ene 311 lecture 6. donors and acceptors we have learned how to find new position of fermi level for...

ENE 311 Lecture 6

Donors and Acceptors

• We have learned how to find new position of Fermi level for extrinsic semiconductors.

• Now let us consider the new electron densit y in case of both donors ND and acceptors NA

are present simultaneously.

• The Fermi level will adjust itself to preserve overall charge neutrality as

(1)A Dn N p N

Donors and Acceptors

• By solving (1) with , t he equilibriu -m electron and hole concentrations in an n t

ype semiconductors yield

2 2

2

14

2n D A D A i

in

n

n N N N N n

np

n

2. in p n

Donors and Acceptors

• - Similarly to p type semiconductors, the elec tron and hole concentrations are expressed

as

2 2

2

14

2p A D D A i

ip

p

p N N N N n

nn

p

Donors and Acceptors

• Generally, in case of all impurities are ionize d, the net impurity concentration ND – NA is l

arger than the intrinsic carrier concentratio n ni ; therefore, we may simply rewrite the a

bove relationship as

if

if n D A D A

p A D A D

n N N N N

p N N N N

Donors and Acceptors

• The figure shows electron density in Si as a function of temperature for a donor concentration of ND 10= 1

5 cm-3.

• At low temperature, not all donor impurities could be ionized and this is called “Freeze-out region” since some electrons are frozen at the donor level.

Donors and Acceptors

• As the temperature increased, all donor impurities are ionized and this remains the same for a wide range of temperature.

• This region is called “Extrinsic region”.

Donors and Acceptors

• Until the temperature is increased even higher and it reaches a point where electrons are excited from valence band.

• This makes the intrinsic carrier concentration becomes comparable to the donor concentration.

• At this region, the semiconductors act like an intrinsic one.

Donors and Acceptors

• If the semiconductors are heavily doped for both n- or p-type, EF will be higher than EC or below EV, respectively.

• The semiconductor is referred to as degenerate semiconductor.

• This also results in the reduction of the ban dgap.

Donors and Acceptors

• This also results in the reduction of the ban dgap. The bandgap reduction Eg for Si at ro

om temperature is expressed by

where the doping is in the unit of cm-3.

1822 meV

10g

NE

Donors and Acceptors

• Ex. Si is doped with 1016 arsenic atoms/cm3 . Find the carrier concentration and the Fermi level at room temperature (300K).

Donors and Acceptors

Soln

At room temperature, complete ionization o f impurity atoms is highly possible, then we

have n = ND = 1016 cm-3.

2923 -3

16

9.65 109.3 10 cm

10i

D

np

N

Donors and Acceptors

Soln

The Fermi level measured from the bottom of the conduction band is

19

10

ln

2.86 100.0259ln

10

0.205 eV

CC F

D

NE E kT

N

Donors and Acceptors

Soln

The Fermi level measur ed from the intrinsic Fe rmi level is

16

9

ln ln

100.0259ln

9.65 10

0.358 eV

DF i

i i

NnE E kT kT

n n

Direct Recombination

• When a bond between neighboring atoms is broken, an electron-hole pair is generated.

• The valence electron moves upward to the conduction band due to getting thermal energy.

• This results in a hole being left in the valence band.

Direct Recombination

• This process is called carrier generation with the generation rate Gth (number of electron-hole pair generation per unit volume per time).

• When an electron moves downward from the conduction band to the valence band to recombine with the hole, this reverse process is called recombination.

• The recombination rate represents by Rth .

Direct Recombination

• Under thermal equilibrium, the generation rate Gth equals to the recombination rate Rth to preserve the condition of

• The direct recombination rate R can be expressed as

where is the proportionality constant.

2ipn n

R np

Direct Recombination

• - Therefore, for an n type semiconductor, wehave

(3)

where nn0

and pn0 represent electron and h ole densities at thermal equilibrium.

0 0th th n nG R n p

Direct Recombination

• If the light is applied on the semiconductor, it pr - oduces electron hole pairs at a rate GL , the carri

er concentrations are above their equilibrium v alues.

• The generation and recombination rates become

where n and p are the excess carrie concentrations

L thG G G

0 0n n n nR n p n n p p

Direct Recombination

n = p to maintain the overall charge neutrality.

• The net rate of change of hole concentratio n is expressed as

(7)

0

0

n n

n n

n n n

p p p

nL th

dpG R G G R

dt

Direct Recombination

• - In steady state, dpn /dt =0 . From (7 ) we have

(8)

where U is the net recombination rate.

Substituting (3 ) and (5 ) into (8 ), this yields

(9)

L thG R G U

0 0n nU n p p p

Direct Recombination

- For low level injection p , pn0 << nn0 , (9 ) becomes

(10)

where p is called excess minority carrier lifetim e .

0 00

01/n n n n

nn p

p p p pU n p

n

0n n p Lp p G

Direct Recombination

We may write pn in the f unction of t as

0( ) exp /n n p L pp t p G t

Direct Recombination

• Ex . A Si sample with nn0 = 1014 cm-3 is illumi nated with light and 1013 - electron hole pairs/

cm3 are created every microsecond. If n = p =2 s, find the change in the minority carr

ier concentration.

Direct Recombination

Soln Before illumination:

After illumination:

2 9 2 14 5 -30 0/ (9.65 10 ) /10 9.31 10 cmn i np n n

135 6 13 -3

0 6

109.31 10 2 10 2 10 cm

10n n p Lp p G

Continuity Equation

• We shall now consider the overall effect when drift, diffusion, and recombination occur at the same time in a semiconductor material.

• Consider the infinitesimal slice with a thickness dx located at x shown in the figure.

Continuity Equation

• The number of electrons in the slice may increase because of the net current flow and the net carrier generation in the slice.

• Therefore, the overall rate of electron increase is the sum of four components: the number of electrons flowing into the slice at x, the number of electrons flowing out at x+dx, the rate of generated electrons, and the rate of recombination.

Continuity Equation

• This can be expressed as

• where A is the cross-section area and Adx is the volume of the slice.

( ) ( )e en n

n J x A J x dx AAdx G R Adxt e e

Continuity Equation

• By expanding the expression for the current at x + dx in Taylor series yields

• Thus, we have the basic continuity equation for electrons and holes as

(14)

( ) ( ) ...ee e

JJ x dx J x dx

x

1

1

en n

hp p

n JG R

t e xp J

G Rt e x

Continuity Equation

• We can substitute the total current density f or holes and electrons and (10) into (14).

e e n

h h p

dnJ e nE eD

dxdn

J e nE eDdx

Continuity Equation

• - For low injection condition, we will have the continuity equation for minority carriers as

(15)

2

2

2

2

p p p p pop e e n n

n

n n n n non h h p p

p

n n n n nEn E D G

t x x x

p E p p p pp E D G

t x x x

The Haynes-Shockley Experiment

• This experiment can be used to measure the carrier mobility μ.

• The voltage source establishes an electric field in the n-type semiconductor bar. Excess carriers are produced and effectively injected into the semiconductor bar at contact (1).

• Then contact (2) will collect a fraction of the excess carriers drifting through the semiconductor bar.

The Haynes-Shockley Experiment

• After the pulse, the transport equation given by equation (15) can be rewritten as

• If there is no applied electric field along the bar, the solution is given by

(16)

2

2n n n n no

h pp

p p p p pE D

t x x

2

( , ) exp44

n nop pp

N x tp x t p

D tD t

The Haynes-Shockley Experiment

• N is the number of electrons or holes genera ted per unit area. If an electric field is applie

d along the sample, an equation (16 ) will becomes

2( , ) exp

44

p

n nop pp

x EtN tp x t p

D tD t

The Haynes-Shockley Experiment

Ex. - - In Haynes Shockley experiment on n type Ge semiconductor, given the bar is1 cm lon

g, L = 0.95 cm, V1 =2 V, and time for pulse

arrival = 0.25 ns. Find mobility μ.

The Haynes-Shockley Experiment

Soln

1

2

0.953800 cm/s

0.25

2 V/cm1

3800 cm/s1900 cm /V.s

2 V/cm

D

D

L cmv

t ns

VE

cm

v

E

The Haynes-Shockley Experiment

Ex. - In a Haynes Shockley experiment, the max imum amplitudes of the minority carriers at

t1 = 100 μs and t

2 = 200 μs differ by a factor

of5 . Calculate the minority carrier lifetime.

Soln

2

2

2 11

2 1 2

( , ) exp44

exp44

The maximum amplitude

exp4

Therefore,

exp /( ) 200 200 100exp

( ) 100exp /

p

n nop pp

p

n nop pp

pp

p

pp

x EtN tp x t p

D tD t

x EtN tp p p

D tD t

N tp

D t

t tp t

p t t t

5

79 sp

Thermionic emission process

• It is the phenomenon that carriers having hi gh energy thermionically emitted into the v

acuum.

• In other words, electrons escapes from the hot or high temperature surface of the mate rial.

• This is called “ thermionic emission process”.

Thermionic emission process

• Electron affinity qχ is th e energy difference betwe

en the conduction band e dge and the vacuum.

• Work function q is the energy between the Ferm

i level and the vacuum lev el in the semiconductor.

Vn

Thermionic emission process

• It is clearly seen that an electron can thermio nically escape from the semiconductor surfac

e into the vacuum if its energy is above qχ.

• The electron density with energies above qχ can be found by

where Vn is the difference between the botto m of the conduction band and the Fermi level.

( ) exp n

th C

q

q Vn n E dE N

kT

Thermionic emission process

• If escaping electrons with velocity normal to the surface and having energy greater than

EF + q , the thermionic current density is eq ual to

3/ 2( ) /

3

( ) ( )

4 (2 )e FE E kT

J qvN E F E dE

mqv dE

h

Thermionic emission process

• Then we use p = mv and , and after integration, it yields

2 2

28

h kE

m

* 2

2*

3

6 2 2 2 2

Richardson

exp

4

1.2

constant

10 A/(m .K ) 120 A/(cm .K )

qJ A T

kT

qmkA

h

Thermionic emission process

Ex. Calculate the thermionically emitted electr on density nth -at room temperature for an n

type silicon sample with an electron affinityqχ = 4.05 eV and qVn = 0.2 eV. If we reduce

the effective qχ to 0.6 eV, what is nth?

Thermionic emission process

Soln

As we clearly see that at the room temperature, th ere is no thermionic emission of electrons into the

vacuum. This thermionic emission process is impor tant for - metal semiconductor contacts.

19 52

19 6 -3

4.05 0.2(4.05 eV) 2.86 10 exp 10 0

0.0259

0.6 0.2(0.6 eV) 2.86 10 exp 1 10 cm

0.0259

th

th

n

n

Tunneling Process• The figure shows the energy

band when two semiconduct or samples are brought close to each other.

• The distance between them(d ) is sufficient small, so that

- the electrons in the left side semiconductor may transpor

t across the barrier and mov - e to the right side semicond

uctor even the electron ener gy is much less than the barr ier height.

• This process is called “quant um tunneling process”.

Tunneling Process

• The transmission coefficient can be expressed as

• This process is used in tunne l diodes by having a small tu

nneling distance d , a low pot ential barrier qV

0 , and a smal

l effective mass.

2 *0

2 2

2 ( )exp 2 e

C m qV ET d

A