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ECE-656 Fall20171
NAME:______________________________________ PUID:______________________________________
ECE656Exam3SOLUTIONS:Fall2017
December12,2017MarkLundstromPurdueUniversity
Thisisaclosedbookexam.Youmayuseacalculatorandtheformulasheetattheendofthisexam.OnlytheTexasInstrumentsTI-30XIISscientificcalculatorisallowed.Therearefourequallyweightedquestions.Toreceivefullcredit,youmustshowyourwork(scratchpaperisattached).Theexamisdesignedtobetakenin75minutes.BesuretofillinyournameandPurduestudentIDatthetopofthepage.DONOTopentheexamuntiltoldtodoso,andstopworkingimmediatelywhentimeiscalled.Thelasttwopages,whichlistequationsmayberemoved,ifyouwish.100pointspossible,25perquestion
1)25points 2a)5points 3a)5points 4a)10points
2.5ptseach 2b)5points 3b)10points 4b)5points 2c)10points 3c)5points 4c)10points
2d)5points 3d)5points
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ECE-656 Fall20172
Exam3Solutions:ECE656Fall20171a) Whenwewritetherecombinationterminthevariousbalanceequationsas
Rφ = nφ − nφ
0( ) τφ ,sometimesatermcorrespondingto nφ appearsandaterm
correspondingtoitsequilibriumvalue, nφ
0 ,doesnotappear.Whendoesitnotappear?
a) Understeady-stateconditions.b) Underspatiallyuniformconditions.c) Whenthebalanceequationcorrespondstoamomenthigherthan2.d)Whenthebalanceequationcorrespondstoamomenthigherthan3.e)Whenthequantityinthebalanceequationisaflux.
1b) Whenwewriteadrift-diffusionequationintheform,
Jnj = nqµnE j + 2 3( )µn ∂W ∂x j ,whatassumptionarewemaking?
a) Non-degeneratecarrierstatistics.b) Thetemperaturedoesnotvarywithposition.c) Theelectrontemperatureisequaltothelatticetemperature.d) Thekineticenergyisequallydistributedbetweenthethreedegreesof
freedom.e) OnlythattheBTEisvalid.
1c) Whatdoesmomentequationdoes
φ !p( ) =υx p2 2m*( ) giveus?
a) Thecarriercontinuityequation.b) Thecarrierfluxequation.c) Thecarrierenergybalanceequation.d) Thecarrierenergyfluxequation.e) Thecarrierenergysquaredcontinuityequation.
1d) Whensimulating
r t( ), p t( )( ) ,inphasespacebytheMonteCarloapproach,whichofthefollowingistrue?
a) r t( )
iscontinuousand
p t( ) iscontinuous.b) r t( )
isdiscontinuousand
p t( ) iscontinuous.c) r t( )
iscontinuousand
p t( ) isdiscontinuous.d) r t( )
isdiscontinuousand
p t( ) isdiscontinuous.e) Noneoftheabove
ECE-656 Fall20173
Exam3Solutions:ECE656Fall20171e) Whatis“selfscattering”inaMonteCarlosimulation?
a) Amanybodyeffectinwhichanelectroninteractswithitself.b) Anelectron-electronscatteringeventinwhichanelectronscattersfromanother
electron.c) Anelectron-electronscatteringeventinwhichanelectronscattersfromthe
entireplasmaofalltheelectrons.d) Amathematicaltechniquethatsimplifiesthecomputationoffree-flight
times.e) Amathematicaltechniquethatsimplifiesthecomputationofthefinalscattering
state.
1f) Inpractice,onecommonlyextendsthenear-equilibriumdrift-diffusionequationto
high-fieldsbyreplacingthemobilityanddiffusioncoefficientsbyelectricfielddependentquantities,asin Jnx = nqµn E( )E x + qDn E( )dn dx .WhatassumptionisnecessarytowritetheDDequationinthisform?
a) Parabolicenergybands.b) Non-degeneratecarrierstatistics.c) Themicroscopicrelaxationtimeapproximation.d) Thattheenergyrelaxationtimeisshorterthanthemomentumrelaxationtime.e) Thattheshapeofthedistributionattheparticularlocation,whateveritis,
dependsonlyontheelectricfieldatthatlocation.1g) Intheclassicdescriptionofthevelocityvs.electricfieldcharacteristicinbulkSi,
υd =µn0E 1+ E E C( )2
,approximatelywhatisthemagnitudeof E C ?
a) ≈ 0.1kV cm .b) ≈1kV cm .c) ≈10 kV cm .d) ≈100 kV cm .e) ≈1000 kV cm .
ECE-656 Fall20174
Exam3Solutions:ECE656Fall20171h) Whatismeantbytheterm,“non-local”semiclassicaltransport.
a) TransportthatcannotbedescribedbyaDDequationwithafield-dependent
mobilityanddiffusioncoefficient.b) Steady-statetransportinanelectricfieldthatvariesmorerapidlyinspacethan
theenergyrelaxationlength,where Te istheelectrontemperature.c) Transienttransportinanelectricfieldthatvariesmorerapidlyintimethanthe
energyrelaxationtime.d) Alloftheabove.e) Noneoftheabove.
1i) Underwhatconditionsdoesvelocityovershootoccurforarapidlyvaryingelectric
field?a) Whentransportisballistic.b) Whentransportisquasi-ballistic.c) Whenthemomentumrelaxationtimeismuchshorterthantheenergy
relaxationtime.d) Whenthemomentumrelaxationtimeismuchlongerthantheenergyrelaxation
time.e) Whenthemomentumrelaxationtimeisnearlyequaltotheenergyrelaxation
time.1j) Whichofthefollowingistrueaboutaballisticdevicewithtwo,idealLandauer
contactsatdifferentvoltages?
a) ThedistributionfunctioninthedeviceisaFermi-DiracdistributionwiththeaverageFermilevelofthetwocontacts.
b) ThedistributionfunctioninthedeviceisaFermi-DiracdistributionwiththeFermilevelofthecontactwiththemorepositivepotential.
c) ThedistributionfunctioninthedeviceisaFermi-DiracdistributionwiththeFermilevelofthecontactwiththemorenegativepotential.
d) Eachstateinthedeviceisinequilibriumwithoneofthetwocontacts.e) Eachstateinthedeviceisinequilibriumwithboththetwocontacts
ECE-656 Fall20175
Exam3Solutions:ECE656Fall20172) MonteCarlosimulationsofhigh-fieldtransportinbulksiliconwithanelectricfieldof
100,000V/cmshowthefollowingresultsfortheaveragevelocityandkineticenergy:
υd = υ = 1.04×107 cm/s
u = KE = 0.364 eV Forthisproblem,youmayassumeasimpleparabolicbandwithaneffectivemassof
m* = mC
* = 0.26m0 .2a) Estimatetheaveragemomentumrelaxationtime, τ m .Anumericalansweris
required.Solution:
µn = υ E = 104 cm2 /V-s
µn =
q τ m
mc*
,where mc
* istheconductivityeffectivemass.UseMKSunitsforthecalculation:
τ m = q
µnmc* =
µnmc*
q= 104×10−4 × 0.26× 9.11×10−31
1.6×10−19 1.54×10−14
τ m = 1.54×10−14 s
2b) Estimatetheaverageenergyrelaxationtime, τ E .Anumericalansweris
required.Solution:
Fromtheenergybalanceequation:
JxE x = nq υ E x =
n u − u0( )τ E
τ E =
u − u0( )q υ E x
u0
q= 1.5
kBTL
q= 0.039
τ E =
u − u0( )q υ E x
=0.364− 0.039( )
1.04×107 ×105 = 0.313×10−12
τ E = 0.3 ps
ECE-656 Fall20176
Exam3Solutions:ECE656Fall20172c) Estimatethedriftenergyinelectronvolts.Anumericalanswerisrequired.Solution:
Edr =
12
m*υd2 = 0.5 0.26( ) 9.11×10−31( ) 1.04×105( )2
= 1.28×10−21 J
Edr eV( ) = 1.28×10−21
q= 8.01×10−3 eV
Edr eV( ) = 8.01×10−3 eV
2d) Estimatethethermalenergyinelectronvolts.Anumericalanswerisrequired.Solution:
Fromtheformulasheet: u = 1
2m*υd
2 + 32
kBTe
32
kBTe = u − 12
m*υd2 = 0.364− 0.008 = 0.356
32
kBTe = 0.356 eV
Virtuallyalloftheenergyisthermalenergy.
ECE-656 Fall20177
Exam3Solutions:ECE656Fall2017 3) Whenderivingthemomentumbalanceequationin3D,atensor,
Wij =
1Ω
υi p j2
f r , p,t( )p∑
occurs.Thisquestionisaboutthediagonalcomponentofthetensor, Wxx .Youmayassumeasimple,parabolicenergybandandspatialvariationinthex-directiononly.
3a) Usethebalanceequationprescriptiontoobtainanexpressionforthex-directedflux
associatedwith Wxx .Solution:
Fφx ≡
1Ω
12
m*υx2⎛
⎝⎜⎞⎠⎟υx f x, !p,t( )
p∑ = FWxx
.
FWxx
isafluxofthequantity, Wxx .Itcanalsobewrittenas
F
Wxx
≡ 1Ω
12
m*υx2⎛
⎝⎜⎞⎠⎟υx f
p∑ = n
12
m* υx3
3b) Usethebalanceequationprescriptiontoobtainanexpressionforthe“generation
rate”associatedwith Wxx .Solution:
Gφ = −qE x
1Ω
∂φ∂px
fp∑
⎧⎨⎪
⎩⎪
⎫⎬⎪
⎭⎪
Theterminsidethesumis
∂φ∂px
=∂ 12m*υx
2⎛⎝⎜
⎞⎠⎟
∂px= 12m*
∂ px2( )
∂px= υx
sowefind
Gφ = −qE i
1Ω
υx f!p∑
⎧⎨⎪
⎩⎪
⎫⎬⎪
⎭⎪= JnxE x
ECE-656 Fall20178
Exam3Solutions:ECE656Fall20173c) Writedownanexpressionforthe“recombinationrate”associatedwith Wxx .Solution:
Rφ ≡nφ − nφ
0
τφ
=Wxx −Wxx
0
τWxx Wxx
0 = n0
kBTL
2
3d) Writedownthetotalbalanceequationfor Wxx ,andexplainwhatwouldneedtobe
doneinordertosolvethisequationincombinationwiththefirstbalanceequation(thecontinuityequation)andthesecondbalanceequation(themomentumbalanceequation).
Solution:
∂Wxx
∂t= − d
dxFWxx( ) + JnxE x −
Wxx −Wxx0
τWxx
Tosolvethisequation,wewouldneedtoterminatethehierarchyanddevelopanapproximationfor
FWxx
intermsofthequantitiesinthepreviousbalanceequations.
Wewouldalsoneedtodevelopanexpressionfortherelaxationtime,
τWxx
.
ECE-656 Fall20179
Exam3Solutions:ECE656Fall20174) ThisproblemconcernstheballisticSchottkybarrierdiodeshowninthefigure
below.Notethatitisforwardbiased.Assumethatthesemiconductorissiliconwithaneffectivedensityofstatesof NC = 3.23×1019 cm-3 andthatthetemperatureis300K.Thequantity, Fn ,istheelectronquasi-Fermilevel(elecrochemicalpotential)inthesemiconductorand EFM istheFermilevelinthemetal.Notshownontheleftisanidealcontactinequilibrium.
Answerthefollowingquestions.Numericalanswersarerequired.
4a) Atthemetal-semiconductorjunction,whatisthedensityofballisticelectronsinthe
semiconductorwithpositivevelocities, n+ x = 0−( )?
Solution:Inabulk,diffusivesemiconductor,
n = NCe Fn−EC( ) kBT At x = 0− onlythepositivevelocitystatesarefilledbythecontactattheleft.Halfofthestateshavepositivevelocities,so
n+ x = 0−( ) = NC
2e Fn−EC( ) kBT = 3.23×1019
2e−0.1/0.026 = 3.45×1017 cm-3
n+ x = 0−( ) = 3.45×1017 cm-3
Thenon-degenerateassumptionwearemakingisclearlyOK.Exam3Solutions:ECE656Fall2017
ECE-656 Fall201710
4b) Atthemetal-semiconductorjunction,whatisthedensityofballisticelectronsinthe
semiconductorwithnegativevelocities, n− x = 0−( ) ?
Solution:Inabulk,diffusivesemiconductor,
n = NCe Fn−EC( ) kBT
At x = 0− onlythenegativevelocitystatesarefilledbythecontactattheright.Halfofthestateshavenegativevelocities,so
n− x = 0−( ) = NC
2e EFM −EC( ) kBT = 3.23×1019
2e−0.2/0.026 = 7.37 ×1015 cm-3
n− x = 0−( ) = 7.37 ×1015 cm-3
Thenon-degenerateassumptionwearemakingisclearlyOK.
4c) WhatisthecurrentdensityinAmperespercm2?Youwillneedaneffectivemassfor
thisquestion;youmayassumeasimpleparabolicbandwithaneffectivemassof
m* = mC
* = 0.26m0 .Solution:
Jn = qυT n+ 0−( )− n− 0−( )⎡
⎣⎤⎦
Fromtheformulasheet:
υx
+ =υT = 2kBT π m* ηF << 0( )
υT =2kBTπ m* =
2 1.38×10−23( )300
3.14 0.26( ) 9.11×10−31( ) = 1.06×105 m/s
Jn = qυT n+ 0−( )− n− 0−( )⎡
⎣⎤⎦ = 1.6×1019 1.06×107( ) 3.45×1017 − 7.37 ×1015⎡⎣ ⎤⎦ A/ cm2
Jn = 5.73×105 A/ cm2
ECE-656 Fall201711
ECE-656KeyEquations(Weeks1-15)
Physicalconstants:
= 1.055 ×10−34 J-s[ ] m0 = 9.109 ×10
−31 kg[ ] kB = 1.380 ×10
−23 J/K[ ] q = 1.602 ×10−19 C[ ] ε0 = 8.854 ×10−14 F/cm[ ]
--------------------------------------------------------------------------------------------------------------------Densityofstatesink-space:1D:
Nk =2 × L 2π( ) = L π 2D: Nk =2 × A 4π 2( ) = A 2π 2 3D:
Nk =2 × Ω 8π 2( ) = Ω 4π 3 Densityofstatesinenergy(parabolicbands,perlength,area,orvolume):
D1D E( ) = gv
π2m*
E − ε1( ) D2D E( ) = gV
m*
π2 D3D E( ) = gv
m* 2m* E − EC( )π 23
--------------------------------------------------------------------------------------------------------------------FermifunctionandFermi-DiracIntegrals/Bose-Einsteinfunction
f0 E( ) = 1
1+ e E−EF( ) kBT N0 =
1e!ω0 kBT −1
F j ηF( ) = 1
Γ( j +1)η jdη
1+ eη−ηF0
∞
∫ F j ηF( )→ eηF ηF << 0
dF j
dηF
=F j−1
Γ(n) = n −1( )! (naninteger) Γ(1 / 2) = π Γ( p +1) = pΓ( p) --------------------------------------------------------------------------------------------------------------------Scattering:
S !p, ′!p( ) = 2π
"H !′p , !p
2δ ′E − E − ΔE( )
H ′p , p =
1Ω
e− i ′p r
−∞
+∞
∫ US (r )ei pr dr
1τ p( ) == S p, ′p( )
′p ,↑∑
1τ mp( ) = S p, ′p( )Δpz
pz0′p ,↑∑
1τ Ep( ) = S p, ′p( )ΔE
E0′p ,↑∑
ADP: Kq
2= q2DA
2 ODP: Kq
2= D0
2 PZ: Kq
2= eePZ κ Sε0( )2 POP: Kβ
2= ρe2ω 0
2
q2κ 0ε0κ 0
κ∞
−1⎛⎝⎜
⎞⎠⎟
S !p, ′!p( ) = π
Ωρ ωKq
2Nω + 1
2∓
12
⎛⎝⎜
⎞⎠⎟δ ′!p − !p ∓ #!q( )δ ′E − E ∓ #ω( )
δ ′!p − !p ∓ #!q( )δ ′E − E ∓ #ωβ( )→ 1
#υqδ ±cosθ + #q
2 p∓ω q
υq⎛
⎝⎜
⎞
⎠⎟
1τ= 1τ abs
+ 1τ ems
= 2π
DA2kBTcl
⎛
⎝⎜⎞
⎠⎟D3D E( )
2(ADP)
LD =
κ Sε0kBTq2n0
1τ= 1τ m
= 2πDO
2
2ρω0
⎛
⎝⎜⎞
⎠⎟N0 +
12
12
⎛⎝⎜
⎞⎠⎟
D3D E ± ω0( )2
N0 =
1eωo kB T −1
(ODP)
ECE-656 Fall201712
BoltzmannTransportEquation:
∂ f∂t
+!υ i∇r f +
!Fe i∇ p f = df
dt coll
!Fe = −q
!E − q
!υ ×!B
dfdt coll
= −f − fS( )τ m
= −δ fτ m
(RTA)
WithasmallB-field,theresulting2Dcurrentequationis !Jn =σ S
!E -σ SµH
!E ×
!B( )
ω cτ m <<1( )
µH = µnrH rH ≡ τ m
2 τ m
2
• ≡ •( )E E
IsothermalNear-EquilibriumTransport:SummaryofLandauerApproach
I = 2q
hT E( )∫ M E( ) f1 − f2( )dE àsmallbias,isothermal: f1 E( )− f2 E( ) = − ∂ f0
∂E⎛⎝⎜
⎞⎠⎟ qV( )
Linearresponse(alsocalledlowbias,near-equilibrium):
I = GV G = 2q2
hT E( )∫ M E( ) −
∂ f0
∂E⎛⎝⎜
⎞⎠⎟
dE G = 2q2
hT E( ) M E( )
T E( ) ≡T E( )∫ M E( ) −∂ f0 ∂E( )dE
M E( ) −∂ f0 ∂E( )dE∫⎧⎨⎪
⎩⎪
⎫⎬⎪
⎭⎪
M E( ) = M E( ) −∂ f0 ∂E( )dE∫
Rball =
1M EF( )
h2q2 =
12.8kΩM
Modes/channels(general):
M E( ) ≡ h4
υx+ E( ) D1D E( ) υx
+ E( ) =υ E( )
M E( ) =WM 2D ≡W h4
υx+ E( ) D2D E( ) υx
+ E( ) = 2πυ E( )
M E( ) = AM 3D ≡ A h4
υx+ E( ) D3D E( ) υx
+ E( ) = 12υ E( )
Modes(Parabolicbands):
E k( ) = EC + !2k 2 2m*( ) Modes(graphene):
M E( ) = M1D E( ) = gv
M E( ) =W M2D E( ) = gvW
2m* E − EC( )π!
M (E) =W 2 E π!υF
M E( ) = A M3D E( ) = gv A m*
2π!2 E − EC( )
Transmission: T E( ) = λ E( )
λ E( ) + L
ECE-656 Fall201713
Mean-free-pathforbackscattering:
1D: λ E( ) = 2υ E( )τ m E( ) 2D: λ E( ) = π2υ E( )τ m E( ) 3D: λ E( ) = 4
3υ E( )τ m E( )
Diffusioncoefficient:
Dn = υx+ λ 2 Dn =υT λ0 2
Dn E( ) = υx+ E( ) λ E( ) 2
Uni-directionalthermalvelocity:
υx
+ =υT = 2kBT π m* ηF << 0( ) Thermoelectrictransport:Coupledcurrentequations(diffusive): Jx =σE x −σ S dT dx Jx
Q = Tσ SE x −κ 0 dT dx
Coupledcurrentequations(inverted): E x = ρJx + S
dTL
dx
Jx
Q = π Jx −κ e
dTdx
Transportcoefficients:
σ = ′σ E( )∫ dE =
2q2
hM λ
′σ E( ) = 2q2
hλ E( ) M E( )
A−∂ f0
∂E⎛⎝⎜
⎞⎠⎟
M = M E( ) −∂f0
∂E⎛
⎝⎜⎞
⎠⎟∫ dE
S = −
kB
qE − EF
kBT⎛
⎝⎜⎞
⎠⎟′σ E( )∫ dE ′σ E( )∫ dE = −
kB
q⎛
⎝⎜⎞
⎠⎟E − EF
kBT=
kB
−q⎛
⎝⎜⎞
⎠⎟EC − EF( )
kBT+
Δn
kBT
⎧⎨⎪
⎩⎪
⎫⎬⎪
⎭⎪ π = TLS (KelvinRelation)
κ 0 = TkB
q⎛
⎝⎜⎞
⎠⎟
2E − EF
kBT⎛
⎝⎜⎞
⎠⎟
2
′σ E( )∫ dE =κ 0 = σTkB
q⎛
⎝⎜⎞
⎠⎟
2E − EF
kBT⎛
⎝⎜⎞
⎠⎟
2⎧⎨⎪
⎩⎪
⎫⎬⎪
⎭⎪ave κ e =κ 0 − πSσ
κ e
σ=
kB
q⎛⎝⎜
⎞⎠⎟
2E − EF
kBT⎛
⎝⎜⎞
⎠⎟
2
−E − EF
kBT⎛
⎝⎜⎞
⎠⎟
2⎧⎨⎪
⎩⎪
⎫⎬⎪
⎭⎪T = LT (Weidemann-Franz“Law”)
2
kB
q⎛⎝⎜
⎞⎠⎟
2
< L < π 2
3kB
q⎛⎝⎜
⎞⎠⎟
2
(parabolicbandswithenergy-independentscattering)
ThermoelectricmaterialFigureofMerit: zT =
S 2σTκ
Latticethermalconductivity: κ L =
π 2kB2TL
3hλ ph !ω( )M ph !ω( )
AWph !ω( )d !ω( )∫
Generalbalanceequation