiconductor, the net rate of carrier
recombination is:
ni2  np
p
n


t R G
t R G  p (n  n1 )   n ( p  p1 )
where n1  ni e( ET  Ei ) kT
p1  ni e ( Ei  ET ) kT
• ET : energy level of R–G center
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Erwin Sitompul
SDP 4/11
Chapter 3
Carrier Action
Continuity Equation
 Consider carrier-flux into / out of an infinitesimal volume:
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
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Erwin Sitompul
SDP 4/12
Chapter 3
Carrier Action
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

t
1 J N ( x) n
n


q x
t thermal t other
R G
p
1 J P ( x) p


t
q x
t
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processes
p

thermal
t other
R G
Erwin Sitompul
processes
SDP 4/13
Chapter 3
Carrier Action
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.
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Erwin Sitompul
SDP 4/14
Chapter 3
Carrier Action
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
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Erwin Sitompul
n
n

t thermal
n
R G
n
t other
processes
SDP 4/15
 GL
Chapter 3
Carrier Action
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
President University
Erwin Sitompul
SDP 4/16
Chapter 3
Carrier Action
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
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n
p
Erwin Sitompul
0
SDP 4/17
Chapter 3
Carrier Action
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
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Erwin Sitompul
SDP 4/18
Chapter 3
Carrier Action
Minority Carrier Diffusion Length
 2 pn pn
 2 is:
 The general solution to the equation
2
x
LP
pn ( x)  Ae  x LP  Be x LP
 A and B are constants determined by boundary conditions:
pn ()  0
pn (0)  pn0
 B0
 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.
President University
Erwin Sitompul
SDP 4/19
Chapter 3
Carrier Action
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
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Erwin Sitompul
SDP 4/20
Chapter 3
Carrier Action
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  ni e( 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  ni e( Ei  FP ) kT
 p
FP  Ei  kT ln  
 ni 
• The quasi-Fermi levels is useful to describe the carrier
concentrations under non-equilibrium conditions
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Erwin Sitompul
SDP 4/21
Chapter 3
Carrier Action
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  N D  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
President University
Erwin Sitompul
SDP 4/22
Chapter 3
Carrier Action
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
c) Find FN and FP?
FN  Ei  kT ln  n ni 
5
Ei
FP

FN  Ei  8.62 10  300  ln 10 10
 0.417 eV
FP  Ei  kT ln  p ni 
5
17

10

Ev
0.238 eV
Ei  FP  8.62 10  300  ln 10 10
 0.238 eV
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
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Ec
FN
Erwin Sitompul
SDP 4/23
0.238
0.02586
Chapter 2
Carrier Action
Homework 3
 1.
(6.17)
A semiconductor has the following properties:
DN = 25 cm2/s
τn0 = 10–6 s
DP = 10 cm2/s
τp0 = 10–7 s
The semiconductor is a homogeneous, p-type (NA = 1017 cm–3) material in
thermal equilibrium for t ≤ 0. At t = 0, an external source is turned on which
produces excess carriers uniformly at the rate GL = 1020 cm–3 s–1.
At t = 2×10–6 s, the external source is turned off.
(a) Derive the expression for the excess-electron concentration as a
function of time for 0 ≤ t ≤ ∞
(b) Determine the value of the excess-electron concentration at
(i) t = 0, (ii) t = 2×10–6 s, and (iii) t = ∞
(c) Plot the excess electron concentration as a function of time.
 2.
(4.38)
Problem 3.24
Pierret’s “Semiconductor Device Fundamentals”.
 Deadline: 10 May 2012, at 08:00.
President University
Erwin Sitompul
SDP 4/24
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