Chapter4 semiconductor in equilibrium

Microelectronics I

Chapter 4: Semiconductor in Equilibrium

Equilibrium; no external forces such as voltages, electrical

fields, magnetic fields, or temperature gradients are acting on

the semiconductor

Microelectronics I

e

eEc

Ev

Conduction

band

Valence

band

T>0K

eband

�Particles that can freely move and contribute to the current flow (conduction)

1. Electron in conduction band

2. Hole in valence band

carrier

Microelectronics I

�How to count number of carriers,n?

If we know

1. No. of energy states

2. Occupied energy states

Density of states (DOS)

The probability that energy states is

Assumption; Pauli exclusion principle

2. Occupied energy states The probability that energy states is

occupied

“Fermi-Dirac distribution function”

n = DOS x “Fermi-Dirac distribution function”

e

Ec

Conduction

bandCEE

h

mEg −=

3

2/3*)2(4)(

π

No of states (seats) above EC for electron

Microelectronics I

Density of state

E

e

Ec

EvValence

band

EEh

mEg v −=

3

2/3*)2(4)(

π

No of states (seats) below Ev for hole

g (E)

Fermi-Dirac distribution

Microelectronics I

e

Ec

Probability of electron having certain energy

E

Electron (blue line)

−+

=

kT

EEEf

F

F

exp1

1)(

Electron

having energy

above Ec

e

Ec

Ev

f (E)10.5

Fermi energy, EF

EF; the energy below which all states are filled with electron and above

which all states are empty at 0K

kT

hole (red line)

−+

−

kT

EE Fexp1

11

Hole having

energy below

Ev

e

Ec

E

Microelectronics I

No of carrier

No of free electrone

Ecfree electron

e

Ev

g(E) x f (E)1

No of free holee

Ev free hole

Microelectronics I

Thermal equilibrium concentration of electron, no

∫∞

=

CE

o dEEfEgn )()(

CEEh

mEg −=

3

2/3*)2(4)(

π

−−≈=

EE )(1

−−≈

−+

=kT

EE

kT

EEEf F

F

F

)(exp

exp1

1)( Boltzmann approximation

−−=

−−

=

kT

EEN

kT

EE

h

kTmn

FCC

FCno

)(exp

)(exp

22

2/3

2

*π

NC; effective density of states

function in conduction band

Microelectronics I

Ex. 1

Calculate the thermal equilibrium electron concentration in Si at T= 300K.

Assume that Fermi energy is 0.25 eV below the conduction band. The value of Nc for Si

at T=300 K is 2.8 x 1019 cm-3.

Ec

EF

0.25 eV

)25.0(

)(exp

+−−

−−=

EE

kT

EENn FC

Co

Ev

315

19

108.1

0259.0

)25.0(exp108.2

−×=

+−−⋅×=

cm

EE CC

Thermal equilibrium concentration of hole, po

Microelectronics I

[ ]∫∞

−=

Ev

o dEEfEgp )(1)(

EEh

mEg v −=

3

2/3*)2(4)(

π

−−≈=−

EE )(1

−−≈

−+

=−kT

EE

kT

EEEf F

F

F

)(exp

exp1

1)(1 Boltzmann approximation

−−=

−−

=

kT

EEN

kT

EE

h

kTmp

vFv

vFp

o

)(exp

)(exp

22

2/3

2

*π

Nv; effective density of states

function in valence band

Microelectronics I

Ex.2

Calculate the thermal equilibrium hole concentration in Si at T= 300K.

Assume that Fermi energy is 0.27 eV below the conduction band. The value of Nc for Si

at T=300 K is 1.04 x 1019 cm-3.

Ec

EF)27.0(

)(exp

−+−

−−=

EE

kT

EENp vF

vo

Ev

EF

0.27 eV

314

19

1009.3

0259.0

)27.0(exp1004.1

−×=

−+−⋅×=

cm

EE Vv

Microelectronics I

−−=

−−

=

kT

EEN

kT

EE

h

kTmn FC

CFCn

o

)(exp

)(exp

22

2/3

2

*π

−−=

−−

=

kT

EEN

kT

EE

h

kTmp vF

vvFp

o

)(exp

)(exp

22

2/3

2

*π

kTkTh

�NC and Nv are constant for a given material (effective mass) and temperature

�Position of Fermi energy is important

If EF is closer to EC than to Ev, n>p

If EF is closer to Ev than to EC, n<p

Microelectronics I

Consider ex. 1

Ec

Ev

EF

0.25 eV

315

19

108.1

0259.0

)25.0(exp108.2

)(exp

−×=

+−−⋅×=

−−=

cm

EE

kT

EENn

CC

FCCo

Eg-0.25 eV

Hole concentration

Eg=1.12 eV

34

19

1068.2

0259.0

)25.012.1(exp1004.1

)(exp

−×=

−−⋅×=

−−=

cm

kT

EENp vF

vo

Microelectronics I

Intrinsic semiconductor; A pure semiconductor with no impurity atoms

and no lattice defects in crystal

1. Carrier concentration(ni, pi)

2. Position of EFi

1. Intrinsic carrier concentration

Concentration of electron in in conduction band, ni

Concentration of hole in in valence band, pi

−−=

−−==

kT

EEN

kT

EENpn vFi

vFiC

Cii

)(exp

)(exp

−=

−−=

kT

ENN

kT

EENNn

g

vcvC

vCi exp)(

exp2

Independent of Fermi energy

Microelectronics I

Ex. 3; Calculate the intrinsic carrier concentration in gallium arsenide (GaAs)

at room temperature (T=300K). Energy gap, Eg, of GaAs is 1.42 eV. The

value of Nc and Nv at 300 K are 4.7 x 1017 cm-3 and 7.0 x 1018 cm-3,

respectively.

1218172 1009.50259.0

42.1exp)100.7)(107.4( ×=

−××=ni

361026.2

0259.0

−×=

cmni

Microelectronics I

2. Intrinsic Fermi level position, EFi

If EF closer to Ec, n>p

If EF closer to Ev, n<p

Intrinsic; n=p

EF is located near the center of the forbidden bandgap

−− −− )()( EEEE

+=

−−=

−−

*

*

ln4

3

)(exp

)(exp

n

p

midgapFi

vFiv

FiCC

m

mkTEE

kT

EEN

kT

EEN

Ec

Ev

Emidgap

Mp ≠ mn

Mp = mn EFi = Emidgap

EFi shifts slightly from Emidgap

Microelectronics I

Efi is located near the center of Eg

no=po

Efi is located near the center of Eg

Microelectronics I

Dopant atoms and energy levels

adding small, controlled amounts of specific dopant, or impurity, atoms

Increase no. of carrier (either electron or hole)

Alter the conductivity of semiconductor

III IV V

B C

Al Si P

Ga Ge As

In Sb

3 valence

electrons

5 valence

electrons

Consider Phosphorus (P) and boron (B) as

impurity atoms in Silicon (Si)

Microelectronics I

1. P as substitutional impurity (group V element; 5 valence electron)

�In intrinsic Si, all 4 valence electrons contribute to covalent bonding.

�In Si doped with P, 4 valence electron of P contribute to covalent bonding

and 1 electron loosely bound to P atom (Donor electron).

Donor electron

can easily break the bond and freely moves

Microelectronics I

�Energy to elevate the donor electron into conduction band is less than that for

the electron involved in covalent bonding

�Ed(; energy state of the donor electron) is located near Ec

�When small energy is added, donor electron is elevated to conduction band,

leaving behind positively charged P ion

�P atoms donate electron to conduction band� P; donor impurity atom

�No. of electron > no. of hole� n-type semiconductor (majority carrier is electron)

Microelectronics I

2. B as substitutional impurity (group III element; 3 valence electron)

�In Si doped with B, all 3 valence electron of B contribute to covalent

bonding and one covalent bonding is empty

�When small energy is added, electron that involved in covalent bond will

occupy the empty position leaving behind empty position that associated

with Si atom

Hole is created

Microelectronics I

�Electron occupying the empty state associated with B atom does not have

sufficient energy to be in the conduction band� no free electron is created

�Ea (;acceptor energy state) is located near Ev

�When electron from valence band elevate to Ea, hole and negatively

charged B are created

�B accepts electron from valence band� B; acceptor impurity atom

�No. of hole > no. of electron���� p-type material (majority carrier is hole)

Microelectronics I

�Pure single-crystal semiconductor; intrinsic semiconductor

�Semiconductor with dopant atoms; extrinsic semiconductor

p-typen-type

Dopant atom;

Majority carrier;�Donor impurity atom

�electron

�Acceptor impurity atom

�holeMajority carrier; �electron �hole

Ionization Energy

The energy that required to elevate donor electron into the conduction (in

case of donor impurity atom) or to elevate valence electron into acceptor

state (in case of acceptor impurity atom).

Microelectronics I

III-V semiconductors

GaAs

Group III Group V

Dopant atoms;Dopant atoms;

�Group II (beryllium, zinc and cadmium) replacing Ga; acceptor

�Group VI (selenium, tellurium) replacing As; donor

�Group IV (Si and germanium) replacing Ga; donor

As; acceptor

Microelectronics I

Carrier concentration of extrinsic semiconductor

When dopant atoms are added, Fermi energy and distribution of electron and hole

will change.

EF>EFi EF<EFi

Electron> hole

n-typehole> electron

p-type

Microelectronics I

−−=

kT

EENn FC

Co

)(exp

−−=

kT

EENp vF

vo

)(exp

Thermal equilibrium concentration of electron

Thermal equilibrium concentration of hole

Ex. 4

Ec0.25 eV

Band diagram of Si. At T= 300 K, Ec

Ev

EF

1.12 eV

0.25 eVBand diagram of Si. At T= 300 K,

Nc=2.8x1019cm-3 and Nv=1.04x1019cm-3.

Calculate no and po.

31519 108.10259.0

25.0exp)108.2( −

×=

−×= cmno

3419 107.20259.0

)25.012.1(exp)1004.1( −

×=

−−×= cmpo

N-type Si

Microelectronics I

�Change of Fermi energy causes change of carrier concentration.

no and po equation as function of the change of Fermi energy

−=

−−=

kT

EEn

kT

EENn FiF

iFC

Co exp)(

exp

−− −− EEEE )()(

−−=

−−=

kT

EEn

kT

EENp FiF

ivF

vo

)(exp

)(exp

ni; intrinsic carrier concentration

Efi; intrinsic Fermi energy

Microelectronics I

The nopo product

2

exp

)(exp

)(exp

i

g

vC

vFFCvCoo

n

kT

ENN

kT

EE

kT

EENNpn

=

−=

−−

−−=

in=

2

ioo npn =

Product of no and po is always a constant for a given material at a given

temperature.

Microelectronics I

Degenerate and Non degenerate semiconductors

Small amount of dopant atoms (impurity atoms)

�No interaction between dopant atoms

�Discrete, noninteracting energy state.

�EF at the bandgap

E

donor acceptor

Nondegenerate semiconductor

EF

EF

Large amount of dopant atoms (~effective density of states)

Microelectronics I

�Dopant atoms interact with each other

�Band of dopant states widens and overlap the allowed band

(conduction @ valence band)

�EF lies within conduction @ valence band

e

e

e

Ec

Ev

Filled states

EFe

Ec

Ev empty states

EF

Degenerate semiconductor

Microelectronics I

Statistic donors and acceptors

Discrete donor level

donor

+−=

−+

= dd

Fd

dd NN

kT

EE

Nn

exp2

11Density of electron

occupying the

donor levelConcentration of

donors

Concentration of

ionized donors

Microelectronics I

acceptor

Discrete acceptor level

acceptor

−−=

−+

= aa

aF

aa NN

kT

EE

g

Np

exp1

1Concentration of

holes in the

acceptor statesConcentration of

acceptorsConcentration of

ionized acceptor

g; degeneracy factor (Si; 4)

Microelectronics I

from the probability function, we can calculate the friction of total electrons still in

the donor states at T=300 K

−−+

=+

kT

EE

N

Nnn

n

dC

d

Cod

d

)(exp

21

1

Consider phosphorus doping in Si for T=300K at concentration of 1016 cm-3

ionization energy

Consider phosphorus doping in Si for T=300K at concentration of 10 cm

(NC=2.8 x1019 cm-3, EC-Ed= 0.045 eV)

%41.00041.0

0259.0

045.0exp

102

108.21

1

16

19

==

−

×

×+

=+ od

d

nn

n

�only 0.4% of donor states contain electron. the donor states are states

are said to be completely ionized

Microelectronics I

Complete ionization; The condition when all donor atoms are positively

charged by giving up their donor electrons and all acceptor atoms are negatively

charged by accepting electrons

Microelectronics I

At T=0 K, all electron in their lowest possible energy state

Nd+=0 and Na

-=0

EF

EF

Freeze-out; The condition that occurs in a semiconductor when the temperature

is lowered and the donors and acceptors become neutrally charged. The

electron and hole concentrations become very small.

Microelectronics I

Charge neutrality

In thermal equilibrium, semiconductor crystal is electrically neutral

“Negative charges = positive charge”

Determined the carrier concentrations as a function of impurity doping

concentration

Charge-neutrality condition

Compensated semiconductor; A semiconductor that contains both donor and

acceptors at the same region

If Nd > Na � n-type compensated semiconductor

If Na > Nd � p-type compensated semiconductor

If Nd = Na � has the characteristics of an intrinsic semiconductor

concentration

Microelectronics I

Charge-neutrality condition

+−+=+ doao NpNn

Negative charges Positive chargesNegative charges Positive charges

)()( ddoaao nNppNn −+=−+

Microelectronics I

)()( ddoaao nNppNn −+=−+

If we assume complete ionization (pa=0, nd=0)

doao NpNn +=+

From nopo=ni2

2

i Nn

Nn +=+

2

2

0

22i

adado

di

ao

nNNNN

n

Nn

nNn

+

−+

−=

+=+

Electron concentration is given as function of donors and acceptors concentrations

Microelectronics I

Example;

Consider an n-type silicon semiconductor at T=300 K in which Nd=1016 cm-3

and Na=0. The intrinsic carrier concentration is assumed to be ni=1.5x1010

cm-3. Determine the thermal equilibrium electron and hole concentrations.

2101616

2

2

)105.1(1010

22

×+

+=

+

−+

−= n

NNNNn i

adado

Electron,

316

210

10

)105.1(2

10

2

10

−≈

×+

+=

cm

hole, 34

16

2102

1025.210

)105.1( −×=

×== cm

n

np

o

io

Microelectronics I

Redistribution of electrons when donors are added

+ + + +

- - - - - -

Intrinsic electron

+ + +

Intrinsic hole

�When donors are added, no > ni and po < ni

�A few donor electron will fall into the empty states in valence band and

hole concentration will decrease

�Net electron concentration in conduction band ≠ intrinsic electron +

donor concentration

Microelectronics I

Temperature dependence of no

2

2

22i

adado n

NNNNn +

−+

−=

Very strong function of temperature

�As temperature increases, n 2 term will dominate. Shows intrinsic characteristics�As temperature increases, ni2 term will dominate. Shows intrinsic characteristics

0 K Temperature

Freeze-out

Partial ionizationExtrinsic

no=Nd

Intrinsic

no=ni

Microelectronics I

Hole concentration

From charge-neutrality condition and nopo product

)()( ddoaao nNppNn −+=−+

2

ioo npn =

2

i NpNn

+=+

2

2

22i

dadao

doa

o

i

nNNNN

p

NpNp

n

+

−+

−=

+=+

Microelectronics I

Example;

Consider an p-type silicon semiconductor at T=300 K in which Na=1016 cm-3

and Nd=3 x 1015 cm-3. The intrinsic carrier concentration is assumed to be

ni=1.5x1010 cm-3. Determine the thermal equilibrium electron and hole

concentrations.

2

2

22+

−+

−= n

NNNNp i

dadao

Hole,

315

21015161516

107

)105.1(2

10310

2

10310

−×≈

×+

×−+

×−=

cm

electron,34

15

2102

1021.3107

)105.1( −×=

×

×== cm

p

nn

o

io

po=Na-Ndapproximation

Microelectronics I

Position of Fermi Energy Level

As a function of doping concentration and temperature

Equations for position of Fermi level (n-type)

=−

o

CFC

n

NkTEE ln

Compensated semiconductor, n =N -N

−=−

ad

CFC

NN

NkTEE ln

Compensated semiconductor, no=Nd-Na

=−

i

oFiF

n

nkTEE ln

Microelectronics I

Equations for position of Fermi level (p-type)

=−

o

vCF

p

NkTEE ln

Compensated semiconductor, po=Na-Nd

−=−

da

vvF

NN

NkTEE ln

− da NN

=−

i

oFFi

n

pkTEE ln

Microelectronics I

Example;

Silicon at T=300 K contains an acceptor impurity concentration of Na=1016 cm-3.

Determine the concentration of donor impurity atoms that must be added so that

the Silicon is n-type and Fermi energy is 0.20 eV below the conduction band

edge.

)(

ln

−−

−=−

EE

NN

NkTEE

ad

CFC

31619 1024.10259.0

2.0exp108.2

)(exp

−×=

−×=

−−=−

cm

kT

EENNN FC

Cad

316316 1024.21024.1 −−×=+×= cmNcmN ad

Microelectronics I

Position of EF as function of donor concentration (n-type) and acceptor

concentration (p-type)

Microelectronics I

Position of EF as function of temperature for various doping concentration

Microelectronics I

Important terms

Intrinsic semiconductor; A pure semiconductor material with no impurity

atoms and no lattice defects in the crystal

Extrinsic semiconductor; A semiconductor in which controlled amounts

of donors and/or acceptors have been added so that the electron and hole

concentrations change from the intrinsic carrier concentration and a

preponderance of either electron (n-type) or hole (p-type) is created.preponderance of either electron (n-type) or hole (p-type) is created.

Acceptor atoms; Impurity atoms added to a semiconductor to create a p-

type material

Donor atoms; Impurity atoms added to a semiconductor to create n-type

material

Microelectronics I

Complete ionization; The condition when all donor atoms are positively

charged by giving up their donor electrons and all acceptor atoms are

negatively charged by accepting electrons

Freeze-out; The condition that occurs in a semiconductor when the

temperature is lowered and the donors and acceptors become neutrally

charged. The electron and hole concentrations become very small

Fundamental relationship

2

ioo npn =

Microelectronics I

problems

1. The value of po in Silicon at T=300K is 1015 cm-3. Determine (a) Ec-EF and (b) no

2. Determine the equilibrium electron and hole concentrations in Silicon for the

following conditions;

(a) T=300 K, Nd= 2x1015cm-3, Na=0

(b) T=300 K, Nd=Na=1015 cm-3

3. (a) Determine the position of the Fermi level with respect to the intrinsic Fermi 3. (a) Determine the position of the Fermi level with respect to the intrinsic Fermi

level in Silicon at T=300K that is doped with phosphorus atoms at a

concentration of 1015cm-3. (b) Repeat part (a) if the Si is doped with boron

atoms at a concentration of 1015cm-3. (c) Calculate the electron concentration in

the Si for part (a) and (b)