Chapter 2: Carrier and Transport phenomena

Questions provide us the understanding needed to design semiconductor devices

If there is an electric field present in the semiconductor, how do the electrons and holes move?

If there is a concentration gradient in the electron or hole density, how do the carriers respond?

Do electrons in the conduction band fall down into the valence band and recombine with holes?

Is it possible for electrons in the valence band to jump up into the conduction band? What cause such processes?

1. CARRIER DRIFT

Objective:An electric field applied to a semiconductor will produce a force on electrons and holes so that they will experience a net acceleration and net movement, provided there are available energy states in the conduction and valence bands. This net movement of charge due to an electric field is called drift. The net drift of charge gives rise to a drift current .

What is drift velocity? (vd)

Is the velocity component that arises when an electric field, ε is applied to a semiconductor. This electric field causes each electron to experience a force -qε due to the field and each electron will be accelerated along the field (in the opposite direction).

What is mobility? (u)

The mobility is an important parameter of the semiconductor since it describes how well a particle will move due to an electric field.

1. CARRIER DRIFT According to quantum mechanics, electron have a free electron-like behavior and no

scattering occurs. In real semiconductors, due to the imperfections, shown in the table, the electrons scatter,

which affects their transport (, , ) properties.

Important Sources of Scattering in Semiconductors Ionized impurities Due to dopants in the semiconductors Phonons Due to lattice vibrations at finite temperatures Alloy Random potential fluctuations Interface roughness Important in heterostructure Chemical impurities Due to unintentional impurities

Results in the Momentum & Energy of electrons will gradually lose coherence with the initial state values. The average time it takes to lose coherence or memory of the initial state

properties is called (Mathieson’s rule).

sc = Scattering time or

Mean time between collisions for an electron or Relaxation time or Delay time

. 1 2

1 1 1...

tot sc sc sc

3

a) Low Electric Fields (d = immediately before end of collision,

after collision the d = 0)

= d = ave.= drift velocity = ave. gain velocity (either d n or d p ) e = electron F = = Electric Field sc = scattering time m* = effective mass (either mn* or mp*) J = Current density = drift current density n = density of charge carrier (either nn = n or np= p) = conductivity of material = resistivity of material = 1/ = mobility effects (factor) (either n or p )

. .

*SC

d

e Fv

m

2. . .. .

*SC

d

n e FJ n e v

m

From Ohm’s Law, J = .F

2. .

*SCn e

m

From def. of mobility,d = - .F (the e- move in a direction

opposite to the electric field while the holes move in the same direction)

.

*SCe

m

For general :

For holes :For e- :

. .

*SC

pp

e

m

. .

*SC

nn

e

m

4

If both e- and holes are present

. . . .n n p pn e n e

From Ohm’s Law, J = .F

( . . ).n n p pJ e n n F

( . . )n n p pe n n

Example 1The mobility of e- in pure Si at 300 K is 1500 cm2/ V.s. Calculate the relaxation time. Given: m* = 0.26 m0

Ans:The time for pure Si:

*.SC

m

e

30 4 2

1319

(0.26 0.91 10 ).(1500 10 / . )2.2 10

1.6 10

x x kg x m V sx s

x C

5

Example 2The mobility of e- in pure GaAs at 300 K is 8500 cm2/ V.s. Calculate the relaxation time. If the GaAs sample is doped at Nd = 1017 cm-3, the mobility decrease to 5000 cm2/ V.s. Calculate the relaxation time due to ionized impurity scattering. Given: Donor (n-type) doped = m* = 0.067 m0

(Note:the mobility, decreases with the increase in temp. in order to ionized the dopants because as the temp. rises the atoms in the crystal vibrate with greater amplitude. In other words, the electrons scatter from the dopants from these vibrations and mobility decreases)

Ans:The time for pure GaAs

1 *.SC

m

e

30 4 2

1319

(0.067 0.91 10 ).(8500 10 / . )3.24 10

1.6 10

x x kg x m V sx s

x C

The time for ionized impurity in GaAs (pure + impurity)30 4 2

1319

(0.067 0.91 10 ).(5000 10 / . )1.9 10

1.6 10

x x kg x m V sx s

x C

2 *.SC

m

e

2 1

1 1 1imp

SC SC SC Mathieson’s rule:

134.6 10impSC x s

6

Example 3Consider 2 semiconductor samples, Si and GaAs. Both materials are doped n-type at Nd = 1017 cm-3. Assume 50% of the donors are ionized at 300 K. Calculate the conductivity of the samples. Compare this conductivity to the conductivity of undoped samples. Given:Pure or undoped density of state (n) for Si ni = pi = n = p = 1.5 x 1010 cm-3

Pure or undoped density of state (n) for GaAs ni = pi = n = p = 1.84 x 106 cm-3

n (Si) = 1000 cm2/ V.s p (Si) = 350 cm2/ V.s n (GaAs) = 8000 cm2/ V.s p (GaAs) = 400 cm2/ V.sAns:

Conductivity for undoped GaAs:

( . . )undoped i n i pe n n 19 10 3 2(1.6 10 )(1.5 10 )(1000 350) / .x C x cm cm V s

Conductivity for undoped Si:

19 6 3 2(1.6 10 )(1.84 10 )(8000 400) / .undoped x C x cm cm V s

7

6 13.24 10 ( )x cm

9 12.47 10 ( )x cm

Conductivity for doped GaAs:

Cont. Example 3

17 16 3(50%) 10 50% 5 10dopedn x x cm 2i

dopeddoped

np

n

10 23 3

16

(1.5 10 )4.5 10

5 10

xx cm

x

Conductivity for doped Si:

( . . )doped n n p pe n n

17 3(100%) 10doped nn n N cm For Si & GaAs

For Si & GaAs

For Si (very small compared to ndoped)

2i

dopeddoped

np

n

6 25 3

16

(1.84 10 )6.77 10

5 10

xx cm

x For GaAs (very small compared to ndoped)

( . . )doped n n p pe n n

164( )cm

18( )cm

19 16 1 3 1(1.6 10 )(5 10 )(1000 / . ) (4.5 10 )(350 / . )x C x cm V s x cm V s

19 16 1 5 1 1(1.6 10 )(5 10 )(8000 / . ) (6.77 10 )(400 / . ) x C x cm V s x cm V s cm

Almost zero

Almost zero

8

Example 4Consider a Si semiconductor at T = 300 K with an impurity doping concentration of Nd = 1016 cm-3 and Na = 0. Calculate the drift current density, J, for an applied field, = 35 V/cm.Given: n (Si) = 1350 cm2/ V.s p (Si) = 480 cm2/ V.s Ans:Since Nd > Na , the semiconductor is n-type at room temperature, we can assume complete ionization: The n = nn ~ Nd = 1016 cm-3 and p = np = 0

2 10 24 3

16

(1.5 10 )2.25 10

10in x

p x cmn

( . . ).n n p pJ e n n F ( . )n ne n F

19 16 2(1.6 10 )(10 )(1350)(35) 75.6 /x A cm 9

10

Example 5 (Final Sem 1 09/10)

An n-type Silicon sample with a conductivity of 0.1 (cm)-1 at 300 K.Given: a) Calculate the electron and hole carrier density of the material. [4 marks]

b) Calculate the Fermi level for n-type and p-type material with the same conductivity using Joyce-Dixon approximation. [4 marks]

c) Calculate the intrinsic Fermi level for n-type and p-type of the material. [4 Marks]

d) Sketch the flat band diagram, indicating clearly the positions of Ec, Ev, EFn, EFp and Ei. [4 marks]

e) How much is the energy gap has been shifted if compared to the energy gap of 1.1 eV. [2 marks]

f) Describe why the mobility carrier in an extrinsic semiconductor decreases with the increases of temperature. [2

marks]

11

n nq n

p pq p

114 3

19 2

0.1( )6.25 10

(1.6 10 )(1000 / )n

n

cmn x cm

q x C cm Vs

115 3

19 2

0.1( )2.08 10

(1.6 10 )(300 / )p

p

cmp x cm

q x C cm Vs

14 14

19 19

6

1ln

8

(6.25 10 ) 1 (6.25 10 )(0.026) ln

(2.78 10 ) (2.78 10 )8

(0.026) ( 10.70) (7.95 10 )

0.278

Fn C BC C

Fn C

Fn C

Fn C

n nE E k T

N N

x xE E

x x

E E x

E E eV

b)

Solutions:a)

15 15

18 18

5

1ln

8

(2.08 10 ) 1 (2.08 10 )(0.026) ln

(9.84 10 ) (9.84 10 )8

(0.026) ( 8.46) (7.47 10 )

0.22

Fp V BV V

Fp V

Fp V

Fp V

p pE E k T

N N

x xE E

x x

E E x

E E eV

12

c)

14

10

exp

ln

6.25 10ln (0.026) ln 0.276

1.5 10

Fn i

i B

Fn i

i B

Fn i Bi

E En

n k T

E En

n k T

n xE E k T eV

n x

15

10

exp

ln

2.08 10ln (0.026) ln 0.3078

1.5 10

i Fp

i B

i Fp

i B

i Fp Bi

E Ep

n k T

E Ep

n k T

p xE E k T eV

n x

d)

0.22

0.278

0.276

0.3078Ei

EC

EFn

EFp

EV

e) From the diagram (d),

EgapBefore = 1.1 eV

EgapAfter = 0.276 + 0.3078 = 0.5838 eV E = 1.1 - 0.5838 = 0.5162

f) At high temperatures, lattice scattering dominates as the thermal vibrations of lattice atoms increase with T hence increasing the probability of charge carrier-lattice collisions. Hence, the mobility decreases as the sample is heated. orThe mobility, decreases with the increase in temperature (in order to ionized the dopants) because as the temperature rises the atoms in the crystal vibrate with greater amplitude. In other words, the electrons scatter from the dopants from these vibrations and mobility decreases.

13

b) Very High Electric Field Transport:Breakdown Phenomena

(When > 100 kV/cm, the semiconductor suffers a “breakdown” in which current has “runaway” behavior. The breakdown occurs due to carrier multiplication means the number of electrons and holes that can participate in current flow increase. (The total number of electrons conserved))

i) Impact ionization or Avalanche Breakdown♣ In normal case during transportation,

the e-/holes remain in the same band.

♣ At very high this does not hold true.♣ An e- which is ‘very hot’ scatters with

an e- in the valence band via coulombic interaction and knocks it into the conduction band as shown in Figure .♣ Thus the initial e- should have energy slightly larger than the bandgap.♣ In the final state we have 2 e- in

the conduction band 1 hole in the valence band.

♣ Thus the number of current carryingcharges have multiplied. The process is called “Avalanching”

♣ The same process could happen to “hot holes”.

+

--

-

Conduction band

Final state has 2 e- + 1 hole

Initial state has 1 e-

Valence band

Avalanche process14

Energy band diagrams under junction-breakdown conditions-Avalanche multiplication.

( )imp

dI zI

dz

( )( ) exp( )

( ) imp

I zN x x

I O

Once avalanching starts,

• I = current• imp = Average rate of ionization per unit distance (coefficient for e-)imp = Average rate of ionization per unit distance (coefficients for hole)N = number of times an initial electron will suffer impact ionization after travelling a distance x

15

ii) Band-to-band Tunneling or Zener Tunneling

When a strong happens, the e- in the valence band can tunnel into an ‘unoccupied state’ in the conduction band or vice versa. As the e- tunnels, the tunneling probability is:

F = Electric Field in the semiconductor

Example 1Calculate the band to band tunneling probabilty

in GaAs and InAs at an applied = 2 x 105 v/m.Given: m* (GaAs) = 0.065 m0

m* (InAs) = 0.02 m0

Eg (GaAs) =1.5 eVEg (InAs) = 0.4eV

(means Zener Tunneling is important when ~2 x 105 V/m)

Energy band diagrams under junction-breakdown conditions- Tunneling effect

* 3/ 24 2exp( )

3gm E

Te F

Available empty states (holes) in valence band

Electrons in conduction band

30 19 3/ 2

19 34 7

4 2 0.065 0.91 10 )(1.5 1.6 10 )exp( ) 0

3 (1.6 10 )(1.05 10 )(2 10 / )GaAs

x x x x kg x x JT

x x C x Js x V m

63.7 10InAsT x 16

b). HIGH FIELD EFFECTS (Proven thru graph)

• High field transport means Carrier velocity tends to saturate and mobility = v.F starts to decrease (The mobility starts to decrease and becomes independent of the electric field)

2. CARRIER DIFFUSION

Objective:There is a second mechanism, in addition to drift, that can induce a current in a semiconductor. We can consider a classic physics example in which a container as shown is divided into 2 parts by a membrane. The left side contains gas molecules at a particular temperature and the right side is initially empty. The gas molecules are in continual random thermal motion so that, when the membrane is broken, there will be a net flow of gas molecules into the right side of the container. Diffusion is the process whereby particles flow from a region of high concentration toward a region of low concentration. If the gas molecules were electrically charged, the net flow of charge would result in a diffusion current.

What is a diffusion current? (J)

Diffusion current will exist when there is a spatial variation of carrier concentration in the semiconductor material. This will cause the carriers to move from a region of high concentration to a region of low concentration?

x= 0

19

2. CARRIER DIFFUSION Arising from thermodynamics, when there is a gradient in the concentration of a species of

mobile particles, the particles diffuse from the regions of high concentration to the low concentration.

Due to the random motion of the particles’ collision of various scattering processes

(remember, no ) in space. Calculate the electron flux, to the right across x = x0 at any instant of time, t.

= free path to each side of x0 boundary in time , = average carrier densities in L-Region = average carrier densities in R-RegionBut,

Net Flux:

for electron

Where Dn = diffusion coefficient of the electron system

for hole

( , )x t

( )( , )

2L R

SC

n nx t

SCLnRn

.L R

dnn n

dx

2 ( , ) ( , )( , )

2n nSC

dn x t dn x tx t D

dx dx

( , )( , )p p

dp x tx t D

dx

20

The current density, J

( , ) ( , )( ) ( ) ( )tot n p n p

dn x t dp x tJ diff J diff J diff eD eD

dx dx

Note: What is the difference between drift and diffusion?Drift of carriers, driven by an electric field.

Diffusion of carriers due to their random thermal motion.

21

Example 1

Determine the carrier density gradient to produce a given diffusion current density.The hole concentration in silicon at T = 300 K varies linearly from x = 0 to x = 0.01 cm. The hole diffusion coefficient is Dp = 10 cm2/s, the hole diffusion current density is Jdif = 20 A/cm2, and the hole concentration at x = 0 is p = 4 x 1017 cm-3. Determine the hole concentration at x = 0.01 cm. ( , ) ( , )

( ) ( ) ( )tot n p n p

dn x t dp x tJ diff J diff J diff eD eD

dx dx

1719

17 3

( ) (0.01) (0)( )

0.01 0

(0.01) (4 10 )20 (1.6 10 )(10)

0.01 0

(0.01) 2.75 10

dif p p p

p x p pJ J diff eD eD

x

p xx

p x cm

22

Example 2

The electron concentration in silicon decreases linearly from 1016 cm-3 to 1015 cm-3 over a distance of 0.10 cm. The cross-sectional area of the sample is 0.05 cm2. The electron diffusion coefficient is 25 cm2/s. Calculate the electron diffusion current.

16 1519

2

2

10 101.6 10 25

0 0.10

0.36 /

For A = 0.05 cm

0.05 0.36 18

n n

dn nJ eD eD

dx x

J x

J A cm

I AJ I mA

23

Example 3

The hole concentration in silicon decreases linearly from 1015 cm-3 to 2x1014 cm-3 over a distance of 0.10 cm (diffusion length). The cross-sectional area of this cylinder is 0.075 cm2. The hole diffusion coefficient is Dp = 10 cm2/s. i) Calculate the hole diffusion current (Ip).ii)Calculate the hole diffusion current density (Jp).iii)Calculate the radius, ‘r’ of the cross-sectional area of this cylinder.iv)Calculate the scattering time, ‘p’ for the hole.v)How are you going to increase the current without increase the hole concentration?

15 14

19

22

2 2

23

10 (2 10 )) 1.6 10 10 0.075 0.96

0 0.10

0.96) 12.8 /

0.075

0.075) 0.075 0.15

) 1 10

) ' ', . ( 0.25 0.

p p

pP

PP p p p

p

dp xi I eD A x mA

dx

I mAii J mA cm

A cm

iii A r cm r cm

Liv L D s

D

v increase the r i e r instead of

22

0.9615) (0.25 ) 2.56

0.075p

mAI cm mA

cm

24

Example 4

The electron concentration in a sample of n-type silicon varies linearly from 1017 cm-3 at x = 0 to 6 x1016 cm-3 at x = 4 m. The electron current density is experimentally measured to be -400 A/cm2. What is the electron diffusion coefficient?

17 1619

4

2

10 6 10400 1.6 10

0 4 10

400 16

25 /

n n n

n

n

n

dn nJ eD eD

dx x

xx D

x

D

D cm s

25

Example 5 (Final Sem 1 2011/2012)

The electron concentration in a sample of n-type silicon varies linearly as shown below. Find the electron current density, Jn at x = 1 m (1 x 10-4 cm).Given the electron diffusion coefficient = 25 cm/s

0.1

5x m

EF - Ei (eV)

0

0.1( ) (0.1)

5

( ) 200 (0.1)

200 0.1exp exp

200 0.1 200 0.1200exp exp

F i

F i

F ii i

B B

i iB B B

y mx c

eVE E eV x eV

m

eVE E eV x eV

cm

E E xn n n

k T k T

x xdn dn n

dx dx k T k T k T

4

19 10

4 3 2

200 0.1200exp

200(1 10 ) 0.1200(1.6 10 )(25) (1.5 10 )exp

0.026 (0.026)

(4.6 10 )exp(3.077) 9.978 10 /

n n n

n

n

iB B

dnJ eD eD

dx

J

J

xn

k T k T

x A cm

26

3. CARRIER INJECTION

Objective:If electrons and holes are injected into a semiconductor, either by external contacts or by optical excitation, the system is no more equilibrium. Now the system is called Quasi-Fermi levels

27

exp

v Fp

vB

E Ep N

k T

exp Fn c

CB

E En N

k T

Electron Density

Hole Density

exp F i

i

E En

kT

exp i F

i

E En

kT

Joyce-Dixon approximation

1ln

8Fn C B

C C

n nE E k T

N N

1

ln8

V Fp BV V

p pE E k T

N N

Boltzmann approximation

lnFn C BC

nE E k T

N

For n or e-: lnV Fp B

V

pE E k T

N

For holes:

For n or e-:

For holes:

ExampleUsing Boltzmann statistics, calculate the position of the electron and hole quasi-fermi levels when an e-h density of (n = p = 1017cm-3) is injected into pure (undoped) Si. At 300 K.Given: NC = 2.8 x 1019 cm-3, NV = 1.04 x 1019 cm-3, Eg (Si) = EC - EV = 1.12 eV

For example

28

lnFn C BC

nE E k T

N

lnFn B CC

nE k T E

N

17

19

10(0.026) ln

2.8 10 CEx

( 0.146)CE eV

EC

EFn

0.146 eV

[ln ] ( 0.121)Fp V B VV

pE E k T E eV

N EFp

EV

0.121 eV

EFn - EFp = (EC – EV) – (0.146+0.121) = 1.1 – 0.267 = 0.833 eV

If we had injected only n = p = 1015cm-3 , the differences in the quasi-fermi levels would be:

EFn - EFp = (EC – EV) – (0.266 + 0.24) = 1.1 – 0.506 = 0.59 eV

Final Sem II 09/10

An n-type Si semiconductor at T = 300 K with an impurity ionization energy of 0.228 eV above from the intrinsic Fermi level.

a) Find the impurity concentration (i.e the majority and minority carrier density).

10 14 3

2 10 26 3

14

,

0.228exp (1.5 10 )exp 1.01466 10

0.025852

,

(1.5 10 )2.21749 10

1.01466 10

n iF Fi

B

i

Majority carrier density n

E En n x x cm

k T

and

Minority carrier density p

n xp x cm

n x

:

( ) 0.228n i i nF F F F

Note

E E E E eV

b) Find the intrinsic Fermi level for the majority and minority carrierdensity (use the value of n & p from part a)).

14

10

6

10

( ) ln

1.01466 10(0.025852) ln 0.228

1.5 10

( ) ln

2.21749 10(0.025852) ln 0.228

1.5 10

n i i n

i p p i

F F F F Bi

F F F F Bi

nE E E E k T

n

xeV

x

pE E E E k T

n

xeV

x

Note: same value of energy, i.e 0.228 eV, why? Because we are trying to find the same level of energy but using different charge carriers, i.e “n” and “p”.

Majority (from ‘n’)

Minority (from ‘p’)

31

c) Find the Fermi level for the majority and minority carrier density by using Joyce-Dixon approximation.

6 6

10 10

5

1ln

8

(2.21749 10 ) 1 (2.21749 10 )(0.026) ln

(1.5 10 ) (1.5 10 )8

(0.026) ( 8.82) (5.22 10 ) 0.229

n

n

n

C F Bi i

C F

C F

n nE E k T

N N

x xE E

x x

E E x eV

14 14

19 19

6

1ln

8

(1.01466 10 ) 1 (1.01466 10 )(0.026) ln

(2.78 10 ) (2.78 10 )8

(0.026) ( 12.52) (1.29 10 ) 0.325

n

n

n

C F BC C

C F

C F

n nE E k T

N N

x xE E

x x

E E x eV

32

d) Sketch the flat band diagram, indicating the positions of Ec,

Ev, EFn, EFp and Efi.

0.7528 eV

0.3236 eV

0.228 eV

EFi

EC

EFn

EFp

EV

33

4. RECOMBINATION-GENERATION CENTERS PROCESSES

Happens due to Lattice defects or Impurity atoms in lattice disrupt the ideal single-crystal lattice structure. As a result, allowed electronic energy states within the bandgap.

These energy states may now serve as “stepping stones” in the recombination-generation process.

These energy states occur near the midgap energy (as shown).

34

Electron-hole generation and recombination(normal case)

35

Recombination via a trapping center: (a)(1)an electron is trapped and then (2) a hole is trapped; (b)(1) or hole is trapped and then (2) an electron is trapped; (c)(1)or an electron is trapped and then (2) the electron falls into an empty state (hole)

Generation via a trapping center: (1) an electron is elevated from the valence band into the trap creating a free hole and then (2) the electron is elevated into the conduction band creating a free electron

(a) (b) (c)

36

Creation of excess electron and hole densities by photons in Generation process

Recombination of excess carriers reestablishing thermal equilibrium

37

4. GENERATION AND RECOMBINATION PROCESSES

♣ Generation – the process whereby electrons and holes (carriers) are created. a) Thermal Energy By increase in temperature, the electrons will be exited from valence band to conduction band. And the electrons will continuously exited up until there will be a build-up of Free carriers. Free carriers can also be generated if electrons leave a donor and go into the conduction band. In order to reach an equilibrium concentration there has to be recombination as well. At equilibrium:

RG = the carrier generation rate RR = the carrier recombination rate = Total electron (may be functions of time or position) = Total hole (may be functions of time or position) =Thermal equilibrium electron concentration (independent of time and position) = Thermal equilibrium hole concentration(independent of time and position) = Intrinsic concentration

b) Optical Radiation (Optoelectronics) (Excess Carriers)

When light (photon-light energy = hf) shines on semiconductor, it can cause an electron in valence band to go into the conduction band. The photon absorption process, a photon scatters an electron in the valence band, causing the electron to go into conduction band.

RG = RR

20 0 inp n p n

20 0 inp n p n

np

0p0n

in

38

4. GENERATION AND RECOMBINATION PROCESSES

♣ Recombination - the process whereby electrons and holes (carriers) are annihilated.

a) Thermal Energy By decrease in temperature, the electrons will be back to valence band. Free carriers can also be generated from valence band going to an

acceptor, causes a hole.

b) Optoelectronics (Excess Carriers) The photon emmision process, light will be emitted when electron go back to a hole (valence band).

39

Conduction band

Valence band

Photon

Photon+

-

“Vertical in “k”

Photon Absorption Photon Emission

Fig. shows the band-to-band absorption in semiconductors. An electron in the valence band “absorbs” a photon and moves into the conduction band. In the reverse process, the electron recombines with a hole to emit a photon (emission process) . Momentum conservation ensures that only vertical transitions are allowed.

40

The figure shows how the excess carriers decay intothe semiconductor

Excess electron concentration =

x

Carrier Injection

0 L

(0)n

( )n x

( )n x L

Carrier lifetime =

Excess hole concentration =

May be functions of time or position

Example 1Excess electrons have been generated in a semiconductor to a concentration of The excess carrier lifetime is . Theforcing function generating the excess carriersturns off at t = 0 so the semiconductor is allowed to return to an equilibrium condition for t>0. Calculate the excess electron concentration for

a) t=0

b) t=1s

c) t=4s

15 3(0) 10n cm 6

0 10n s

0

15

( ) (0)exp( )

( ) 10 exp( )1

n

tn t n

tn t

s

15 310n cm

14 33.68 10n x cm

13 31.83 10n x cm

0 0n por

0

0

n n n

p p p

41

Example 2Find the recombination rate of the excess electrons for example 1.

a)

b)

c)

Note:1. When excess electron density ( ) is injected in p-type material. This excess minority carriers ( ) will recombine with the majority carriers ( ) with a rate given by:

2. When excess holes density ( ) is injected in n-type material. This excess minority carriers ( ) will recombine with the majority carriers ( ) with a rate given by:

0

( )

n

n tR

1521 3 1

6

1010

10R cm s

1420 3 1

6

3.68 103.68 10

10

xR x cm s

1319 3 1

6

1.83 101.83 10

10

xR x cm s

0

( )

p

p t

n

p

n p

0

( )

n

n tR

p n

0

( )

p

p tR

42

Diffusion length = Average distance = n p n n p pL orL D or D

Diffusion coefficient = D

Example 3Electrons are injected into a p-type silicon sample at 300 K with the carrier lifetime is given by 5x10-9s. The excess electrons have been generated in a semiconductor to a concentration of .a) Calculate the rate of the recombination of the excess electron (excess minority carriers ( ) ) in a p-type silicon at t = 0 s. b) Calculate the diffusion length for the electrons if the diffusion coefficient is 30 cm2/s at t = 0 s.

Ans:

a)

b)

12 3(0) 10n cm

1220 3 1

90

( ) (1 10 )2 10

(5 10 )n

n t xR x cm s

x

n

2 1 9 4(30 )(5 10 ) 3.873 10n n nL D cm s x s x cm

43

Example 4

Consider a semiconductor in which and . Assume that the excess carrier lifetime is . Determine the electron-hole recombination rate if the excess hole concentration is .

10 310in cm610 s

13 35 10p x cm

n-type semiconductor, low-injection so that

13

6

19 3 1

5 10

10

5 10

pO

p xR

R x cm s

15 30 10n cm

44

5. CONTINUITY EQUATION (critical in p-n diode & bjt)(Is created due to alter of the charge transport during:)Generation process – additional of electrons & holesRecombination process – removes of the electrons and holes

• The continuity equation allows us to calculate the carrier distribution in the presence of generation and recombination.

• Continuity equation Conservation of electrons and hole densities

• Total rate of particle flow = Particle flow due to current – (Loss due to recombination rate + Gain due to generation rate)

Particle Current

Loss Gain

45

Figure shows current flow and generation-recombination processes in an infinitesimal slice of thickness dx.

46

Design a uniformly doped p-type Germanium semiconductor bar with circular cross sectional area. The semiconductor should provide a current of 3 mA under a bias voltage of 6V. The maximum current density is to be 200 A/cm2. You are required to provide the following information to the device production team:(a) The required cross sectional area. [3 marks](b) The resistance of the semiconductor bar. [3 marks](c) The chosen doping concentration (refer to Figure 1 below) and the required length. [4 marks]

(a) A = I/Jdrift = 3mA/200 (1) =1.5 x 10-5 cm2 (2)

(b) R = V/I = 6/3mA (1) = 2 kΩ (2)

(c) R = L/eµpNaA (1)

Any reasonable Na is acceptable. If Na is 1016cm-3, µp ≈ 1400 cm2/Vs (from the

table). (1)

L = eµpNaAR=1.6 x 10-19 x1400 x 1016 x 1.5 x 10-5 x 2 k (1) = 6.72 x 10-2 cm (1)

Final Sem II 2010/2011 (10 marks)

47

Figure 1. Electron and hole mobility versus impurity concentrations for germanium, silicon and gallium arsenide at T=300K

48

Final Sem III 2011/2012 (10 marks)

8V

As shown in Figure above, a voltage of 8 V is applied to an N-type semiconductor with length, L of 0.04 m and cross-section area, A of 2 x 10-2 cm2. Let mn = 0.067mo. Given that the average time between collisions is 0.5 x 10-9 s, find out how fast the electron is moving under the influence of the electric field.

19 92

30

2 7

. . . .(2)

.

(1.6 10 ) (0.5 10 )1312 / (3)

0.067 (0.91 10 )

(2)

8(1312 / )( ) 262400 / 2.624 10 / (3)

0.04

d sc scn

e e e

n

drift n n

drift

v e F e q

F m F m m

m Vs

Vv E

L

Vv m Vs m s cm s

m