semiconductor device modeling and characterization ee5342, lecture 3-spring 2002
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Semiconductor Device Modeling and Characterization EE5342, Lecture 3-Spring 2002. Professor Ronald L. Carter [email protected] http://www.uta.edu/ronc/. Classes of semiconductors. Intrinsic : n o = p o = n i , since N a &N dTRANSCRIPT
L3 January 22 1
Semiconductor Device Modeling and CharacterizationEE5342, Lecture 3-Spring 2002
Professor Ronald L. [email protected]
http://www.uta.edu/ronc/
L3 January 22 2
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
L3 January 22 3
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
L3 January 22 4
• For Nd > Na
>Let N = Nd-Na, and (taking the + root)no = (N)/2 + {[N/2]2+ni
2}1/2
• For Nd+= Nd >> ni >> Na we have
>no = Nd, and
>po = ni2/Nd
Equilibrium conc n-type
L3 January 22 5
• 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
L3 January 22 6
Electron Conc. inthe MB approx.• Assuming the MB approx., the
equilibrium electron concentration is
kTEE
expNn
dEEfEgn
Fcco
E
Eco F
max
c
L3 January 22 7
Hole Conc in MB approx• Similarly, the equilibrium hole
concentration ispo = Nv exp[-(EF-Ev)/kT]
• So that nopo = NcNv exp[-Eg/kT]
• ni2 = nopo, Nc,v = 2{2m*n,pkT/h2}3/2
• Nc = 2.8E19/cm3, Nv = 1.04E19/cm3 and ni = 1E10/cm3
L3 January 22 8
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
L3 January 22 9
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)/ni
2]
• 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 10
EF relative to Efi
• Letting ni = no gives Ef = Efi
ni = Nc exp[-(Ec-Efi)/kT], soEc - Efi = kT ln(Nc/ni). ThusEF - Efi = kT ln(no/ni) and for n-typeEF - 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 11
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 12
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 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 13
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
L3 January 22 14
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
L3 January 22 15
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 coll, a collision occurs which “resets” the velocity to <vx(coll)> = 0, then <vx> = qExcoll/m* = Ex
L3 January 22 16
Carrier mobility (cont.)• The response function is the
mobility.• The mean time between collisions,
coll, may has several important causal events: Thermal vibrations, donor- or acceptor-like traps and lattice imperfections to name a few.
• Hence thermal = qthermal/m*, etc.
L3 January 22 17
Carrier mobility (cont.)• If the rate of a single contribution
to the scattering is 1/i, then the total scattering rate, 1/coll is
all
collisions itotal
all
collisions icoll
11
by given is mobility total
the and , 11
L3 January 22 18
Drift Current
• The drift current density (amp/cm2) is given by the point form of Ohm LawJ = (nqn+pqp)(Exi+ Eyj+ Ezk), so
J = (n + p)E = E, where
= nqn+pqp defines the conductivity
• The net current is
SdJI
L3 January 22 19
Drift currentresistance• Given: a semiconductor resistor
with length, l, and cross-section, A. What is the resistance?
• As stated previously, the conductivity,
= nqn + pqp
• So the resistivity, = 1/ = 1/(nqn + pqp)
L3 January 22 20
Drift currentresistance (cont.)• Consequently, since
R = l/AR = (nqn + pqp)-1(l/A)
• For n >> p, (an n-type extrinsic s/c)R = l/(nqnA)
• For p >> n, (a p-type extrinsic s/c) R = l/(pqpA)
L3 January 22 21
Drift currentresistance (cont.)• Note: for an extrinsic semiconductor and
multiple scattering mechanisms, sinceR = l/(nqnA) or l/(pqpA), and
(n or p total)-1 = i-1, then
Rtotal = Ri (series Rs)
• The individual scattering mechanisms are: Lattice, ionized impurity, etc.
L3 January 22 22
Exp. mobility modelfunction for Si1
Parameter As P B
min 52.2 68.5 44.9
max 1417 1414 470.5
Nref 9.68e16 9.20e16 2.23e17
0.680 0.711 0.719
ref
a,d
minpn,
maxpn,min
pn,pn,
N
N1
L3 January 22 23
Exp. mobility modelfor P, As and B in Si
0
500
1000
1500
1.E+13 1.E+15 1.E+17 1.E+19
Doping Concentration (cm̂ -3)
Mob
ility
(cm̂
2/V
-sec
)
P
As
B
L3 January 22 24
Carrier mobilityfunctions (cont.)• The parameter max models 1/lattice
the thermal collision rate
• The parameters min, Nref and model 1/impur the impurity collision rate
• The function is approximately of the ideal theoretical form:
1/total = 1/thermal + 1/impurity
L3 January 22 25
Carrier mobilityfunctions (ex.)• Let Nd
= 1.78E17/cm3 of phosphorous, so min = 68.5, max = 1414, Nref = 9.20e16 and = 0.711. Thus n = 586 cm2/V-s
• Let Na = 5.62E17/cm3 of boron, so
min = 44.9, max = 470.5, Nref = 9.68e16 and = 0.680. Thus n = 189 cm2/V-s
L3 January 22 26
Lattice mobility
• The lattice is the lattice scattering mobility due to thermal vibrations
• Simple theory gives lattice ~ T-3/2
• Experimentally n,lattice ~ T-n where n = 2.42 for electrons and 2.2 for holes
• Consequently, the model equation is lattice(T) = lattice(300)(T/300)-
n
L3 January 22 27
Ionized impuritymobility function• The impur is the scattering mobility
due to ionized impurities
• Simple theory gives impur ~ T3/2/Nimpur
• Consequently, the model equation is impur(T) = impur(300)(T/300)3/2
L3 January 22 28
Net silicon (ex-trinsic) resistivity• Since
= -1 = (nqn + pqp)-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 (Nimpur)
L3 January 22 29
Net silicon extrresistivity (cont.)
1.00E-02
1.00E-01
1.00E+00
1.00E+01
1.00E+02
1.00E+03
1.E+13 1.E+15 1.E+17 1.E+19
Doping Concentration (cm̂ -3)
Res
isti
vity
(oh
m-c
m)
P
As
B
L3 January 22 30
Net silicon extrresistivity (cont.)• Since = (nqn + pqp)-1, and
n > p, ( = q/m*) we have
p > n
• Note that since1.6(high conc.) < p/n < 3(low conc.), so
1.6(high conc.) < n/p < 3(low conc.)
L3 January 22 31
Net silicon (com-pensated) res.• For an n-type (n >> p) compensated
semiconductor, = (nqn)-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 = (nqn)-1 = [Nqn(NI)]-1
L3 January 22 32
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.