Semiconductor Device Modeling and Characterization – EE5342 Lecture 5 – Spring 2011 Professor...
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Transcript of Semiconductor Device Modeling and Characterization – EE5342 Lecture 5 – Spring 2011 Professor...
Semiconductor Device Modeling and
Characterization – EE5342 Lecture 5 – Spring 2011
Professor Ronald L. [email protected]
http://www.uta.edu/ronc/
©rlc L05-28Jan2011
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First Assignment
• e-mail to [email protected]– 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 [email protected], with EE5342 in subject line.
©rlc L05-28Jan2011
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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
©rlc L05-28Jan2011
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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.