EECS130 Integrated Circuit Devices

Professor Ali Javey9/04/2007

Semiconductor FundamentalsLecture 3

Reading: finish chapter 2 and begin chapter 3

Announcements

•HW 1 is due next Tuesday, at the beginning of the class. Late HWs will not be accepted.

•Midterm 1 is on Oct. 4th, in class. It’s closed book, but you may bring and use one sheet of note.

•Check EE130 web site often for lecture notes and announcements. (http://inst.eecs.berkeley.edu/~ee130/fa07/)

Last Lecture: Energy Band Diagram

Ec

Ev

• Energy band diagram shows the bottom edge of conduction band, Ec , and top edge of valence band, Ev .

• Ec and Ev are separated by the band gap energy, Eg .

Ed

Carrier Concentrations

kTEEc

fceNn /)( −−=kTEE

vvfeNp /)( −−=

At E=Ef , f(E)=1/2

From density of states and Fermi function, we obtain:

Example: The Fermi Level and Carrier Concentrations

kTEEc

fceNn /)( −−=

( ) ( ) eV 614.010/108.2ln026.0ln 1719 =×==− nNkTEE cfc

E cEf

E v

0.146 eV

Where is Ef for n =1017 cm-3? Solution:

The np Product and the Intrinsic Carrier Concentration

• In an intrinsic (undoped) semiconductor, n = p = ni .

kTEvci

geNNn 2/−=

2innp =

kTEEc

fceNn /)( −−= kTEEv

vfeNp /)( −−=andMultiply

kTEvc

kTEEvc

gvc eNNeNNnp //)( −−− ==

Question: What is the hole concentration in an N-type semiconductor with 1015 cm-3 of donors?

Solution: n = 1015 cm-3.

After increasing T by 60°C, n remains the same at 1015 cm-3 while p increases by about a factor of 2300 because .

Question: What is n if p = 1017cm-3 in a P-type silicon wafer?

Solution:

EXAMPLE: Carrier Concentrations

3-5315

-3202

cm10cm10cm10

=≈= −nnp i

kTEi

gen /2 −∝

3-3317

-3202

cm10cm10cm10

=≈= −pnn i

EXAMPLE: Complete ionization of the dopant atomsNd = 1017 cm-3 and Ec -Ed =45 meV. What fraction of the donors are not ionized?

Solution: First assume that all the donors are ionized.

Ec

Ef

Ev

146 meVEd

45meV

Probability of non-ionization ≈ 02.01

11

1meV26/)meV)45146((/)( =

+=

+ −− ee kTEE fd

Therefore, it is reasonable to assume complete ionization, i.e., n = Nd .

meV146cm10 317 −=⇒== −cfd EENn

Doped Si and Charge

• What is the net charge of your Si when it is electron and hole doped?

Bond Model of Electrons and Holes (Intrinsic Si)• Silicon crystal in

a two-dimensionalrepresentation.

Si Si Si

Si Si Si

Si Si Si

Si Si Si

Si Si Si

Si Si Si

Si Si Si

Si Si Si

Si Si Si

• When an electron breaks loose and becomes a conduction electron, a hole is also created.

Dopants in Silicon

Si Si Si

Si Si

Si Si Si

Si Si Si

Si Si

Si Si Si

As B

• As (Arsenic), a Group V element, introduces conduction electrons and creates N-type silicon,

• B (Boron), a Group III element, introduces holes and creates P-type silicon, and is called an acceptor.

• Donors and acceptors are known as dopants.

and is called a donor.

N-type Si P-type Si

General Effects of Doping on n and p

Charge neutrality: da NpNn −−+ +_

= 0

da NpNn −−+ = 0

Assuming total ionization of acceptors and donors:

aN_

: number of ionized acceptors /cm3

dN+

: number of ionized donors /cm3

aN : number of ionized acceptors /cm3

dN+

: number of ionized donors /cm3

I. (i.e., N-type)

If ,

II. (i.e., P-type)

If ,

ad NNn −=

nnp i2=

iad nNN >>−

ad NN >> dNn = di Nnp 2=and

ida nNN >>− da NNp −=

pnn i2=

da NN >> aNp = ai Nnn 2=and

General Effects of Doping on n and p

EXAMPLE: Dopant Compensation

What are n and p in Si with (a) Nd = 6×1016 cm-3 and Na = 2×1016 cm-3

and (b) additional 6×1016 cm-3 of Na ?

(a)

(b) Na = 2×1016 + 6×1016 = 8×1016 cm-3 > Nd !

+ + + + + + + + + + + +. . . . . .

. . . . . . . . . . . - - - - - - - -

. . . . . .

. . . . . .

316 cm104 −×=−= ad NNn3316202 cm105.2104/10/ −×=×== nnp i

3161616 cm102106108 −×=×−×=−= da NNp3316202 cm105102/10/ −×=×== pnn i

Nd = 6×1016 cm-3

Na = 2×1016 cm-3

n = 4×1016 cm-3

Nd = 6×1016 cm-3

Na = 8×1016 cm-3

p = 2×1016 cm-3

Carrier Concentrations at Extremely High and Low Temperatures

intrinsic regime

n = Nd

freeze-out regime

lnn

1/Thigh temp.

roomtemperature

cryogenictemperature

Infrared Detector Based on Freeze-out

To image the black-body radiation emitted by tumors requires a photodetector that responds to hν’s around 0.1 eV. In doped Si operating in the freeze-out mode, conduction electrons are created when the infrared photons provide the energy to ionized the donor atoms.

photon Ec

Ev

electron

Ed

Chapter 2 Summary

Energy band diagram. Acceptor. Donor. mn , mp . Fermi function. Ef .

kTEEc

fceNn /)( −−=kTEE

vvfeNp /)( −−=

ad NNn −=

da NNp −=

2innp =

Thermal Motion

• Zig-zag motion is due to collisions or scatteringwith imperfections in the crystal.

• Net thermal velocity is zero.• Mean time between collisions (mean free time) is τm ~ 0.1ps

Thermal Energy and Thermal Velocity

electron or hole kinetic energy 2

21

23

thmvkT ==

kg101.926.0K300JK1038.133

31

123

−

−−

×××××

==eff

th mkTv

cm/s103.2m/s103.2 75 ×=×=

~8.3 X 105 km/hr

Drift

Electron and Hole Mobilities

• Drift is the motion caused by an electric field.

Effective Mass

In an electric field, , an electron or a hole accelerates.

Electron and hole effective massesSi Ge GaAs GaP

m n /m 0 0.26 0.12 0.068 0.82m p /m 0 0.39 0.30 0.50 0.60

electrons

holes

Remember :F=ma=-qE

Remember :F=ma=mV/t = -qE

Electron and Hole Mobilities

p

mpp m

qτμ =

n

mnn m

qτμ =

• μp is the hole mobility and μn is the electron mobility

mpp qvm τ=

p

mp

mq

vτ

=

pv μ=nv μ−=

Electron and hole mobilities of selected semiconductors

Electron and Hole Mobilities

Si Ge GaAs InAsμ n (cm2/V·s) 1400 3900 8500 30000μ p (cm2/V·s) 470 1900 400 500

. sV

cmV/cmcm/s 2

⎥⎦

⎤⎢⎣

⎡⋅

=v =μ

; μ has the dimensions of v/

Based on the above table alone, which semiconductor and which carriers (electrons or holes) are attractive for applications in high-speed devices?

EXAMPLE: Given μp = 470 cm2/V·s, what is the hole drift velocity at = 103 V/cm? What is τmp and what is the distance traveled between

collisions (called the mean free path)? Hint: When in doubt, use the MKS system of units.

Drift Velocity, Mean Free Time, Mean Free Path

EXAMPLE: Given μp = 470 cm2/V·s, what is the hole drift velocity at = 103 V/cm? What is τmp and what is the distance traveled between

collisions (called the mean free path)? Hint: When in doubt, use the MKS system of units.

Solution: ν = μp = 470 cm2/V·s ×

103 V/cm = 4.7×

105 cm/s

τmp = μp mp /q =470 cm2/V ·s ×

0.39 ×

9.1×10-31 kg/1.6×10-19 C

= 0.047 m2/V ·s ×

2.2×10-12 kg/C = 1×10-13s = 0.1 ps

mean free path = τmhνth ~ 1×

10-13 s ×

2.2×107 cm/s

= 2.2×10-6 cm = 220 Å = 22 nm

This is smaller than the typical dimensions of devices, but getting close.

Drift Velocity, Mean Free Time, Mean Free Path

There are two main causes of carrier scattering:1. Phonon Scattering2. Impurity (Dopant) Ion Scattering

2/32/1

11 −∝×

∝×

∝∝ TTTvelocitythermalcarrierdensityphononphononphonon τμ

Phonon scattering mobility decreases when temperature rises:

μ = qτ/mvth ∝

T1/2

Mechanisms of Carrier Scattering

∝

T

_+

- -Electron

Boron Ion Electron

ArsenicIon

Impurity (Dopant)-Ion Scattering or Coulombic Scattering

dada

thimpurity NN

TNN

v+

∝+

∝2/33

μ

There is less change in the direction of travel if the electron zips bythe ion at a higher speed.

Total Mobility

1E14 1E15 1E16 1E17 1E18 1E19 1E20

0

200

400

600

800

1000

1200

1400

1600

Holes

Electrons

Mob

ility

(cm

2 V-1 s

-1)

Tota l Im purity Concenration (atom s cm -3)Na + Nd (cm-3)

impurityphonon

impurityphonon

μμμ

τττ

111

111

+=

+=

Temperature Effect on Mobility

1015

Question:What Nd will make dμn /dT = 0 at room temperature?

Velocity Saturation

When the kinetic energy of a carrier exceeds a critical value, it generates an optical phonon and loses the kinetic energy. Therefore, the kinetic energy is capped and the velocity does not rise above a saturation velocity, vsat , no matter how large is.

Velocity saturation has a deleterious effect on device speed as we will see in the later chapters.