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ENE 311
Lecture 7
p-n Junction
• A p-n junction plays a major role in electronic
devices.
• It is used in rectification, switching, and etc.
• It is the simplest semiconductor devices.
• Also, it is a key building block for other
electronic, microwave, or photonic devices.
Basic fabrication steps
The basic fabrication steps for p-n junction include
• oxidation,
• lithography,
• diffusion or ion implanation,
• and metallization.
Basic fabrication steps
Oxidation
• This process is to make a high-quality
silicon dioxide (SiO2) as an insulator
in various devices or a barrier to
diffusion or implanation during
fabrication process.
• There are two methods to grow SiO2:
dry and wet oxidation, using dry
oxygen and water vapor,
respectively.
•
Generally, dry oxidation is used to
form thin oxides because of its good
Si-SiO2 interface characteristics,
while wet oxidation is used for
forming thicker layers since its higher
growth rate.
Basic fabrication steps
Lithography
• This process is called
photolithography
used to delineate the
pattern of the p-n
junction.
Basic fabrication steps
• (a) The wafer after
the development.
• (b) The wafer after
SiO2 removal.
• (c) The final result
after a complete
lithography process.
Basic fabrication steps
Diffusion &
Ion Implantation
• This is used to put the impurity into
the semiconductor.
• For diffusion method, the
semiconductor surface not protected
by the oxide is exposed to a high
concentration of impurity. The
impurity moves into the crystal by
solid-state diffusion.
• For the ion-implantation method, the
impurity is introduced into the
semiconductor by accelerating the
impurity ions to a high-energy level
and then implanting the ions in the
semiconductor.
Basic fabrication steps
Metallization
• This process is used
to form ohmic
contacts and
interconnections.
• After this process is
done, the p-n
junction is ready to
use.
Thermal equilibrium condition
• The most important
characteristic of p-n
junction is rectification.
• The forward biased voltage
is normally less than 1 V and
the current increases rapidly
as the biased voltage
increases.
• As the reverse bias
increases, the current is still
small until a breakdown
voltage is reached, where
the current suddenly
increases.
Thermal equilibrium condition
• Assume that both pand n-type
semiconductors are
uniformly doped.
• The Fermi level EF is
near the valence
band edge in the ptype material and
near the conduction
band edge in the ntype material.
Thermal equilibrium condition
• Electrons diffuse from n-side
toward p-side and holes diffuse
from p-side toward n-side.
• As electrons leave the n-side,
they leave behind the positive
donor ions (ND+) near the
junction.
•
In the same way, some of
negative acceptor ions (NA-)
are left near the junction as
holes move to the n-side.
Thermal equilibrium condition
Space-charge
region
neutral
neutral
• This forms 2 regions
called “neutral”
regions and “spacecharge” region.
• The space-charge
region is also called
“depletion region”
due to the depletion
of free carriers.
Thermal equilibrium condition
• Carrier diffusion induces
an internal electric field
in the opposite direction
to free charge diffusion.
• Therefore, the electron
diffusion current flows
from left to right,
whereas the electron
drift current flows from
right to left.
Thermal equilibrium condition
• At thermal equilibrium,
the individual electron
and hole current flowing
across the junction are
identically zero.
• In the other words, the
drift current cancels out
precisely the diffusion
current. Therefore, the
equilibrium is reached as
EFn = EFp.
Thermal equilibrium condition
• The space-charge density distribution and the
electrostatic potential  are given by Poisson’s
equation as
d 2
dE

e

     ND  N A  p  n
2
dx
dx


(1)
• Assume that all donor and acceptor atoms are
ionized.
Thermal equilibrium condition
• Assume NA = 0 and n >> p for n-type neutral
region and ND = 0 and p >> n for p-type neutral
region.
Thermal equilibrium condition
• The electrostatic potential  in of the n- and ptype with respect to the Fermi level can be
found with the help of n  ni exp  EF  Ei  / kT and
p  ni exp  Ei  EF  / kT
as
kT  N D 
 n  EF  Ei   ln 

e  ni 
(2)
kT  N A 
 p    Ei  EF    ln 

e  ni 
(3)
Thermal equilibrium condition
• The total electrostatic potential difference
between the p-side and the n-side neutral
region is called the “built-in potential” Vbi. It is
written as
kT  N A N D 
Vbi   n   p 
ln 

e  ni2 
(5)
• a) A p-n junction with
abrupt doping changes at
the metallurgical
junction.
• (b) Energy band diagram
of an abrupt junction at
thermal equilibrium.
• (c) Space charge
distribution.
• (d) Rectangular
approximation of the
space charge distribution.
Thermal equilibrium condition
Ex. Calculate the built-in potential for a silicon p-n
junction with NA = 1018cm-3 and ND = 1015 cm-3 at
300 K.
Thermal equilibrium condition
Ex. Calculate the built-in potential for a silicon p-n
junction with NA = 1018cm-3 and ND = 1015 cm-3 at
300 K.
Soln
 N AND 
Vbi  kT ln 

2
 ni 

18
15 
10

10

 0.0259ln 
2
  9.65  109  


Vbi  0.774 eV
Depletion Region
The p-n junction may
be classified into two
classes depending on
its impurity
distribution:
• the abrupt junction
and
• the linearly graded
junction.
Depletion Region
• An abrupt junction can be
seen in a p-n junction that
is formed by shallow
diffusion or low-energy ion
implantation.
• The impurity distribution
in this case can be
approximated by an abrupt
transition of doping
concentration between the
n- and the p-type regions.
Depletion Region
• In the linearly graded
junction,
the
p-n
junction may be formed
by deep diffusions or
high-energy
ion
implantations.
• The impurity distribution
varies linearly across the
junction.
Abrupt junction
• Consider an abrupt junction
as in the figure above,
equation (1) can be written
as
d 2 eN A

2
dx

d 2
eN D

2
dx

for -x p  x  0
for 0  x  xn
• The charge conservation is
expressed by the condition
Q = 0 or
N A x p  N D xn
Abrupt junction
• To solve equation (5), we need to solve it separately for
p- and n-type cases.
p-side:
Integrate eq.(4) once, we have
d
dx

eN A x

c
We know that E  
d
dx
eN A x


E p ( x)   
 c
 

Abrupt junction
Apply boundary condition: E p ( x  x p )  0
 eN A (  x p ) 
Ep (xp )   
 c  0



c
eN A x p

E p ( x)  
eN A ( x  x p )

(7)
Abrupt junction
n-side:
• Similarly, we can have
En ( x) 
eN D ( x  xn )

  Em 
eN D x

(8)
Abrupt junction
• Let consider at x = 0
E p (0)  En (0)  
eN A x p


eN D ( xn )

 Em
(9)
We may relate this electric field E to the potential
over the depletion region as
Vbi  
xn
0
 xp
 xp
xn
 E ( x)dx    E ( x)dx
  E ( x)dx
p  side
0
n  side
Abrupt junction
2
A p
eN x
2
D n
eN x
Vbi 

2
2
(10)
• From (6), we have
xn 
N Axp
ND
N D xn
xp 
NA
(11)
Abrupt junction
• Substitute (11) into (10), this yields
2Vbi
ND
xp 

e  N A  ND  N A
2Vbi
NA
xn 

e  N A  ND  ND
(12)
Abrupt junction
• Hence, the space-charge layer width or
depletion layer width can be written as
2Vbi
W  x p  xn 
e
 N A  ND 


N
N
 A D 
(13)
Abrupt junction
Ex. Si p-n diode of NA = 5 x 1016 cm-3 and ND = 1015
cm-3. Calculate
(a) built-in voltage
(b) depletion layer width
(c) Em
Abrupt junction
Soln (a)
kT  N A N D 
Vbi 
ln 

2
e  ni 

16
15 
5  10  10 

 0.0259ln
 1.45  1010 2 


Vbi  0.679 eV
Abrupt junction
Soln (b)
From (13)
2Vbi
W
e
 N A  ND 


N
N

A D 
2  8.85  1014  11.8  0.679  5  1016  1015 


19
16
15 
1.6  10
5

10

10


W  0.95  m
Abrupt junction
Soln (c)
Emax  E ( x  0)  
eN A x p


eN D ( xn )

2Vbi
N
xn 
 A
e N A  ND  ND
2  8.85  1014  11.8  0.679 5  1016


19
16
15
1015
1.6  10  5  10  10 
xn  9.299  105 cm.
Emax 
Emax
eN D ( xn )

1.6  1019  1015  9.299  105

8.85  1014  11.8
 1.431 104 V/cm
Abrupt junction
• If one side has much
higher impurity doping
concentration than
another, i.e. NA >> ND or
ND >> NA, then this is
called “one-sided
junction”.
• Consider case of p+-n
junction as in the figure
(NA >> ND),
W  xn 
2 Vbi
eN D
Abrupt junction
• Similarly, for n+-p junction of ND >> NA
W  xp 
2Vbi
eN A
• The electric-field distribution could be written as
E ( x )   Em 
eN B x

where NB = lightly doped bulk concentration
(i.e., NB = ND for p+-n junction)
Abrupt junction
• The maximum electric field Em at x = 0 can be
found as
Em 
eN BW

• Therefore, the electric-field distribution E(x)
can be re-written as
x

E ( x) 
 W  x    Em 1  

 W
eN B
(16)
Abrupt junction
• The potential distribution can be found from
integrating (16) as
Vbi x 
x
 ( x) 
2 
W 
W
(17)
Abrupt junction
Ex. For a silicon one-sided abrupt junction with NA
= 1019 cm-3 and ND = 1016 cm-3, calculate the
depletion layer width and the maximum field at
zero bias.
Abrupt junction
Soln

19
16 
10

10
  0.895 V
Vbi  0.0259ln 
2
  9.65  109  


2Vbi
W
 3.41 105  0.343  m
eN D
Em 
qN BW

 0.52  104 V/cm
Linearly Graded Junction
Linearly Graded Junction
• In this case, the Possion equation (1) is
expressed by
d 2
dE

e
W
W

    ax for -  x 
2
dx
dx


2
2
(18)
where a is the impurity gradient in cm-4 and W is
the depletion-layer width
Linearly Graded Junction
• By integrating (18) with the boundary conditions
that the electric-field is zero at W/2, E(x) can
be found as
2
2

ea W / 2   x 
E ( x)   

 
2

(19)
• The maximum field Em at x = 0 is
eaW 2
Em 
8
(20)
Linearly Graded Junction
• The built-in potential is given by
eaW 3
Vbi 
12
(21)
and
kT   aW / 2  aW / 2   2kT  aW 
Vbi 
ln 
ln 


2
e 
ni
e
2
n

 i
(22)
Linearly graded junction in
thermal equilibrium.
(a) Impurity distribution.
(b) Electric-field
distribution.
(c) Potential distribution
with distance.
(d) Energy band diagram.
Linearly Graded Junction
Ex. For a silicon linearly graded junction with an
impurity gradient of 1020 cm-4, the depletionlayer width is 0.5 μm. Calculate the maximum
field and built-in voltage.
Linearly Graded Junction
Soln
1.6  10  10   0.5  10 
eaW
3
Em 


4.75

10
V/cm
14
8
8  11.9  8.85  10
 1020  0.5  104 
2kT  aW 
Vbi 
 0.645 V

  2  0.0259 
9 
e  2ni 
 2  9.65  10 
2
19
20
4 2
Note: Practically, the Vbi is smaller than that
calculated from (22) by about 0.05 to 0.1 V.