donors and acceptors in semiconductors

8/12/2019 Donors and Acceptors in Semiconductors

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Donors and acceptors in

Semiconductors

Prof. Dr. Mohammed GulamAhamad


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CARRIER CONCENTRATIONS INSEMICONDUCTORS

Donors and Acceptors

Fermi level , Ef Carrier concentration equations

Donors and acceptors both present


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Donors and Acceptors

The conductivity of a pure(intrinsic) s/c is low due tothe low number of freecarriers.

For an intrinsic semiconductor

n = p = ni

n = concentration of electrons per unit volume

p = concentration of holes per unit volume

ni = the intrinsic carrier concentration of the semiconductor under consideration.

The number of carriers aregenerated by thermally orelectromagnetic radiationfor a pure s/c.


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n.p = ni2

n = p

number of e-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 behindin VB. We have a bipolar (two carrier)

conduction and the number of holes and e- sare equal.

n.p = ni2This equation is called as mass-action

law.


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The intrinsic carrier concentration ni depends on;

the semiconductor material, and

the temperature.

n.p = ni2

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 = ni2

This equation is valid for extrinsic as well as

intrinsic material.


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To increase the conductivity, one can

dope pure s/c with atoms from column lllor V of periodic table. This process iscalled as doping and the added atomsare called as dopantsimpurities.

What is doping and dopants impurities ?

There are two types of doped or extrinsic s/cs;

n-type

p-type

Addition of different atoms modify the conductivity ofthe intrinsic semiconductor.


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p-type doped semiconductor

Si + Column lll impurity atoms

Boron (B)has three valance

e-s

Havefour

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


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Boron(column III) atoms have three valance

electrons, there is a deficiency of electron ormissing electron to complete the outer shell.

This means that each added or doped boron

atom introduces a single holein the crystal.

There are two ways of producing hole

1) Promote e

-

sfrom VB to CB,2) Add column lll impurities to the s/c.


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Energy Diagram for a p-type s/c

Ec = CB edge energy level

Ev = VB edge energy level

EA= Acceptor energ level

Eg

CB

VB

acceptor(Column lll) atoms

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

Electron

Hole


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The impurity atoms from column lll occupy at an energy level

within Eg . These levels can be1. Shallow levels which is close to the band edge,

2. 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 addedboron atom accepts an e- from VB and have a fullconfiguration of e-sat the outer shell.

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

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

p-type semiconductor


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Holes = p = majority carriers

Electrons = n = minority carriers

Majority and minority carriers in a p-type semiconductor

t2

t1

t3

Electric field direction

Holes movement as afunction of appliedelectric field

Hole movement direction

Electron movementdirection


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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


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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 donorcentre

onized (+ve)donor centre

Neutralacceptorcentre

onized (-ve)acceptorcentre

Ec

Ev

Ea

Ec

Ev

Ea


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

n-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


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np, pn

n-type , n >> p; n is the majority carrier

concentration nnp is the minority carrierconcentration pn

p-type , p >> n; p is the majority carrierconcentration pp

n is the minority carrierconcentration np

np

pn

Type of semiconductor


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calculation

Calculate the hole and electron densities in a piece of p-type siliconthat has been doped with 5 x 1016acceptor atoms per cm3 .

ni= 1.4 x 10

10

cm

-3

(

at room temperature)

Undoped

n = p = ni

p-type ; p >> n

n.p = ni2 NA= 5 x 10

16 p = NA = 5 x 1016 cm-3

3

316

23102

109.3105

)104.1(xcmx

cmx

p

nn i

electrons per cm3

p >> ni and n


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Fermi level , EF

This is a reference energy level at which the probability of occupationby an electron is .

Since Ef is a reference level therefore it can appear anywhere in theenergy level diagram of a S/C .

Fermi energy level is not fixed.

Occupation probability of an electron and hole can be determinedby Fermi-Dirac distribution function, FFD ;

EF = Fermi energy level

kB = Boltzman constant

T = Temperature

)exp(1

1

Tk

EEF

B

FFD


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E is the energy level under investigation.

FFDdetermines the probability of the energy level E beingoccupied by electron.

determines the probability of not finding anelectron at an energy level E; the probability of finding ahole .

)exp(1

1

Tk

EEF

B

FFD

FD

FDF

f

fEEif

1

2

1

0exp1

1

Fermi level , EF

C i t ti ti


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Carrier concentration equations

The number density, i.e., the number of electronsavailable 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 E n

h kT

E E E En N n nkT kT

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


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Donors and acceptors both present

Both donors and acceptors present in a s/c in

general. However one will outnumber the otherone.

In an n-type material the number of donorconcentration is significantly greater than that

of the acceptor concentration. Similarly, in a p-type material the number of

acceptor concentration is significantly greaterthan that of the donor concentration.

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

W k d l


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How does the position of the Fermi Level change with

(a) increasing donor concentration, and

(b) increasing acceptor concentration?

Worked example

(a) We shall use equation

fn is increasing then the quantity EC-EF must be decreasing

i.e. as the donor concentration goes up the Fermi level movestowards the conduction band edge Ec.

exp ( )C FCE E

n NkT

W k d l


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Worked example

But the carrier density equations such as;

arent 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

223

2

*


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Worked example

Eg/2

Eg/2

Eg/2

Eg/2

Eg/2

Eg/2

EC

EV

EF1

EC

EV

EF1EF2

EF2

EF3

EF3

n3n1 n2

p1 p3p2

p3 > p2 > p1

n3 > n2 > n1

W k d l


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Worked 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 VFvexp


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Donors and acceptor both present

n D An N N 2) Similarly, when the number of shallow acceptor concentration is signicantly

greater than the shallow donor concentration in a piece of a s/c, it can be

considered as a p-type s/c and

In general, both donors and acceptors are present in a piece of a

semiconductor although one will outnumber the other one.

The impurities are incorporated unintentionally during the growth of the

semiconductor crystal causing both types of impurities being present in a piece

of a semiconductor.

How do we handle such a piece of s/c?

1) Assume that the shallow donor concentration is significantly greater thanthat of the shallow acceptor concentration. In this case the material behaves as

an n-type material and

D d t b th t


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For the case NA>ND , i.e. for p-type material


2

2

2 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

D d t b th t


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2

2 21,2

12 22

2

4( ) 0 , solving for ;2

1

42

p D A p i p

p A D A D i

i

pp

b b acp N N p n p xa

p N N N N n

n

n p

majority

minority

D d t b th t


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For the case ND>NA, i.e. n-type material

22

22 2

2

1,2

12 22

2

.

0

0 ( ) 0

4solving for n ;

2

14

2

i

n n i nn

n A D n n A n D

i

n n A D n A D n in

n

n D A D A i

in

nn p n p

n

n N N P n N p N

n

n n N N n N N n nn

b b acx

a

n N N N N n

np

donors and acceptors in semiconductors

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