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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 density in
case of both donors ND and acceptors NA are
present simultaneously.
• The Fermi level will adjust itself to preserve
overall charge neutrality as
n  N A  p  N D
(1)
Donors and Acceptors
2
• By solving (1) with n. p  ni , the equilibrium
electron and hole concentrations in an n-type
semiconductors yield
1 
nn   N D  N A 
2
ni2
pn 
nn
N

D
N

 2
A

 4n 

2
i
Donors and Acceptors
• Similarly to p-type semiconductors, the electron
and hole concentrations are expressed as
1 
p p   N A  N D 
2
ni2
np 
pp
N  N

D

 2
A

 4ni2 

Donors and Acceptors
• Generally, in case of all impurities are ionized,
the net impurity concentration ND – NA is larger
than the intrinsic carrier concentration ni;
therefore, we may simply rewrite the above
relationship as
nn  N D  N A if N D  N A
p p  N A  N D if N A  N D
Donors and Acceptors
• The figure shows electron
density in Si as a function of
temperature for a donor
concentration of ND = 1015 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
bandgap.
Donors and Acceptors
• This also results in the reduction of the
bandgap. The bandgap reduction Eg for Si at
room temperature is expressed by
N
Eg  22
meV
18
10
where the doping is in the unit of cm-3.
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 of
impurity atoms is highly possible, then we have
n = ND = 1016 cm-3.
9.65 10

n
p

ND
1016
2
i

9 2
 9.3 103 cm-3
DonorsE and
Acceptors
N 
 E  kT ln


N
 D
Soln
 2.86 1019 
 0.0259 ln 

10
10
The Fermi level measured from the bottom of
the conductionband
0.205iseV
C
C
F
Donors and Acceptors
Soln
The Fermi level measured
from the intrinsic Fermi
level is
n
EF  Ei  kT ln 
 ni

 ND 
  kT ln 

n

 i 
 1016 
 0.0259 ln 
9 
9.65

10


 0.358 eV
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
pn  ni2
• The direct recombination rate R can be expressed as
R   np
where  is the proportionality constant.
Direct Recombination
• Therefore, for an n-type semiconductor, we
have
Gth  Rth   nn 0 pn 0
(3)
where nn0 and pn0 represent electron and hole
densities at thermal equilibrium.
Direct Recombination
• If the light is applied on the semiconductor, it
produces electron-hole pairs at a rate GL, the
carrier concentrations are above their equilibrium
values.
• The generation and recombination rates become
G  GL  Gth
R   nn pn    nn0  n  pn0  p 
where n and p are the excess carrie concentrations
Direct Recombination
n  nn  nn 0
p  pn  pn 0
• n = p to maintain the overall charge
neutrality.
• The net rate of change of hole concentration is
expressed as
dpn
 G  R  GL  Gth  R
dt
(7)
Direct Recombination
• In steady-state, dpn/dt = 0. From (7) we have
GL  R  Gth  U
(8)
where U is the net recombination rate.
Substituting (3) and (5) into (8), this yields
U    nn0  pn0  p  p
(9)
Direct Recombination
For low-level injection p, pn0 << nn0, (9) becomes
U   nn 0 p 
pn  pn 0 pn  pn 0

1/  nn 0
p
(10)
where p is called excess minority carrier lifetime .
pn  pn 0   p GL
Direct Recombination
We may write pn in the
function of t as
pn (t )  pn 0   p GL exp  t /  p 
Direct Recombination
• Ex. A Si sample with nn0 = 1014 cm-3 is illuminated
with light and 1013 electron-hole pairs/cm3 are
created every microsecond. If n = p = 2 s, find
the change in the minority carrier
concentration.
Direct Recombination
Soln Before illumination:
pn 0  ni2 / nn 0  (9.65 109 )2 /1014  9.31105 cm-3
After illumination:

1013 
6
pn  pn 0   pGL  9.3110   2 10  6   2 1013 cm-3
10 

5
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
n
 J e ( x) A J e ( x  dx) A 
Adx  

  Gn  Rn  Adx

t
e
 e

• where A is the cross-section area and Adx is the volume of the
slice.
Continuity Equation
• By expanding the expression for the current at x
+ dx in Taylor series yields
J e
J e ( x  dx)  J e ( x) 
dx  ...
x
• Thus, we have the basic continuity equation for
electrons and holes as
n 1 J e

  Gn  Rn 
t e x
p
1 J h

 Gp  Rp 
t
e x
(14)
Continuity Equation
• We can substitute the total current density for
holes and electrons and (10) into (14).
dn
dx
dn
J h  eh nE  eD p
dx
J e  ee nE  eDn
Continuity Equation
• For low-injection condition, we will have the
continuity equation for minority carriers as
n p
n p
 2n p
n p  n po
E
 n p e
 e E
 Dn 2  Gn 
t
x
x
x
n
pn
E
pn
 pn
pn  pno
  pn h
 h E
 Dp
 Gp 
2
t
x
x
x
p
2
(15)
The Haynes-Shockley Experiment
• This experiment can be used to measure the carrier
mobility μ.
• The voltage source establishes an electric field in the ntype 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
pn
pn
 2 pn pn  pno
 h E
 Dp

2
t
x
x
p
• If there is no applied electric field along the
bar, the solution is given by
 x2
N
t 
pn ( x, t ) 
exp  
   pno

 (16)
4
D
t

4 Dpt
p
p


The Haynes-Shockley Experiment
• N is the number of electrons or holes generated
per unit area. If an electric field is applied along
the sample, an equation (16) will becomes
2


x   p Et 

N
t
pn ( x, t ) 
exp  
   pno

4 D pt
p 
4 D pt


The Haynes-Shockley Experiment
Ex. In Haynes-Shockley experiment on n-type Ge
semiconductor, given the bar is 1 cm long, L =
0.95 cm, V1 = 2 V, and time for pulse arrival =
0.25 ns. Find mobility μ.
The Haynes-Shockley Experiment
Soln
L 0.95 cm
vD  
 3800 cm/s
t 0.25 ns
V1
E
 2 V/cm
1 cm
vD 3800 cm/s
 
 1900 cm 2 /V.s
E
2 V/cm
The Haynes-Shockley Experiment
Ex. In a Haynes-Shockley experiment, the
maximum amplitudes of the minority carriers at
t1 = 100 μs and t2 = 200 μs differ by a factor of 5.
Calculate the minority carrier lifetime.
Soln
2


x   p Et 

N
t
pn ( x, t ) 
exp  
   pno

4 D pt
p 
4 D pt


2


x   p Et 

N
t
p  pn  pno 
exp  
 

4 D pt
p 
4 D pt


The maximum amplitude
 t 
N
p
exp   
  
4 D pt
 p
Therefore,
t2 exp  t1 /  p 
 200  100 
p(t1 )
200


exp 
  5

p(t2 )
100
p
t1 exp  t2 /  p 


 p  79 s
Thermionic emission process
• It is the phenomenon that carriers having high
energy thermionically emitted into the vacuum.
• In other words, electrons escapes from the hot
or high temperature surface of the material.
• This is called “thermionic emission process”.
Thermionic emission process
Vn
• Electron affinity qχ is the
energy difference between
the conduction band edge
and the vacuum.
• Work function q is the
energy between the Fermi
level and the vacuum level
in the semiconductor.
Thermionic emission process
• It is clearly seen that an electron can
thermionically escape from the semiconductor
surface into the vacuum if its energy is above
qχ.
• The electron density with energies above qχ
can be found by
 q    Vn  
nth   n( E )dE  NC exp 

kT


q

where Vn is the difference between the bottom
of the conduction band and the Fermi level.
Thermionic emission process
• If escaping electrons with velocity normal to the
surface and having energy greater than EF + q,
the thermionic current density is equal to
J   qvN ( E ) F ( E )dE
 4 (2m)3/ 2   ( E  EF ) / kT
  qv 
dE
e
3
h


Thermionic emission process
h2k 2
• Then we use p = mv and E  2 , and after
8 m
integration, it yields
 q 
J  A T exp 

kT


2
4

qmk
A* 
 Richardson constant
3
h
 1.2  106 A/(m 2 .K 2 )  120 A/(cm 2 .K 2 )
*
2
Thermionic emission process
Ex. Calculate the thermionically emitted electron
density nth at room temperature for an n-type
silicon sample with an electron affinity qχ = 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
 4.05  0.2 
52
nth (4.05 eV)  2.86  10 exp  

10
0

 0.0259 
 0.6  0.2 
6
-3
nth (0.6 eV)  2.86  1019 exp  

1

10
cm

0.0259


19
As we clearly see that at the room temperature,
there is no thermionic emission of electrons into
the vacuum. This thermionic emission process is
important for metal-semiconductor contacts.
Tunneling Process
• The figure shows the energy
band when two semiconductor
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 transport
across the barrier and move to
the right-side semiconductor
even the electron energy is
much less than the barrier
height.
• This process is called “quantum
tunneling process”.
Tunneling Process
• The transmission coefficient
can be expressed as

2me* (qV0  E ) 


T  2  exp 2d

2
A




C
2
• This process is used in tunnel
diodes by having a small
tunneling distance d, a low
potential barrier qV0, and a
small effective mass.