CHAPTER 3CHAPTER 3CARRIER CONCENTRATIONS IN CARRIER CONCENTRATIONS IN

SEMICONDUCTORSSEMICONDUCTORS

CHAPTER 3CHAPTER 3CARRIER CONCENTRATIONS IN CARRIER CONCENTRATIONS IN

SEMICONDUCTORSSEMICONDUCTORS

Prof. Dr. Beşire GÖNÜL

CARRIER CONCENTRATIONS IN CARRIER CONCENTRATIONS IN SEMICONDUCTORSSEMICONDUCTORS

• Donors and Acceptors

• Fermi level , Ef

• Carrier concentration equations

• Donors and acceptors both present

Donors and AcceptorsDonors and Acceptors

The conductivity of a pure The conductivity of a pure (intrinsic) s/c is low due to (intrinsic) s/c is low due to the low number of free the low number of free carriers.carriers.

For an intrinsic semiconductor

n = p = ni

n = concentration of electrons per unit volumep = concentration of holes per unit volumeni = the intrinsic carrier concentration of the semiconductor under

consideration.

The number of carriers are generated by thermally or electromagnetic radiation for a pure s/c.

n.p = nn.p = nii22

n = pn = p

number of enumber of e--’s in CB = number of holes in VB’s in CB = number of holes in VB

This is due to the fact that when an e-

makes a transition to the CB, it leaves a hole behind in VB. We have a bipolar (two carrier) conduction and the number of holes and e- ‘s are equal.

n.p = ni2

This equation is called as mass-action law.

The intrinsic carrier concentration nThe intrinsic carrier concentration n i i depends on; depends on; the the semiconductor materialsemiconductor material, and, and the the temperaturetemperature. .

n.p = nn.p = nii22

For silicon at 300 K, ni has a value of 1.4 x 1010 cm-3.

Clearly , equation (n = p = ni) can be written as

n.p = nn.p = nii22

This equation is valid for extrinsic as well as intrinsic material.

To increase the conductivity, one can To increase the conductivity, one can dope pure s/c with atoms from column dope pure s/c with atoms from column lll or V of periodic table. This process is lll or V of periodic table. This process is called as called as dopingdoping and the added atoms and the added atoms are called as are called as dopantsdopants impurities.impurities.

What is doping and dopants impurities What is doping and dopants impurities ??

There are two types of doped or extrinsic s/c’s; n-type p-type

Addition of different atoms modify the conductivity of the intrinsic semiconductor.

p-type doped semiconductorp-type doped semiconductor

Si + Column lll impurity atoms

Boron (B)has three valance

e-’ s

Have four

valance e-’s

Si

Si Si

Si

B

Electron

Hole

Bond with missingelectron

Normal bond with two electrons

Boron bonding in Silicon Boron sits on a lattice side

p >> n

Boron(column III)Boron(column III) atoms have three valance atoms have three valance electrons, there is a deficiency of electron or electrons, there is a deficiency of electron or missing electron to complete the outer shell.missing electron to complete the outer shell.

This means that each added or doped This means that each added or doped boronboron atom introduces a atom introduces a single holesingle hole in the crystal. in the crystal.

There are two ways of producing hole1) Promote e-’s from VB to CB,2) Add column lll impurities to the s/c.

Energy Diagram for a p-type s/cEnergy Diagram for a p-type s/c

Ec = CB edge energy level

Ev = VB edge energy level

EA= Acceptor energ level

Eg

CBCB

VBVB

acceptor(Column lll) atoms

The energy gap is forbidden only for pure material, i.e. Intrinsic material.

Electron

Hole

The impurity atoms from column lll occupy at an The impurity atoms from column lll occupy at an energy level within Eenergy level within Eg g . These levels can be. These levels can be

1.1. Shallow levels which is close to the band edge,Shallow levels which is close to the band edge,

2.2. Deep levels which lies almost at the mid of the band gap.Deep levels which lies almost at the mid of the band gap.

If the EA level is shallow i.e. close to the VB edge, each added boron atom accepts an e- from VB and have a full configuration of e-’s at the outer shell.

These atoms are called as acceptor atoms since they accept an e- from VB to complete its bonding. So each acceptor atom gives rise a hole in VB.

The current is mostly due to holes since the number of holes are made greater than e-’s.

p-type semiconductor

Holes Holes = = pp = majority carriers = majority carriersElectrons Electrons = = nn = minority carriers = minority carriers

Majority and minority carriers in a p-type semiconductorMajority and minority carriers in a p-type semiconductor

t2

t1

t3

Electric field direction

Holes movement as a function of applied electric field

Hole movement directionElectron movement direction

Ec

Ev

Ea

Eg

Electron

Hole

Shallow acceptor in silicon

Si

Si Si

Si

P

Electron

Weakly bound electron

Normal bond with two electrons

Phosporus bonding in silicon

Ec

Ev

Ed

Eg

Electron

Valance band

Conduction band

Band gap is 1.1 eV for silicon

Shallow donor in silicon

Donor and acceptor charge states

Electron

Hole

Neutral donor centre

İonized (+ve)donor centre

Neutral acceptor centre

İonized (-ve)acceptor centre

Ec

Ev

Ea

Ec

Ev

Ea

Extra e- of column V atom is weakly attached to its host atom

n-type semiconductorn-type semiconductor

Si

Si

Si

As

Si

Si + column V (with five valance e- )

ionized (+ve)donor centre

Ec

Ev

ED = Donor energy level (shallow)

Band gap is 1.1 eV for silicon

Hole

Electron

Eg

n - type semiconductor

nnpp , p , pnn

n-type , n-type , n >> pn >> p ; n is the majority carrier ; n is the majority carrier concentration concentration nnnn

p is the minority carrier p is the minority carrier concentration concentration ppnn

p-type , p-type , p >> np >> n ; p is the majority carrier ; p is the majority carrier concentration concentration pppp

n is the minority carrier n is the minority carrier concentration concentration nnpp

np pn

Type of semiconductor

calculationcalculation Calculate the hole and electron densities in a piece of p-type Calculate the hole and electron densities in a piece of p-type

silicon that has been doped with 5 x 10silicon that has been doped with 5 x 101616 acceptor atoms per cm acceptor atoms per cm3 3 . .

nnii = 1.4 x 10= 1.4 x 101010 cm cm-3 -3 (( at room temperature)at room temperature)

UndopedUndoped

n = p = nn = p = ni i

p-type ; p >> np-type ; p >> n

n.p = nn.p = nii2 2 NNAA = 5 x 10 = 5 x 1016 16 p = Np = NA A = 5 x 10= 5 x 1016 16 cmcm-3-3

3316

23102

109.3105

)104.1(x

cmx

cmx

p

nn i

electrons per cm3

p >> ni and n << ni in a p-type material. The more holes you put in the less e-’s you have and vice versa.

Fermi level , EFermi level , EFF

This is a reference energy level at which the probability of This is a reference energy level at which the probability of occupation by an electron is ½.occupation by an electron is ½.

Since ESince Ef f is a reference level therefore it can appear anywhere in is a reference level therefore it can appear anywhere in the energy level diagram of a S/C .the energy level diagram of a S/C .

Fermi energy level is not fixed.Fermi energy level is not fixed. Occupation probability of an electron and hole can be determined Occupation probability of an electron and hole can be determined

by Fermi-Dirac distribution function, Fby Fermi-Dirac distribution function, FFD FD ; ;

EEF F = Fermi energy level= Fermi energy level

kkBB = Boltzman constant= Boltzman constantT T = Temperature= Temperature

)exp(1

1

TkEE

F

B

FFD

E is the energy level under investigation.E is the energy level under investigation. FFFDFD determines the probability of the energy level E determines the probability of the energy level E

being occupied by electron.being occupied by electron.

determines the probability of not finding an determines the probability of not finding an electron at an energy level E; the probability of finding electron at an energy level E; the probability of finding a hole .a hole .

)exp(1

1

TkEE

F

B

FFD

FD

FDF

f

fEEif

1

2

1

0exp1

1

Fermi level , EFermi level , EFF

Carrier concentration equationsCarrier concentration equations

The number density, i.e., the number of electrons The number density, i.e., the number of electrons available for conduction in CB is available for conduction in CB is

3 / 2*

2

22 exp ( )

exp ( ) exp( )

p F V

F V i FV i

m kT E Ep

h kT

E E E Ep N p n

kT kT

3 / 2*

2

22 exp ( )

exp ( ) exp( )

n C F

C F F iC i

m kT E En

h kT

E E E En N n n

kT kT

The number density, i.e., the number of holes The number density, i.e., the number of holes available for conduction in VB is available for conduction in VB is

Donors and acceptors both presentDonors and acceptors both present

Both donors and acceptors present in a s/c in Both donors and acceptors present in a s/c in general. However one will outnumber the other general. However one will outnumber the other one.one.

In an n-type material the number of donor In an n-type material the number of donor concentration is significantly greater than that of concentration is significantly greater than that of the acceptor concentration.the acceptor concentration.

Similarly, in a p-type material the number of Similarly, in a p-type material the number of acceptor concentration is significantly greater than acceptor concentration is significantly greater than that of the donor concentration.that of the donor concentration.

A p-type material can be converted to an n-type A p-type material can be converted to an n-type material or vice versa by means of adding proper material or vice versa by means of adding proper type of dopant atoms. This is in fact how p-n type of dopant atoms. This is in fact how p-n junction diodes are actually fabricated.junction diodes are actually fabricated.

How does the position of the Fermi Level change withHow does the position of the Fermi Level change with

(a)(a) increasing increasing donor concentrationdonor concentration, and, and

(b)(b) increasing increasing acceptor concentrationacceptor concentration ??

Worked exampleWorked example

(a) We shall use equation

İf n is increasing then the quantity EC-EF must be decreasing i.e. as the donor concentration goes up the Fermi level moves towards the conduction band edge Ec.

exp ( )C FC

E En N

kT

Worked exampleWorked example

But the carrier density equations such as;

aren’t valid for all doping concentrations! As the fermi-level comes

to within about 3kT of either band edge the equations are no longer

valid, because they were derived by assuming the simpler Maxwell

Boltzmann statics rather than the proper Fermi-Dirac statistic.

kT

EEnp

andkT

EE

h

kTmn

Fii

Fcn

exp

exp2

22

3

2

*

Worked example Worked example

EEgg/2/2

EEgg/2/2

EEgg/2/2

EEgg/2/2

EEgg/2/2

EEgg/2/2

EECC

EEVV

EEF1F1

EECC

EEVV

EEF1F1

EEF2F2

EEF2F2

EEF3F3

EEF3F3

n3n1 n2

p1 p3p2

p3 > p2 > p1

n3 > n2 > n1

Worked exampleWorked example

(b) Considering the density of holes in valence band;

It is seen that as the acceptor concentration increases, Fermi-level

moves towards the valance band edge. These results will be used in

the construction of device (energy) band diagrams.

kT

EENp VFv exp

Donors and acceptor both presentDonors and acceptor both present

n D An N N 2) 2) Similarly, when the number of shallow acceptor concentration is signicantly Similarly, when the number of shallow acceptor concentration is signicantly greater than the shallow donor concentration greater than the shallow donor concentration in in a piece of a s/c, it can be a piece of a s/c, it can be considered as a p-type s/c andconsidered as a p-type s/c and

• In general, both donors and acceptors are present in a piece of a In general, both donors and acceptors are present in a piece of a semiconductor although one will outnumber the other one.semiconductor although one will outnumber the other one.• The impurities are incorporated unintentionally during the growth of the The impurities are incorporated unintentionally during the growth of the semiconductor crystal causing both types of impurities being present in a piece semiconductor crystal causing both types of impurities being present in a piece of a semiconductor.of a semiconductor.• How do we handle such a piece of s/c?How do we handle such a piece of s/c?

1) Assume that the shallow donor concentration is significantly greater 1) Assume that the shallow donor concentration is significantly greater than than that of the shallow acceptor concentration. In this case the material behaves as that of the shallow acceptor concentration. In this case the material behaves as an n-type material andan n-type material and

For the case For the case NNAA>N>ND D , i.e. , i.e. for p-type materialfor p-type material

Donors and acceptor both present Donors and acceptor both present

2

22 2

.

0

0 ( ) 0

p p i

p A D P p D p A

ip p D A p D A p i

p

n p n

n N N p p N n N

np p N N p N N p n

p

Donors and acceptor both presentDonors and acceptor both present

22 2

1,2

12 22

2

4( ) 0 , solving for ;

2

14

2

p D A p i p

p A D A D i

ip

p

b b acp N N p n p x

a

p N N N N n

nn

p

majority

minority

Donors and acceptor both presentDonors and acceptor both present For the case For the case NNDD>N>NAA , , i.e. n-type materiali.e. n-type material

22

22 2

2

1,2

12 22

2

.

0

0 ( ) 0

4solving for n ;

2

14

2

in n i n

n

n A D n n A n D

in n A D n A D n i

n

n

n D A D A i

in

n

nn p n p

n

n N N P n N p N

nn n N N n N N n n

n

b b acx

a

n N N N N n

np

n