Transcript Semiconductor Device Modeling and Characterization – EE5342 Lecture 5 – Spring 2011 Professor Ronald L.

Semiconductor Device Modeling
and Characterization – EE5342
Lecture 5 – Spring 2011
Professor Ronald L. Carter
[email protected]
http://www.uta.edu/ronc/
First Assignment
• e-mail to [email protected]
– In the body of the message include
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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 of
semiconductors
• 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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Equilibrium
concentrations
• 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 + N d = n o + N a
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Equilibrium conc
n-type
• 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
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Equilibrium conc
p-type
• For Na > Nd
> Let N = Nd-Na, and (taking the + root)
po = (-N)/2 + {[-N/2]2+ni2}1/2
• For Na-= Na >> ni >> Nd we have
> po = Na, and
> no = ni2/Na
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Position of the
Fermi 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 midband
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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], so
Ec - Efi = kT ln(Nc/ni). Thus
EF - Efi = kT ln(no/ni) and for ntype
EF - Efi = kT ln(Nd/ni)
• Likewise
Efi - EF = kT ln(po/ni) and for ptype
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 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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Sample
calculations
• 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 electron
conc. and energies
 no 
 Ef  Ec 
no
 exp
, or Ef  Ec  kT ln ;
Nc
 kT 
 Nc 
 no 
 Ef  Efi 
no
 exp
, or Ef  Efi  kT ln ;
ni
 kT 
 ni 
 noNv 
 Nv 
and Ef  Ev  kT ln 2   kT ln 
 po 
 ni 
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Equilibrium hole
conc. and energies
po
 po 
 Ev  Ef 
 exp
, or Ev  Ef  kT ln ;
Nv
 kT 
 Nv 
 po 
po
 Efi  Ef 
 exp
, or Efi  Ef  kT ln ;
ni
 kT 
 ni 
 poNc 
 Nc 
and Ec  Ef  kT ln 2   kT ln 
 no 
 ni 
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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
1
all
1

, and the



tcoll collisions  ti 
total mobility m is given by
1
all
1


m total collisions  mi 
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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
I   J  dS
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Drift current
resistance
• 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 current
resistance (cont.)
• Consequently, since
R = rl/A
R = (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.
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