chenming hu ch2 slides

8/10/2019 Chenming Hu Ch2 Slides

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Modern Semiconductor Devices for Integrated Circuits (C. Hu) Slide 2-1

Chapter 2 Motion and Recombination

of Electrons and Holes

2.1 Thermal Motion

Average electron or hole kinetic energy 22

1

2

3thmvkT

kg101.926.0

K300JK1038.13331

123

eff

thm

kTv

cm/s103.2m/s103.2 75


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2.1 Thermal Motion

Zig-zag motion is due to collisions or scatteringwith imperfections in the crystal. Netthermal velocity is zero. Mean time between collisions ism ~ 0.1ps


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Modern Semiconductor Devices for Integrated Circuits (C. Hu)Slide 2-3

Hot-point Probe can determine sample doing type

Thermoelectr ic Generator

(f rom heat to electricity )

and Cooler (fromelectr icity to refr igeration)

Hot-point Probe

distinguishes Nand P type

semiconductors.


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2.2 Drif t

2.2.1 Electron and Hole Mobilities

Drift is the motion caused by an electric field.


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p is the hole mobility andn is the electron mobility

mpp qvm

p

mp

m

qv

p

mp

pm

q

n

mn

n

m

q

Epv Env


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Electron and ho le mob i l it ies o f selected

semiconductors


Si Ge GaAs InAs

n (cm2/Vs) 1400 3900 8500 30000

p (cm

2

/Vs) 470 1900 400 500

.sV

cm

V/cm

cm/s 2

v =E; has the dimensions ofv/E

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


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There are two main causes of carrier scattering:1. Phonon Scattering2. Ionized-Impurity (Coulombic) Scattering

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2/1

11

TTTvelocitythermalcarrierdensityphonon

phononphonon

Phonon scatter ingmobility decreases when temperature rises:

= q/m

vthT1/2

2.2.2 Mechanisms of Carrier Scattering

T


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_+

- -Electron

Boron IonElectron

ArsenicIon

Impurity (Dopant)-I on Scatter ing or Coulombic Scatter ing

dada

thimpurity

NN

T

NN

v

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


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Total Mobil i ty

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

0

200

400

600

800

1000

1200

1400

1600

Holes

Electrons

Mobility(cm2

V-1s

-1)

Total Impurity Concenration (atoms cm-3)N

a+ N

d(cm-3)

impurityphonon

impurityphonon

111

111


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Temperature Effect on Mobil i ty

101 5

Question:WhatNdwill makedn/dT= 0 at room

temperature?


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Velocity Saturation

When the kinetic energy of a carrier exceeds a critical value, itgenerates an optical phonon and loses the kinetic energy.

Therefore, the kinetic energy is capped at large E and the

velocity does not rise above a saturation velocity, vsat.Velocity saturationhas a deleterious effect on device speed asshown in Ch. 6.


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2.2.3 Drift Current and Conductivity

Jp

n

E

unitarea

Jp= qpv A/cm2or C/cm2sec

Ifp = 1015cm-3andv = 104 cm/s, thenJp= 1.610

-19C 1015cm-3104cm/s

= 22 A/cm1.6cmC/s6.1

EXAMPLE:

Hole current density


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Jp,drift = qpv = qppE

Jn,drift =qnv = qnnE

Jdrift = Jn,drift + Jp,drift = E=(qnn+qpp)E

conductivity(1/ohm-cm) of a semiconductor is= qnn+ qpp

2.2.3 Drift Current and Conductivity

1/=is resistivity (ohm-cm)


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N-type

P-type

Relationship between Resistivity and Dopant Density

= 1/

DOPANTDE

NSITYcm-3

RESISTIVITY (cm)


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EXAMPLE: Temperature Dependence of Resistance

(a) What is the resistivity () of silicon dopedwith 1017cm-3 of arsenic?

Solution:

(a) Using the N-type curve in the previous

figure, we find that= 0.084 -cm.

(b) What is the resistance (R) of a piece of this

silicon material 1m long and 0.1 m2in cross-

sectional area?

(b) R =L/A = 0.084 -cm 1 m / 0.1 m2

= 0.084 -cm 10-4cm/ 10-10cm2= 8.4 10-4


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EXAMPLE: Temperature Dependence of Resistance

By what factor will R increase or decrease from

T=300 K to T=400 K?

Solution:The temperature dependent factor in (and

therefore) is n. From the mobility vs. temperature

curve for 1017cm-3, we find that n decreases from 770

at 300K to 400 at 400K. As a result, R increases by

93.1400770


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2.3 Diffusion Current

Particles diffuse from a higher-concentration locationto a lower-concentration location.


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2.3 Di ffusion Current

dx

dn

qDJ ndiffusionn , dx

dp

qDJ pdiffusionp ,

D is called the diffusion constant. Signs explained:

n p

x x


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Total Current Review of Four Cur rent Components

Jn= Jn,drift+ Jn,diffusion= qnnE+dx

dnqDn

Jp= Jp,drift+ Jp,diffusion= qppEdx

dpqDp

JTOTAL = Jn+ Jp


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2.4 Relation Between the Energy

Diagram and V, E

E(x)=dx

dE

q

1

dx

dE

q

1

dx

dV vc

Ecand Evvary in the opposite

direction from the voltage. That

is, Ecand Evare higher where

the voltage is lower.

V(x)

0.7V

x0

x

Ef(x)

E

0.7V

-

+

Ec(x)

Ev(x)

N-+

0.7eV

N type Si


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2.5 Einstein Relationship between D and

dx

dEe

kT

N

dx

dn ckT/)EE(c fc

dx

dE

kT

n c

kT/)EE(

cfceNn

Consider a piece of non-uniformly doped semiconductor.

Ev(x)

Ec(x)

Ef

n-type semiconductor

Decreasing donor concentration

N-type semiconductor

EqkT

n


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2.5 Einstein Relationship between D and

EqkT

n

dx

dn

0dx

dnqDqnJ nnn E at equilibrium.

EkTqDqnqn nn 0

nn

q

kTD

These are known as the Einstein relationship.

pp

q

kTD Similarly,


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EXAMPLE: Di ffusion Constant

What is the hole diffusion constant in a piece of

silicon with p = 410 cm2 V-1s-1?

Solution:

Remember: kT/q = 26 mV at room temperature.

/scm11sVcm410)mV26( 2112

pp

q

kTD


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2.6 Electron-Hole Recombination

The equilibrium carrier concentrations are denoted with

n0andp0.

The total electron and hole concentrations can be different

fromn0andp0.

The differences are called theexcess carr ier

concentrationsnandp.

'0 nnn

'0 ppp


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Charge Neutrali ty

Charge neutrality is satisfied at equilibrium(n=p= 0).When a non-zeron is present, an equalp may

be assumed to be present to maintain charge

equality and vice-versa.If charge neutrality is not satisfied, the net chargewill attract or repel the (majority) carriers throughthe drift current until neutrality is restored.

'p'n


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Recombination L ifetime

Assume light generatesn andp. If the light issuddenly turned off, n andp decay with timeuntil they become zero.

The process of decay is calledrecombination.The time constant of decay is therecombinationtime orcarr ier li fetime, .Recombination is natures way of restoring

equilibrium (n= p= 0).


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ranges from 1ns to 1ms in Si anddepends on

the density of metal impurities (contaminants)such as Au and Pt.Thesedeep trapscapture electrons and holes tofacilitate recombination and are calledrecombination centers.

Recombination L ifetime

Ec

Ev

DirectRecombinationis unfavorable insilicon

Recombinationcenters


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Direct and Indirect Band Gap

Direct band gapExample: GaAs

Direct recombination is efficientas k conservation is satisfied.

Indirect band gap

Example: Si

Direct recombination is rare as kconservation is not satisfied

Trap

Modern Semiconductor Devices for Integrated Circuits (C. Hu)


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n

dt

nd

dt

pdpn

dt

nd

Rate of recombination (s-1cm-3)

pn


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EXAMPLE: Photoconductors

A bar of Si is doped with boron at 1015cm-3. It is

exposed to light such that electron-hole pairs are

generated throughout the volume of the bar at the

rate of 1020/scm3. The recombination lifetime is

10s. What are (a) p0, (b) n0, (c) p,(d) n,(e) p ,

(f) n, and (g) the np product?


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Solution:

(a) What is p0?

p0= Na = 1015 cm-3

(b) What is n0 ?n0= ni

2/p0= 105 cm-3

(c) What is p?

In steady-state, the rate of generation is equal to the

rate of recombination.1020/s-cm3= p/

p= 1020/s-cm3 10-5s= 1015 cm-3


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(d) What is n?n= p= 1015 cm-3

(e) What is p?

p= p0+ p= 1015cm-3 + 1015cm-3= 21015cm-3

(f) What is n?

n = n0+ n= 105cm-3 + 1015cm-3~ 1015cm-3 since n0> ni2 = 1020 cm-6.

The np product can be very different from ni2.


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2.7 Thermal Generation

Ifn is negative, there are fewerelectrons than the equilibrium value.

As a result, there is a net rate ofthermal generation at the rate of |n|/.


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2.8 Quasi-equi l ibrium and Quasi-Fermi Levels

Whenevern = p 0, np ni2. We would like to preserve

and use the simple relations:

But these equations lead tonp = ni2. The solution is to introduce

twoquasi-Fermi levelsEfnandEfp such that

kTEEc

fceNn /)(

kTEE

vvfeNp

/)(

kTEE

cfnceNn

/)(

kTEE

vvfpeNp

/)(

Even when electrons and holes are not at equilibrium, within

each groupthe carriers can be at equilibrium. Electrons areclosely linked to other electrons but only loosely to holes.


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Now assume n= p= 1015cm-3.(b) Find Efnand Efp.

n = 1.011017cm-3=

EcEfn= kT ln(Nc/1.011017cm-3)

= 26 meV ln(2.81019cm-3/1.011017cm-3)= 0.15 eV

Efnis nearly identical toEfbecause nn0 .

kTEE

cfnceN

/)(

EXAMPLE: Quasi-Fermi Levels and Low-Level I njection


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EXAMPLE: Quasi-Fermi Levels

p = 1015 cm-3=

EfpEv= kT ln(Nv/1015cm-3)

= 26 meV ln(1.041019cm-3/1015cm-3)= 0.24 eV

kTEEv

vfpeN /)(

Ec

Ev

Efp

Ef Efn


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2.9 Chapter Summary

dx

dnqDJ ndiffusionn ,

dx

dpqDJ pdiffusionp ,

nnq

kTD

ppq

kTD

Eppv

Ennv

Epdriftp qpJ ,

Endriftn qnJ ,

-


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Modern Semiconductor Devices for Integrated Circuits (C. Hu)Slid 2 40

2.9 Chapter Summary

is the recombination lifetime.n andp are the excess carr ier concentrations.

n = n0+ n

p = p0+ p

Charge neutrality requiresn= p.

rate of recombination = n/= p/

Efn

andEfp

are the quasi-Fermi levels of electrons andholes.

kTEE

vvfpeNp

/)(

kTEE

cfnceNn

/)(

chenming hu ch2 slides

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