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EE 5340Semiconductor Device TheoryLecture 4 - Fall 2003

Professor Ronald L. Carterronc@uta.edu

http://www.uta.edu/ronc

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Web Pages

* If you have not already done so:• R. L. Carter’s web page

– www.uta.edu/ronc/

• EE 5340 web page and syllabus– www.uta.edu/ronc/5340/syllabus.htm

• University and College Ethics Policies– www2.uta.edu/discipline/– www.uta.edu/ronc/5340/COE_EthicsStatement_Fall02.htm

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Assignment 1: If you have not received response from me ...• Send e-mail to ronc@uta.edu

– On the subject line, put “5340 e-mail”– In the body of message include

• email address: ______________________• Your Name*: _______________________• Last four digits of your Student ID: _____

* Your name as it appears in the UTA Record - no more, no less

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Assignment 2: 930AM stu-dents only*• IF you have a class OTHER than 5340

at 8 AM, e-mail to ronc@uta.edu– Subject line: “5340 8 AM Class”– In the body of message include

• email address: ______________________• Your Enrollment Name*: ______________• Class you are taking at 8:00 AM: ________

* If you don’t do this, and don’t take Test 1 at 8:00 AM, you will not get credit for Test 1

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Classes ofsemiconductors• Intrinsic: no = po = ni, since Na&Nd << ni

=[NcNvexp(-Eg/kT)]1/2, (not easy to get)

• n-type: no > po, since Nd > Na, noNd-Na

• p-type: no < po, since Nd < Na, poNa-Nd

• Compensated: no=po=ni, w/ Na- = Nd

+ > 0

• Note: n-type and p-type are usually partially compensated since there are usually some opposite- type dopants

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Equilibriumconcentrations• Charge neutrality requires

q(po + Nd+) + (-q)(no + Na

-) = 0

• Assuming complete ionization, so Nd

+ = Nd and Na- = Na

• Gives two equations to be solved simultaneously

1. Mass action, no po = ni2, and

2. Neutrality po + Nd = no + Na

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Equilibriumconc (cont.)• For Nd > Na (taking the + root) no

= (Nd-Na)/2 + {[(Nd-Na)/2]2+ni2}1/2

• For Nd >> Na and Nd >> ni, can use the binomial expansion, giving

no = Nd/2 + Nd/2[1 + 2ni2/Nd

2 + … ]

• So no = Nd, and po = ni2/Nd in the limit

of Nd >> Na and Nd >> ni

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Equilibriumconc (cont.)• For Na > Nd (taking the + root) po

= (Na-Nd)/2 + {[(Na-Nd)/2]2+ni2}1/2

• For Na >> Nd and Na >> ni, can use the binomial expansion, giving

po = Na/2 + Na/2[1 + 2ni2/Na

2 + … ]

• So po = Na in the limit of Na >> Nd and Na >> ni

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Examplecalculations• For Nd = 3.2E16/cm3, ni = 1.4E10/cm3

no = Nd = 3.2E16/cm3

po = ni2/Nd , (po is always ni

2/no)

= (1.4E10/cm3)2/3.2E16/cm3

= 6.125E3/cm3 (comp to ~1E23 Si)

• For po = Na = 4E17/cm3,

no = ni2/Na = (1.4E10/cm3)2/4E17/cm3

= 490/cm3

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Position of theFermi Level• Efi is the Fermi level

when no = po

• Ef shown is a Fermi level for no > po

• Ef < Efi when no < po

• Efi < (Ec + Ev)/2, which is the mid-band

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EF relative to Ec and Ev• Inverting no = Nc exp[-(Ec-EF)/kT] gives

Ec - EF = kT ln(Nc/no) For n-type material: Ec - EF =kTln(Nc/Nd)=kTln[(Ncpo)/ni

2]

• Inverting po = Nv exp[-(EF-Ev)/kT] gives EF - Ev = kT ln(Nv/po) For p-type material: EF - Ev = kT ln(Nv/Na)

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EF relative to Efi

• Letting ni = no gives Ef = Efi

ni = Nc exp[-(Ec-Efi)/kT], soEc - Efi = kT ln(Nc/ni). ThusEF - Efi = kT ln(no/ni) and for n-typeEF - Efi = kT ln(Nd/ni)

• Likewise Efi - EF = kT ln(po/ni) and for p-type Efi - EF = kT ln(Na/ni)

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Locating Efi in the bandgap • Since Ec - Efi = kT ln(Nc/ni),

and Efi - Ev = kT ln(Nv/ni)

• The 1st equation minus the 2nd givesEfi = (Ec + Ev)/2 - (kT/2) ln(Nc/Nv)

• Since Nc = 2.8E19cm-3 > 1.04E19cm-3 = Nv, the intrinsic Fermi level lies below the middle of the band gap

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Samplecalculations• Efi = (Ec + Ev)/2 - (kT/2) ln(Nc/Nv), so at

300K, kT = 25.86 meV and Nc/Nv = 2.8/1.04, Efi is 12.8 meV or 1.1% below mid-band

• For Nd = 3E17cm-3, given that Ec - EF = kT ln(Nc/Nd), we have Ec - EF

= 25.86 meV ln(280/3), Ec - EF = 0.117 eV =117meV ~3x(Ec - ED) what Nd

gives Ec-EF =Ec/3

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Equilibrium electronconc. and energies

o

v2i

vof

i

ofif

fif

i

o

c

ocf

cf

c

o

pN

lnkTn

NnlnkTEvE and

;nn

lnkTEE or ,kT

EEexp

nn

;Nn

lnkTEE or ,kT

EEexp

Nn

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Equilibrium hole conc. and energies

o

c2i

cofc

i

offi

ffi

i

o

v

ofv

fv

v

o

nN

lnkTn

NplnkTEE and

;np

lnkTEE or ,kT

EEexp

np

;Np

lnkTEE or ,kT

EEexp

Np

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Carrier Mobility

• In an electric field, Ex, the velocity (since ax = Fx/m* = qEx/m*) is

vx = axt = (qEx/m*)t, and the displ

x = (qEx/m*)t2/2

• If every coll, a collision occurs which “resets” the velocity to <vx(coll)> = 0, then <vx> = qExcoll/m* = Ex

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Carrier mobility (cont.)• The response function is the

mobility.• The mean time between collisions,

coll, may has several important causal events: Thermal vibrations, donor- or acceptor-like traps and lattice imperfections to name a few.

• Hence thermal = qthermal/m*, etc.

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Carrier mobility (cont.)• If the rate of a single contribution

to the scattering is 1/i, then the total scattering rate, 1/coll is

all

collisions itotal

all

collisions icoll

11

by given is mobility total

the and , 11

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Drift Current

• The drift current density (amp/cm2) is given by the point form of Ohm LawJ = (nqn+pqp)(Exi+ Eyj+ Ezk), so

J = (n + p)E = E, where

= nqn+pqp defines the conductivity

• The net current is SdJI

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Net silicon extrresistivity (cont.)• Since = (nqn + pqp)-1, and

n > p, ( = q/m*) we have

p > n

• Note that since1.6(high conc.) < p/n < 3(low conc.), so

1.6(high conc.) < n/p < 3(low conc.)