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Chapter 5
Carrier transport phenomena
W.K. Chen
„
„
1
Transport
The net flow of electrons and holes in material is called transport
Two basic transport mechanisms
‰
‰
„
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Drift: movement of charged due to electric fields
Diffusion: the flow of charges due to density gradient
We implicitly assume the thermal equilibrium during the carrier
transport is not substantially disturbed
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Outline
„
„
„
„
Carrier drift
Carrier diffusion
Graded impurity distribution
The Hall effect
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5.1 Carrier drift
„
„
Drift: the net movement of charge due to electric fields is called drift
Drift current: The net drift of charge gives rise to a drift current
l
l
N: total number of flow charge
n: volume density of flow charge
A: cross-sectional area
υ: average drift velocity
l: traveling length of carrier per Δt
A
N nAl nA(υΔt )
#
= nAυ ( )
=
=
Δt
Δt
Δt
sec
Φ
#
)
Flux density φ = = nυ (
A
sec ⋅ cm 2
Flux Φ =
Drift current density J = qφ = qnυ d
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A
( 2)
cm
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Ampere
4
mobility
The average drift velocity for low electric fields is directly
proportional to the electric field, similar to the terminal velocity case
in “Fundamental Physics”
r
Fext
Hole: F = m*p a = (+ e)E
E: electric field
r
υ dp = μ p E
r
υdp: average drift velocity for holes
μp: hole mobility, proportionality factor
„
crystal
The mobility describe how well a particle will move due to an
electric field
r
Fext = m*a Drift current due to holes J
p , drf = ( + e) pυ dp = ( + e) pμ p E
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J = qφ = qnυ d (
A
)
cm 2
F = (−e) E
Flux φ p
r
υ dn = − μ n E
r
Flux φn
J n ,drf = qnυ = (−e)nυ dp = (−e) p (− μ n E )
Drift current due to electrons
Total drift current
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F = (+ e) E
E
J p ,drf
J n ,drf
J n ,drf = enμ n E
J = J n ,drf + J p ,drf = (enμ n + epμ p )E
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Example 5.1 drift current density
GaAs sample at 300K, N a = 0, N d = 1016 cm −3
Assume complete ionization
⇒ Calulate the drift current density if applied electric field E = 10 V/cm
Solution:
complete ionization ⇒ n ≈ N d = 1016 cm -3
ni2 (1.8 ×1016 ) 2
=
= 3.24 × 10 − 4 cm -3
p=
16
10
n
⇒ Calulate the drift current density if applied electric field E = 10 V/cm
J dft = J n ,drf + J p ,drf = (enμ n + epμ p )E
J dft = (1.6 × 10 −19 )(8500)(1016 )(10) = 136A/cm 2
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5.1.2 Mobility effect
F = m*p a = m*p
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F
eE
dυ
= ( + e) E ⇒ υ = * t
mp
dt
t
Under thermal equilibrium
Assume the net drift velocity is a small perturbation on the random thermal
velocity, so the time between collision will not be altered appreciably
⎛ eτ cp ⎞
mean peak velocity υ d , peak = ⎜ * ⎟E τcp: mean time between collisions
⎜ m ⎟
⎝ p ⎠
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The average drift velocity is one half the peak value
1 ⎛ eτ cp ⎞
average velocity υ d = ⎜ * ⎟E
2 ⎜⎝ m p ⎟⎠
Due to the statistic nature, the factor of ½ does not appear in a more
accurate model
⎛ eτ cp ⎞
average velocity υ dp = ⎜ * ⎟E = μ p E
⎜ m ⎟
⎝ p ⎠
μp =
μn =
υ dp
E
υ dn
E
=
=
eτ cp
m*p
,
eτ cn
mn*
The less the collisions, the longer the mean collision time and the higher the
mobility
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Scattering (collision) mechanisms
„
Two major scattering mechanisms
‰
Phonon (lattice) scattering
A perfect periodic potential in a solid allows electrons to move
unimpeded, or with no scattering, through the crystal
The thermal vibrations of lattice atoms cause a disruption in the perfect
periodic potential, resulting the interactions between the electrons or
holes and the vibrating lattice atoms
μ L ∝ T −3 / 2
lattice
‰
Ionized impurity scattering
The impurites in semiconductor at higher temperatures. The coulomb
interactions between the electrons or holes and the ionized impurities
produce scattering or collisions.
impurity
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T −3 / 2
μI ∝
NI
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total ionized impurity
N I = ( N d+ + N a− )
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Mobilitis versus temperarure
„
„
Inserts show the temperature dependence for “almost” intrinsic silicon
The inserts show that the parameter n is not equal to 3/2, but is 2.2, as
the first-order scattering theory predicted. However , the mobilites do
increase as the temperature increases
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Mobilties versus impurity concentrations at 300K
Ge
Si
GaAs
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Net mobility
The probability of a scattering even in the differential time dt is the sum of
individual events
dt
τ
Qμp =
υ dp
E
=
eτ cp
m
*
p
=
dt
τI
, μn =
υ dn
E
+
dt
τL
=
eτ cn
⇒ μ ∝τ
mn*
The net mobility due to the ionized and lattice scattering processes
1
μ
„
„
=
1
μI
+
1
μL
High effective mass of carrier results in low mobility
The mobility will increase with the increasing collision time
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5.1.3 Conductivity
J = e(nμ n + epμ p )E = σE
V = IR, R =
J=
ρL
A
=
L
σA
I
V / R EL
1
=
=
⋅
= σE
A
A
A ⎛ L ⎞
⎜ ⎟
⎝ σA ⎠
J = σE
σ = (enμ n + epμ p )
„
ρ=
1
σ
=
1
e(nμ n + epμ p )
The conductivity and resistivity of an extrinsic semiconductor are a
function primarily of the majority carrier parameters, such carrier
concentrations and mobilities
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Resistivity versus impurity concentration at 300K
Si
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Ge, GaAs and GaP
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σ i = (eμ n + eμ p )ni
„
„
In the midtemperature range (extrinsic range)
We have complete ionization, the electron concentration remains
essentially constant, However, the mobility decreases with increasing
temperature
At higher temperatures
The intrinsic carrier concentration begins to dominate the electron
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Chen
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concentration
as well as the conductivity
Example 5.2 mobility
Compensated n - type Si at 300K,
σ = 16 (Ω - cm) −1 , Na = 1016 cm −3
⇒ Determine the donor concentration and electron mobility
Solution:
compenstaed semiconductor and complete ionization at 300K
⇒ n ≈ (Nd − Na )
σ ≈ eμ n n = eμ n ( N d − N a )
16 = (1.6 × 10 −19 ) μ n ( N d − 1017 )
Use the left figure with trial and error
For example if we choose
N I = N d+ − N a− = 3 × 1017 cm -3 (i.e., N d = 2 × 1017 cm -3 )
Then μ n = 510 cm 2 /V - s ⇒ σ = 8.16 (Ω - cm)-1
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If we choose
N I = N d+ − N a− = 6 ×1017 cm -3 (i.e., N d = 5 × 1017 cm -3 )
Then μ n = 325 cm 2 /V - s ⇒ σ = 20.8 (Ω - cm)-1
Further tial and error yields
N I = N d+ + N a− = 4.5 ×1017 cm -3
N d = 3.5 ×1017 cm -3 )
μ n = 400 cm 2 /V - s
⇒ σ = 16 (Ω - cm)-1 (agree with given value)
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5.1.4 Velocity saturation
1
3
mυth2 = kT
2
2
at T = 300 K
1
3
mυth2 = (0.0259) = 0.03885eV
2
2
2(0.03885eV)
2(0.03885)(1.6 ×10 −19 )
υth =
=
≈ 105 m/s
−31
m
(9.11×10 )
υth ≈ 107 cm/s
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Figure 5.8
„
For Si
‰
‰
At low electric fields, there is linear variation of velocity with electric field
At high electric fields, the velocity saturated at approximately 107cm/s
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GaAs : mn* = 0.067mo
„
Si : mn* = 1.08mo
For GaAs
Due to low effective mass, the low-electric field electron velocity in GaAs is
much larger than in Si.
‰ At high electric fields, negative differential mobility occurs due to the scattering
of electrons into upper valley. Because of larger effective mass in the upper
valley (0.55 mo vs. 0.067 mo), the intervalley transfer mechanism results in
decreasing average drift velocity
of electrons
with electric field.
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‰
5.2 Carrier diffusion
(1)
xo
xo − l
n(x)
xo
(2)
xo
xo + l
x
Fn = −υthl
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l = υthτ cn
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dn
dx
22
„
During electron travel between collisions
In a mean free time, One half of electrons at
segment (1) will move to the right and cross the xo
plane into segment (2)
One half of electrons at segment (2) will move to
the left and cross the xo plane into (segment (1))
n1
Net rate of electron flow to the right (electron flux)
xo − l
l
n2
(1)
1
1
1
n1υth − n2υth = (n1 − n2 )υth
2
2
2
dn
n1 − n2 ≈ − l
dx
1
dn
⇒ Fn ( x) = υth ⋅ (− l )
2
dx
Fn ( xo ) =
(2)
xo
xo + l
l = υthτ cn
1
dn
net electron flux Fn = − υthl
2
dx
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Electron diffusion current
1
dn
J n = (−e) Fn = + eυthl
2
dx
J n ,dif = eDn
dn
dx
electron diffusion coefficient
1
Dn = υthll = υthτ cn
2
Hole diffusion current
J p ,dif = −eD p
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dp
dx
hole diffusion coefficent
l = υthτ cp
1
D p = υthl
2
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Example 5.4
n - type GaAs at 300K,
The electron concentration varies linearly from 1× 1018 to 7 × 1018 cm −3 over a
distance of 0.10 cm
⇒ Calculate the diffusion current if diffusion coefficient Dn = 225 cm 2 /s
Solution:
Δn
dn
≈ eDn
Δx
dx
⎛ 1×1018 − 7 ×1017 ⎞
−19
⎟⎟ = 108 A/cm2
= (1.6 ×10 )(225)⎜⎜
0.10
⎝
⎠
J n ,dif = eDn
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5.2.2 Total current density
J = enμ n E + epμ p E + J n ,dif + eDn
dp
dn
− eD p
dx
dx
J = enμ n E + epμ p E + J n ,dif + eDn∇n − eD p ∇p
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5.3 Graded impurity distribution
„
„
In nonuniform doped semiconductor, there will be a diffusion of majority
electrons from the region of high concentration to the region of low
concentration.
The flow of electrons leave behind positively charged donor ions. The
separation of positive and negative charges induces a electric field
The electric field is defined as
The electric potential is related to the electric potential energy by the charge
( -e)
φ=
Ec
( − e)
1
⇒ φ = ( EF − EFi )
e
dφ 1 dEFi
=
Ex = −
dx e dx
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⎛ N d ( x) ⎞
⎡ E f − E fi ⎤
⎜⎜
⎟⎟
≈
⇒
−
=
no = ni exp ⎢
N
(
x
)
E
E
kT
ln
d
f
fi
⎥
n
kT
⎦
⎣
i
⎝
⎠
−
dE fi
dx
=
kT dN d ( x)
N d ( x) dx
The induced electric field due to the nonuniform doping
⎛ kT ⎞ 1 dN d ( x)
E x = −⎜ ⎟
⎝ e ⎠ N d ( x) dx
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Example 5.5
N d ( x) = 1016 − 1019 x (cm -3 ) (0 ≤ x ≤ 1μm)
Solution:
(0.0259)(1019 )
⎛ kT ⎞ 1 dN d ( x)
E x = −⎜ ⎟
=−
(1016 − 1019 x)
⎝ e ⎠ N d ( x) dx
At x = 0, we find
E x ( x = 0) = 25.9 V/cm
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5.3.2 The Einstein relation
In thermal equilibrium, the individual electron and hole current must be zero
dn
dx
dN d ( x)
J n = 0 = enμ n E + eDn
dx
J n = 0 = enμ n E + eDn
⎧ ⎛ kT ⎞ 1 dN d ( x) ⎫
dN d ( x)
0 = enμ n ⎨− ⎜ ⎟
⎬ + eDn
dx
⎩ ⎝ e ⎠ N d ( x) dx ⎭
Q n( x ) ≈ N d ( x )
Dn
μn
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=
kT
e
Dn
μn
=
Dp
μp
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=
kT
e
30
The hole current must also be zero
Einstein relation
μ
D
Dn
μn
=
Dp
μp
=
kT
e
0.0256 at T = 300 K
≈ 40 at T = 300K
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5.4 The Hall effect
The Hall effect is used
„ to distinguish whether a
semiconductor is n-type or ptype
„ To measure the majority
carrier concentration and
„ Majority carrier mobility
FB = qυ × B
F = q[E + υ × B] = 0
qE y = qE H = qυ x Bz
The induced electric field in the y-direction
is called the Hall field
The induced electric field produce a voltage in the y-direction is called the
Hall Voltage
VH = + E H W
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VH = υ xWBz
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υ x : drift velocity
32
Semiconductor type & concentration
For p-type semiconductor, the drift velocity is related to material parameters
υ dx =
Jx
Ix
=
ep (ep)(Wd )
⇒ VH =
I x Bz
epd
p=
I x Bz
edVH
⇒ VH =
I x Bz
ned
n=
I x Bz
edVH
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mobility
Once the majority carrier concentration is determined, we can calculate the
low-field majority carrier mobility
Q J x = epμ p E x
Ix
V
= epμ p ⋅ x
Wd
L
μp =
W.K. Chen
IxL
epVxWd
μn =
IxL
enVxWd
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Example 5.7 carrier concnetration & mobility
L = 0.1 cm, W = 10 −2 cm, d = 10 −3 cm
I x = 1.0 mA, Vx = 12.5 V, Bz = 500 gauss = 5 ×10 − 2 tesla, and VH = −6.25 mV
Solution:
I B
n= x z
edVH
− (10 −3 )(5 ×10 −2 )
n=
= 5 ×10 21 m 3 = 5 × 1015 cm 3
−19
−5
−3
(1.6 ×10 )(10 )(−6.25 ×10 )
(10 −3 )(10 −3 )
IxL
= 0.10 m 2 /V - s
μn =
μn =
−19
−4
−5
21
(1.6 ×10 )(5 ×10 )(12.5)(10 )(10 )
enVxWd
μ n = 1000 cm 2 /V - s
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Figure 5.14 Figure for Problem 5.22
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