Semiconductor Device Modeling and

Characterization – EE5342 Lecture 5 – Spring 2011

Professor Ronald L. Carterronc@uta.edu

http://www.uta.edu/ronc/

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First Assignment

• e-mail to listserv@listserv.uta.edu– In the body of the message include

subscribe EE5342 • This will subscribe you to the

EE5342 list. Will receive all EE5342 messages

• If you have any questions, send to ronc@uta.edu, with EE5342 in subject line.

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Second Assignment

• Submit a signed copy of the document that is posted at

www.uta.edu/ee/COE%20Ethics%20Statement%20Fall%2007.pdf

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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

• p-type: no < po, since Nd < Na

• 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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• For Nd > Na > Let N = Nd-Na, and (taking the + root)

no = (N)/2 + {[N/2]2+ni2}1/2

• For Nd+= Nd >> ni >> Na we have

> no = Nd, and

> po = ni2/Nd

Equilibrium conc n-type

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• For Na > Nd > Let N = Nd-Na, and (taking the + root)

po = (-N)/2 + {[-N/2]2+ni

2}1/2

• For Na-= Na >> ni >> Nd we have

> po = Na, and

> no = ni2/Na

Equilibrium conc p-type

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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)/ni2]

• 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).

Thus EF - Efi = kT ln(no/ni) and for n-type EF - 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), andEfi - Ev = kT ln(Nv/ni)

• The sum of the two equations gives Efi = (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 thatEc - 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 tcoll, a collision occurs which “resets” the velocity to <vx(tcoll)> = 0, then <vx> = qExtcoll/m* = mEx

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

mobility.• The mean time between collisions,

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

• Hence mthermal = qtthermal/m*, etc.

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

to the scattering is 1/ti, then the total scattering rate, 1/tcoll 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 Law

J = (nqmn+pqmp)(Exi+ Eyj+ Ezk), so

J = (sn + sp)E = sE, where

s = nqmn+pqmp defines the conductivity

• The net current is SdJI

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Drift currentresistance• Given: a semiconductor resistor

with length, l, and cross-section, A. What is the resistance?

• As stated previously, the conductivity,

s = nqmn + pqmp

• So the resistivity, r = 1/s = 1/(nqmn +

pqmp)

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Drift currentresistance (cont.)• Consequently, since

R = rl/AR = (nqmn + pqmp)-1(l/A)

• For n >> p, (an n-type extrinsic s/c)

R = l/(nqmnA)• For p >> n, (a p-type extrinsic s/c)

R = l/(pqmpA)

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References *Fundamentals of Semiconductor Theory and

Device Physics, by Shyh Wang, Prentice Hall, 1989.

**Semiconductor Physics & Devices, by Donald A. Neamen, 2nd ed., Irwin, Chicago.

M&K = Device Electronics for Integrated Circuits, 3rd ed., by Richard S. Muller, Theodore I. Kamins, and Mansun Chan, John Wiley and Sons, New York, 2003.

• 1Device Electronics for Integrated Circuits, 2 ed., by Muller and Kamins, Wiley, New York, 1986.

• 2Physics of Semiconductor Devices, by S. M. Sze, Wiley, New York, 1981.