Semiconductor Device
Modeling and Characterization
EE5342, Lecture 3-Spring 2002
Professor Ronald L. Carter
[email protected]
http://www.uta.edu/ronc/
L3 January 22
1
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
L3 January 22
2
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 + Nd = no + Na
L3 January 22
3
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
L3 January 22
4
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
L3 January 22
5
Electron Conc. in
the MB approx.
• Assuming the MB approx., the
equilibrium electron concentration is
no 
Emax
 gc E fF E dE
Ec
  Ec  EF  
no  Nc exp 

kT


L3 January 22
6
Hole Conc in
MB approx
• Similarly, the equilibrium hole
concentration is
po = Nv exp[-(EF-Ev)/kT]
• So that nopo = NcNv exp[-Eg/kT]
• ni2 = nopo, Nc,v = 2{2pm*n,pkT/h2}3/2
• Nc = 2.8E19/cm3, Nv = 1.04E19/cm3
and ni = 1E10/cm3
L3 January 22
7
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
L3 January 22
8
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)
L3 January 22
9
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 p-type
Efi - EF = kT ln(Na/ni)
L3 January 22
10
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
L3 January 22
11
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
L3 January 22
12
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 
L3 January 22
13
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 
L3 January 22
14
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
L3 January 22
15
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.
L3 January 22
16
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
L3 January 22

all
1


m total collisions  mi 
17
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
L3 January 22
18
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)
L3 January 22
19
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)
L3 January 22
20
Drift current
resistance (cont.)
• Note: for an extrinsic semiconductor
and multiple scattering mechanisms,
since
R = l/(nqmnA) or l/(pqmpA), and
(mn or p total)-1 = S mi-1, then
Rtotal = S Ri (series Rs)
• The individual scattering mechanisms
are: Lattice, ionized impurity, etc.
L3 January 22
21
Exp. mobility model
function for Si1
min
mn, p  mn,
p 
Parameter
mmin
mmax
Nref
a
L3 January 22
max
min
mn,

m
p
n, p
 Nd, a 

1  

 Nref 
As
52.2
1417
9.68e16
0.680
a
P
68.5
1414
9.20e16
0.711
B
44.9
470.5
2.23e17
0.719
22
Mobility (cm^2/V-sec)
Exp. mobility model
for P, As and B in Si
1500
1000
P
As
500
B
0
1.E+13
1.E+15
1.E+17
1.E+19
Doping Concentration (cm^-3)
L3 January 22
23
Carrier mobility
functions (cont.)
• The parameter mmax models 1/tlattice
the thermal collision rate
• The parameters mmin, Nref and a model
1/timpur the impurity collision rate
• The function is approximately of the
ideal theoretical form:
1/mtotal = 1/mthermal + 1/mimpurity
L3 January 22
24
Carrier mobility
functions (ex.)
• Let Nd = 1.78E17/cm3 of phosphorous,
so mmin = 68.5, mmax = 1414, Nref =
9.20e16 and a = 0.711. Thus mn = 586
cm2/V-s
• Let Na = 5.62E17/cm3 of boron, so mmin
= 44.9, mmax = 470.5, Nref = 9.68e16 and
a = 0.680. Thus mn = 189 cm2/V-s
L3 January 22
25
Lattice mobility
• The mlattice is the lattice scattering
mobility due to thermal vibrations
• Simple theory gives mlattice ~ T-3/2
• Experimentally mn,lattice ~ T-n where n =
2.42 for electrons and 2.2 for holes
• Consequently, the model equation is
mlattice(T) = mlattice(300)(T/300)-n
L3 January 22
26
Ionized impurity
mobility function
• The mimpur is the scattering mobility
due to ionized impurities
• Simple theory gives mimpur ~ T3/2/Nimpur
• Consequently, the model equation is
mimpur(T) = mimpur(300)(T/300)3/2
L3 January 22
27
Net silicon (extrinsic) resistivity
• Since
r = s-1 = (nqmn + pqmp)-1
• The net conductivity can be obtained
by using the model equation for the
mobilities as functions of doping
concentrations.
• The model function gives agreement
with the measured s(Nimpur)
L3 January 22
28
Resistivity (ohm-cm)
Net silicon extr
resistivity (cont.)
1.00E+03
1.00E+02
P
1.00E+01
As
1.00E+00
B
1.00E-01
1.00E-02
1.E+13
1.E+15
1.E+17
1.E+19
Doping Concentration (cm^-3)
L3 January 22
29
Net silicon extr
resistivity (cont.)
• Since
r = (nqmn + pqmp)-1, and
mn > mp, (m = qt/m*) we have
rp > rn
• Note that since
1.6(high conc.) < rp/rn < 3(low conc.), so
1.6(high conc.) < mn/mp < 3(low conc.)
L3 January 22
30
Net silicon (compensated) res.
• For an n-type (n >> p) compensated
semiconductor, r = (nqmn)-1
• But now n = N = Nd - Na, and the
mobility must be considered to be
determined by the total ionized
impurity scattering Nd + Na = NI
• Consequently, a good estimate is
= (nqmn)-1 = [Nqmn(NI)]-1
L3 January 22
r
31
References
• 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.
L3 January 22
32