carrier transport:drift and diffusionece.uprm.edu/~mtoledo/5209/2012/transport.pdf · semiconductor...
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Carrier transport: Drift and DiffusionINEL 5209 - Solid State Devices - Spring 2012
Manuel Toledo
April 10, 2012
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Outline
.. .1 DriftDrift currentMobilityResistivityResistanceHall Effect
.. .2 DiffusionHaynes-Shockley Experiment
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Drift
Drift: carrier motion due to applied electric field
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Drift current
Drift current
drift velocity: vd
distance traveled by hole in time t: vdt
holes crossing the plane in time t: pvdtA
drift current for holes: ip,drift = qpvdA
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Drift current
Drift current
In vector notation:
Jp,drift = qpvd Jn,drift = qnvd
mobility: constant of proportionality between vd and E
Jp,drift = qpµpE Jn,drift = qnµnE
< µ >: cm2/V − sec;
For Si: µn = 1360cm2/V − sec and µp = 460cm2/V − sec atT = 300K and ND = 1014/cm3, NA = 1014/cm3, respectively.
For GaAs: µn = 8000cm2/V − sec and µp = 320cm2/V − sec atT = 300K and ND, NA < 1014/cm3.
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Drift current
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Mobility
Mobility is affected by scattering events with:
i Phonons,
ii Ionized impurity atoms,
iii neutral impurity atoms and defects,
iv other carriers,
v internal electric field created by the piezoelectric effect.
Most important are i and ii.
Mobility if only i is present: µL ∝ T−3/2
Mobility if only ii is present: µI ∝ T3/2/NI
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Mobility
(1µn
= 1µLn
+ 1µIn
+ · · ·1µp
= 1µLp
+ 1µIp
+ · · ·
)
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Mobility
Experimental fit
µ = µmin +µ0
1 + (N/Nref)α
N is either ND or NA. Other parameters exhibit a temperaturedependence of the form
A = A0(T/300)ν
where A0 is a constant, T is the absolute temperature, and ν is thetemperature exponent.
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Mobility
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Mobility
Small-device effects
velocity saturation of electrons on Si:
vdsat =v0dsat
1 + AeT/Td
where A = 0.8, v0dsat = 2.4 × 107cm/sec, Td = 600K, and T istemperature in degrees Kelvin.
Inter-valley carrier transfer
Ballistic transport
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Mobility
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Resistivity
Resistivity
Resistivity
ρ =1
q(µnn + µpp
) = 1σ
For n-type let p → 0 and set n = ND. For p-type let n → 0 andset p = NA.
Ohm’s Law:
J = σE = Eρ → E = ρJ
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Resistivity
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Resistance
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Resistance
Ohm’s Law: I = V/R
Integrated Circuit Resistor (see previous slide)
R = ρL
xjW
ρ is the material’s resistivity
conductivity = σ = 1ρ
Sheet resistance: RS = 1σxj
R = RSLW
Sheet resistance is expressed in Ω/.
Example: For R = 3.5kΩ, and a sheet resistance of RS = 200Ω/ and a feature sizeW = 1µm, use L = 17.5µm.
The smaller W the better.
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Resistance
Integral = Average → like above form of Ohm’s Law
Differential = local → more accurate model.
Conductivity varies with depth. See next slide.
Average value of conductivity is used as an approximation.
σ =1xj
∫ ∞
0σ(x)dx
Average conductivity → no information about currentdistribution in resistor.
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Resistance
Typical conductivity profile.
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Resistance
Using x0 =√
2, we can find the average conductivity:
σ(x) = σ(0)e−(x/x0)2
σ =10xj
∫ ∞
0e−(x2/2)dx
=10(Ω − cm)−1
3µm×√
π/2µm
= 4.18(Ω − cm)−1
Integration was performed using what is known as Laplaceintegral.
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Resistance
Using this device, to build a 100kΩ resistor
RS =1
σxj
=1
4.18(Ω − cm)−1 × 3µm= 797.4Ω/
LW
=RRS
=100, 100Ω797.4Ω
= 125
For a feature size of W = 1µm, L = 125µm is required.
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Hall Effect
Hall Effect
Hall effect setup (from http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/hall.html).
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Hall Effect
Hall Effect
Lorentz Force (due to magnetic field): Fm = qvB
For n-type, drift velocity = vx = −µnEx.
Force due to Hall Effect’s electric field: FH = qEH
Equilibrium: FH = Fm and (ignoring sign)
qvB = −qµExB = qEH → EH = µExB
Vx = logitudinal voltage = Ex × L
VH = Hall’s voltage = EH × W = µExB × W
VH = µ
(WL
)VxB
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Hall Effect
Hall Effect
A Silicon sample contains 1016 phosphorous atoms per cc. A 1mAcurrent is flowing through the sample, which is 1cm long. Atransversal magnetic field Bz = 10−4Wb/cm(1Wb = 1V − s = 1T − m2 = 108G − cm2) is applied. Find thesample’s dimensions if a Hall voltage VH = 20mV is measured.
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Diffusion
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Diffusion
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F1 =12n(−l) × l
τc=
12
n(−l) × vth F2 =12
n(l) × vth
Net flow from left to right:
F = F1 − F2 =12
vth (n(−l) − n(l))
≃ 12
vth
((n0 − l
dndx
)−(
n0 + ldndx
))= −vthl
dndx
≡ −Dndndx
Dn = diffusion coefficient = diffusivity. The diffusion current due to agradient in electron concentration is:
Jn = −qF = qDndndx
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Example
Assume that, in an n-type semiconductor at T = 300K, the electronconcentration varies linerly from 1 × 1018cm−3 to 7 × 1017cm−3
over a distance of 0.1 cm. Find the diffusion current if the electrondiffusion coefficient is Dn = 22.5cm2/s.
Jn,diff ≃(1.6 × 10−19
) (22.5cm2/s
)(0.3 × 1018
0.1
)= 10.8 A/cm2
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Example
Assume that, in an n-type semiconductor at T = 300K, the electronconcentration varies linerly from 1 × 1018cm−3 to 7 × 1017cm−3
over a distance of 0.1 cm. Find the diffusion current if the electrondiffusion coefficient is Dn = 22.5cm2/s.
Jn,diff ≃(1.6 × 10−19
) (22.5cm2/s
)(0.3 × 1018
0.1
)= 10.8 A/cm2
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Einstein Relationship
Under equilibrium conditions, the Fermi level inside a material doesnot change with position. A nonzero electric field is established insidea non-uniformly doped semiconductor under equilibrium conditions.
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Jn,drift + Jn,diff = qµnnE + qDndndx
= 0
n = NCF1/2(ηc) where ηc = (EF − Ec)/kT,under equilibrium dEF/dx = 0,E = 1
qdEcdx
dndx
= − 1kT
dndηc
dEc
dx= − q
kTdndηc
E
qE(
µnn −qkT
Dndndηc
)= 0 ⇒ µn =
qkT
Dn1n
dndηc
non-degenerate limit: n → NC exp ηc, dndηc
→ n and
µn =qkT
Dn ⇒ Dn =kTq
µn
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Example
Minority carriers (holes) are injected into a homogeneous n-typesemiconductor sample at one point. An electric field of 50 V/cm isapplied across the sample, and the field moves these minoritycarriers a distance of 1 cm in 100 µs Find the drift velocity and thediffusivity of the minority carriers.
vp =1 cm
100 µs= 104 cm/s
µp =vp
E=
104
50= 200 cm2/V − s
Dp =kTq
µp = 0.0259 × 200 = 5.18 cm2/s
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Example
Minority carriers (holes) are injected into a homogeneous n-typesemiconductor sample at one point. An electric field of 50 V/cm isapplied across the sample, and the field moves these minoritycarriers a distance of 1 cm in 100 µs Find the drift velocity and thediffusivity of the minority carriers.
vp =1 cm
100 µs= 104 cm/s
µp =vp
E=
104
50= 200 cm2/V − s
Dp =kTq
µp = 0.0259 × 200 = 5.18 cm2/s
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Haynes-Shockley Experiment
Haynes-Shockley Experiment
Light pulse creates excess minoritycarriers - in this case, excess holes,above thermal equilibrium. Excesscarriers drift due to the electric field.
Excess carriers drift due to theelectric field. A second kind ofcurrent due to diffusion takes place:
jdiff = −qDp∂p∂x
Excess carriers are collected at theterminals and a voltage pulse isobserved.
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Haynes-Shockley Experiment
Haynes-Shockley Experiment
Measure the time it takes the pulse to arrive to figure out carrier drift velocity andmobility:
vd = L/tmax ⇒ µp =vd
E=
L/tmax
V/L=
L2
V × tmax
Due to diffusion, the width of the pulse widens with time. The hole distribution can berepresented by a Gaussian:
p = pmaxe− (x−xmax)2
4Dpt
For (x − xmax)2 = 4Dpt, p = pmaxe. Identify this level in the measured pulse todetermine
∆x =p
4Dpt =q
4Dp(tmax + ∆t)
Using vd = ∆x∆t , vd = L
tmax, and solving for Dp:
Dp =((∆t × L)/tmax)
2
4(tmax + ∆t)
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