ECE-305 Spring20161

NAME:_________________________________________________PUID:_____________________________________

SECTION:Circleone: Alam Lundstrom

SOLUTIONS:ECE305Exam4:Spring2016March28,2016

M.A.AlamandM.S.LundstromPurdueUniversity

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ECE-305 Spring20162

Exam4Solutions:ECE305 Spring2016

Answerthefivemultiplechoicequestionsbelowbydrawingacirclearoundtheone,bestanswer.1a) WhichofthefollowingistrueoftheQFL’s, Fn x( ) and Fn x( ) ,inaforward-biased

NPjunction?a)

Fn x( ) = Fp x( ) everywhere.

b) Fn x( ) ≥ Fp x( ) everywhere.

c) Fn x( ) ≥ Fp x( ) onlyontheN-side.

d) Fn x( ) ≥ Fp x( ) onlyinthetransitionregion.

e) Fn x( ) ≥ Fp x( ) onlyontheP-side.

1b) Assumethat np = ni

2eqVA kBT holdsatthebeginningoftheP-region( x = xp )andthat

low-levelinjectionapplies.Whichofthefollowingistrue?a) n = n0 = ni

2 N A , p = p0 ≈ N A .

b) n = ni

2 N A( )eqVA kBT , p = p0 ≈ N A .

c) n = n0 = ni2 N A , p = p0e

qVA kBT ≈ N AeqVA kBT .d) n = N A, p = p0 ≈ ni

2 N A .

e) n = N A, p = ni

2 N A( )eqVA kBT .1c) Ionimplantationisusedto:

a) Dopeasemiconductor.b) Depositdielectricthinfilmsonasemiconductor.c) Depositmetallicthinfilmsonasemiconductor.d) Growcrystallinelayersonasemiconductor.e) Patternthinfilmsonasemiconductor.

1d) ConsideranNPjunctionwith N A >> N D .Whichoneofthefollowingistrue?

a) MostofthedepletionlayerisontheP-side.b) Mostoftheforward-biasedcurrentisduetoelectrons.c) Mostofthereverse-biasedcurrentisduetoelectrons.d) MostofthebandbendingisontheP-side.e) TheelectricfieldpeaksattheNPjunction.

ECE-305 Spring20163

Exam4Solutions:ECE305 Spring20161e) Thelithographyprocesstypicallyconsistsofseveralsteps-firstresistapplication,

thenprebake,thenexposure,andthenwhat?a) Etching.b) Hardbake.c) Resiststripping.d) Epitaxy.e) Develop

2) AnswerthefollowingquestionsaboutMetal-Silicon(MS)junctions.2a) IfweformanMSdiodefromthemetalandsemiconductorshownbelow,whatisthe

magnitudeofthebuiltinvoltageinvolts?(Anumericalanswerisrequired.)

Solution:

qVbi = ΦM −ΦS = 4.5− 4.05+ 0.10( ) = 0.35 eV

Vbi = 0.35 V

ECE-305 Spring20164

Exam4Solutions:ECE305 Spring2016Considerametal-semiconductordiode.Theequilibriumelectricfieldvs.positioninthesemiconductor, E x( ) ,issketchedbelowwhere x1 = 0.05 ×10−4 cm and xn = 0.20 ×10

−4 cm .

2b) Sketchthespacechargedensityinthesemiconductor, ρ x( ) ,ontheaxesbelow.Explainyoursketch.

Solution:BeginwiththePoissonequation:

dEdx

= ρKSε0

0 < x < x1,E x( ) constant ⇒ ρ x( ) = 0

x1 < x < xn ,E x( ) constant, positive slope ⇒ ρ x( ) > 0 and constant

x > xn ,E x( ) constant ⇒ ρ x( ) = 0

ECE-305 Spring20165

Exam4Solutions:ECE305 Spring20162c) Whatisthenumericalvalueofthedopingdensityfor 0 < x < x1?

Solution:

Nocharge–impliesnodopingandnegligiblecarrierdensity(i.e.depletionapprox.).

dEdx

= 0 = ρKSε0

⇒ ρ = 0⇒ N D = 0

N D = 0

2d) Whatisthenumericalvalueofthedopingdensityfor x1 < x < xn ?(Hint.Thelinearity

oftheelectricfieldallowsyoutocalculatethisquantityeasily).

Solution:Positivecharge–impliesN-typedopingandnegligiblecarrierdensity(i.e.depletionapproximation).

dEdx

= ρKSε0

=qN D

KSε0

N D =

KSε0

qdEdx

= 11.8×8.854×10−14

1.6×10−19 ×0− −7 ×104( )( )

0.15×10−4 = 3.05×1016

N D = 3.05×1016 cm-3

2e) Whatisthenumericalvalueofthebuilt-inpotential, Vbi ?Notethatyoucannotusethe

p-njunctionformula, Vbi = kBT q( )ln N AN D ni

2( ) forMSdiodes.

Solution:

Vbi = E 0( ) x1 +

E 0( )2

xn − x1( )

Vbi =

E 0( )2

x1 + xn( ) = 7 ×104

20.05+ 0.20( )×10−4 = 0.875

Vbi = 0.875 V

ECE-305 Spring20166

Exam4Solutions:ECE305 Spring2016

3) Asilicondiodeisasymmetricallydopedat N D = 1019 cm-3and N A = 1017 cm-3.Answerthefollowingquestionsassumingroomtemperature( kBT q = 0.026 V ).AssumethatthelengthsoftheneutralNandPregionsare L = 1000 µm .TwoidealmetalcontactsontheendsoftheN-andP-regionsmaintainequilibriumcarrierdensitiesunderallconditions.

TheminorityelectronconcentrationontheP-sideissketchedbelowforacertainforwardbias.

3a) IstheP-sideinlow-levelinjection?Explainwhyorwhynot.

Solution:YES

Δn x > xp( ) ≤ 7.2×1013 << p0 = N A = 1017

Excessminorityelectronconcentrationismuchlessthanthemajorityconcentration.

ECE-305 Spring20167

Exam4Solutions:ECE305 Spring2016

3b) Determinetheforwardbiasinvoltsusingthelawofthejunction.

Solution:

Δn xp( ) = ni

2

N A

eqVA kBT −1( ) ≈ ni2

N A

eqVA kBT = 7.2×1013

eqVA kBT = 7.2×1013

ni2 N A

= 7.2×1013

103 = 7.2×1010

VA = 0.026ln 7.2×1010( ) = 0.65

VA = 0.65 V

3c) Determinetheminoritycarrierdiffusionlength(incm)basedonthepairofcarrierconcentrationsat

xp and

xp +10×10−4 cmasindicatedabove.

Solution:

Δn x( ) = Δn xp( )e− x−xp( ) Ln

Δn xp +10×10−4cm( ) = Δn xp( )

2= Δn xp( )e− 10×10−4( ) Ln

Ln =

10×10−4

ln2= 14.4×10−4

Ln = 14.4×10−4 cm = 14.4 µm

3d) Determineanumericalvaluefortheholequasi-Fermilevel, Ei x( )− Fp x( ) ,at

x = xp +10 ×10−4 cm .(Hint.Findtheformulathatrelatescarrierconcentrationto

theintrinsicFermilevelofthesemiconductor).

Solution:

p x( ) ≈ p0 = nieEi−Fp x( )( ) kBT (lowlevelinjection)

Ei − Fp x( ) = 0.026ln 1017 1010( ) = 0.419 eV

Ei − Fp xp +10×10−4cm( ) = 0.419 eV

ECE-305 Spring20168

Exam4Solutions:ECE305 Spring2016

3e) Determineanumericalvaluefortheelectronquasi-Fermilevel, Fn x( )− Ei x( ) ,atx = xp +10 ×10

−4 cm .Solution:

n x( ) ≈ Δn x( ) = nieFn x( )−Ei( ) kBT (lowlevelinjection)

Fn x( )− Ei = 0.026ln 3.6×1013 1010( ) = 0.213eV

Fn xp +10×10−4cm( )− Ei = 0.213eV

Notealsothat

Fn xp +10×10−4cm( )− Fp xp +10×10−4cm( ) = 0.632 eV islessthanthe

appliedbias,sotheQFL’saresplitbylessthantheyareat xp .