L3 January 27 1

Semiconductor Device Modeling and CharacterizationEE5342, Lecture 3-Spring 2004

Professor Ronald L. Carterronc@uta.edu

http://www.uta.edu/ronc/

L3 January 27 2

Web Pages* You should be aware of information

at• R. L. Carter’s web page

– www.uta.edu/ronc/

• EE 5342 web page and syllabus– www.uta.edu/ronc/5342/syllabus.htm

• University and College Ethics Policies– www2.uta.edu/discipline/– www.uta.edu/ronc/5342/Ethics.htm

• Submit a signed copy to Dr. Carter

L3 January 27 3

First Assignment

• e-mail to listserv@listserv.uta.edu– 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 ronc@uta.edu, with EE5342 in subject line.

L3 January 27 4

Semiconductor Electronics - concepts thus far• Conduction and Valence states due to

symmetry of lattice• “Free-elec.” dynamics near band edge• Band Gap

– direct or indirect– effective mass in curvature

• Thermal carrier generation• Chemical carrier gen (donors/accept)

L3 January 27 5

Counting carriers - quantum density of states function• 1 dim electron wave #s range for n+1

“atoms” is 2/L < k < 2/a where a is “interatomic” distance and L = na is the length of the assembly (k = 2/)

• Shorter s, would “oversample”• if n increases by 1, dp is h/L• Extn 3D: E = p2/2m = h2k2/2m so a vol

of p-space of 4p2dp has h3/LxLyLz

L3 January 27 6

QM density of states (cont.)• So density of states, gc(E) is

(Vol in p-sp)/(Vol per state*V) =4p2dp/[(h3/LxLyLz)*V]

• Noting that p2 = 2mE, this becomes gc(E) = {42mn*)3/2/h3}(E-Ec)1/2

and E - Ec = h2k2/2mn*

• Similar for the hole states whereEv - E = h2k2/2mp*

L3 January 27 7

Fermi-Diracdistribution fctn• The probability of an electron having

an energy, E, is given by the F-D distr fF(E) = {1+exp[(E-EF)/kT]}-1

• Note: fF (EF) = 1/2

• EF is the equilibrium energy of the system

• The sum of the hole probability and the electron probability is 1

L3 January 27 8

Fermi-DiracDF (continued)• So the probability of a hole having

energy E is 1 - fF(E)

• At T = 0 K, fF (E) becomes a step function and 0 probability of E > EF

• At T >> 0 K, there is a finite probability of E >> EF

L3 January 27 9

Maxwell-BoltzmanApproximation• fF(E) = {1+exp[(E-EF)/kT]}-1

• For E - EF > 3 kT, the exp > 20, so within a 5% error, fF(E) ~ exp[-(E-EF)/kT]

• This is the MB distribution function

• MB used when E-EF>75 meV (T=300K)

• For electrons when Ec - EF > 75 meV and for holes when EF - Ev > 75 meV

L3 January 27 10

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 27 11

Electron and HoleConc 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 27 12

Calculating theequilibrium no• The ideal is to calculate the

equilibrium electron concentration no for the FD distribution, where

fF(E) = {1+exp[(E-EF)/kT]}-1

gc(E) = [42mn*)3/2(E-Ec)1/2]/h3

dEEfEgnF

max

c

E

Eco

L3 January 27 13

Equilibrium con-centration for no• Earlier quoted the MB approximation no

= Nc exp[-(Ec - EF)/kT],(=Nc exp F)

• The exact solution is no = 2NcF1/2(F)/1/2

• Where F1/2(F) is the Fermi integral of order 1/2, and F = (EF - Ec)/kT

• Error in no, is smaller than for the DF: = 31%, 12%, 5% for -F = 0, 1, 2

L3 January 27 14

Equilibrium con-centration for po• Earlier quoted the MB approximation po =

Nv exp[-(EF - Ev)/kT],(=Nv exp ’F)

• The exact solution is po = 2NvF1/2(’F)/1/2

• Note: F1/2() = 0.678, (1/2/2) = 0.886

• Where F1/2(’F) is the Fermi integral of order 1/2, and ’F = (Ev - EF)/kT

• Errors are the same as for po

L3 January 27 15

Degenerate andnondegenerate cases• Bohr-like doping model assumes no

interaction between dopant sites• If adjacent dopant atoms are within 2

Bohr radii, then orbits overlap

• This happens when Nd ~ Nc (EF ~ Ec), or when Na ~ Nv (EF ~ Ev)

• The degenerate semiconductor is defined by EF ~/> Ec or EF ~/< Ev

L3 January 27 16

Donor ionization• The density of elec trapped at donors

is nd = Nd/{1+[exp((Ed-EF)/kT)/2]}

• Similar to FD DF except for factor of 2 due to degeneracy (4 for holes)

• Furthermore nd = Nd - Nd+, also

• For a shallow donor, can have Ed-EF >> kT AND Ec-EF >> kT: Typically EF-Ed ~ 2kT

L3 January 27 17

Donor ionization(continued)• Further, if Ed - EF > 2kT, then

nd ~ 2Nd exp[-(Ed-EF)/kT], < 5%

• If the above is true, Ec - EF > 4kT, sono ~ Nc exp[-(Ec-EF)/kT], < 2%

• Consequently the fraction of un-ionized donors is nd/no = 2Nd exp[(Ec-Ed)/kT]/Nc = 0.4% for Nd(P) = 1e16/cm3

L3 January 27 18

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 27 19

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 27 20

• 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 27 21

• 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 27 22

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 27 23

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 27 24

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 27 25

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 27 26

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 27 27

Locating Efi in the bandgap • Since

Ec - Efi = kT ln(Nc/ni), andEfi - Ev = kT ln(Nv/ni)

• The sum of the two equations givesEfi = (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 27 28

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 27 29

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 27 30

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 27 31

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 27 32

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 27 33

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 27 34

Drift Current

• The drift current density (amp/cm2) is given by the point form of Ohm Law

J = (nqn+pqp)(Exi+ Eyj+ Ezk), so

J = (n + p)E = E, where

= nqn+pqp defines the conductivity

• The net current is

SdJI

L3 January 27 35

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 27 36

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 27 37

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 27 38

Exp. mobility modelfunction for Si1

Parameter As P Bmin 52.2 68.5 44.9

max 1417 1414470.5

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 27 39

Exp. mobility modelfor P, As and B in Si

L3 January 27 40

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 27 41

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 27 42

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 27 43

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 27 44

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 27 45

Net silicon extrresistivity (cont.)

L3 January 27 46

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 27 47

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 27 48

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.