EE 5340
Semiconductor Device Theory
Lecture 05 – Spring 2011
Professor Ronald L. Carter
ronc@uta.edu
http://www.uta.edu/ronc
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 carrier
conc. (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 of
semiconductors
• 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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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 (cont.)
• For Nd > Na (taking the + root)
no = (Nd-Na)/2 + {[(NdNa)/2]2+ni2}1/2
• For Nd >> Na and Nd >> ni, can use the
binomial expansion, giving
no = Nd/2 + Nd/2[1 + 2ni2/Nd2 + … ]
• So no = Nd, and po = ni2/Nd in the limit
of Nd >> Na and Nd >> ni
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n-type equilibrium
concentrations
• N ≡ Nd - Na , n type  N > 0
• For all N,
no = N/2 + {[N/2]2+ni2}1/2
• In most cases, N >> ni, so
no = N, and
po = ni2/no = ni2/N,
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(Law of Mass Action is always true in equilibrium)
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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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p-type equilibrium
concentrations
• N ≡ Nd - Na , p type  N < 0
• For all N,
po = |N|/2 + {[|N|/2]2+ni2}1/2
• In most cases, |N| >> ni, so
po = |N|, and
no = ni2/po = ni2/|N|,
(Law of Mass Action is always true in equilibrium)
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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 n-type
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 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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Example
calculations
• For Nd = 3.2E16/cm3, ni = 1.4E10/cm3
no = Nd = 3.2E16/cm3
po = ni2/Nd , (po is always ni2/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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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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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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m  m min 
m max  m min

1  N i N ref

a
Parameter
Arsenic
Phosphorus
Boron
μmin
μmax
Nref
α
52.2
1417
9.68 X 1016
0.680
68.5
1414
9.20 X 1016
0.711
44.9
470.5
2.23 X 1017
0.719
Figure 1.16 (cont. M&K)
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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
M&K and 1Device
Electronics for
Integrated Circuits, 2 ed., by Muller
and Kamins, Wiley, New York, 1986.
– See Semiconductor Device Fundamentals, 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.
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