carrier transport phenomena.ppt
DESCRIPTION
carrier in semi condutorTRANSCRIPT
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Carrier Transport Phenomena
Transport; the process by which charged particles (electrons and holes) move
By Dr. Ghanshyam Kumar Singh
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In this section, we will
•Describe the mechanism of carrier drift due to an applied electric field
•Describe the characteristics of carrier mobility
•Describe the mechanism of carrier diffusion
•Describe the effects of a nonuniform impurity doping concentration
•Discuss and analyze the Hall Effect
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Understanding of electrical properties (I-V characteristics)
Basic current equation;
EneI
e; electronic charged (constant, 1.6 x 10-19 C); mobility ( figure of merit that reflect the speed)n; carrier concentrationE; Electric field
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Carrier Transport
“Drift”The movement of carrier due
to electric field (E)
“Diffusion”The flow of carrier due to density
gradients (dn/dx)
V
+ -E
electron
dividerelectron
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Carrier Drift
Drift current density
Consider a positively charged hole,
When electric field, E, is applied, the hole accelerates
eEamF p *
m*p; effective mass of hole, a; acceleration, e; electronic charge
However, hole collides with ionized impurity atoms and with thermally vibrating lattice atom
hole
Lattice atom
Ionized impurity atomE
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hole
Lattice atom
Ionized impurity atomE
Holes accelerates due to E
Involves in collision (“Scattering Process”) Loses most of energy
Gain average drift velocity, vdp
Ev pdp
µp; Hole mobility (unit; cm2/Vs)Describes how well a carrier move due to E
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Drift current density, Jdrf (unit; A/cm2) due to hole
dpdrfp epvJ |
pEeJ pdrfp |
Drift current density due to electron
nEeJ ndrfn |
Total drift current;
EpneJ pndrf )(
The sum of the individual electron and hole drift current densities
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Mobility effects
*p
cpp m
e
Mobility is important parameter to determine the conductivity of material
*n
cnn m
e
; mean time between collisions
If =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
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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, µL
2/3 TL
As temperature decreases, the probability of a scattering event decreases. Thus, mobility increases
Temperature , Scattering Probability , Mobility Temperature , Scattering Probability , Mobility
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electron hole
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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, µI
Total ionized impurity concentration
• 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.
Temperature , Thermal velocity, Time around ionized impurity, Mobility
• If the number of ionized impurity centers increases, then the probability of a carrier encountering an ionized impurity centers increases, thus reducing mobility
Ionized Impurity , Scattering Probability , Mobility
IN
TI
2/3
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The net mobility is given by
IL 111
Due to phonon scattering Due to ionized ion scattering
Normally, more than one source of scattering is present, for example both impurities and lattice phonons.
It is normally a very good approximation to combine their influences using "Matthiessen's Rule" (developed from work by Augustus Matthiessen in 1864):
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Conductivity
EEpneJ pndrf )(
Drift current
σ; conductivity [(Ω.cm)-1]
)( pne pn
electron
hole
Function of electron and hole concentrations and mobolities
Ρ; resistivity [Ω.cm]
)(
11
pne pn
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L
+- V
I
Area, A
Bar of semiconductor
Current density,A
IJ Electric field,
L
VE
IRIA
LI
A
LV
L
V
A
I
EJ
Resistance, R is a function of resistivity, or conductivity, as well as the geometry of the semiconductor
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Consider p-type semiconductor with an acceptor doping Na (Nd=0) in which Na>>ni
pepne npn )(
Assume complete ionization
1
an Ne
Function of the majority carrier
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ex.;
Consider compensated n-type Silicon at T=300 K with a conductivity of σ=16 (Ωcm)-1 and an acceptor doping concentration of 1017 cm-3. Determine the donor concentration and the electron mobility.
Solution;
At T=300 K, we can assume complete ionization. (Nd-Na >>ni)
)10()106.1(16
)(1719
dn
adnn
N
NNene
To determine µn and Nd, we can use figure mobility vs. impurity concentration with trial and error
)10(10 1720 dn N
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If Nd=2 x 1017 cm-3, impurity concentration, NI= Nd
++Na-=3 x 1017
cm-3. from the figure, µn= 510 cm2/Vs. so σ=8.16 (Ωcm)-1.
If Nd=5 x 1017 cm-3, impurity concentration, NI= Nd
++Na-=6x 1017
cm-3. from the figure, µn= 325 cm2/Vs. so σ=20.8 (Ωcm)-1.
Nd should be between 2 x 1017 and 5 x 1017 cm-3. after trial and error.
Nd= 3.5 x 1017 cm-3
µn=400 cm2/Vsσ= 16 (Ωcm)-1
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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 Nd=5 x 1015 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/cm2 when 5 V is applied.
Solution;
When 5 V is applied to 10 kΩ resistor, the current, I
mAR
VI 5.0
10
54
If the current density, J is limited to 50 A/cm2, the cross-sectional area, A is
253
1050
105.0cm
J
IA
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Consider that electric field, E is limited to 100 V/cm. Then the length of the resistor, L is
The conductivity, σ of the semiconductor is
cmE
VL 2105
100
5
154
2
)(5.01010
105
cm
RA
L
The conductivity of the compensated p-type semiconductor is
)( dapp NNepe
Here, the mobility is function of total ionized impurity concentration Na+Nd
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Using trial and error, if Na=1.25x1016cm-3 , then Na+Nd=1.75x1016cm-3, and the hole mobility, from figure mobility versus impurity concentration, is approximately µp=410 cm2/Vs. The conductivity is then,
492.0)10)55.12((410106.1)( 1519 dap NNe
This is very close to the value we need. From the calculation
L=5x10-2 cmA=10-5cm2
Na=1.25x1016cm-3
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Velocity Saturation
Evd 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
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Carrier diffusion
Diffusion; process whereby particles flow from a region of high concentration toward a region of low concentration.
dividerCarrier
Ele
ctro
n co
ncen
trat
ion,
n
Position x
Electron diffusion current density
Electron flux
dx
dneDJ
dx
dnDeJ
ndifnx
ndifnx
|
| )(
Dn; electron diffusion coefficient
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Hol
e ce
ntra
tion,
p
Position x
Hole diffusion current density
Hole flux
dx
dpeDJ
dx
dpeDJ
pdifpx
pdifpx
|
|
Dp; hole diffusion coefficient
Diffusion coefficient; indicates how well carrier move as a result of density gradient.
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Total Current Density
Total Current Density
Electron drift current
hole drift current
Electron diffusion current
hole diffusion current
difpxdrfpdifnxdrfn JJJJJ ||||
dx
dpeD
dx
dneDEepEenJ pnxpxn
1-D
3-D
peDneDEepEenJ pnpn
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Graded impurity distribution
Mobility,µ; indicates how well carrier moves due to electrical fieldDiffusion coefficient, D; how well carrier moves due to density gradient
Here, we derive relationship between mobility and diffusion coefficient using nonuniformly doped semiconductor model
“Einstein relation”
Non-uniformly doped semiconductor
electron
x
EC
EF
Ev
x
Energy-band diagram
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EC
EF
Ev
x
Doping concentration decreases as x increases Electron diffuse in +x directionThe flow of electron leaves behind positively charged donor
Induce electrical field, Ex, given by
dx
xdN
xNe
kTE d
dx
)(
)(
1
Since there are no electrical connections, there is no current(J=0)
0)(
)( dx
xdNeDExNeJ d
nxdn
…eq.1
…eq.2Electron current
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From eq.1 and 2,
e
kTD
n
n
Hole current must also be zero. We can show that
e
kTD
p
p
e
kTDD
p
p
n
n
Diffusion coefficient and mobility are not independent parameters.The relationship between this 2 parameter “Einstein relation”
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Exercise 1
Assume the mobility of a particular carrier is 1000 cm2/V-s at T=300K.Determine the diffusion coefficient given the carrier mobility.
Using the Einstein relation we have that
e
kTDD
p
p
n
n
20.0259 1000 25.9 cm /sD kT kT
De e
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Exercise 2
Assume that electron diffusion coefficient of a semiconductor at T=300K is Dn=215 cm2/s. Determine the electron mobility.
Using the Einstein relation we have that
e
kTDD
p
p
n
n
22158301cm /V-s
0.0259
D kT DkTee
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The Hall effect
Using the effect, we can determine
The type of semiconductorCarrier concentrationmobility
Magnetic field
Applied electrical field
Force on charged particle in magnetic field (“Lorentz force”)
BqvF
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the Lorentz force on electron and hole is in –y directionThere will be buildup of negative charge (n-type) or positive charge (p-type) at y=0As a results, an electrical field called “Hall field, EH” is induced. Hall field produces “Hall voltage, VH”
In y-direction, Lorentz force will be balanced by force due to Hall field
zxH
Hzx
WBvVW
VqBqv
(p-type)
Polarity of VH is used to determine the type of semiconductor
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For p-type
))(( Wdep
Iv x
x
deV
BIp
epd
BIV
H
zx
zxH
Can calculate the hole concentration in p-type
For n-type
deV
BIn
end
BIV
H
zx
zxH
Note that VH is negative for n-type
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When we know the carrier concentration, we can calculate carrier mobility
xpx EepJ
Similar with n-type, mobility is determined from
WdenV
LI
x
xn
WdepV
LIL
Vep
Wd
I
x
xp
xpx
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Exercise 3
Silicon at T = 300 K is uniformly doped with phophorus atoms at a concentration of 2 1016 cm. A Hall effect device has been fabricated with the following geometry: d = 10-3 cm, W = 10-2 cm, and L = 10-1 cm. The electrical parameters measured are: Ix = 1.2 mA, and Bz = 500 gauss = 5 10-2 Tesla. Determine a) The Hall voltageb) The Hall field
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Exercise 3
Determine a) The Hall voltage
3 2
22 19 5
3
1.2 10 5 10
2 10 1.6 10 10
1.875 10 1.875
X ZH
I BV
ned
V mV
b) The Hall Field
3
2
1.875 100.1875 V cm
10
HH H H
HH
VV W
W
V
W
, n=2 1016 cm, Ix = 1.2 mA, Bz = 500
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Quiz 1
Consider a silicon at T = 300 K. A Hall effect device has been fabricated with the following geometry: d = 5 10-3 cm, W = 5 10-2 cm, and L = 0.50 cm. The electrical parameters measured are: Ix = 0.50 mA, Vx = 1.25 V, and Bz = 650 gauss = 6.5 10-2 Tesla. The Hall voltage is EH = -16.5 mV/cm. Determine a) The Hall voltageb) The conductivity typec) The majority carrier concentrationd) The majority carrier mobility
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Quiz 1
Determine a) The Hall voltage
3 216.5 10 5 10
0.825 mV
H HV W
b) The conductivity type
negative n-typeHV
c) The majority carrier concentration
3 2
15 -3
19 5 3
0.5 10 6.5 104.924 10 cm
1.6 10 5 10 0.825 10x z
H
I Bn
edV
d) The majority carrier mobility
3 2
19 21 4 5
0.5 10 0.5 10
1.6 10 4.924 10 1.25 5 10 5 10
1015 cm / V-s
xn
x
I L
enV Wd
d = 5 10-3 cm, W = 5 10-2 cm, and L = 0.50 cm. Ix = 0.50 mA, Vx = 1.25 V, Bz = 650 gauss = 6.5 10-2 TeslaEH = -16.5 mV/cm