EE 5340Semiconductor Device TheoryLecture 05 – Spring 2011

Professor Ronald L. Carterronc@uta.edu

http://www.uta.edu/ronc

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Review the Following• R. L. Carter’s web page:

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

all EE 5340 pages when downloading to assure the latest version.) All links at:– www.uta.edu/ronc/5340/syllabus.htm

• University and College Ethics Policies– www.uta.edu/studentaffairs/conduct/

• Makeup lecture at noon Friday (1/28) in 108 Nedderman Hall. This will be available on the web.

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First Assignment

• 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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Second Assignment

• Submit a signed copy of the document posted at

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

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Schedule Changes Due to University Weather Closings• Make-up class will be held Friday, February 11 at 12 noon in 108 Nedderman Hall.

• Additional changes will be announced as necessary.

• Syllabus and lecture dates postings will be updated in the next 24 hours.

• Project Assignment will be posted in the next 36 hours.

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Intrinsic carrierconc. (MB limit)

• ni2 = no po = Nc Nv e-Eg/kT

• Nc = 2{2pm*nkT/h2}3/2

• Nv = 2{2pm*pkT/h2}3/2

• Eg = 1.17 eV - aT2/(T+b) a = 4.73E-4 eV/K b = 636K

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

<< ni, ni2 = NcNve-Eg/kT, ~1E-13

dopant level !• 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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Equilibriumconc (cont.)• For Nd > Na (taking the + root)

no = (Nd-Na)/2 + {[(Nd-Na)/2]2+ni

2}1/2

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

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

2/Nd2 + … ]

• So no = Nd, and po = ni2/Nd in the

limit of Nd >> Na and Nd >> ni

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n-type equilibriumconcentrations• N ≡ Nd - Na , n type N > 0• For all N,

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

• In most cases, N >> ni, sono = N, and

po = ni2/no = ni

2/N, (Law of Mass Action is al-

ways true in equilibrium)

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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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p-type equilibriumconcentrations• N ≡ Nd - Na , p type N < 0 • For all N,

po = |N|/2 + {[|N|/2]2+ni2}1/2

• In most cases, |N| >> ni, sopo = |N|, and

no = ni2/po = ni

2/|N|, (Law of Mass Action is al-

ways true in equilibrium)

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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 1st equation minus the 2nd 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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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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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

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by given is mobility total

the and , 11

Figure 1.16 (p. 31 M&K) Electron and hole mobilities in silicon at 300 K as functions of the total dopant concentration. The values plotted are the results of curve fitting measurements from several sources. The mobility curves can be generated using Equation 1.2.10 with the following values of the parameters [3] (see table on next slide).

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Figure 1.16 (cont. M&K)

Parameter Arsenic Phosphorus Boronμmin 52.2 68.5 44.9

μmax 1417 1414 470.5

Nref 9.68 X 1016 9.20 X 1016 2.23 X 1017

α 0.680 0.711 0.719

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refi NN

1minmax

min

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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

M&K and 1Device Electronics for Integrated Circuits, 2 ed., by Muller and Kamins, Wiley, New York, 1986.– See Semiconductor Device

Fundamen-tals, by Pierret, Addison-Wesley, 1996, for another treatment of the m model.

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

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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.