SOLUTIONS: ECE 656 Homework (Week 4) Mark Lundstrom …

MarkLundstrom

ECE-656 Fall20171

SOLUTIONS:ECE656Homework(Week4)MarkLundstromPurdueUniversity

1) Repeattheelectron-phononenergy-momentumconservationargumentsdiscussedin

Sec.2.5ofFundamentalsofCarrierTransport,butthistimeassumeelectronsingraphene.

Solution:Beginwithenergyandmomentumconservation: ′E = E ± ω (*) !′p = !p ± "!q (**)Usethedispersionofgraphenetowrite:

E = υF k =υF p

E2 = υF p( )2

E = p2

E υF2( ) =

p2

m* E( ) m* E( ) ≡ E υF

2( ) Usingthisdefinitionofaneffectivemassforgraphene,(*)becomes

′p 2

m* = p2

m* ± ω (*’)

whichissimilartotheparabolicbandresult.

Nowtakethedotproductof(**)withitselftofind:

!′p i!′p = ′p 2 = p2 ± 2"!p i

!q + "2q2 nowdividethroughby m*

′p 2

m* = p2

m* ±2!"p i

"qm* + !

2q2

m*

anduse(*’)

±!ω = ± 2!"p i

"qm* + !

2q2

m* = ± 2!pqcosθm* + !

2q2

m*

MarkLundstrom

ECE-656 Fall20172

ECE656Homework(Week4)Solutions(continued)Solvingfor !

2q2 m* wefind

!2q2

m* = ∓ 2!pqcosθm* ± !ω

!β = 2 p ∓cosθ ± m*ω

2 pq⎡

⎣⎢

⎤

⎦⎥ ,

whichisastatementofenergyandmomentumconservationforelectronsingraphene.Nowusethedispersionofgrapheneandourdefinitionofeffectivemasstowrite

m*

p=

E υF2

E υF

= 1υF

whichcanbeusedtoexpressourrelationforenergy-momentumconservationas

!q =2 p ∓cosθ ± ω

2qυF

⎡

⎣⎢

⎤

⎦⎥

Notethatthisresultisalmostthesameasfortheparabolicbandresult(exceptforthefactorof2downstairs).Theconclusionswillbesimilar.2) Assumeatransitionrateoftheform:

S!p→ !′p( ) = Cδ E − ′E( )δ !′p − !p ∓ #!q( )

whereCisaconstant.Answerthefollowingquestionsassumingparabolicenergybands.

a) Deriveanexpressionfor

!q = q ,whichexpressesconservationofenergyandmomentum.

b) Usingtheresultsofa),determinetheminimumandmaximummagnitudeof !q .

MarkLundstrom

ECE-656 Fall20173

ECE656Homework(Week4)Solutions(continued)

Solution:a)Beginwithenergyandmomentumconservation:

′E = E (*) !′p = !p ± "!q (**)Sincethebandareparabolic,(*)canbewrittenas:

′p 2

2m* =p2

2m* ± ω (*’)

Nowtakethedotproductof(**)withitselftofind:

!′p i!′p = ′p 2 = p2 ± 2"!p i

!q + "2q2 nowdividethroughby 2m*

′p 2

2m* =p2

2m* ±2!"p i

"q2m* + !

2q2

2m*

anduse(*’)

′p 2

2m* −p2

2m* = 0 = ± 2!"p i"q

2m* + !2q2

2m* = ± 2!pqcosθ2m* + !

2q2

2m*

Solvingfor !q wefind

!q = ∓2 pcosθ ,whichisastatementofenergyandmomentumconservationonforelasticscattering.b)Notethat q isthemagnitudeof

!q ,soitmustalwaysbegreaterthanorequaltozero.Thelargest q willoccurfor

Absorption: cosθ = −1, θ = π Emission: cosθ = 1, θ = 0

Ineithercase, !qmax = 2 p Thesmallest q occursfor θ = π 2 forbothabsorptionandemission.Ineithercase, !qmin = 0 Weconcludethatforelasticscattering:

0 ≤ !q ≤ 2 p

MarkLundstrom

ECE-656 Fall20174

ECE656Homework(Week4)Solutions(continued)3) Thisproblemconcernselectronscatteringinbulk(3D)GaAs.Assumethattheoptical

phononenergyis ω0 = 35 meV.RecallthatGaAsisadirectgapsemiconductorandthattheLvalleys(along<111>)haveenergyminimathatare0.3eVabovetheΓ valleyminimum.ThefourXvalleys(along<100>)haveenergyminima0.5eVabovetheΓ valleyminimum.RecallthatΓ valleyelectronshavealighteffectivemassandthattheLandXvalleyelectronshavealarge(Si-like)effectivemass.Answerthefollowingquestions.

Solution:

a) Sketchthetotalelectronscatteringratevs.energyforelectronsintheΓ valley.

Labelallcriticalenergiesandgiveabriefexplanation(labelabsorptionandemissionprocessesseparately).AllenergiesshouldbereferredtothebottomoftheΓ valley,i..e. ECΓ = 0 .

Thescatteringrateisthesumof6differentprocesses;thelastfiveofthesehavethresholdsfortheironset.

1) POPAbsorption(ABS).Γ − Γ intravalleyscattering.2) POPEmission(EMS).Γ − Γ intravalleyscatteringplus1)3) OPABS. Γ − L intervalleyscatteringplus1)and2).4) OPEMS. Γ − L intervalleyscatteringplus1),2),and3).5) OPABS. Γ − X intervalleyscatteringplus1),2),3),and4).6) OPEMS. Γ − X intervalleyscatteringplus1),2),3),4),and5).

MarkLundstrom

ECE-656 Fall20175


b) SketchtheΓ toΓ electronscatteringratevs.energy.Labelallcriticalenergiesandgiveabriefexplanation(labelabsorptionandemissionprocessesseparately).

1) POPAbsorption(ABS).Γ − Γ intravalleyscattering.2) POPEmission(EMS).Γ − Γ intravalleyscattering

NotethatPOPABSisproportionalto N0 andPOPEMSisproportionalto N0 +1 ,soPOPEMSislargerinmagnitude.

c) SketchtheΓ toLelectronscatteringratevs.energy.Labelallcriticalenergiesand

giveabriefexplanation(labelabsorptionandemissionprocessesseparately).

3) OPABS. Γ − L intervalleyscattering.4) OPEMS. Γ − L intervalleyscattering

MarkLundstrom

ECE-656 Fall20176


NotethatEMSis,again,strongerthanABS.Alsonotethatthescatteringratesincreaseasthesquarerootofenergybecausetheygoesasthedensityoffinalstates(noPOPintervalleyscatteringbecauselargephononwavevectorsareneededforthistypeofscattering).

d) SketchoftheLtoΓ electronscatteringratevs.energy.Labelallcriticalenergies

andgiveabriefexplanation(labelabsorptionandemissionprocessesseparately).

1) OPABS. L − Γ intervalleyscattering.2) OPEMS. L − Γ intervalleyscattering

NotethatbothABSandEMSbeginatthesameenergybecauseelectronsatthebottomoftheL-valleycanscatterdowntotheΓ valleybyeitherabsorbingoremittingaphonon.Alsonotethatthemagnitudeofthescatteringratesissmallerthaninc)becausethefinalstatesinthiscasehaveavalleydegeneracyof1insteadof4asinc).Mostimportantly,notethatthescatteringratesdonotbeginatzero,becausetheyareproportionaltoaDOSinthefinalvalley,wheretheΓ valleyDOSisquitehigh. 1 τΓ−L ∝ DΓ E ± ω0( ) beginsatanenergy E = 0.3± ω0 ,wherethedensity-of-finalstates(intheΓ valley)isquitehigh.

e) SketchtheLtoLelectronscatteringratevs.energy.Labelallcriticalenergiesand

giveabriefexplanation(labelabsorptionandemissionprocessesseparately).

MarkLundstrom

ECE-656 Fall20177


AnelectronatthebottomofanL-valleycanabsorbaphononandscattertoanotherL-valley,butanelectronnearthebottomofanL-valleycannotscattertoanotherL-valleybyemission,becausetherearenoL-valleystatesavailableatthatenergy.

4) Thedeformationpotentialscatteringrateforopticalphononemission(ODPemission)

isdescribedby:

1τ= 2πD0

2

2ρω0

⎛

⎝⎜⎞

⎠⎟N0 +1( ) D3D E − ω0( )

2 .

Obtaintheenergyfluxrelaxationrateforthisscatteringprocess.

Solution:

Beginwiththedefinitionoftheenergyfluxrelaxationrate:

1τ FW

= S p, ′p( ) ′p ,↑∑

FWp( )− FW

′p( )FWp( )

⎡

⎣⎢⎢

⎤

⎦⎥⎥ (*)

Workontheterminbracketsandassumethattheinitialfluxisorientedalongthez-axis:

FWp( )− FW

′p( )FWp( )

⎡

⎣⎢⎢

⎤

⎦⎥⎥=

Eυz − E − ω0( ) ′υz

Eυz

= 1−E − ω0( )

E′υz

υz

⎛

⎝⎜⎞

⎠⎟= 1−

′υz

υz

⎛

⎝⎜⎞

⎠⎟+ω0

E′υz

υz

⎛

⎝⎜⎞

⎠⎟

MarkLundstrom

ECE-656 Fall20178


FWp( )− FW

′p( )FWp( )

⎡

⎣⎢⎢

⎤

⎦⎥⎥= 1−

′υz

υz

⎛

⎝⎜⎞

⎠⎟+ω0

E′υz

υz

⎛

⎝⎜⎞

⎠⎟= 1−

′υz

υz

⎛

⎝⎜⎞

⎠⎟−ω0

E1−

′υz

υz

⎛

⎝⎜⎞

⎠⎟+ω0

E

Ifweassumethattheeffectivemassisconstant(parabolicenergybands),then

m*υz = pz andwefind:

FWp( )− FW

′p( )FWp( )

⎡

⎣⎢⎢

⎤

⎦⎥⎥= 1−

′pz

pz

⎛

⎝⎜⎞

⎠⎟+ω0

E−ω0

E1−

′pz

pz

⎛

⎝⎜⎞

⎠⎟

Nowinsertthisexpressionin(*)

1τ FW

= S p, ′p( )′p ,↑∑ 1−

′pz

pz

⎛

⎝⎜⎞

⎠⎟+ω0

E−ω0

E1−

′pz

pz

⎛

⎝⎜⎞

⎠⎟⎡

⎣⎢⎢

⎤

⎦⎥⎥

= S p, ′p( ) 1−′pz

pz

⎛

⎝⎜⎞

⎠⎟+ S p, ′p( ) ω0

E⎛⎝⎜

⎞⎠⎟′p ,↑

∑′p ,↑∑ − S p, ′p( )

′p ,↑∑ ω0

E1−

′pz

pz

⎛

⎝⎜⎞

⎠⎟

1τ FW

= 1τ m

+ω0

E⎛⎝⎜

⎞⎠⎟

1τ−ω0

E⎛⎝⎜

⎞⎠⎟

1τ m

1τ FW

= 1τ m

1−ω0

E⎛⎝⎜

⎞⎠⎟+ω0

E⎛⎝⎜

⎞⎠⎟

1τ

Sinceweareassumingthatphononemissionoccurs,thisonlyappliesfor E > ω0 .

5) TheODPscatteringratefor2Delectronsis:

1τ n, ′n

⎞

⎠⎟

a,e

= πD0

2

ρω0

⎛

⎝⎜⎞

⎠⎟N0 +

12

12

⎛⎝⎜

⎞⎠⎟

D2 D E ± ω0( )2

2 +δn, ′n

2⎛

⎝⎜

⎞

⎠⎟

Let ω0 = 1.1kBT ,assumetwosubbands,andplottheabsorptionandemissionscatteringratesvs.energyforanelectroninsubbandone.

MarkLundstrom

ECE-656 Fall20179

ECE656Homework(Week4)Solutions(continued)Solution:Let’swritethescatteringrateas

1τ n, ′n

⎞

⎠⎟

a,e

= Γ0 N0 +12

12

⎛⎝⎜

⎞⎠⎟

2 +δn, ′n

2⎛

⎝⎜

⎞

⎠⎟

where

Γ0 =

πD0

2

ρω0

⎛

⎝⎜⎞

⎠⎟m*

2π2

N0 =

1eω0 kBT −1

= 1e1.1 −1

= 13−1

= 12

Nowworkoutthevariousscatteringrates:

1τ n, ′n

⎞

⎠⎟

a,e

= Γ0

12+ 1

2

12

⎛⎝⎜

⎞⎠⎟

2 +δn, ′n

2⎛

⎝⎜

⎞

⎠⎟ = Γ0 1 1

2⎛⎝⎜

⎞⎠⎟

2 +δn, ′n

2⎛

⎝⎜

⎞

⎠⎟

1τ1,1

⎞

⎠⎟

a

= Γ0

12

⎛⎝⎜

⎞⎠⎟

32

⎛⎝⎜

⎞⎠⎟= 3

4Γ0

1τ1,1

⎞

⎠⎟

e

= Γ0

32

⎛⎝⎜

⎞⎠⎟

32

⎛⎝⎜

⎞⎠⎟= 9

4Γ0

1τ1,2

⎞

⎠⎟

a

= Γ0

12

⎛⎝⎜

⎞⎠⎟

1( ) = 24Γ0

1τ1,2

⎞

⎠⎟

e

= Γ0

32

⎛⎝⎜

⎞⎠⎟

1( ) = 64Γ0

Totalscatteringratefor E > ε2

1τ1

⎞

⎠⎟ tot

= Γ0

34+ 9

4+ 2

4+ 6

4⎛⎝⎜

⎞⎠⎟= 5Γ0

MarkLundstrom

ECE-656 Fall201710


Thetotalscatteringrateforelectronsinsubband1isplottedbelow.Inthisplot,wetakethezeroofenergytobethebottomofthefirstsubband, E = ε1 = 0 .

6) UseargumentssimilartothoseinSec.2.2ofFCTandevaluatethemomentumrelaxationtimeforpiezoelectricscattering.

a) Showthatthescatteringpotentialis

U PZ =

eePZ

κ Sε0

u (2.29)ofFCT

where ePZ isthepiezoelectricconstantand u istheelasticwavedisplacement.Wehavereplacedthechargeonaelectron,q,withetoavoidconfusionwiththephononwavevector,

!q .

HINT:Beginwith D =κ Sε0E + ePZ

∂u∂x

b) Usethescatteringpotentialofparta)andevaluatethematrixelementforPZ

scattering.Showthattheresultis

H !′p , !p

2=

eePZ

κ Sε0

⎛

⎝⎜⎞

⎠⎟

2kBT

2cℓq2Ω

δ !′p − !p ∓ $!q( ) = Kq

2Aq

2δ !′p − !p ∓ $!q( )

c) Writeanexpressionforthetransitionrate, S

p, ′p( ) ,anddetermine Cq .

d) Evaluate 1 τ m assumingthatthescatteringiselastic.

MarkLundstrom

ECE-656 Fall201711


Solution:

a)Beginwith D =κ Sε0E + ePZ

∂u∂x (*)

Assumespacechargeneutrality: ∇iD =κ Sε0E + ePZ

∂u∂x

= 0

If, D ∝ eiqx ,thisimpliesthatD=0,sowefindtheelectricfieldas:

E = −

ePZ

κ Sε0

∂u∂x

= −iqePZ

κ Sε0

u

Thescatteringpotentialis

US = −e E dx∫ = e

ePZ

κ Sε0

u

U PZ = e

ePZ

κ Sε0

u

b)Evaluatethematrixelement

H ′p , p =

e−i ′p ir

Ω∫ US

ei ′p ir

Ωd 3r

Usetheresultsfromparta)

U PZ =

eePZ

κ Sε0

u = Kq Aqei!qi!r

Kq =

eePZ

κ Sε0 (eqn.2.59dofFCT)

H !′p , !p = Kq Aq

1Ω

ei !p− !′p ±"!qi!r( ) "∫ d 3r

Followingthetext,FCT,wefind

MarkLundstrom

ECE-656 Fall201712


H !′p , !p

2= Kq

2Aq

2δ !p − !′p ∓ #!q( ) eqn.(2.60ofFCT) (**)

Nowquantizethelatticevibrationsaccordingtoeqn.(2.71c)

Aq

2→ !

2ρΩω q

Nω + 12∓

12

⎛⎝⎜

⎞⎠⎟

andwefinallywritethematrixelement(**)as

H !′p , !p

2= Kq

2Aq

2δ !p − !′p ∓ #!q( ) = eePZ

κ Sε0

⎛

⎝⎜⎞

⎠⎟

2#

2ρΩω q

Nω + 12∓

12

⎛⎝⎜

⎞⎠⎟δδ !p − !′p ∓ #!q( ) (***)

c)Beginwiththetransitionrate

S !p, !′p( ) = 2π

"H !′p , !p

2δ !p − !′p ∓ "!q( )δ ′E − E ∓ "ω( )

Nowuse(***)forthematrixelementandeqn.(2.66)FCTforthetwodeltafunctionstofind

S !p, !′p( ) = Cq Nω + 1

2∓

12

⎛⎝⎜

⎞⎠⎟δ ′E − E ∓ #ω( )

where

Cq =

2π!2

eePZ

κ Sε0

⎛

⎝⎜⎞

⎠⎟

2!

2ρΩω1!υq

= 2π!2

eePZ

κ Sε0

⎛

⎝⎜⎞

⎠⎟

2!

2ρΩωm*

!pq× q

q

Cq =

eePZ

κ Sε0

⎛

⎝⎜⎞

⎠⎟

2πm*

!ρυSq2 pΩ (eqn.(2.73d)ofFCT

Here,wehaveused: ω q =υS and p = m*υ forthemomentumoftheelectron.

MarkLundstrom

ECE-656 Fall201713

ECE656Homework(Week4)Solutions(continued)d)Beginwitheqn.(2.80)inFCT.Assume

ωυq

=υS

υ<<1 (averageelectronvelocitymuchlessthanthephononvelocity,so

(2.80)becomes

1τ m

= Ω4π 2 Nω + 1

2∓

12

⎛⎝⎜

⎞⎠⎟qmin

qmax

∫ Cq

"q2 p"q3

pdq

Nowassumeequipartition:

N ≈ N +1=

kBTω

1τ m

=πm*e2ePZ

2 kBTcℓκ Sε0 2 p3( ) q dq

qmin

qmax

∫

Acousticphononscatteringiselastic: qmax = 2 p and qmin = 0

q dq

qmin

qmax

∫ =qmax

2

2−

qmin2

2= 2 p2

Finally,thescatteringratebecomes:

1τ m

=πm*e2ePZ

2 kBTcℓκ Sε0 p

∝ 1E

Themomentumrelaxationratedecreaseswithenergy–asexpectedforanelectrostaticscatteringmechanisms.

7) Foralloysofcompoundsemiconductorslike AlxGa1-xAs ,microscopicfluctuationsinthe

alloycomposition,x,produceperturbationsinthebandedges.Thetransitionrateforallowscatteringis

S p, ′p( ) = 2π

2

3π 2

16⎛⎝⎜

⎞⎠⎟ΔU

2

NΩδ ′E − E( )

whereNistheconcentrationofatomsand

MarkLundstrom

ECE-656 Fall201714


ΔU = x 1− x( ) χGaAs − χAlAs( ) with χ beingtheelectronaffinity.a) Explainwhythealloyscatteringratevanishesatx=0andatx=1.

b) Deriveanexpressionfor τ m foralloyscattering.

Solution:a)Forx=0,wehaveGaAs,whichisnotanalloy,sothereisnoalloyscattering.Forx=1,wehaveAlAs,whichusnotanalloy,sothereisnoalloyscattering.Thestrongestallowscatteringwilloccurwhenx=½.b)

S p, ′p( ) = C

Ωδ ′E − E( )

where

C = 2π2

3π 2

16⎛

⎝⎜⎞

⎠⎟ΔU

2

N

Sincethereisnodependenceonphononwavevector,thescatteringrateandmomentumrelaxationratesshouldbeequal.Sincethereisnodependentonβ ,wedonotneedtoworryaboutenergy-momentumconservationandcandothesumsimply.

1τ= 1τm

= S p, ′p( ) ′p ,↑∑ = C 1

Ωδ ′E − E( )

′p ,↑∑

Werecognizethesumasone-halfofthedensity-of-states,

1τ E( ) =

1τm E( ) = C

D3D E( )2

τm E( )∝ E−1/2

Whichispowerlawscatteringwithacharacteristicexponentof s = −1/ 2 .

MarkLundstrom

ECE-656 Fall201715


8) Acousticphononscatteringwasassumedtobeelasticwhenworkingoutthemomentumrelaxationrateineqn.(2.84)ofFCT.Repeatthecalculationbutdonotassumeelasticscattering.Showthattheresultisnearlyequaltoeqn.(2.84)nearroomtemperature.

Solution:Thetransitionrateforphononscatteringis:

S !p, !′p( ) = 2π

"Kq

2Aq

2δ !p − !′p ∓ "!q( )δ ′E − E ∓ "ω( )

Thetwodelta-functionscanbereplacedbyonethatexpressesenergyandmomentumconservation(eqn.(2.66)ofFCT):

δ !p − !′p ∓ #!q( )δ ′E − E ∓ #ω( )→ 1

#υqδ ±cosθ + #q

2 p∓ω q

υq⎛

⎝⎜

⎞

⎠⎟

sothetransitionratebecomes:

S !p, !′p( ) = 2π

"2υqKq

2Aq

2δ ±cosθ + "q

2 p∓ω q

υq⎛

⎝⎜

⎞

⎠⎟,

whichiseqn.(2.67)ofFCT.ForADPscattering:

Kq

2= q2DA

2 (eqn.(2.59a)ofFCT)

Aq

2→ !

2ρΩω q

Nωq+ 1

2∓

12

⎛⎝⎜

⎞⎠⎟ (eqn.(2.71c)ofFCT)

Sowefind:

S !p, !′p( ) = πm*DA

2

"ρ pυS

1Ω

Nωq+ 1

2∓

12

⎛⎝⎜

⎞⎠⎟δ ±cosθ + "q

2 p∓ω q

υq⎛

⎝⎜

⎞

⎠⎟

or

MarkLundstrom

ECE-656 Fall201716


S !p,

!′p( ) = Cq Nωq+ 1

2∓

12

⎛⎝⎜

⎞⎠⎟δ ±cosθ + #q

2 p∓ω q

υq⎛

⎝⎜

⎞

⎠⎟ (eqn.(2.72FCT)

Cq =

πm*DA2

!ρ pυS

1Ω (eqn.(2.73a)FCT)

FollowingFCT,wewritethemomentumrelaxationrateas:

1τ m

= S!p,!′p( )

!q ,↑∑ ∓#qcosθ( )

p= Ω

8π 3 dφo

2π

∫ S!p,!′p( ) ∓#qcosθ( )

p−1

+1

∫0

∞

∫ d cosθ( )q2dq

or

1τ m

= Ω4π 2 Cq Nωq

+ 12∓

12

⎛⎝⎜

⎞⎠⎟δ ±cosθ + "q

2 p∓ω q

υq⎛

⎝⎜

⎞

⎠⎟∓"qcosθ( )

p−1

+1

∫0

∞

∫ d cosθ( )q2dq

Nowassumeequipartition: Nωq

≈ Nωq+1≈

kBT!ω q

andusetheexpressionfor Cβ .

1τ m

=m*DA

2kBT4π!ρ p2υS

2 δ ±cosθ + !q2 p∓ω q

υq⎛

⎝⎜

⎞

⎠⎟ ∓cosθ( )

−1

+1

∫0

∞

∫ d cosθ( )q2dq

Integrateoverβ first:

1τ m

=m*DA

2kBT4π!ρ p2υS

2

!q2 p∓ω q

υq⎛

⎝⎜

⎞

⎠⎟

qmin

qmax

∫ q2dq =m*DA

2kBT4π!ρ p2υS

2

!q2 p∓υS

υ⎛⎝⎜

⎞⎠⎟qmin

qmax

∫ q2dq

wherewehaveusedtheacousticphonondispersionforthelaststep.

1τ m

=m*DA

2kBT4π!ρ p2υS

2

!2 p

qmax4

4−

qmin4

4

⎛

⎝⎜⎞

⎠⎟∓υS

υqmax

3

3−

qmin3

3

⎛

⎝⎜⎞

⎠⎟⎡

⎣⎢⎢

⎤

⎦⎥⎥

MarkLundstrom

ECE-656 Fall201717

ECE656Homework(Week4)Solutions(continued)Assume qmax >> qmin and qmax

4 >> qmax3

1τ m

=m*DA

2kBT8πρ p3υS

2

qmax4

4

⎛

⎝⎜⎞

⎠⎟

Energyandmomentumconservationgives(seeFCT,eqn.2.54)

!qmax = 2 p 1±

υS

υ⎛⎝⎜

⎞⎠⎟

qmax4

4= 4 p4

!4 1±υS

υ⎛⎝⎜

⎞⎠⎟

4

sothemomentumrelaxationratebecomes:

1τm

=m* pDA

2kBT2π4ρυS

2 1±υS

υ⎛

⎝⎜⎞

⎠⎟

4

=m* pDA

2kBT2π4c

1±υS

υ⎛

⎝⎜⎞

⎠⎟

4

Recallthatthe3Ddensity-of-statesis:

D3D E( ) = 2m*( )3/2

4π 23 E1/2

whichcanbeusedtore-expressthemomentumrelaxationrateas:

1τ m E( ) =

m*DA2kBTc

D3D E( )2

1±υS

υ⎛⎝⎜

⎞⎠⎟

4

,

whichisalmostexactlytheresultforelasticscattering(eqn.(2.84)ofFCT)sinceforatypicalelectron υ <<υS .

SOLUTIONS: ECE 656 Homework (Week 4) Mark Lundstrom …

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