Chapter 5: Carrier Transport Phenomena - FKE

Microelectronics I
Chapter 5: Carrier Transport
Phenomena
Transport; the process by which charged particles (electrons
and holes) move

Microelectronics I
Understanding of electrical properties ( I-V characteristics)
Basic current equation;
I ∝ e⋅µ
⋅n⋅
E
e; electronic charged (constant, 1.6 x 10-19 C)
u; mobility ( figure of merit that reflect the speed)
n; carrier concentration
E; Electric field
Carrier concentration (electron, n o and hole, p o)
Chapter 4
Carrier transport (current)
This chapter

Microelectronics I
+
“Drift”
The movement of carrier due
to electric field (E)
E
V
electron
-
Carrier Transport
“Diffusion”
The flow of carrier due to density
gradients (dn/dx)
electron
divider

Microelectronics I
5.1 Carrier Drift
Drift current density
Consider a positively charged hole,
When electric field, E, is applied, the hole accelerates
F = mp
a =
*
m*p; effective mass of hole, a; acceleration, e; electronic charge
However, hole collides with ionized impurity atoms and with thermally vibrating
lattice atom
Lattice atom
hole
Ionized impurity atom
eE
E

Microelectronics I
hole
Ionized impurity atom
Holes accelerates
due to E
Involves in collision
(“Scattering Process”)
Loses most of energy
Lattice atom
Gain average drift velocity, v dp
vdp p
= µ E
µ p; Hole mobility (unit; cm 2 /Vs)
Describes how well a carrier move due to E
E

Microelectronics I
Drift current density, J drf (unit; A/cm 2 ) due to hole
J =
p|
drf
epv
J p | drf = p
dp
eµ
pE
Drift current density due to electron
Total drift current;
J n | drf = n
eµ
nE
J drf = n p
e(
µ n + µ p)
E
The sum of the individual electron and hole drift current densities

Microelectronics I
Mobility is important parameter to determine the conductivity of material
Mobility effects
eτ
cp
p *
mp
µ =
µ =
n
eτ
m
τ; mean time between collisions
If τ=10 -15 s, in average, every 10 -15 s, carrier involves in collision @ scattering
Two dominant scattering mechanism
1. Phonon or lattice scattering
2. Ionized scattering
cn
*
n

Microelectronics I
1. Lattice scattering or phonon scattering
At temperature, T > 0 K, atoms randomly vibrate. This thermal vibrations cause a
disruption of the periodic potential function. This resulting in an interaction
between carrier and the vibrating lattice atoms.
Mobility due to lattice scattering, µ L
µ µ
L
∝ T
− 3 / 2
As temperature decreases, the probability of a scattering event decreases. Thus,
mobility increases

Microelectronics I
electron hole

Microelectronics I
2. Ionized Ion scattering
Coulomb interaction between carriers and ionized impurities produces scattering
or collusion. This alter the velocity characteristics of the carriers.
Mobility due to ionized ion scattering, µ I
µ
L
∝
T 2 / 3
N Total ionized impurity concentration
I
If temperature increases, the random thermal velocity of a carrier increases,
reducing the time the carrier spends in the vicinity of the ionized impurity center.
This causes the scattering effect decreases and mobility increases.
If the number of ionized impurity centers increases, then the probability of a
carrier encountering an ionized impurity centers increases, thus reducing
mobility

Microelectronics I

Microelectronics I
The net mobility is given by
1 1
=
µ µ
L
+
1
µ
Due to phonon scattering Due to ionized ion scattering
I

Microelectronics I
Conductivity
Drift current
electron
J drf n p
σ; conductivity [(Ω.cm) -1 ]
= e(
µ n + µ p)
E = σE
hole
σ = ( µ n + µ p)
e n p
Function of electron and hole concentrations and mobolities
Ρ; resistivity [Ω.cm]
1
ρ =
=
σ
1
( µ n + µ p)
e n p

Microelectronics I

Microelectronics I
Current density,
J
I
A
V
Area, A
Bar of semiconductor
L
+
- V
I
V
J = Electric field, E =
A
L
= σE
V
= σ
L
⎛ L ⎞
= ⎜ ⎟I
⎝σA
⎠
⎛ ρL
⎞
= ⎜ ⎟I
⎝ A ⎠
Resistance, R is a function of resistivity, or
conductivity, as well as the geometry of the
semiconductor
=
IR
I

Microelectronics I
Consider p-type semiconductor with an acceptor doping N a (N d=0) in which N a>>n i
Assume complete ionization
σ = ( µ n + µ p)
≈ eµ
p
e n p
n
σ ≈ eµ
nN a
1
≈
ρ
Function of the majority carrier

Microelectronics I
ex.;
Consider compensated n-type Silicon at T=300 K with a conductivity of σ=16
(Ωcm) -1 and an acceptor doping concentration of 10 17 cm -3 . Determine the donor
concentration and the electron mobility.
Solution;
At T=300 K, we can assume complete ionization. (N d-N a >>n i)
σ ≈ eµ
n = eµ
( N
16
=
n
( 1.
6
× 10
n
−19
d
− N
) µ ( N
n
d
a
)
−10
To determine µn and Nd, we can use figure mobility vs. impurity concentration with
trial and error
10
20
= µ
n( Nd
−10
17
)
17
)

Microelectronics I
If N d=2 x 10 17 cm -3 , impurity
concentration, N I= N d + +Na - =3 x 10 17
cm -3 . from the figure, µn= 510
cm 2 /Vs. so σ=8.16 (Ωcm) -1 .
If Nd=5 x 1017 cm-3 , impurity
concentration, NI= N +
d +Na - =6x 1017 cm-3 . from the figure, µn= 325
cm2 /Vs. so σ=20.8 (Ωcm) -1 cm .
2 /Vs. so σ=20.8 (Ωcm) -1 .
N d should be between 2 x 10 17 and 5 x
10 17 cm -3 . after trial and error.
Nd= 3.5 x 10 17 cm -3
µn=400 cm 2 /Vs
σ= 16 (Ωcm) -1

Microelectronics I
Ex 2.; designing a semiconductor resistor with a specified resistance to
handle a given current density
A Si semiconductor at T=300 K is initially doped with donors at a concentration of
N d=5 x 10 15 cm -3 . Acceptors are to be added to form a compensated p-type
material. The resistor is to have a resistance of 10 kΩ and handle a current
density of 50 A/cm 2 when 5 V is applied.
Solution;
When 5 V is applied to 10 kΩ resistor, the current, I
V 5
I = = = 0.
5mA
4
R 10
If the current density, J is limited to 50 A/cm 2 , the cross-sectional area, A is
A =
I
J
=
0.
5×
10
50
−3
= 10
−5
cm
2

Microelectronics I
Consider that electric field, E is limited to 100 V/cm. Then the length of the
resistor, L is
L
V
E
5 −2
= = 5×
10
100
The conductivity, σ of the semiconductor is
=
cm
L
σ = =
cm
RA 10 × 10
−2
5× 10
−1
= 0.
5(
Ω )
4 −5
The conductivity of the compensated p-type semiconductor is
σ ≈ eµ p = eµ
−
p
p ( Na
Nd
Here, the mobility is function of total ionized impurity concentration Na+Nd
)

Microelectronics I
Using trial and error, if Na=1.25x10 16 cm -3 , then Na+Nd=1.75x10 16 cm -3 , and the
hole mobility, from figure mobility versus impurity concentration, is approximately
µp=410 cm 2 /Vs. The conductivity is then,
σ = eµ ( N
p
a
− N
d
)
=
1.
6
× 10
−19
× 410×
(( 12.
5
−
5)
× 10
This is very close to the value we need. From the calculation
L=5x10 -2 cm
A=10 -5 cm 2
Na=1.25x10 16 cm -3
15
)
=
0.
492

Microelectronics I
Velocity Saturation
v d
= µ E
Drift velocity increase linearly with applied electric field.
At low electric field,
vd increase linearly
with applied E.
slope=mobility
At high electric field,
vd saturates
Constant value

Microelectronics I
Carrier diffusion
Diffusion; process whereby particles from a region of high concentration toward
a region of low concentration.
Electron concentration, n
Carrier
Electron flux
divider
Electron diffusion
current density
Position x
J
J
nx|
dif
nx|
dif
⎛ dn ⎞
= ( −e)
Dn⎜
− ⎟
⎝ dx ⎠
dn
= eDn
dx
D n; electron diffusion coefficient

Microelectronics I
Hole centration, p
Hole flux
Hole diffusion
current density
Position x
J
J
px|
dif
px|
dif
=
eD
p
= −eD
⎛
⎜−
⎝
p
dp
dx
dp
dx
D p; hole diffusion coefficient
Diffusion coefficient; indicates how well carrier move as a result of
density gradient.
⎞
⎟
⎠

Microelectronics I
Total Current Density
Electron drift
current
1-D
3-D
Total Current
Density
hole drift
current
Electron diffusion
current
J = J n|
drf + J nx|
dif + J p|
drf + J px|
dif
dn
J = enµ
nE
x + epµ
pE
x + eDn
− eDp
dx
dp
dx
J = enµ
nE
+ epµ
pE
+ eDn∇n
− eDp∇p
hole diffusion
current

Microelectronics I
Mobility,µ; indicates how well carrier moves due to electrical field
Diffusion coefficient, D; how well carrier moves due to density gradient
Here, we derive relationship between mobility and diffusion
coefficient using nonuniformly doped semiconductor model
Graded impurity distribution
electron
“Einstein relation”
x
nonuniformly doped semiconductor
Energy-band diagram
x
E C
E F
E v

Microelectronics I
x
E C
E F
E v
Doping concentration decreases as x increases
Electron diffuse in +x direction
The flow of electron leaves behind positively
charged donor
Induce electrical field, E x, given by
⎛ kT ⎞ 1 dN d ( x )
E Ex
= − ⎜ ⎟
⎝ e ⎠ N ( x)
dx
Since there are no electrical connections, there is no current(J=0)
Electron current
J
=
eµ
N
n
d
dNd
( x)
( x)
Ex
+ eDn
=
dx
0
d
…eq.2
…eq.1

Microelectronics I
From eq.1 and 2,
D
µ
n =
n
kT
e
Hole current must also be zero. We can show that
D
µ
n
Dp kT
=
µ µ e
p
Dp
=
µ
n =
p
kT
e
Diffusion coefficient and mobility are not independent parameters.
The relationship between this 2 parameter “Einstein relation”

Microelectronics I
The Hall effect
Using the effect, we can determine
The type of semiconductor
Carrier concentration
mobility
Applied electrical field
Magnetic field
Force on charged particle
in magnetic field (“Lorentz
force”)
F =
qv×
B

Microelectronics I
the Lorentz force on electron
and hole is in –y direction
There will be buildup of negative
charge (n-type) or positive charge
(p-type) at y=0
As a results, an electrical field
called “Hall field, E EH” H” is induced.
Hall field produces “Hall voltage,
V H”
Polarity of V H is used to determine the type of semiconductor
In y-direction, Lorentz force will be balanced by force due to Hall field
qv
V
H
V
B q
W
= v WB
H
x × z = (p-type)
x
z

Microelectronics I
For p-type
For n-type
V
p
H
v
=
x =
I xB
epd
eV H
I x
( ep)(
Wd)
z
I xBz
=
eV d
I xBz
VH
= −
end
I xBz
n = −
eV d
H
Can calculate the hole concentration in p-type
Note that V H is negative for n-type

Microelectronics I
When we know the carrier concentration, we can calculate carrier mobility
J = epµ
E
x
p
x
Similar with n-type, mobility is determined from
n = µ
I xL
enV Wd
x
I epµ
E
x
p
=
Wd L
I xL
µ p =
epV Wd
x
x