L3 January 22 1

Semiconductor Device Modeling and CharacterizationEE5342, Lecture 3-Spring 2002

Professor Ronald L. Carterronc@uta.edu

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.