EE 5340Semiconductor Device TheoryLecture 05 – Spring 2011

Professor Ronald L. Carterronc@uta.edu

http://www.uta.edu/ronc

©rlc L05-08Feb2011

2

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.

©rlc L05-08Feb2011

3

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

©rlc L05-08Feb2011

4

Second Assignment• Submit a signed copy of the

document posted at

www.uta.edu/ee/COE%20Ethics%20Statement%20Fall%2007.pdf

©rlc L05-08Feb2011

5

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.

©rlc L05-08Feb2011

6

Intrinsic carrierconc. (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

©rlc L05-08Feb2011

7

Classes ofsemiconductors• 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

©rlc L05-08Feb2011

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 + Na8

©rlc L05-08Feb2011

9

Equilibriumconc (cont.)• For Nd > Na (taking the + root)

no = (Nd-Na)/2 + {[(Nd-Na)/2]2+ni

2}1/2

• For Nd >> Na and Nd >> ni, can use the binomial expansion, giving

no = Nd/2 + Nd/2[1 + 2ni

2/Nd2 + … ]

• So no = Nd, and po = ni2/Nd in the

limit of Nd >> Na and Nd >> ni

©rlc L05-08Feb2011

10

n-type equilibriumconcentrations• N ≡ Nd - Na , n type N > 0• For all N,

no = N/2 + {[N/2]2+ni2}1/2

• In most cases, N >> ni, sono = N, and

po = ni2/no = ni

2/N, (Law of Mass Action is al-ways true in

equilibrium)

©rlc L05-08Feb2011

11

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

©rlc L05-08Feb2011

12

p-type equilibriumconcentrations• N ≡ Nd - Na , p type N < 0 • For all N,

po = |N|/2 + {[|N|/2]2+ni2}1/2

• In most cases, |N| >> ni, sopo = |N|, and

no = ni2/po = ni

2/|N|, (Law of Mass Action is al-ways true in

equilibrium)

©rlc L05-08Feb2011

13

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

©rlc L05-08Feb2011

14

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)

©rlc L05-08Feb2011

15

EF relative to Efi• Letting ni = no gives Ef = Efi

ni = Nc exp[-(Ec-Efi)/kT], soEc - 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)

©rlc L05-08Feb2011

16

Locating Efi in the bandgap • Since

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

©rlc L05-08Feb2011

17

Examplecalculations• For Nd = 3.2E16/cm3, ni =

1.4E10/cm3

no = Nd = 3.2E16/cm3

po = ni2/Nd , (po is always ni

2/no)=

(1.4E10/cm3)2/3.2E16/cm3

= 6.125E3/cm3 (comp to ~1E23 Si)

• For po = Na = 4E17/cm3,no = ni

2/Na = (1.4E10/cm3)2/4E17/cm3 = 490/cm3

©rlc L05-08Feb2011

18

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 thatEc - 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

©rlc L05-08Feb2011

19

Equilibrium electronconc. and energies

ov

2i

vof

io

fiffif

io

cocf

cfco

pNlnkT

nNnlnkTEvE and

;nnlnkTEE or ,kT

EEexpnn

;NnlnkTEE or ,kT

EEexpNn

©rlc L05-08Feb2011

20

Equilibrium hole conc. and energies

oc

2i

cofc

io

ffiffi

io

vo

fvfv

vo

nNlnkT

nNplnkTEE and

;nplnkTEE or ,kT

EEexpnp

;NplnkTEE or ,kT

EEexpNp

©rlc L05-08Feb2011

21

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

©rlc L05-08Feb2011

22

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.

©rlc L05-08Feb2011

23

Carrier mobility (cont.)• If the rate of a single contribution

to the scattering is 1/ti, then the total scattering rate, 1/tcoll is

m

m

m

t

t

all

collisions itotal

all

collisions icoll

11by given is mobility total

the and , 11

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).

©rlc L05-08Feb2011

24

Figure 1.16 (cont. M&K)

Parameter Arsenic Phosphorus Boronμmin 52.2 68.5 44.9μmax 1417 1414 470.5Nref 9.68 X 1016 9.20 X 1016 2.23 X 1017

α 0.680 0.711 0.719

©rlc L05-08Feb2011

25

ammmmrefi NN

1minmax

min

©rlc L05-08Feb2011

26

Drift Current• The drift current density (amp/cm2)

is given by the point form of Ohm LawJ = (nqmn+pqmp)(Exi+ Eyj+ Ezk), so

J = (sn + sp)E = sE, wheres = nqmn+pqmp defines the

conductivity• The net current is SdJI

©rlc L05-08Feb2011

27

Drift currentresistance• 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)

©rlc L05-08Feb2011

28

Drift currentresistance (cont.)• Consequently, since

R = rl/AR = (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)

©rlc L05-08Feb2011

29

ReferencesM&K and 1Device Electronics for

Integrated Circuits, 2 ed., by Muller and Kamins, Wiley, New York, 1986.– See Semiconductor Device

Fundamen-tals, by Pierret, Addison-Wesley, 1996, for another treatment of the m model.

2Physics of Semiconductor Devices, by S. M. Sze, Wiley, New York, 1981.

©rlc L05-08Feb2011

30

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.