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Ionization Energies in Si, Ge, GaAs

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Page 1: Ionization Energies in Si, Ge, GaAs - MyBlog...Complete Ionization Ex: we want to determine the fraction of electrons are still in the donor state at T=300oK in the case of P donor

Ionization Energies in Si, Ge, GaAs

Page 2: Ionization Energies in Si, Ge, GaAs - MyBlog...Complete Ionization Ex: we want to determine the fraction of electrons are still in the donor state at T=300oK in the case of P donor

Extrinsic Semiconductors An extrinsic semiconductors is defined as a semiconductor in which controlled amounts of specific dopant or impurity atoms have been added so that the thermal-equilibrium electron and hole concentration are different from the intrinsic carrier concentration. One type of carrier will be predominant.

Adding impurities will change the distributions of electrons and holes in the material and since the Fermi level is related to the distribution function EF will also change. In the figure: EF>Eg/2 no>po n-type

⎥⎦

⎤⎢⎣

⎡ −−=

kTEE

Nn FcCo

)(exp

⎥⎦

⎤⎢⎣

⎡ −−=

kTEE

Np vFVo

)(exp

Page 3: Ionization Energies in Si, Ge, GaAs - MyBlog...Complete Ionization Ex: we want to determine the fraction of electrons are still in the donor state at T=300oK in the case of P donor

Carrier distribution of a p-type semiconductor

EF<Eg/2 po>no p-type semiconductor

Page 4: Ionization Energies in Si, Ge, GaAs - MyBlog...Complete Ionization Ex: we want to determine the fraction of electrons are still in the donor state at T=300oK in the case of P donor

Extrinsic Semiconductors The expressions previously derived for the thermal-equilibrium concentrations of electrons and holes are general:

⎥⎦

⎤⎢⎣

⎡ −−=⎥

⎤⎢⎣

⎡ −−=

kTEE

NpkTEE

Nn VFvo

FCCo

)(exp

)(exp

for n-type semiconductor electrons are majority carriers and holes minority for p-type semiconductor holes are majority carriers and electrons minority nopo

2

exp)(

exp )(

exp

ioo

GVC

VFFCVCoo

npnkTENN

kTEE

kTEENNpn

=

⎥⎦

⎤⎢⎣

⎡−=⎥

⎤⎢⎣

⎡ −−⎥⎦

⎤⎢⎣

⎡ −−=

The product is always a constant for a given semiconductor at a given temperature (only when the Boltzmann approximation is valid) .

Page 5: Ionization Energies in Si, Ge, GaAs - MyBlog...Complete Ionization Ex: we want to determine the fraction of electrons are still in the donor state at T=300oK in the case of P donor

Non validity of the Boltzmann approximation

If (Ec-EF) >>kT then (E-EF) >> kT the Fermi probability function reduces to the Boltzmann approximation

kTEE

kTEE

Ef F

FF

)([exp)(exp1

1)( −−≅

−+

=

If the Boltzmann approximation does not hold (E-EF) ∼kT (EF very close to the conduction band) :

dE

kTEE

EEhm

nF

c

En

oc

⎥⎦

⎤⎢⎣

⎡ −−+

−= ∫

)(exp1

)2(43

2/3*π

If we again make a change of variable kTEE C−

and also define kTEE CF

F−

Page 6: Ionization Energies in Si, Ge, GaAs - MyBlog...Complete Ionization Ex: we want to determine the fraction of electrons are still in the donor state at T=300oK in the case of P donor

Non validity of the Boltzmann approximation

[ ]

[ ]η

ηηη

η

ηηη

ηπ

dF

dhm

n

FF

F

no

−+=

−+=

(exp1)(

as defined is integral The(exp1

)2(4

2/1

02/1

2/1

03

2/3*

This function is called Fermi-Dirac integral and is a tabulated function of the variable ηF.

( )[ ] kT

EEkTEEd

hm

p FVF

V

F

po

−=

−=

−+= ∫

∞ ''''

2/1'

03

2/3*

and (exp1

)2(4ηηη

ηηηπ

The same general method can be used to calculate the thermal equilibrium hole concentration

Page 7: Ionization Energies in Si, Ge, GaAs - MyBlog...Complete Ionization Ex: we want to determine the fraction of electrons are still in the donor state at T=300oK in the case of P donor

Example

Ex: Si at 300oK

ηF= 2 EC -EF=52meV kT=25meV EC -EF≈KT

1008.2 3200

−×= cmn

319190

2/1319

2/10

1027.7)3.2)(108.2(2

3.2)2( and 108.2 )(2

×=×=

=×==

cmn

FcmNFNn CFC

π

ηπ

Note if we would used the Boltzmann approximation (not valid in this case)

Page 8: Ionization Energies in Si, Ge, GaAs - MyBlog...Complete Ionization Ex: we want to determine the fraction of electrons are still in the donor state at T=300oK in the case of P donor

Degenerate and Nondegenerate semiconductors In the previous discussion of adding dopant atoms to a semiconductor we have implicitly assumed that the concentration of dopant atoms added is small when compared to the semiconductor atoms (~1024 cm-3). The impurity atoms are spread far enough apart so that there is not interations between donor electrons. We have assumed that the impurities introduce discrete, noninteracting donor (acceptor) energy states in n-type (p-type) semiconductor (nondegenerate semiconductor). If the impurity concentration increases the distance between the impurity atoms decreases and at one point the donor (acceptor) electrons will begin to interact with each other. When this occurs the single descrete donor (acceptor) energy state will split into a band of energies and may overlap the bottom of the conduction band. When the electron (hole) concentrations exceeds Nc then EF lies within the conduction band (degenerate semiconductor).

Page 9: Ionization Energies in Si, Ge, GaAs - MyBlog...Complete Ionization Ex: we want to determine the fraction of electrons are still in the donor state at T=300oK in the case of P donor

Nondegenerate and Degenerate Energy-bands Nondegenerate semiconductor Degenerate semiconductor

Small Nd or Na and EF in the gap High Nd or Na

Page 10: Ionization Energies in Si, Ge, GaAs - MyBlog...Complete Ionization Ex: we want to determine the fraction of electrons are still in the donor state at T=300oK in the case of P donor

Statistics of donors and acceptors

The derivation of the Fermi-Dirac probability function was derived considering the Pauli exclusion principle. This principle also applies to the donor and acceptor states. The statistics of filling band gap levels differs slightly from the statistics of filling energy band levels. A donor site can accomodate only one electron however the donor-site electron can have either spin-up and a spin-down (g=2 spin degeneracy). This fact increases the ways of arranging electrons on the donor-level sites The probability function of electrons occupying the donor state is:

2g

exp11

=

⎟⎠

⎞⎜⎝

⎛ −+

=

kTEE

g

NnFd

dd

where nd is the electron density occupying the donor level and Ed is the donor energy level

+−= ddd NNn Nd= density of the donor atoms Nd

+= density of ionized donors

Page 11: Ionization Energies in Si, Ge, GaAs - MyBlog...Complete Ionization Ex: we want to determine the fraction of electrons are still in the donor state at T=300oK in the case of P donor

Statistics of acceptors

The same for acceptor atoms:

4g

exp11

=

−=

⎟⎠

⎞⎜⎝

⎛ −+

= −aa

aF

aa NN

kTEE

g

Np

g=2x2=4 for Si and GaAs due to two-fold (heavy and light-hole) degeneracy of the valence band

where pa is the hole density occupying the acceptor level and

Ea is the acceptor energy level

Na= density of the acceptor atoms Na

-= density of ionized acceptor atoms

Page 12: Ionization Energies in Si, Ge, GaAs - MyBlog...Complete Ionization Ex: we want to determine the fraction of electrons are still in the donor state at T=300oK in the case of P donor

Number of electrons in the donor state

( )⎥⎦

⎤⎢⎣

⎡ −−=

⎟⎠

⎞⎜⎝

⎛ −≈

kTEEN

kTEE

Nn Fdd

Fd

dd exp2

exp21

If we assumed ( ) kTEE Fd >>−

In this condition the Boltzmann approximation is also valid then for the electron in the conduction band we have:

)(

exp ⎥⎦

⎤⎢⎣

⎡ −−=

kTEE

Nn FCCo

We would like to determine the relative number of electrons in the donor state compared to the total number of electrons:

( )

( )⎥⎦

⎤⎢⎣

⎡ −−+

=

⎥⎦

⎤⎢⎣

⎡ −−+⎥

⎤⎢⎣

⎡ −−

⎥⎦

⎤⎢⎣

⎡ −−

=+

kTEE

NN

kTEEN

kTEEN

kTEEN

nnn

dC

d

CFCC

Fdd

Fdd

od

d

)(exp

21

1

)(

expexp2

exp2

Page 13: Ionization Energies in Si, Ge, GaAs - MyBlog...Complete Ionization Ex: we want to determine the fraction of electrons are still in the donor state at T=300oK in the case of P donor

Complete Ionization

Ex: we want to determine the fraction of electrons are still in the donor state at T=300oK in the case of P donor atoms in Si (Nd=1016cm-3).

%4.00041.0

0259.0)045.0(exp

)10(2108.21

1

16

19 ==

⎥⎦

⎤⎢⎣

⎡−×+

=+ od

d

nnn

This example shows that at RT only few electrons are in the donor state compared with the conduction band. Only 0.4% of the donor states contain electrons: the donor states are said to be completely ionized. The same is valid for the acceptor states. Therefore at RT all donor (acceptor) atoms have donated an electron to the conduction band.

Page 14: Ionization Energies in Si, Ge, GaAs - MyBlog...Complete Ionization Ex: we want to determine the fraction of electrons are still in the donor state at T=300oK in the case of P donor

Freeze-out The opposite occurs at T=0oK. All electrons are in their lowest possible energy state

0or == +ddd NNn

dF

Fd

Fd

dd

EEkTEE

kTEE

Nn

>⇒

=−∞=⎟⎠

⎞⎜⎝

⎛ −

⎟⎠

⎞⎜⎝

⎛ −+

=

0)exp( ; 0exp

exp

211

Fig. 4-13

Page 15: Ionization Energies in Si, Ge, GaAs - MyBlog...Complete Ionization Ex: we want to determine the fraction of electrons are still in the donor state at T=300oK in the case of P donor

Compensated semiconductors

In thermal equilibrium the semiconductor crystal is electrically neutral. This charge-neutrality condition is used to determine the thermal-equilibrium electron and hole concentrations as a function of the impurity doping concentration. A compensated semiconductor is one that contains both donor and acceptor impurity atoms in the same region. n-type compensated semiconductor occurs when Nd>Na p-type compensated semiconductor occurs when Nd<Na If Nd=Nawe have a completely compensated semiconductor (characteristics similar to intrinsic).

Page 16: Ionization Energies in Si, Ge, GaAs - MyBlog...Complete Ionization Ex: we want to determine the fraction of electrons are still in the donor state at T=300oK in the case of P donor

Energy-band diagram

Compensated semiconductor

Page 17: Ionization Energies in Si, Ge, GaAs - MyBlog...Complete Ionization Ex: we want to determine the fraction of electrons are still in the donor state at T=300oK in the case of P donor

no for compensated semiconductors In thermal equilibrium the semiconductor crystal is electrically neutral (the net charge density is zero).

( ) ( )ddoaao

doao

nNppNnNpNn

−+=−+

+=+ +−

If we assumed complete ionization (at RT)

22

22

2

2)(

2)(

0)(

0 and 0

iadad

o

ioado

do

idoao

ad

nNNNNn

nnNNn

NnnNpNn

pn

+⎟⎠

⎞⎜⎝

⎛ −+

−=

=−−−

=+=+=+

==

For an intrinsic semiconductor Na=Nd=0 io nn =

for Nd>Na

Page 18: Ionization Energies in Si, Ge, GaAs - MyBlog...Complete Ionization Ex: we want to determine the fraction of electrons are still in the donor state at T=300oK in the case of P donor

Determination of no in compensated semiconductor

Ex: n-type Si at T=300oK with Nd=1016 cm-3 and Na= 1012 cm-3 Assumed ni=1.5x1010 cm-3

( ) 3162101616

22

10105.12

102

10

2)(

2)(

−≅×+⎟⎟⎠

⎞⎜⎜⎝

⎛+=

=+⎟⎠

⎞⎜⎝

⎛ −+

−=

cm

nNNNNn iadad

o

majority carrier concentration

( ) 3416

2102

1025.2101105.1 −×=

×

×== cm

nnpo

io minority carrier concentration

no>ni po<ni electron redistribution along the available energy states =>

NOTE:

Page 19: Ionization Energies in Si, Ge, GaAs - MyBlog...Complete Ionization Ex: we want to determine the fraction of electrons are still in the donor state at T=300oK in the case of P donor

n0 in compensated semiconductor

The electron concentration in the conduction band increases above ni as we add donor atoms and the minority carrier hole concentration will decrease.

Page 20: Ionization Energies in Si, Ge, GaAs - MyBlog...Complete Ionization Ex: we want to determine the fraction of electrons are still in the donor state at T=300oK in the case of P donor

nd versus T

Between T=0oK (freeze-out) and T=300oK (complete ionization), we have partial ionization of donor and acceptor atoms (n=Nd).

0or == +ddd NNn

High T the semiconductor shows an intrinsic behaviour

Page 21: Ionization Energies in Si, Ge, GaAs - MyBlog...Complete Ionization Ex: we want to determine the fraction of electrons are still in the donor state at T=300oK in the case of P donor

no vs T

ni strong depends from T. By increasing T electron and hole pairs are thermally generated and then the ni

2 term start to dominate. The semiconductor will lose the extrinsic characteristics.

22

2)(

2)(

iadad

o nNNNNn +⎟⎠

⎞⎜⎝

⎛ −+

−=

Page 22: Ionization Energies in Si, Ge, GaAs - MyBlog...Complete Ionization Ex: we want to determine the fraction of electrons are still in the donor state at T=300oK in the case of P donor

Position of the Fermi energy level for n-type

Determination of the position of the Fermi level as a function of doping concentration and as a function of temperature. We assume that the Boltzmann appoximation is valid.

⎟⎟⎠

⎞⎜⎜⎝

⎛=−

⎟⎟⎠

⎞⎜⎜⎝

⎛=−⇒⎥

⎤⎢⎣

⎡ −−=

d

cFC

Do

o

cFC

FCco

NNkTEE

NnnNkTEE

kTEENn

ln

tor semiconduc type-nan consider weIf

ln )(

exp

EF moves to Ec if the donor concentration (Nd) increases. For a compensated semiconductor Nd term is replaced with Nd-Na.

all donor atoms are ionized

Page 23: Ionization Energies in Si, Ge, GaAs - MyBlog...Complete Ionization Ex: we want to determine the fraction of electrons are still in the donor state at T=300oK in the case of P donor

po for compensated semiconductors

22

22

2

2)(

2)(

0)(

0 and 0

idada

o

ioado

dodo

iao

ddaaad

nNNNNp

ppNNp

NpNpn

Nn

NNandNNpn

+⎟⎠

⎞⎜⎝

⎛ −+

−=

=−−−

=+=+=+

==== +−

If we assumed complete ionization

Na>Nd

Thermal-equilibrium majority carrier hole concentration in a p-type semiconductor or when Na>Nd.

Minority carrier concentration :

da

i

o

io NN

npn

n−

==22

Page 24: Ionization Energies in Si, Ge, GaAs - MyBlog...Complete Ionization Ex: we want to determine the fraction of electrons are still in the donor state at T=300oK in the case of P donor

Position of the Fermi energy level for n-type

We develop a different expression for EF for n-type semiconductor

( ) with lnexp ⎟⎟⎠

⎞⎜⎜⎝

⎛=−⇒⎥

⎤⎢⎣

⎡ −=

i

oFiF

FiFio n

nkTEEkTEEnn

22

2)(

2)(

iadad

o nNNNNn +⎟⎠

⎞⎜⎝

⎛ −+

−=

For a semiconductor with Nd-Na=0 (completely compensated) then no=ni and EF=EFI => same characteristics of an intrinsic semiconductor.

EX: Si T=300oK Nd=1016 cm-3 Na=0 ni=1.5x1010 cm-3 n-type semiconductor

appr.)Boltzmann the(valid 206.010

108.2ln0259.0ln 16

19

kTeVNNkTEEd

cFC >>=⎟⎟

⎞⎜⎜⎝

⎛ ×=⎟⎟

⎞⎜⎜⎝

⎛=−

Page 25: Ionization Energies in Si, Ge, GaAs - MyBlog...Complete Ionization Ex: we want to determine the fraction of electrons are still in the donor state at T=300oK in the case of P donor

Position of the Fermi energy level of p-type

For a p-type semiconductor

⎟⎟⎠

⎞⎜⎜⎝

⎛=−

⎟⎟⎠

⎞⎜⎜⎝

⎛=−⇒⎥

⎤⎢⎣

⎡ −−=

a

VVF

Ao

o

vVF

VFpvo

NNkTEE

NppNkTEE

kTEEexNp

ln

tor semiconduc type-pan consider weIf

ln )(

as the Na increases EF moves closer to the valence band or

( )

ln

exp

⎟⎟⎠

⎞⎜⎜⎝

⎛=−⇒

⎥⎦

⎤⎢⎣

⎡ −−=

i

oFFi

FiFio

np

kTEE

kTEE

np

Page 26: Ionization Energies in Si, Ge, GaAs - MyBlog...Complete Ionization Ex: we want to determine the fraction of electrons are still in the donor state at T=300oK in the case of P donor

EF vs EFi

For n-type no>ni EF >EFi For p-type po>ni EFi> EF

Page 27: Ionization Energies in Si, Ge, GaAs - MyBlog...Complete Ionization Ex: we want to determine the fraction of electrons are still in the donor state at T=300oK in the case of P donor

EF vs T

⎟⎟⎠

⎞⎜⎜⎝

⎛=−

⎟⎟⎠

⎞⎜⎜⎝

⎛=−

i

oFFi

i

oFiF

npkTEE

nnkTEE

ln

ln

⎥⎦

⎤⎢⎣

⎡−=

kTE

NNn gVCi exp2

ni strongly depends from T EF is a function of T also. As T increases ni increases too and EF=>EFi

Fig. 4-19

At high T the semiconductor loses the extrinsic characteristics and behaves as intrinsic.

At low T freeze-out occurs, Boltzmann approximation is not longer valid. The equations we derived the Fermi level position no longer apply. EF->Ed for n-type and EF->Ea for p-type.

Page 28: Ionization Energies in Si, Ge, GaAs - MyBlog...Complete Ionization Ex: we want to determine the fraction of electrons are still in the donor state at T=300oK in the case of P donor

EF vs doping

⎟⎟⎠

⎞⎜⎜⎝

⎛=−

a

VVF N

NkTEE ln⎟⎟⎠

⎞⎜⎜⎝

⎛=−

d

cFC N

NkTEE ln

n-type semiconductor p-type semiconductor

Page 29: Ionization Energies in Si, Ge, GaAs - MyBlog...Complete Ionization Ex: we want to determine the fraction of electrons are still in the donor state at T=300oK in the case of P donor

Relevance of EF

EF and his relations are significant in p-n juntions or other semiconductor devices. In thermal-equilibrium EF is constant throughout a system.

Fig. 4-20