SEMICONDUCTOR DEVICES
p-n junction diode

Basic Structure of p-n Junction

Zero Applied Bias
o
o
o
o
o

Reverse Applied Bias
o
o
o
o

Energy band diagram
Built – in potential barrier
Electric field
Potential
Space charge width
Energy band diagram
Space charge width
Electric field
Junction capacitance
Forward Biasing of p-n junction
o
o
o
o
o
o
o
o
o
Boundary condition
Minority carrier distribution
Ideal p-n junction current
Temperature effect of p-n junction diode current
Diffusion resistance
Diffusion capacitance
Small signal Equivalent circuit
Generation Current
Recombination Current

Formula to remember

Numerical

Question Bank
P-N Junction Diode
P – N Junction( No Biasing)
Basic structure of the P-N junction
1.
p-n junction is made on the single piece of semiconductor.
2.
The interface separating the n and p regions
is referred to as the metallurgical junction.
3.
In case of a step junction, doping
concentration is uniformly distributed in
each region and there is an abrupt change in
doping at the junction.
4.
Initially at the metallurgical junction, there is
a very large density gradient in both the electron and hole concentrations.
5.
Electron of n – type side diffuse into p – type side. Holes of p –type side diffuse
into n – type side.
6.
As one electron diffuse from n – region, a positively charged donor atom left
behind near the junction. These are immobile.
7.
Similarly negatively charged acceptor atoms are formed in the junction in
type side. These are also immobile.
8.
These immobile charges are called space charge region or depletion region.
9.
Due to the presence of these immobile charges near the junction, an induced
electrical field is created in a direction from n to p region.
10.
Density gradient still exist in the majority carrier concentration at the edge of the
space charge region. This density gradient provides a diffusion force that acts on
the majority carriers.
11.
At thermal equilibrium
12.
Force due to space charge = Diffusion
force
13.
The space charge region produced due to
the presence of immobile charge acts as a
fictitious battery whose + ve terminal is
connected to
N – type side and – ve
terminal connected to
P – type side.
The corresponding potential is called
built – in potential barrier and is denoted
Vbi.
14.
Due to the potential barrier the conduction band of N- region is lower by eVbi
with respect the conduction band of P – region.
15.
This potential difference across the junction can not be measured with a
voltmeter. This potential Vbi maintains equilibrium, so this voltage produces no
current
16.
Energy band diagram during no biasing.
p-
2|P age
Dr. RajanikantaParida/Phy/ Semiconductor Devices /ITER
P-N Junction Diode
Built – in Potential Barrier
During no biasing of p – n junction, a potential is developed across the metallurgical
junction due to the presence of the space charge near the junction. This potential is
called built – in potential Barrier. It is denoted by Vbi.
Due to this built – in potential barrier the conduction band of n – region is lowered by
eVbi with respect to the conduction band of p- region and the Fermi level gets disturbed.
The energy band diagram of a p – n junction in thermal equilibrium at zero baising is
shown below.
In the figure
eVbi =
=
Conduction band energy difference between p and n - region
Intrinsic Fermi energy difference between p and n - region
=
 E Fi P   E Fi n
Thus,
eVbi = eφ FP  eφ Fn

Vbi =
φ FP + φ Fn
----------------- (1)
From the figure we have,
and
eφ Fn = E Fn - E Fi
For n – region
-------------------(2)
eφ FP = E Fi - E Fp
For p – region
-------------------(3)
where φFn and φFP are the potential.
We know that
n 0  Nc e
=>
=>
n 0  Nc e
 E  E Fn 
 c

 kT 
 E  E Fn 
 c


kT


n0  ni e
 E Fn  E Fi 


kT


.e
 E  E Fi 
  Fn


kT


Where ni = intrinsic carrier concentration
EFi = intrinsic Fermi energy
3|P age
P-N Junction Diode
 eφ Fn

 kT



=>
n 0  ni e
=>

n0
 e
ni
=>
n
ln  0
 ni
 eφ Fn

kT

=>
φ Fn 
n 
kT
 ln  0 
e
 ni 
=>
φ Fn 
N 
kT
 ln  d 
e
 ni 
 eφ Fn 

kT 
-------------------------- (4)
where no = Nd = Net donor concentration
For p-region we have,
p0  N a = n i e
 E Fi  E Fp

kT




eφ Fp
=>
Na = n i e
=>
Na
= e
ni
=>
ln
=>
 kT 
φFp =  
 e 
kT
eφ Fp
kT
Na
=
ni
e φ Fp
kT
N 
ln  a 
 ni 
------------------------------(5)
Now from equation (1) we have,
Vbi 
φ Fn
+ φ Fp
=>
 kT  N
N 
Vbi    ln d  ln a  
n i  
 e  n i
=>
Vbi 
=>
N N 
Vbi  Vt ln  a 2 d 
 ni 
kT  N a N d 
ln 

e  ni2 
------------------------------- (6)
kT
, thermal voltage.
e
Equation (6) represents the built – in – potential. It depends on the Na and Nd.
where, Vt 
Electric Field in the space charge region
Near the metallurgical junction space charge is created due to the presence of positively
charged immobile donor ions in the n-region and negatively charged immobile acceptor
ions in the p – region. These positive and negative charges create an electric field (E).
4|P age
P-N Junction Diode
p
Let
W
= Space charge width
n
Wp
Wn
Wn = Width of the neutral part of n-region
Wp
= Width of the neutral part of p region
Let the p-n junction is uniformly doped and let us
assume the abrupt junction approximation. Then the
variation of space charge density with distance is as
shown in the graph given below.
x = - xp
x = 0 x = xn
Wp
 = Volume charge density
Let
Here it is assumed that the space charge region abruptly
ends in n -region at x = + xn and p-region at x = -xp
Here
 = - e Na
-xp < x < 0
and
=
0 < x < xn
e Nd
From Poisson’s equation we have,
dE  x  ρ  x 
=
, ε s  Permitivity of semiconductor
dx
εs
dE  x  =

1
ρ  x  dx
εs
----------------------------- (1)
Following assumptions are made :
1.
Electric field is zero in the neutral
current flows in thermal equilibrium)
p – region for x < - xp ( Since no
2.
Electric field is a continuous function ( since there is no surface charge
density within the p-n junction structure)
3.
E=0
at
x = xp
E=0
at
x = xn
Now integration (1) and applying the above assumptions we can get Electrical field for
p and n-region
Electric field in p – region is
 dE x 


At
Ep = 
e N
a
E p = - 
ε
 s
x = -xp
= 
1
ρ  x  dx
εs


1
- e Na dx
εs

x+C
1


E =0
------------------------------- (2)
 e Na 
Thus, 0  - 
   x p   C1
 s 
5|P age
P-N Junction Diode
 e Na 
C1  - 
 xp
 s 
Putting this value in equation (2) we get


Ep
eN
a
- 
ε
 s


 x -  e Na

 ε

 s

x
 p

 eN a
 x + xp 
s
Electric field in n – region is
1
 dE  x  =  εs ρ  x  dx

Ep 

En =
1
 ε  e N  dx
d
for  x p  x  0

s
At
----------------(3)
 e Nd 
E = 
 x + C 2 --------------- (4)
 εs 
x = xn
E =0
 e Nd 
We have, 0 = 
 x n + C2
 εs 
 e Nd 
C2 = - 
 xn
 εs 
Putting this value in equation (4) we get
 e Nd 
 e Nd 
E = 
 x -
 xn
 εs 
 εs 
 eN d
En 
 xn  x 

s
At x = 0 , Electric field is continuous
for 0  x  x n
So,
Ep = En

 - eN a

 - eN d

x + x p  = 
 x n -x  


 εs
 x=0  ε s
 x=0

Na x p  Nd x n
--------------(5)
at the metallurgical junction
------------------ (4)
 No. of -ve charge per   No. of +ve charge per 



 unit area in p-region   unit area in n-region 
Now the variation of electric field (E) with distance inside the space charge region
(depletion region) for a uniformly doped p-n junction is shown in the figure.

From the figure we have



E – field is a linear function of distance through the junction.
Emax occurs at x = 0 (i.e at metallurgical junction)
E – field exists in the space charge region even no voltage is applied
between p and n region.
6|P age
P-N Junction Diode
Potential across the junction
p – region
We know that
dφ
dx
d φ = - Edx
E = -

Where,  = potential difference
And
E = electric field in the space charge region
φ p = -  E p dx

 eN a 
  x + x p  dx
εs 

φp = -
  

φp =
eN a
 x dx +

εs

φp =

eN a  x 2
+ x p .x   C1

εs
 2

At
x = -xp
x
- eN a
x + xp  ]
εs
dx 

------------------------------------(1)
 =0
eN a  x p

+ x p .  -x p 
εs
 2
2

 eN a   x p
2
- xp 
C1 = - 
 

 ε s   2
2
Thus, 0 =

p
[ Since E p 

  C1

2
 eN a  x p
C1 = 


 εs  2
Putting this value in equation in equation (1) we get
2

  eN a  x p
eN a  x 2
φp =
+ x p .x   


εs
 2
  s  2
xp
eN a  x 2

φp =
+ x p .x+
εs
2
 2
2
eN a  2
φp =
x + 2x p .x  x p

2ε s
2



φp =
eN a
2ε s
x + x 





2
p
for
 x p  x  0 -------------------(2)
n – region
We know that
E = -
dφ
dx
d φ = - Edx

Where,  = potential difference
7|P age
P-N Junction Diode
and
E = electric field in the space charge region

φn = -
E
n
dx
 eN d 
  x n -x  dx
εs 
eN d
 x n dx - x dx 
φn =



εs
φn =


φn =

 
eN d 
x2
x
.x
 n
εs
2

[ Since E n 

 + C2

- eN d
xn  x  ]
εs
----------------------------------(3)
As potential is a continuous function we have
Eqn (2) = Eqn (3)
at x = 0
eN a
2ε s

x + x 
p
2
=
eN d 
x2
x
.x
 n
εs
2


 + C2

at
x=0
eN a 2
xp = C2
-----------------------------------(4)
2ε s
Putting the value of equation (4) in equation (3) we get
eN a 2
eN d 
x2 
φn =
x
.x
xp
for 0  x  x n ----------(5)
 n
+
2ε s
εs
2 


The variation of potential through the junction with the distance is shown in the figure
given below
From the figure it is clear that
 Potential through the junction is a
quadratic function of the distance

Energy = - e 
As  is a quadratic function of
distance, the energy is also a quadratic function of distance. It is shown in the
energy band diagram.

At x = xn
 = Vbi
Where Vbi = built-in-potential barrier
eN d 
x 2n  eN a 2
xp
 x n .x n +
εs 
2  2ε s
2
e 
Vbi =
N d x 2n + N a x p 

-----------------------(6)


2ε s
Equation (6) is another form of the built-in-potential barrier.

Thus Vbi =
8|P age
P-N Junction Diode
Space Charge Width
The positive donor immobile ion in the n-region and negative acceptor immobile ion in
p-region near the metallurgical junction are called space charge region or depletion
region.
xn = space charge width in n-region
Let
And xp = space charge width in p-region
Then the space charge width (W) is given by
W
=
xn + xp
--------------------------(1)
We know that
Na x p = Nd x n
N 
xp   d  xn
 Na 
Again the built-in-potential barrier is given by

Vbi =
2
e 
N d x n2 + N a x p 

2ε s 
--------------------------(2)
---------------------------(3)
Putting the value of equation (2) in equation (3) we get
Vbi =
2
e 
N d x n2 + N a x p 


2ε s

N d2 2 
e 
2
Vbi =
 Nd x n + Na 2 x n 
2ε s 
Na 

Vbi =

Nd 
e
N d x 2n  1 +

2ε s
Na 


Vbi =
 N + Nd 
e
N d x 2n  a

2ε s
Na 


 2ε
x 2n =  s
 e

 2ε   N  


1
x n =  s   a  
 Vbi 
 e   N d   N a + N d 


1
  N a 
  N  N + N  Vbi
  d  a
d 
1
2
-------------------------(4)
From equation (2) we have
N 
xp =  d  xn
 Na 


 2ε   N  


1
x p =  s   a  
 Vbi 

 e   N d   N a + N d 
1
2
 2ε   N 


1
x p =  s   d 
 Vbi 
 e   N a  N a + N d 

We know that
W
=
 Nd 


 Na 
1
2
------------------------(5)
xn + xp
9|P age
P-N Junction Diode



1

 2ε V
W =  s bi
 e
1

 2ε s Vbi  

1
W = 



 e   N a + N d  

 2ε V
W =  s bi
 e

1

 N + N 
 a
d 
  N a +Nd  
 N N  
  a d  
1
2
2
 N  1 2  N  1 2 
 a  +  d  
 N d 
 N a  

 N +N
 a1 1 d
 N a 2 N d 2



2
-------------------------(6)
Equation (6) represents the space charge width of the p-n junction at zero biasing.
p-n Junction with Reverse Biasing
During reverse biasing +ve terminal of the source is connected to the n – type side and –
ve terminal is connected to the p – type side. As a result of which space charge region is
increased, potential barrier at the junction is increased, the energy band diagram
changes accordingly.
 Energy band diagram for reverse biasing:
The energy band diagram for reverse
biasing is shown below
Let VR = Reverse biased voltage applied
to the p-n junction
Vtotal = Fn  Fp + VR

Vtotal = Vbi + VR
Where Vbi = built-in potential barrier
 Space charge width for reverse biasing:
Due to the reverse bias voltage (VR) applied to the p-n junction, the space charge
width can be obtained by replacing Vbi by (Vbi + VR) and is given by
 2
W =  s V V
bi
R
 e



 N N
d
 a
 N N
 a d




1
 2



From the above equation it is clear that by increasing VR the space charge width
(W) increases
 Electrical field
Maximum electric field (Emax) is obtained at the metallurgical junction. It is given
by
 eNa x p 
 eN x n 
E max = -  d
=



 εs 
 εs 
Emax can also be written in the form of total space charge width (W)
We know that
 eN 
dx
 εs  n


Emax = - 
10 | P a g e
P-N Junction Diode
 eN
d
E
=-
max
 ε
 s
E
m ax
=

- 
 




2e
ε
s
  2ε   N
s  a

  e   N d




N
N
a
m ax

= -
N
a d
+ N
-2
E

1

 N + N
d
 a
d




1
2

V
bi
+ V

R

1
 2


1
2
Vb i + VR 
 2 e  N + N
s 
a
d


e
N
N
 

a d
E max  





 V +V
R
 bi


 Vb i + VR



2  Vbi  VR





1
2
W
11 | P a g e
P-N Junction Diode
ε
Graphical representation
During reverse biasing the space charge width is
given by
 2ε
 N +N d  
W =  s  Vbi + VR   a

 N a N d  
 e
1
2
P – N Junction Forward biasing
1.
If p-side is connected to the +ve terminal of the battery and n- side is connected to
the -ve terminal of the battery then p-n junction is said to be forward biased.
2.
Applied electrical field is in opposite direction
to the electrical field due the space charge
region.
3.
As a result of which the barrier potential
decreases to ( Vbi - Va ). Where Va = applied
potential.
4.
The Fermi energy level in n-region will be at
higher level compared to the Fermi energy level
in p-region
5.
The space charge width (W) decreases.
6.
Electron of n – type side diffuse into p – type side. Holes of p –type side diffuse
into n – type side.
7.
As electron of n-region diffuses into p-region it became minority charge carrier
for p-region. And as hole of p-region diffuses into n-region it became minority
charge carrier for n-region.
Therefore Ambipolar transport equation is required to study the behaviour of these
minority carriers. Thus by applying forward bias, excess minority carriers are
created in each region of the p-n junction.
8.
As the diffusion of carrier takes place across the junction means there exists a
diffusion current.
12 | P a g e
P-N Junction Diode
Energy band diagram:
Energy band diagram for pn-junction during forward is shown in the figure
Boundary condition
During thermal equilibrium at zero biasing the built-in potential barrier maintains
equilibrium between the carrier distributions on either side of the junction.
It is given by
Vbi 
 N N 
kT
ln  a 2 d 
e
 ni


eV b i
kT
 N N 
= ln  a 2 d 
 ni


Na Nd
n i2
 eV 
= exp  bi 
 kT 

ni2
Na Nd
 - eVbi 
= exp 

 kT 
--------------------------------(1)
Let
nn0 = thermal equilibrium concentration of majority carrier electron in n-region
And
np0 = thermal equilibrium concentration of minority carrier electron in p-region
Then we can write
ni 2
Na
Putting these values in equation (1) we get
n n0  Nd
n p0
n n0

n p0
and
np0 
-------------------------------(2)
 - eVbi 
= exp 

 kT 
 - eVbi 
= n n0 exp 

 kT 
-------------------------------(3)
During forward biasing the voltage across the p-n junction is
13 | P a g e
P-N Junction Diode
Vtotal
 Vbi  Va
Where
Va  Forward voltage
And equation (3) can be written as
 - e (Vbi  Va ) 
n p  n n0 exp 

kT


Where n p  Total minority carrier concentration of electron in p-region
 - e Vbi 
e V 
exp  a 
 n n0 exp 

 kT 
 kT 
e V 
n p  n p0 exp  a 

------------------------------(4)
 kT 
Equation (4) represents the expression for minority carrier electron concentration at the
edge of the space charge region in p-region ( x = - xp )
Similarly, minority carrier hole concentration pn at the edge of the space charge region
in n-region ( x = xn ) is given by
e V 
p n  p n0 exp  a 
-----------(5)
 kT 
np

Graphical representation for excess minority carrier concentration at the edge of the
space charge, generated due to forward bias is shown in the figure.
Minority carrier distribution
During forward biasing electron of n-region
diffuses into p-region and it becomes minority
charge carrier for p-region. Similarly hole of
p-region diffuses into n-region and it becomes
minority charge carrier for n-region.
Therefore Ambipolar transport equation is
required to study the behaviour of these
minority carrier. Thus by applying forward
bias, excess minority carriers are created in
each region of the p-n junction.
Ambipolar transport equation for excess minority carrier hole in an n-region is
  p n 
  p n 
 2 ( p n )
 pn
---(1)
 Dp
 μp E
+ g/ 
2
t
x
x
τ p0
Where
 p n  pn  pno
p n = T otal m inority carrier concentration o f holes in n-region
14 | P a g e
P-N Junction Diode
p no = Thermal equilibrium concentration of minority carrier hole in n-region
In neutral part of n-region we have
x > xn
E = 0
g/ = 0
  δp n 
t
( assuming no generation )
 0
( assuming steady state )
Then equation (1) becomes
 2 (δp n ) δp n
Dp

 0
x 2
τ p0


 2 ( p n )
 pn

 0
2
 p0 Dp
x
 2 ( pn )  pn
 2  0
Lp
x 2
-----------------------(2)
L2p = τ p0 Dp
Where
-----------------------(3)
Lp = Distance within which the minority carrier hole diffuses inside the
neutral part of the n-region
= Minority carrier diffusion length for hole
In neutral part of p-region we have
x  xp ,
E=0,
g/ = 0
( assuming no generation )
  n p 
 0
t
Then we have
Dn


 2 ( n p )
x
2
 2 ( n p )
x
2
 2 ( n p )
x 2


( assuming steady state )

 np
 n0
 np
 n0 Dn
 np
L2n
 0
 0
 0
-----------------(4)
L2n = D n τ n0
-----------------(5)
Ln = Minority carrier diffusion length for electron
General solution for equation (2) is
Where
x
-x
δp n = A e p + B e p
General solution for equation (4) is
 np  C e
x
L
L
------------------(6)
x
 D e Ln
------------------(7)
The values of A, B, C and D can be obtained by using the boundary condition.
The boundary conditions are:Ln
15 | P a g e
P-N Junction Diode
e V 
 p n0 exp  a 
 kT 
e V 
 n p0 exp  a 
 kT 
i)
pn
ii)
np
iii)
iv)
p n ( x   )  p n0
n p ( x   )  n p0
( for long diode, i.e
( for long diode, i.e
Wn >> Lp )
Wp >> Ln )
Putting the boundary conditions (iii) and (iv) in equations (6) and (7) respectively we
get
A = 0 and D = 0
So, equations (6) and (7) become
δp n = B e
-x
Lp
----------------------(8)
x
And
 np  C e L
From the boundary condition (i) we have
e V 
 p n  x  xn  p n0 exp  a 
 kT 
----------------------(9)
n
e V 
 p n0 exp  a 
 kT 

eV  

 δpn  x  xn  pn0 exp  a   1
 kT  

n
At x = xn, eq (8) become
x 
 δpn  x  xn = B exp  n 
 Lp 

Now
 δp n
+ p n0  x  x
n
---------------------(11)
eqn (10) = eqn (11)
x
B exp  n

 Lp

--------------------(10)
B  p n0


 e Va  
  p n0 exp 
  1
 kT  


 xn 

 e Va  
 exp  kT   1 exp  L 

 

 p 
---------------------(12)
Putting this value in equation (8) we get
 x 
δp n = B exp  
 L p 



δp n =
x -x

 e Va  
p n0  exp 
- 1 exp  n

 L
 kT  

 p



---------------------(13)
Similarly eqn (9) can be written as

eV
 n p  n po exp  a
 kT

 xp  x 
 
  1 exp  L 
 
n


-------------(14)
16 | P a g e
P-N Junction Diode
Graphical representation of equations (13) and (14) is
shown below
From the graph it is clear that

The graph shows the steady – state minority
carrier concentrations in a p-n junction under
forward biasing.

Minority carrier concentrations decay
exponentially with distance away from the
junction and reach the thermal equilibrium
values.
Ideal p-n junction current
Assumption made to derive current – voltage relationship is given below:_
1. The abrupt depletion layer approximation applies. The space charge regions
have abrupt boundaries and the semiconductor is neutral outside of the depletion
layer.
2. The Maxwell – Boltzmann approximation applies to carrier statistics.
3. The concept of low injection applies.
4. (i)
(ii)
The total current is constant throughout the entire p-n junction.
The individual electron and hole currents are continuous function
through the p-n junction.
The individual electron and hole currents are constant throughout the
depletion layer.
Let JP (xn) = Minority carrier hole diffusion current density at x = xn
d
J p (x n ) = - e D p
p n (x)
at x=x n
dx
d

----------------(1)
at x = x n
J p (x n ) = - e D p
 δp n (x) 
dx
Since p no is constant we have

d

 p n  x  = d  p no + δp n  x   = d δp n  x  
 dx

dx
dx
But we know that the excess minority carrier hole is
x -x

 eV  
δp n  x  = p no  exp  a  - 1 exp  n
 L 
 kT  

p


(iii)

 x - x   -1

d
 eV  
δp n  x   = p no  exp  a  - 1 exp  n
.

 L  L
dx
 kT  

p
p


At x = xn we have
-p
d
δpn  x   = no

dx
Lp

 eVa
exp  kT


 
 - 1 ------------- (2)
 
Putting the value of equation (2) in equation (1)
17 | P a g e
P-N Junction Diode
e Dp p no 
 eV
exp  a

Lp 
 kT
J p (x n ) =
 
 -1
 
--------------- (3)
Similarly the minority carrier electron diffusion current density at x = - xp is given by
J n (-x p ) =
e Dn n po 
 eV
exp  a

Ln
 kT

 
 -1
 
---------------- (4)
As the individual electron and hole currents were continuous functions and constant
through the space charge region, we have
 Total current 

 =
 density 
 electron diffusion 


 current density 

J = J p (x n ) + J n (-x p )

 e D p p no
e D n n po  
 eVa
J= 
+
 exp 

Lp
Ln
 kT




 eV  
J = J s exp  a  -1
 kT  

Where, J s =
e D p p no
Lp
+
+
 hole diffusion 


 current density 
 
 -1
 
--------------- (5)
--------------- (6)
e D n n po
---------------- (7)
Ln
Equation (5) and (6) are the ideal current – voltage
relationship of a p-n junction. This is also called as ideal
– diode equation.
Equation (7) is called reverse saturation current density.
The variation of forward current density with applied
forward voltage is shown in the figure.
If Va >> kT/e , then equation (6) will be
 eVa 
J = Js exp 

 kT 
eVa
kT

ln J = ln J s +

 e 
ln J = 
 V a + ln J s
 kT 
----------------(8)
The ideal current – a voltage characteristic of a p-n
junction diode with current is plotted on a log scale and
is shown in the figure.

ln Js
log(J) ~ Va graph is a straight when Va is greater than a few
kT
e
volt and –1
term in equation(6) can be neglected.

Forward bias current is an exponential function of the forward bias voltage.

As we have assumed the electric field to be zero at the space charge edge, we
have neglected minority carrier drift current.
18 | P a g e
P-N Junction Diode
Different form of Js
We know that the reverse bias saturation current density is
e D p p no
e D n n po
---------------------(i)
Js 

Lp
Ln
But we know that
n i2
n i2
and
L n = Dn τ no
p no =
, n po =
,
Lp = D p τ po
Nd
Na
Putting these values in equation (i) we get
e D p n i2
e D n n i2
Js =
+
D p τ po N d
D n τ no N a

Js
=
e
Dp
τ po
n i2
Nd
+
e
Dn
τ no
n i2
Na
 1 Dp
1 Dn 
= e n i2 
+
-------------------(ii)

N a τ no 
 N d τ po
This equation represents the reverse bias saturation current density in terms of minority
carrier lifetime.

Js
Temperature Effect of Current – Voltage relationship during Forward Biasing
The Forward bias current is given by
ID = A Jf
Where
A = Area of cross-section
J f  Current density

 eV  
I D = A J s exp  a  -1
 kT  


 eV  
I D = Is exp  a  -1
 kT  



--------------------(1)
Where, J s  Reverse saturation current density
e D p p no
e D n n po
Js =
+
Lp
Ln
------------------- (2)
From equation (2) it is clear that

Js depends on pno and npo ( thermal equilibrium minority charge carrier
concentration). We know that pno and npo depends on temperature. So Js
depends on the temperature.

Equation (1) contains the term
 eV 
exp  a  .
 kT 
So ID is function temperature. As
temperature increases the diode current also increases.

The factor Js is more sensitive to temperature than
 eV 
exp  a  .
 kT 
19 | P a g e
P-N Junction Diode
Formula to remember
1.
p-n junction with no biasing

N N 
Vbi  Vt ln  a 2 d 
 ni 
Where Nd = Net donor concentration
Na = Net acceptor concentration
kT
e
Vt 
, thermal voltage

Ep 
 eN a
 x + xp 
s
for
 xp  x  0
Ep = Electric field in p – region

En 
 eN d
 xn  x 
s
for
0  x  xn
En = Electric field in n – region

Na xp  Nd xn

p =
eN a
2 s
 No. of -ve charge per   No. of +ve charge per 



 unit area in p-region   unit area in n-region 

x + x 
2
for
p
 xp  x  0
p = Potential in p – region

n =
eN d
s

x2
 x n .x 2

2
eN a

xp

2 s

for 0  x  x n
n = Potential in n – region

Vbi =
2
e 
N d x n2 + N a x p 


2 s


1
  Na  
  N   N + N  Vbi 
 d  a

d 
1

  2 s
x n = 
 e
1

  2 s   N d  


1
x p = 


 Vbi 


 e   N a   N a + N d 

 2 s Vbi
W = 
 e
2.
  N a N d  
 N N  
  a d  
1
2
2
2
 Space charge width in the n-region
 Space charge width in the p-region

Total Space charge width
p-n junction with Reverse biasing
 2
 N N d  
W =  s  Vbi  VR   a

 e
 N a N d  
1
2

Space charge width

Max. electric field



e s N a N d
Capacitance at metallurgical junction C  

 2  Vbi  VR   N a + N d  
3..
2  Vbi  VR
 eNa x p 
 eN x 
E max    d n    
 
W
 s 
 s 
1

2
p-n junction with forward biasing

 - eVbi 
n p0  n n0 exp 

 kT 
nn0 = thermal equilibrium concentration of majority carrier electron in n-region
And np0 = thermal equilibrium concentration of minority carrier electron in p-region
24 | P a g e
P-N Junction Diode

The boundary conditions are:i)

e V 
p n  p n0 exp  a 
 kT 
iii)
p n  x     p n0
iv)
n p  x     n p0
Minority carrier distribution
 pn

 xp  x 
 

  1 exp 
 
 Ln 
The graph for the steady – state minority
carrier concentrations in a p-n junction
under forward biasing.
Ideal current – voltage relationship of a pn junction

 eV  
J = J s exp  a   1
 kT  


( for long diode, i.e Wn >> Lp )
( for long diode, i.e Wp >> Ln )
 xn x 

 e Va  
 p n0 exp 

  1 exp 

 kT  

 Lp 
 e V
 n p  n po exp  a
 kT

e V 
n p  n p0 exp  a 
 kT 
ii)
 1
J s = e n i2 
 N d
Dp
τ po
Where
+
1
Na
Dn
τ no
Js 
e D p pno
Lp

e D n n po
Ln



no

25 | P a g e
Dr. RajanikantaParida/Phy/ Semiconductor Devices /ITER
P-N Junction Diode
Numericals
1.
Calculate the built-in potential barrier in a silicon pn junction at T = 300 K for
N a = 5× 1017 cm -3 , N d = 2 × 1016 cm -3 and n i = 1.5× 1010 cm -3 .
N N 
Vbi = kT ln  a 2 d 
 ni 
 (5 x 1017 ) x (2 x 1016 ) 
= 0.0259 ln 
 = 0.796 V
(1.5 x 1010 ) 2


Answer:
2.
A silicon pn junction at T = 300 K with zero applied bias has doping concentration
of N a = 5× 1016 cm -3 and N d = 5 × 1015 cm -3 . Determine x a , x p , W and E max .
Answer:
N N 
Vbi = kT ln  a 2 d 
 ni 
 (5 x 1016 ) x (5 x 1015 ) 
= 0.0259 ln 
 = 0.718 V
(1.5 x 1010 ) 2


  2 s
x n = 
 e


1
  Na  
  N   N + N  Vbi 
 d  a

d 
1
2
 2(11.7) x ( 8.85 x 10-14 )   ( 5 x 1015 )  


1
x n = 
( 0.718) 

-19
16  
15
16 
( 1.6 x 10 )
  ( 5 x 10 )   ( 5 x 10 ) + ( 5 x 10 ) 


1
2
x n = 4.11x10-6 cm
 Nd 
-5

 x n = 4.11 x 10 cm
 Na 
W = x n + x p = 4.52 x 10 -5 cm
xp =
E max =
3.
e Nd x n
= 3.18 x 104 V/cm
εs
A silicon p-n junction at T = 300 K is reversed biased at VR = 8.0V. The doping
concentrations are
and
Determine
N a = 5× 1016 cm -3
N d = 5 × 1015 cm -3 .
x a , x p , W and E max .
Answer:
N N 
 (5 x 1016 ) x (5 x 1015 ) 
Vbi = kT ln  a 2 d  = 0.0259 ln 
 = 0.718 V
(1.5 x 1010 ) 2


 ni 
  2 s   N a  


1
x n = 

 (Vbi  VR ) 

 e   N d   N a + N d 

1
2
 2(11.7) x ( 8.85 x 10-14 )   ( 5 x 1015 )  


1
x n = 
( 0.718 + 8.0) 

-19
16  
15
16 
( 1.6 x 10 )
  ( 5 x 10 )   ( 5 x 10 ) + ( 5 x 10 ) 


1
2
x n = 1.43 x10-4 cm

-5
 x n = 1.43 x 10 cm

W = x n + x p = 1.57 x 10 -4 cm
xp =
E max =
 Nd

 Na
2( Vbi + VR ) 2( 0.718 + 8.0)
=
 1.11 x 105 V/cm
4
W
1.57 x 10
26 | P a g e
P-N Junction Diode
-3
5.
An abrupt p-n junction at zero bias has dopant concentrations of N a = 1017 cm -3 and
N d = 5 × 1015 cm -3 at T = 300 K.
a) Calculate the Fermi level on each side of the junction with respect to the
intrinsic Fermi level.
b) Sketch the equilibrium energy band diagram for the junction and determine
Vbi from the diagram and the result of part (a).
c) Calculate Vbi from the equation and compare the result to part (b)
d) Determine x a , x p , and E max .
Answer:
n-region:
a)
p-region:
Vbi = 0.3294 + 0.4070 = 0.7364 volts
c)
N N 
Vbi = kT ln  a 2 d   0.7363 volts
 ni 
d)
  2 s
x n = 
 e
E max
Ans.:
EF
b)
xp =
6.
E Fi -
 N 
E F i = k T ln  d   0 .3 2 9 4 e V
 ni 
 N 
= k T ln  a   0 .4 0 7 0 e V
 ni 
EF -


1
  Na  

 (Vbi ) 

  Nd   Na + Nd 

 Nd

 Na
e
=
1
2
 0.426 μm

 x n = 0.0213 μm

Nd x n
= 3.29 x 104 V/cm
εs
A silicon abrupt junction in thermal equilibrium at T = 300 K is doped such that
pE c - E F = 0 .2 1 e V in the n-region and E F - E v = 0 .1 8 e V in the
19
-3
19
-3
region. N c = 2.8x 10 cm and N v = 1.04 × 10 cm at T = 300 K.
a) Determine the impurity doping concentrations in each region
b) Determine Vbi
   Ec  EF  
n 0 = N n exp 

kT



0.20


15
-3
n 0 = 2.8 x 1019 exp 
  8.43 x 10 cm
0.0259


15
-3
n 0 = N d = 8.43 x 10 cm
a)
  EF  Ev  
p 0 = N v exp 

kT


27 | P a g e
P-N Junction Diode
 0.18 
15
-3
p0 = 1.04 x 1019 exp 
  9.97 x 10 cm
 0.0259 
N N 
Vbi = kT ln  a 2 d   0.690 volts
 ni 
b)
7.
Consider the uniformly doped GaAs junction at T = 300 K. At zero bias only 20%
of the total space charge is to be in the p-region. The built –in potential barrier is
Vbi = 1.20 Volts. Determine N a , N d , x a , x p , and E max .
Ans. : a)
Ga As : Vbi = 1.20 volts
ni = 1.8 × 106 cm -3
x p = 0.2 W = 0.2 (x n + x p ) = 0.2x n + 0.2 x p
 x p Na = Nd x n
xp

0.8 x p =0.2 x n
xn
xp

xn
=
=
1
=0.25
4
Na
=0.25
Nd
kT  N a N d 
kT N a  0.25N a 
ln 
ln
 =
2
e
e
n i2
 ni 
0.25N 2a  kT


 N a2  
kT
1.20 =
ln
ln
0.25


 2 
e
e
n i2
 n i  


 N2 
 N2 
1.20
1.20
= ln 0.25  2a  
)
 0.25  2a  = exp (
0.0259
0.0259

 n i  
 ni 
ni
1.20
Na =
exp (
)  4.14 x 1016 cm-3
0.25
2 x 0.0259
b) Na = 0.25 Nd  1.04 x 1016 cm-3
Vbi =
c) Calculate x a , x p , and E max using the formula.
8.
A silicon p-n junction at 300 K is doped with impurity concentration of
Na=5x016cm-3 and Nd = 2 x 1016 cm-3. The junction is forward biased at Va =
0.610V. Determine the minority carrier concentrations at the edge of the space
charge region.
Ans: p
n0
=
n i 2 ( 1.5 x 1010 ) 2

 1.125 x 10 4 cm -3
Nd
5 x 1016
Then p n (x n ) = p n0 exp (
Again,
n p0 =
Va
0.610
) =  4.5 x 103  exp (
) = 1.90 x 1014 cm-3
Vt
0.0259
n i 2 ( 1.5 x 1010 ) 2

 4.5 x 103 cm -3
16
Na
2 x 10
Then n p (-x p ) = n p0 exp (
9.
Va
0.610
) = 1.125 x 104  exp (
) = 7.62 x 1013 cm -3
Vt
0.0259
The impurity doping concentration in a Silicon p-n junction at T = 300 K are Na =
5 x 1016 cm-3, Nd = 5 x 1015 cm-3. The minority carrier concentration at either
space charge edge is to be no larger than 10% of the respective majority carrier
28 | P a g e
P-N Junction Diode
concentration. Calculate the forward bias voltage that can be applied to this
junction and still meet the required specification.
Ans.:
The maximum carrier concentration will occur on the low – doped side.
p n (max) = 10% of N d = 5 x 1014 cm -3
p n0 =
n i 2 ( 1.5 x 1010 )2

 4.5 x 104 cm -3
Nd
5 x 1015
Then p n = p n0 exp (

10.
 5 x 1014 =  4.5 x 10 4  exp (
Va
)
Vt
Va
)
0.0259
Va = 0.599 Volts
A silicon p-n junction at 300 K has the following parameters: N a = 5 x 1016 cm -3 ,
D p = 10 cm 2 / s ,  n0 = 5 x 10 -7 s ,  p0 = 1 x 10-7 s ,
N d = 1 x 1016 cm -3 ,
D n = 25 cm 2 / s ,
A = 10-3cm2 , Va = 0.625Volts . Calculate the
a) Minority electron diffusion current at the edge of the space charge.
b) Minority hole diffusion current at the edge of the space charge.
c) Total current in the p-n junction diode.
Ans.: a)
11.
Ans.:
I n = A J sn exp (
Va
)
Vt

In = A
eD n n p0
Ln
exp (
b) I n =
Aen i2
Na
Dn
V
exp ( a )  1.54 x 10-4 Amp = 0.154 mA
 n0
Vt
Ip =
Aen i2
Nd
Dp
 p0
exp (
Va
)
Vt
Va
)  1.09 x 10-3 Amp = 1.09 mA
Vt
c) I = I n + I p = 0.154 +1.09 = 1.244 mA
Consider the silicon p-n junction using data given in Q.3. Calculate the electron
and hole current at:
a)
x = xn and x = - xp ( i.e at the edge of the space charge region )
b)
a)
x = xn + Lp
Already solved in Q3. I n = 0.154 mA and I p = 1.09 mA
b)
I p = I p (x = x n ) exp (
I p = 1.09 exp (
xn - x
)
Lp
x n - x n - Lp
Lp
) = 1.09 exp (
-Lp
Lp
) = 1.09 exp (-1) = 0.401mA
I n = I total  I p  (0.154  1.09)  0.401  0.843mA
Consider an ideal p-n junction diode at T 300 K operating in the forward bias
region. Calculate the change in diode voltage that will cause a factor of 10
increase in current.
Ans.:
T = 300 K
12.
If1
= 10
If2

eV1
)
kT = 10
eV
Is exp( 2 )
kT
Is exp(

 e

exp   V1  V2   = 10
 kT

29 | P a g e
P-N Junction Diode
V  V 
e
 V1  V2  = ln 10  1 2 = ln 10
kT
kT/e
  V1  V2  = 0.0259 x ln 10  59.9 mV

13.
Ans.:
Calculate the applied reverse bias voltage at which the ideal reverse current in a pn junction diode at T 300 K reaches 90% of its reverse saturation current value.
VR = ?
I
=  0.90 (reverse bias current is -ve )
Is


V
I = Is exp ( a ) 1
Vt


V
I
= exp ( a ) 1
Is
Vt
V
I
+ 1 = exp ( a )
Is
Vt

 I
 V
ln  + 1  = a
I
 s
 Vt
 Va  0.0259 ln  -0.90 + 1
 I

 Va  Vt ln  + 1 
I
 s

 Va  -59.6 mV


For a silicon p-n junction at T = 300 K, assume  p0 = 0.1 n0 and μ n = 2.4 μ p . The
ratio of electron current crossing the depletion region to the total current is defined
as the electron injection efficiency. Determine the expression for the electron
injection efficiency as a function of : (a) (N a / N d ) (b)
the ratio of n-type
conductivity to p-type conductivity.
Ans.:
T = 300 K
14.
 p0 = 0.1  n0
μ n = 2.4 μ p


 Electron injection 

 =
 efficiency

 Electron injection 

 =
 efficiency

 n0
= 10
 p0
μp
μn
=
Dp
1

2.4
Dn
 Electron current crossing 


 the depletion region

 Total current 
Jn
Jn  Jp
 Electron injection 

 =
 efficiency

1
eD p p n0
1+
=
Lp
eDn n p0
Ln
=
 Electron injection 

 =
 efficiency


1
1
Jp
Jn
1
 en 2
 i
N
1   d


1

1  n0
 p0
Dp
 p0
Dp Na
Dn Nd
en i2
Na
=



Dn 

 n0 

1
1  Na 
1  10


2.4  N d 
=
1
N 
1  2.04  a 
 Nd 
30 | P a g e
P-N Junction Diode
15.
Consider
an
ideal
p-n
junction
diode
with
following
parameters:
 n0 =  p0 = 0.1x 10-6 s , D n = 25 cm 2 / s , D p = 10 cm 2 / s . What must be the ratio of
(N a / N d ) so
that 95% of the current in the depletion region is carried by electron?
eD n n p0
Ln
Jn
= 0.95
Jn  Jp
Ans.:
eD n n p0
Ln
+
= 0.95
eD p p n0
Lp
eD n n p0
Ln
= 0.95
D p p n0 L n
1+
D n n p0 L p



16.

1
3
 Dp  2  N a   n0
1+
 

 Dn   N d   p0
3
 25  2  N 
1+   a  1
 10   N d 
3
2
= 0.95
 D p   Na  Dn  n0
1+


 Dn  N d  D p  p0
= 0.95
3
1
 2.5
1
= 0.95
 Na 
1
0.05
1
-1=
=

 =
0.95
0.95 19
 Nd 

1
 25  2  N 
1+   a  =
0.95
 10   N d 

 1
 Na 
1

 =  
3
 19    2.5  2
 Nd 


 = 0.083


A silicon p-n junction with a cross-sectional area of 10-4 cm2 has following
parameters: N a = 5 x 1015 cm -3 , N d = 1017 cm -3 , n i2 = 1.5 x 1010 cm -3 . Sketch the thermal
equilibrium energy band diagram of the p-n junction, including the values of the
Fermi level with respect to the intrinsic level on each side of the junction.
Ans:
p-side
N 
 5x1015 
E Fi  E F = kT ln  a  = 0.0259 ln 
 0.329 eV
10 
 1.5x10 
 ni 
n-side
N 
 1017
E F  E Fi = kT ln  d  = 0.0259 ln 
10
 1.5x10
 ni 
17.

  0.407 eV

A germanium p+ n diode at T = 300K has the following parameters: N a = 1018 cm -3 ,
-4
2
N d = 1016 cm -3 , D p = 49 cm 2 / s , D n = 100 cm 2 /s ,  n0 =  n0 = 5 x 10 -6 s and A = 10 cm .
Determine the diode current for (a) a forward bias voltage of 0.2 V (b) a reverse
bias voltage of 0.2 V.


V
I = Is exp ( a ) - 1 
Vt


Ans.
For p+ n diode

Is = 10
 D n2 
 eD n n p0 
n
i

Is = A 
 = A  e

τ
N
p0
d 
 Ln 

-4
2

2.4 x 1013 
 1.6 x 10 -19  10 

10-6
1016






I s = 2.91 x 10-9 A
(a) for Va = + 0.2 V
31 | P a g e
P-N Junction Diode
I=
 2.91 x 10 
-9

 0.2  
exp  0.0259  - 1

 


I = 6.55 μA

I =  2.91 x 10 -9 A
(b) for Va = - 0.2 V

 - 0.2  
exp  0.0259  - 1

 

I = - Is   2.91 nA
I=

(c) I n =
 2.91 x 10 
-9
Aen i2
Na
Dn
V
exp ( a )  1.54 x 10-4 Amp = 0.154 mA
 n0
Vt
Question Bank
1.
Which is the metallurgical junction in a p-n junction?
2.
The net +ve and –ve charges in the n & p regions around the metallurgical
junction constitute to form the --------------------
3.
The electric field generated in a p-n junction is always --------------- to that of the
diffusion current.
4.
Define the term contact potential.
5.
Define the term depletion region.
6.
Show the dependence of contact potential on doping concentration for a
semiconductor.
7.
Where does the maximum electric field occur in the space charge region?
8.
Write the expressions for the electric field in the space charge region for p-side
and n-side
9.
Write the boundary condition for the electric field and the potential in the space
charge region.
10.
Does the space charge width in p and n-region same? Justify your answer.
11.
What is the value of electric field and potential at x = -xp and at x = xn?
12.
Plot and explain the behavior of electric field in the depletion region.
13.
What the behavior of electrical field at the metallurgical junction?
14.
Where do you get maximum electric field in the space charge region?
15.
How the built-in- potential- barrier maintains thermal equilibrium?
16.
Why is an electric field formed in the space charge region?
17.
Draw the energy band diagram of a zero bias and reverse biased p-n junction.
18.
Write the expression for the built-in potential barrier voltage.
19.
Draw a curve to show the variation of the potential through the space charge
region with distance for a uniformly doped p-n junction and hence find the builtin potential voltage. How does the potential vary with distance?
32 | P a g e
P-N Junction Diode
20.
Write the expression for the space charge region at zero biasing and reverse
biasing.
21.
What do you mean by junction capacitance?
22.
Express junction capacitance in terms of space charge width.
23.
Why does the potential barrier decreases in a forward biasing?
24.
Describe the mechanism of charge flow across the space charge region of a p-n
junction when forward bias voltage is applied.
25.
Write the boundary conditions for the excess minority carriers in a p-n junction
at the edge of the space charge region under forward bias and under reverse bias.
26.
Sketch the steady – state minority carrier concentrations in a forward biased p-n
junction.
27.
Explain the procedure that is used in deriving the ideal current – voltage
relationship in a p-n junction diode.
28.
Sketch the electron and hole current through a forward biased p-n junction
diode.
29.
What do you mean by carrier injection?
30.
What is meant by short diode?
31.
What is meant by long diode?
32.
Explain the physical mechanism of diffusion capacitance
33.
What is diffusion resistance?
34.
Explain the physical meaning of generation current.
35.
Explain the physical meaning of recombination current.
36.
Describe the Avalanche breakdown mechanism in a p-n junction.
37.
Describe the Zener breakdown mechanism in a p-n junction.
38.
Plot a graph for the electron and hole current components through the space
charge region during avalanche multiplication.
39.
Sketch the energy band diagram for p-n junction at zero biasing, reverse biasing
and forward biasing.
40.
Compare the position of the Fermi energy level for p-region and n-region of a pn junction diode at zero biasing, reverse biasing and forward biasing.
41.
Write the expression for the reverse bias saturation current density in terms of
mean lifetime of minority carriers.
42.
How does the reverse bias saturation current density depend on temperature?
43.
Write the expression for the forward current density when the forward voltage
applied is more than a few kT .
e
33 | P a g e
P-N Junction Diode
Long Questions
1.
Derive built-in potential voltage.
2.
Derive an expression for electrical field in the space charge region
3.
Derive an expression for potential in the space charge region and find an
expression for the built-in potential.
4.
Derive the expression for the space charge width of p-n junction.
5.
Derive the boundary conditions for the excess minority carriers in a p-n junction
at the edge of the space charge region under forward bias and under reverse bias.
6.
Derive the expression for the steady – state minority carrier concentrations in a
forward biased p-n junction.
7.
Derive the ideal current – voltage relationship for a p-n junction diode.
34 | P a g e