Chapter 2: Carrier (n & p) and

Transport Phenomena (v, , or , I or J and related)

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? (Carrier Drift)
If there is a concentration gradient in the electron or hole density,
how do the carriers respond? (Carrier Diffusion)
Do electrons in the conduction band fall down into the valence band
and recombine with holes? (Generation and Recombination process)
Is it possible for electrons in the valence band to jump up into the
conduction band? What cause such processes?
(Generation and Recombination process)

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, or F is
applied to a semiconductor. This electric field causes each electron
to experience a force -q 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
Alloy
Interface roughness
Chemical impurities

Due to lattice vibrations at finite temperatures

Random potential fluctuations
Important in heterostructure
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 (Mathiesons rule).

tot .sc

sc1 sc 2

...

sc = Scattering time or
Mean time between collisions for an electron or
Relaxation time or
Delay time
3

 = d = ave.= drift velocity

= ave. gain velocity
(either d n or d p )
e = electron

a) Low Electric Fields (When < 100 kV/cm)

(d = immediately before end of collision, after
collision the d = 0)

e.F . SC
vd
m*

F = = Electric Field
sc = scattering time
m* = effective mass
(either mn* or mo 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 )

n.e2 . SC .F
J n.e.vd
m*
From Ohms Law,
J = .F

n.e2 . SC

n.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)

For general :

e. SC

m*

For e- :

e. .SC
n
mn *

For holes :

e. .SC
p
mp *

If both e- and holes are present

nn .e.n np .e. p e(nn .n n p . p )

From Ohms Law,
J = .F

J e(nn .n n p . p ).F
Example 1
The 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 *. (0.26 x0.91x1030 kg ).(1500 x104 m2 / V .s)

13

2.2
x
10
s
19
e
1.6 x10 C

Example 2
The 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 *. (0.067 x0.91x1030 kg ).(8500 x104 m2 / V .s)

13

3.24
x
10
s
19
e
1.6 x10 C

The time for ionized impurity in GaAs (pure + impurity)

2
SC

m *. (0.067 x0.91x1030 kg ).(5000 x104 m2 / V .s)

13

1.9
x
10
s
19
e
1.6 x10 C

Mathiesons rule:

1
2
SC

1
1
SC

1
imp
SC

imp
SC
4.6 x1013 s
6

Example 3
Consider 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.s
Ans:

Conductivity for undoped Si:

undoped e(ni .n ni . p )
(1.6 x1019 C )(1.5x1010 cm3 ){(1000 350)}cm2 / V .s
3.24 x106 (cm)1
Conductivity for undoped GaAs:

undoped (1.6 x1019 C )(1.84 x106 cm3 ){(8000 400)}cm2 / V .s

2.47 x109 (cm)1

Cont. Example 3

ndoped (100%) nn N 1017 cm3

For Si & GaAs

ndoped (50%) 1017 x50% 5x1016 cm3

For Si & GaAs

pdoped
pdoped

ni2
ndoped
ni2
ndoped

(1.5 x1010 )2
3
3 For Si (very small compared to ndoped)

4.5
x
10
cm
5 x1016
(1.84 x106 )2
5
3 For GaAs (very small compared to ndoped)

6.77
x
10
cm
5 x1016

Conductivity for doped Si:

Almost zero
doped e(nn .n np . p )
(1.6 x1019 C){(5x1016 )(1000cm1 / V .s) (4.5x103 )(350cm1 / V .s)}

8(cm)1
Conductivity for doped GaAs:
Almost zero
doped e(nn .n np . p )
(1.6 x1019 C){(5x1016 )(8000cm1 / V .s) (6.77 x105 )(400cm1 / V .s)cm1}
64(cm)1
8

Example 4
Consider 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

ni2 (1.5 x1010 )2

4
3
p

2.25
x
10
cm
n
1016

J e(nn .n n p . p ).F e(nn .n ) F

19

(1.6 x10

)(10 )(1350)(35) 75.6 A / cm

16

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]

10

Solutions:
a)

n q n n

p q p p
b)

n
0.1(cm)1
14
3
n

6.25
x
10
cm
qn (1.6 x1019 C )(1000cm2 / Vs)
p
0.1(cm)1
15
3
p

2.08
x
10
cm
q p (1.6 x1019 C )(300cm2 / Vs)

n
1 n
EFn EC k BT ln

N
N
8
C
C

EFn

(6.25 x1014 ) 1 (6.25 x1014 )

EC (0.026) ln

19
19
(2.78
x
10
)
(2.78
x
10
)
8

EFn EC (0.026) (10.70) (7.95 x106 )

EFn EC 0.278eV

p
1 p
EFp EV k BT ln

N
N
8
V
V

EFp

(2.08 x1015 ) 1 (2.08 x1015 )

EV (0.026) ln

18
18
8 (9.84 x10 )
(9.84 x10 )

EFp EV (0.026) (8.46) (7.47 x105 )

EFp EV 0.22eV

11

c) n exp EFn Ei

ni

k BT

d)

n E Ei
ln Fn
k BT
ni
n
6.25 x1014
EFn Ei k BT ln (0.026) ln
0.276eV
10
n
1.5
x
10

EC
0.278
EFn

0.276
Ei
0.3078
EFp
0.22
EV

Ei EFp
p
exp

ni
k
T
B

p Ei EFp
ln
k BT
ni
p
2.08 x1015
Ei EFp k BT ln (0.026) ln
0.3078eV
10
1.5 x10
ni

e) From the diagram (d),

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

12

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.
or
The 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 carrying
charges have multiplied.
The process is called Avalanching
The same process could happen
to hot holes.

Conduction
band

Initial state
has 1 e-

Final state
has 2 e+ 1 hole

+
Valence
band

Avalanche process
14

Once avalanching starts,

dI ( z )
imp I
dz
I ( z)
N ( x)
exp( imp x)
I (O)
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

Energy band diagrams under junctionbreakdown conditions-Avalanche

multiplication.
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:

T exp(

4 2m* Eg3/ 2
3e F

Electrons in
conduction
band

)
Available
empty states
(holes) in
valence
band

F = Electric Field in the semiconductor

Example 1
Calculate the band to band tunneling probabilty

Energy band diagrams under

in GaAs and InAs at an applied = 2 x 105 v/m.

junction-breakdown
Given:
conditions- Tunneling effect
m* (GaAs) = 0.065 m0
m* (InAs) = 0.02 m0
Eg (GaAs) =1.5 eV
4 x 2 x0.065 x0.91x1030 kg )(1.5 x1.6 x1019 J )3/ 2
Eg (InAs) = 0.4eV

TGaAs exp(

3x(1.6 x10

19

C )(1.05 x10

34

Js)(2 x10 V / m)

)0

TInAs 3.7 x106 (means Zener Tunneling is important when ~2 x 105 V/m)
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)

Example 6 (Test 1 Sem 3 2011/2012)

Figure 1

L
V

p-type semiconductor

Figure 2

18

(a) Refer to the Figure 1 (graph) above, determine and explain whether Zener Tunneling can
happen for the GaAs (electrons) at the time scattering between the range of:
5.00 x 10-8 s and 4.50 x 10-8 s.

[3 marks]

(ii) 1.50 x 10-9 s and 1.00 x 10-9 s.

[3 marks]

(i)

(b) What assumption can you made as an electronic engineer from the graph above related to
the carrier drift velocity and the electric field for GaAs (Electrons)?
[3marks]

(c) Figure 2 shows a p-type semiconductor with applied voltage, V of 6.0 V and the length, L of
10.0 cm. Find the mobility of the holes in this semiconductor if the drift velocity is given as
2 x 107 cm/s.
[4 marks]
(d) An n-type Silicon sample with a resistivity of 10 cm at 300 K
(i)

Calculate the electron carrier density of the Si Sample.

[4 marks]

(ii) Calculate the Fermi level for n-type Si sample from the intrinsic Fermi level

[4 marks]

(iii) Sketch the flat band diagram for part (ii)

[4 marks]

(iv) How much is the energy gap has been shifted?

[3 marks]
19

Answers:

2.1 X 107

Figure 1

3.4 X 103

First you need to find the scattering time of electrons at the maximum value of
carrier drift velocity with respect to the electric field for the GaAs (Electron). The
scattering time before this point will be considered at the low electric field region.
The scattering time after this point will be considered under high electric field and
can be solved thru Zener Tunneling and Avalanching breakdown theories only.
20

vd

SC

e.F . SC
me

vd .me (2.1107 cm / s)(0.911030 kg )

3.52 108 s
19
3
e.F
(1.6 10 C )(3.4 10 V / cm)

(2.0 106 cm / s)
me vd 1
12
1 .
(5.7 10 ).
4.39 10 8 s
2
(2.6 10 V / cm)
e F1
6
me vd 2
12 (4.0 10 cm / s )
2 .
(5.7 10 ).
4.56 108 s
2
(5 10 V / cm)
e F2

(1.0 107 cm / s)
me vd 3
12
3 .
(5.7 10 ).
1.9 10 9 s
4
(3.0 10 V / cm)
e F3
(7.0 106 cm / s )
me vd 4
12
4 .
(5.7 10 ).
0.32 10 9 s
5
(1.25 10 V / cm)
e F4

Note: d the e- move in a direction opposite to the electric field while the holes move
in the same direction
3
(a i)No. From the calculation we can see that the scattering time is about 3.52 x 10 -8 s. Therefore any range that is bigger than
this value will consider that the Zener Tunneling is not happen yet.
3
(a ii)Yes. From the calculation we can see that the scattering time is 3.52 x 10 -8 s. Therefore any range that is smaller than
this value will consider that the Zener Tunneling will happen.
(b) After a certain point, or F around 3.4 x 103 V/cm, the vd start to slow down, showing that the voltage breakdown will
happen therefore we have to use Zener Tuneling or Avalanching breakdown theories to solve the problem.

3
21

(c ) p

v
v
10.0cm
L

v. (2 107 cm / s).
3.33 107 cm / V .s
F V L
6.0V
V

1 n
n
0.1(cm) 1
14
3
(d )(i )n

6.25
x
10
cm
q n q n (1.6 x1019 C )(1000cm 2 / Vs)
n
6.25 x1014
(d )(ii ) EFn Ei k BT ln (0.026) ln
0.276eV
10
1.5 x10
ni
(d )(iii )

(d )(iv)E Egbefore Eg After 1.124 [(1.124 / 2) 0.276] 0.286eV

additional : % shifted [(1.124 0.838) /(1.124)].100% 25.44%

22

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)

x= 0

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?

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, ( x, t ) to the right across x = x0 at any instant of time, t.

( x, t )

But,

(nL nR )
2 SC

= free path to each side of x0 boundary in time SC ,

nL = average carrier densities in L-Region
n = average carrier densities in R-Region
R

nL nR

dn
.
dx

Net Flux:

n ( x, t )

2 SC

dn( x, t )
dn( x, t )
Dn
dx
dx

for electron

Where Dn = diffusion coefficient of the electron in the system

Where Dp = diffusion coefficient of the hole in the system

dp( x, t )
p ( x, t ) Dp
dx

for hole
24

The current density, J

J tot (diff ) J n (diff ) J p (diff ) eDn

dn( x, t )
dp( x, t )
eDp
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.

25

Example 7
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.

dn( x, t )
dp( x, t )
J tot (diff ) J n (diff ) J p (diff ) eDn
eDp
dx
dx

J dif

p( x)
p(0.01) p(0)
J p (diff ) eD p
eD p
x
0.01 0

17

p
(0.01)

(4
x
10
)
19
20 (1.6 x10 )(10)

0.01

p(0.01) 2.75 x1017 cm 3

26

Example 8
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.

dn
n
J eDn
eDn
dx
x
15
16

10

10
19
J 1.6 x10 25

0.10 0
J 0.36 A / cm 2
J 0.36 A / cm 2
For A = 0.05 cm 2
I AJ 0.05 0.36 I 18 mA
27

Example 9
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?

(2 x1014 ) 1015
dp
19
i ) I p eD p A 1.6 x10 10 0.075
0.96 mA
dx
0.10

Ip

0.96mA
2

12.8
mA
/
cm
A 0.075cm 2
0.075
iii ) A r 2 0.075cm 2 r
0.15cm

ii ) J P

L
iv) LP D p p p P 1103 s
Dp
v)increase the ' r ', i.e (r 0.25 instead of 0.15) I p

0.96 mA
(0.25cm 2 ) 2.56mA
2
0.075cm
28

Example 10
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?

J n eDn

dn
dx

eDn

400 1.6 x10

n
x

19

400 Dn 16

(6 x10 ) 10
Dn (4 x104 ) 0

16

17

Dn 25 cm / s
2

29

Example 11 (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
E - E (eV)
F

y mx c

0.1
0.1 eV
( EF Ei )eV
x

(0.1)
eV

5 m
eV

( EF Ei )eV 200
x (0.1)eV

cm

0
E Ei
200 x 0.1 Boltzman equation
n ni exp F

n
exp

i
k BT
k BT

x m

200 x 0.1
200 x 0.1
dn
d

200

n
exp

n
exp
i
i

dx dx
k BT
k BT
k BT

J n eDn

dn
dx

eDn

200 x 0.1
200

n
exp
i

k BT
k BT

200(1 104 ) 0.1

200

19
10

J n (1.6 10 )(25)
(1.5

10
)
exp

(0.026)
0.026

J n (4.6 10

) exp(3.077) 9.978 x10 3 A / cm 2

30

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

31

Example 12
Using 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

Electron Density
Hole Density

E Ec
E Ei
n NC exp Fn
ni exp F

kBT

kT
Ev EFp
p N v exp

k
T

Ei EF
ni exp

kT

Boltzmann approximation
For n or e-:

For holes or p:

n
EFn EC k BT ln

NC

p
EV EFp k BT ln

N
V

32

 n
EFn EC k BT ln

N
C

EC
0.146 eV

EFn

1017
EFn k BT ln
EC ( EC 0.146)eV
EC (0.026) ln
19
N
2.8
x
10
C

EFp

p
EV kBT [ln
] ( EV 0.121)eV
NV

EFp
0.121 eV

EV
EFn - EFp = (EC EV) (0.146+0.121) = 1.12 0.267 = 0.853 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) = 1.12 0.506 = 0.614 eV

33

1.

2.

3.

Why does Fermi level shift toward conduction or valence band with doping?
The Fermi level is the energy separating occupied states (or levels) of the valence band from empty states
(levels) of the conduction band at the absolute temperature T=0 Kelvin. The Fermi level is an energy level
characteristic of the statistics (distribution law) which controls the occupation of any energy state by a
given particle: an electron or a hole in semiconductors. Electrons and holes are both fermion particles (
with half values of the spins) and both of them obey the Fermi-Dirac statistics which becomes
asymptotically the Boltzman statistics in dilute systems (with few electrons and or holes) or nondegenerate
systems. The position of the Fermi level with respect to valence and or conduction bands depends on
various parameters as the temperature, the effective masses of electrons and holes, and the number of
free electrons and holes. This variation of the Fermi level obeys two condictions : - the ''mass action law''
which states that the number of particles of each type as well as the overall number of the particles must
conserve whatever is their distribution on the available energy levels. - The neutrality equation which states
that the electrical neutrality has to be fulfilled, i.e. the number of negative charges must be
counterbalanced excatly by the same number of positive charges. 2.
In an intrinsic semiconductor (with no doping at all), the Fermi level is lying exactly at the middle of the
energy bandgap at T=0 Kelvin. With increasing temperature T>0 Kelvin the Fermi energy remains at this
midgap position if conduction and valence bands have exactly the same dispersion energy or more simply
the same effective masses for electrons and holes. If not then the Fermi level will shift away from midgap
position with increasing temperature and will move towards the band with the smaller effective mass
In an extrinsic semiconductor (with added doping), in order to conserve the number of particles (mass
action law) and to fulfill the overall electrical charge neutrality (neutrality equation), the Fermi level has to
move away from the midgap position. It has to shift towards conduction band in an n-type
semiconductor (extrinsic semiconductors with added doping impurities which are donors i.e.
impurities which give additional electrons to the system) where the number of electrons n is higher than the
number of holes (n>p). It shifts towards valence band in a p-type semiconductor (extrinsic semiconductors
with added doping impurities which are acceptors i.e. impurities which trap electrons from the system
giving rise to a deficit of electrons or an excess of holes ) where the number of electrons n is lower than the
number of holes (n<p)
34

Example 13 (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).
Majority carrier density, n
EFn EFi
n ni exp
k BT
and

0.228
10
14
3

(1.5
x
10
)
exp

1.01466
x
10
cm

0.025852

Minority carrier density, p

ni2 (1.5 x1010 ) 2
p

2.21749 x106 cm 3
14
n 1.01466 x10

Note :
( EFn EFi ) EFi EFn 0.228eV

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

n
EFn EFi ( EFi EFn ) k BT ln
ni
1.01466 x1014
(0.025852) ln
0.228eV
10
1.5 x10

Majority (from n)

p
EFi EFp ( EFp EFi ) k BT ln
ni
2.21749 x106
(0.025852) ln
0.228eV
10
1.5 x10

Minority (from p)

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.

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

n
1 n
EC EFn k BT ln

N
N
8
C
C

(1.01466 x1014 ) 1 (1.01466 x1014 )

EC EFn (0.026) ln

19
19
(2.78
x
10
)
(2.78
x
10
)
8

EC EFn (0.026) (12.52) (1.29 x106 ) 0.325eV

n
1 n
EC EFn k BT ln

N
N
8 i
i

(2.21749 x106 ) 1 (2.21749 x106 )

EC EFn (0.026) ln

10
10
(1.5
x
10
)
(1.5
x
10
)
8

EC EFn (0.026) (8.82) (5.22 x105 ) 0.229eV

37

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

Ev, EFn, EFp and Efi.

EC
0.3236 eV
EFn
0.228 eV

EFi
EFp
0.7528 eV
EV

38

4. a) RECOMBINATION-GENERATION CENTERS PROCESSES

(JFYI)
Happens due to Lattice defects or Impurity atoms in lattice disrupt the ideal singlecrystal lattice structure.
As a result, allowed electronic energy states within the bandgap.
These energy states may now serve as stepping stones in the recombinationgeneration process.
These energy states occur near the midgap energy (as in Figure 10 & 11).

Electron-hole generation and recombination (normal case)

39

Figure 10

Figure 11

(a)

(b)

(c)

Now, we have
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

or,
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)
40

4.b)GENERATION AND RECOMBINATION PROCESSES (the one concern)

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:
2
RG = the carrier generation rate
np

n
p

n
R
=
R
0 0
i
G
R
RR = the carrier recombination rate
n = Total electron (may be functions of time or position)
p = Total hole (may be functions of time or position)
n0 =Thermal equilibrium electron concentration (independent of time and position)
p0 = Thermal equilibrium hole concentration(independent of time and position)
ni = Intrinsic concentration

b) Optical Radiation (Optoelectronics)

(Excess Carriers)

np n0 p0 ni2

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.

41

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

Recombination of
excess carriers
reestablishing thermal
equilibrium

42

4. GENERATION AND RECOMBINATION PROCESSES (recall)

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).

43

Conduction
band
-

Photon

Vertical in k

Photon
+

Valence
band

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.
44

Carrier lifetime = n 0 or p 0
Excess electron concentration = n

n n0

p p p0

Excess hole concentration =

May be functions
of time or position

Example 1
Excess electrons have been generated in a
semiconductor to a concentration of n(0) 1015 cm3
6
The excess carrier lifetime is n 0 10 s . The
forcing function generating the excess carriers
turns off at t = 0 so the semiconductor is
allowed to return to an equilibrium condition
for t>0. Calculate the excess electron
concentration for
t
n(t ) n(0) exp( )
a) t=0
n0

n 1015 cm3

b) t=1s

t
n(t ) 10 exp(
)
1 s

n 3.68x1014 cm3

c) t=4s

n 1.83x1013 cm3

15

n(0)

Carrier
Injection

n( x )

n( x L)

The figure shows how the

excess carriers decay into
the semiconductor
45

Example 14
Find the recombination rate of the excess electrons for example 1.

n(t ) p(t )

R
p0
n0
1015
a) R 6 1021 cm3 s 1
10
14
3.68
x
10
20
3 1
b) R

3.68
x
10
cm
s
6
10

1.83x1013
c) R
1.83x1019 cm3 s 1
6
10
Note:
1. When excess electron density ( n ) is injected in p-type material. This excess minority
carriers (n ) will recombine with the majority carriers (p ) with a rate given by:

n(t )
n0

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

p(t )
p0

46

Diffusion length = Average distance = Ln orLp

Diffusion coefficient =

Dn n or Dp p

Example 15
Electrons 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 n(0) 1012 cm3 .
a) Calculate the rate of the recombination of the excess electron (excess minority carriers
( n) ) 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
i) t = 0 s, ii) t = 3 x 10-9 s, t = 4 x 10-9 s
Ans:
a)

n(t ) (1x1012 )
20
3 1
R

2
x
10
cm
s
9
n 0 (5 x10 )

b) Same answer for i), ii), iii)

Ln Dn n (30cm2 s 1 )(5x109 s) 3.873x104 cm

47

Example 16
Consider a semiconductor in which n0 1015 cm3 and ni 1010 cm3 . Assume
that the excess carrier lifetime is 106 s . Determine the electron-hole
recombination rate if the excess hole concentration is p 5x1013 cm3 .
n-type semiconductor, low-injection so that

p
pO

13

5 x10
10

R 5 x10 cm s
19

3 1

48

5. CONTINUITY EQUATION (critical in p-n diode & bjt) (JFYI)

(Is created due to alter of the charge transport during:)
Generation process additional of electrons & holes
Recombination 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

49

Figure shows current flow and generation-recombination

processes in an infinitesimal slice of thickness dx.
50

Example 17 (Final Sem II 2010/2011 (10 marks))

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 = 3 mA / 200 A/cm2 (1) =1.5 x 10-5 cm2 (2)
(b) R = V/I = 6 V / 3 mA (1) = 2 k (2)
(c) R = L / e p Na A (1)
[Any reasonable Na is acceptable. If Na is 1016cm-3, p 1400 cm2/Vs (from Figure 1). (1)]

L = e.p.Na.A.R = (1.6 x 10-19) . (1400) . (1016) . (1.5 x 10-5) . (2 k) = 6.72 x 10-2 cm (2)

(1400)

(1016)

Figure 1. Electron and hole mobility versus impurity concentrations for germanium,
silicon and gallium arsenide at T=300K
52

Final Sem I 2014/2015

QUESTION 2 [10 marks]
A uniformly doped n-type Si semiconductor devices of 2 x 1018 cm-3 with a current density of 2.56 x 107 A/cm2 is
given to the system.
(a) Based on Figure 1, you are required to examine whether this semiconductor devices will or will not exceed the
high Electric Field, E of 500 kV/cm.

[6 marks]

(b) As a design engineer, what can you conclude if the value of E exceed 500 kV/cm and also what is the value of
breakdown voltage, VB at the doping concentration of 3x1012 cm-3.

[4 marks]

53

J
(2.56 x107 Acm 2 )
7
1
(a )vd

8
x
10
cm
.
s
(3)
18
3
19
n.e (2 x10 cm ).(1.6 x10 C )
(8 x107 cm.s 1 )
5
E

(4
x
10
V / cm)(2)
2
2
(2 x10 cm / V .s )
E is not exceeded 500kV/cm.(1)
vd

(b) If E exceed 500kV/cm, it means that the breakdown voltage will happen
therefore we have to use Zener Tunneling or Avalanching Breakdown
theories to solve the problem.(2)

S .E 2

(11.9).(8.85 x1014 Fcm 1 ).(5 x105Vcm 1 ) 2

5
VB

2.7

10
V
19
12
-3
2.e.N B
2.(1.6 x10 C ).(3 10 cm )
note: the value of VB 2.7 105V (2)

54

Example 18 (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 (velocity/speed) the electron is moving under the influence of the electric
field.

vd e.F . sc e. sc q.

F
me .F
me
me

(2)

(1.6 1019 ) (0.5 109 )

2
n

1312
m
/ Vs (3)
30
0.067 (0.9110 )
V
vdrift n E n (2)
L
8V
vdrift (1312 m 2 / Vs )(
) 2624 00 m / s 2.624 107 cm / s (3)
0.04m

55

Example 19 (Final Sem I 2012/2013 (10 marks))

A uniformly doped n-type Ge semiconductor with the current of 0.5 A and charge densities
of 5.0 x 1018 cm-3. This semiconductor has electric field and bias voltage of 8.0 V/m and
5.0 V respectively. Find the cross-sectional area, A and then comments on the required
dopant if you increase the radius of this semiconductor.
[10 marks]

Ld
V /F
(5V ) /(8V / cm)

d .R.N d .q V
(3900cm 2 / Vs). 5V (5.0 1018 cm 3 )(1.6 1019 C )

.
.
N
.
q

dI d

0.5 A

5+2
0.625cm
5
2
A
2.0 10 cm
1
31200cm
Ld
A r 2 (if r )
d .R.N d ().q
N d will decrease

2+1

56