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

Manuel Toledo Transport 1/ 32

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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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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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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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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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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)


Carrier transport:Drift and Diffusionece.uprm.edu/~mtoledo/5209/2012/transport.pdf · semiconductor...

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Transcript of Carrier transport:Drift and Diffusionece.uprm.edu/~mtoledo/5209/2012/transport.pdf · semiconductor...