Consider the formation of a p-n junction by placing two ndoped and p-doped crystals side-by-side:
 (doping
density)
n-type
p-type
e
Nd
Na
Positive charges occur when donors on
the n-side are ionized during the electron
transfer to the p-side.
0
h+
x
Negative charges form when holes are ionized by the
capture of electrons. Electrons (e-) flow to the p-side
holes (h+) flow to the n-side. Consider a p-n junction in
equilibrium: Assume that non-degenerate conditions hold
for all x.
Before:
EC

EV
There is a transfer of charge between
the n- and p-type regions in order to
equalize the Fermi-level (or chemical
potential) on both sides.
p-side
 N , x  0
N d ( x)   d

 0, x  0 
 0, x  0 
N a ( x)  

N
,
x

0
 a

After:
EC
n-side

 (EF)
EV
p-side
“Abrupt junction
model” (1)
n-side
In order to describe the spatial variation in the band edges, EC and
EV, we introduce a potential function, (x), such that each level is
shifted by -e(x). That is, EC  EC - e(x) and EV  EV - e(x).
Therefore, the carrier concentrations can be expressed by
 E  e ( x)   
n( x)  N C (T ) exp  C

k
T
B


   EV  e ( x) 
p ( x)  PV (T ) exp 

k BT


Consider limits at x = 
 E  e ()   
N d  n()  N C (T ) exp  C

k BT


   EV  e () 
N a  p()  PV (T ) exp 

k
T
B


E  e ()   
N
 ln d   C
NC
k BT
 ln
  EV  e ()
Na

PV
k BT
(2)
We can solve for the difference of the two potentials at 
N N 
 e ()  e ()  EC  EV  k BT ln  d a 
 N C PV 
N N 
 e  E g  k BT ln  d a 
 N C PV 
The electric field is related to the potential by Gauss’s Law:
E 
4(r )

E  
d 2 4( x)
     2 
dx

2
for our 1D problem.
We assume that donor and acceptor impurities are fully ionized at all x.
  ( x)  eN d ( x)  N a ( x)  nc ( x)  pv ( x)
(3)
We can substitute Eqs. (1) and (2) into Eq.(3) to get a difficult nonlinear
differential equation.
We can make some important approximations to simplify and solve:
Assume that the potential change  occurs in a finite region near the junction that is
defined by -dp x  dn. This is referred to as the depletion region. Also n << Nd
and p << Na in this region since  is closer to the middle of the gap (i.e., midgap).
  ( x)  eN d ( x)  N a ( x)
We can solve this problem assuming an abrupt junction model which will give a
simple linear second order ordinary differential equation:
x  dn 
 0,


4

eN
d

dn  x  0 
d 2 





dx 2   4eN a 0  x   d 
p




 d p  x 
0

 (),
x  dn 



2

eN
2
d
  ( ) 
(x  dn )
dn  x  0 



  ( x)  

2eN a
2
0

x


d
 () 
(x  d p )
p




 d p  x 
 ()

Note that we have first used the boundary conditions at the edges of the depletion region (see
graph on next slide) to solve for (x):
The electric field must
The
potential
must
be
E
|



'
(
d
)

0
xd n
n
 ( d n )   ( )
be zero outside the
its limiting value at the
E | x   d p   ' (d p )  0 depletion region for
 (d p )   () boundaries of the
equilibrium to hold.
depletion region.
Graphical illustration of the
Abrupt-junction model.
p-type
n-type
Nd
Carrier density
Na
nc(x)
pv(x)
-dp
0
dn
x
Depletion layer
eNd
Charge density
(x)
-dp
dn
x
-eNa
()
(x)
-dp
dn
(-)

x
Comparison of the potential and electric
field in the abrupt-junction model:
()
(x)
-dp
dn

x
(-)
E
d
dx
x
 Emax
There are important relations that we get by imposing continuity of (x) and ’(x).
Continuity of the derivative gives:
 4eN d
4eN a
d (0  ) d (0  )


(d n ) 
(d p )
dx
dx


 Nd dn  Nad p
Secondly, continuity of (x) gives
 (0 )   (0 ) 


2e

N d
d
2
n

 N a d p2   ()   ()  
Together with the derivative condition: Nddn=Nadp this gives
d n, p 
N a N d 1
N a  N d 2e
For Na,d ranging from 1014 to 1018 cm-3
 dn,p ranges from 104 to 102 Å
The total depletion width is LD = dn +dp
We can also write this in a numerically convenient form:
d n , p  105
N a
N d  e eV o

18
10  N a  N d 
1
The maximum electric f ield within the depletion layer is of the order of / (dn+adp) and
ranges from 105 to 107 volts/m.
Consider the effect of an applied voltage on
the depletion width of a p-n junction:
Charge density,
p
(x)
(x)
o
-dp
V = 0 “zero-bias”
dn
The potential simply
changes according to
=o|V|
x
-eNa
p
p
n
n
1(x)
o- |V1|
- sign: “Forward bias”
+ sign: “Reverse bias”
n
-dp
V1 > 0 “Forward-Bias”
dn
x
-eNa
2(x)
We are changing the size of
the depletion region by
applying an external voltage.
o + |V2|
-dp
dn
-eNa
p
n
x
V2 < 0 “Reverse-Bias”
The formula for the depletion width can easily be modified to include the
effects of a voltage bias:
d n, p 
N a
N d    o  V 
V
 d no, p 1 
N a  N d 2e
 o
1
Zero-bias depletion length
Consider rectification by a p-n junction. The symbol J is used for # of
carriers /(area·time). The lower case symbol j is more commonly used for
current density and has units j = coulombs/ (area·time). Note j e= -e J e and
j h= e J h .
Consider electron and hole generation currents.
These carriers are generated by thermal excitation
as we saw before according to
p-side
n-side
EC
 ( EF )
 E  e ( x)   
n( x)  N C (T ) exp  C

k BT


e-
   EV  e ( x) 
p ( x)  PV (T ) exp 

k
T
B


EV
Note: The generation currents involve
minority carriers.
h+
In a band diagram, electrons fall down hill and holes “float” uphill (both towards lowest
energy).
In the process of the generation currents, holes (from n-side)  p-side, electrons (from
p-side)  n-side.
Secondly, consider another kind of current, Recombination current. With this kind of
current holes (from p-side)  n-side. Electrons from the n-side  p-side. This current
involves majority carriers, and it is made possible by thermal excitation over the barrier,
as shown below:
p-side
n-side
o- |V1|
e-
EC
 (EF)
EV
h+
Consider the hole recombination
current. The current is approximated well
by assuming thermionic emission over the
barrier:
J hrec  exp  e o  V  / k BT 
J hrec |V 0  J hgen
 J hrec  J hgen exp V / k BT 
The middle equation is a statement that no net
current can flow during equilibrium when V=0.
The total hole current from the p to the n-side is given by the recombination
current minus the generation current:
Note: In our notation Jh and
Jp will mean the same thing.
  eV  
rec
gen
gen
  1
J h  J h  J h  J h exp 
  k BT  
  eV  
  1
 j  eJ hgen  J egen exp 
  k BT  
p
n
The last equation includes the currents of both holes and electrons, since
the same analysis will also apply to electrons. It is obvious to understand
the rectifying properties of a p-n junction (diode) from this equation.
Reverse bias (V<0)
j
V
Forward bias (V>0)
Rectifying behavior of a p-n
junction (diode).
Saturation

e J hgen  J egen

V
We still need to calculate this term in terms of
fundamental parameters describing transport.
In a more general treatment, it is not necessary to separate into generation and
recombination currents. Note that to solve for the following five fundamental
quantities (Je(x), Jh(x), n(x), p(x), and (x)), we need five equations. In the equilibrium
case with V=0, Je(x)= Jh(x)=0, we need three equations to solve the three unknowns,
including Poisson’s equation for the potential.
In the presence of a field E and n and  p, we can write the electron and hole current
densities as:
dn
dx
dp
J h   p pE  D p
dx
When dn/dx=0, j  E  eJ e  e n nE
J e    n nE  Dn
ne 2 col

m
Therefore,
e ncol
n 
mn
p 
e col
p
mp
Where n and p are the electron and hole
mobilities in units of cm2/V·s, and Dn and Dp
are the electron and hole diffusion constants
in units of cm2/s
From the Dude theory
(mv=Ft) recall
v  eE col / m
ne 2 col
j  nev 
E  E
m
Which expresses  in terms of the
collision time and masses, based on the
Drude model.
Now, recall that
At equilibrium,
 E  e ( x)   
n( x)  N C (T ) exp  C

k
T
B


 E  e ( x)    e d ne( E )
dn
 N C (T ) exp  C


dx
k BT
k BT

 k BT dx
 J e    n nE  Dn
eD
 n  n ;
k BT
dn
1
   n nE  Dn (neE )
0
dx
k BT
p 
eD p
These last two relations are known
as the Einstein relations.
k BT
Consider continuity equations for transport of charge:
J
n
 e;
t
x
J
p
 h
t
x
If V0 and carriers are conserved.
We have to include two other processes which act as a source and drain for carriers:
EC
(i) Generation by thermal excitation
(ii) Recombination (electron  hole)
(i)
(ii)
EV
As a result the continuity equations
need to include these additional terms:
J
n  dn 
   e ;
t  dt  g r x
J
p  dp 
   h
t  dt  g r x
The g-r terms act to restore the system to equilibrium when the system deviates
from equilibrium. These terms can be further described by electron and hole
lifetimes (n and p):
nn
 dn 


;
 
n
 dt  g r
p  po
 dp 
  
p
 dt  g r
o
n o p  ni2 ;
np o  ni2
where no and po are equilibrium concentrations as
determined by the law of mass action.
Note that n >>ncol and p >>pcol ; typically n ~10-9 s and ncol ~10-13 s
Very often we deal with a steady state condition in which
This gives
dJ e n  n o

 0;
dx
n
n p

0
t t
dJ h p  p o

0
dx
p
Note that Je = Jh = 0 for V = 0  n =
no
and p =
po.
Now suppose that E0, i.e., the electric field is negligible. Then
J e    n nE  Dn
dn
dn
  Dn
dx
dx
Fick’s Law for Diffusion
p
n
V
For V0, we are not in
equilibrium but we have a
steady state for V = const.
dJ e n  n o

 0;
dx
n
dJ h p  p o

0
dx
p
d 2n n  no
 Dn 2 
;
dx
n
d 2 p p  po
Dp 2 
dx
p
Therefore,
further give the following 2nd order
ordinary differential equations which
are referred to as diffusion equations:
Note no and po are
equilibrium values.
The solutions are easily written as:
 ( x  xo ) 
 ( x  xo ) 
p  p  A exp 
  B exp 
;
L p 

 L p 
o
 ( x  xo ) 
 ( x  xo ) 
n  n  A exp 
  B exp 

L
L
n
n




o
The diffusion lengths Lp
and Ln are given by
L p  D p p ;
Ln  Dn n
Note that when E0, the majority carrier density is constant and the carriers in the
diffusion equation are for the minority carriers. For example if we have a p-type
material pn = ni2 and p >> n so that electrons (np) are minority carriers. The notation is
slightly changed to be more precise:
p-type
Dn
d 2n p
dx 2

np  np
n
o
;
d 2 p n pn  pn
Dp

dx 2
p
o
n-type
Examine the meaning of the diffusion lengths, Ln and Lp.
Since
k T
Dn  B  n ;
e
Ln  Dn n ;
 Ln 
e ncol
n 
mn
k BT ncol n
mn
From the equipartion theorem and Drude approximation: ½ mvth2 = 3/2 kBT
Further, the mean free path ln is given by ln=vthcol where vth is the thermal velocity
of a carrier.
2
1  ln 
 k BT  mn  col
 ;
3  
Ln 
 1  ln    n 
k BT  n
 Ln   mn  col


mn
3

m



n 

2
col
col
1/ 2
 ln
n
3 col
The interpretation here is as follows: When a carrier undergoes N collisions before

recombination, the net displacement will be
Nl ;
N n
n
The factor of 3 comes from the number of degrees of
freedom (3D). This is basically a “random walk” problem.
1
4
2
3
3 ncol
Note that n >>ncol and p >>pcol ; typically n ~10-9 s and
ncol ~10-13 s  N3000 as the number of collisions before
recombination in GaAs.
We need to improve our understanding of transport across the p-n junction. Consider an
equivalent description of Boltzmann statistics at thermal equilibrium:
Remember that
 EF  Ei  where ni is the intrinsic carrier concentration EF = 
 is the Fermi-level (or chemical potential), and Ei is
n  ni exp 
 k BT  the position of the chemical potential for the intrinsic
case.
Let us define potentials  = -Ei/e and  = -EF/e. At thermal equilibrium pn=ni2 (V=0 across
the junction). For the case of V0, we can write the carrier concentrations as
 e(   n ) 
n  ni exp 
;
 k BT 
 e( p  ) 
p  pi exp 

k
T
B


Where n and p are called quasi-Fermi levels or imrefs (imaginary references).
k BT  n 
k T  p
ln    p    B ln  
e
e
 ni 
 ni 
 e( p   n ) 
2
 pn  ni exp 

 k BT

 n  
The advantage of this formalism is that the potential difference V across the junction is
just the difference in the quasi-Fermi levels, i.e.
V   p | x 0  n | x 0   p | x    n | x  
Forward-Bias Conditions
p-side
n-side
E (eV)
Reverse-Bias Conditions
p-side
n-side
EC
-en
E (eV)
-ep
EF ()
-ep
EC
-en
EV
-dp 0 dn
Potential
-dp 0 dn

V
n
-dp 0 dn
EV
x
p
x

Potential
n
-V
p
x
-dp 0 dn
log(n,p) ppo
nno
log(n,p) ppo
nno
ni
pn p
no
pno
npo
-dp 0 dn
x
Note that (dp+dn)
for reverse bias is
greater than (dp+dn)
for forward bias.
x
ni
npo np
EF ()
pn
np
-dp 0 dn
x
 e( p   n ) 
 eV 
2


pn  ni2 exp 

n
exp
 i
 k BT 
 k BT 
For a forward bias V = p-n > 0 and pn > ni2 in the junction. The opposite is the case for
a reverse bias V = p-n < 0 and pn < ni2 .
Now consider the current density and write in terms of  and :

k T n 
k T  en    n 

  
J e  e n  nE  B


  e n n n
  e n n 
  e n B 
e x 
e  k BT  x
x 
x

 x 
 1 Ei

Note that
x e x
 p
Similarly, J p  e p p
x
E
For p and n constant, Je and Jp
are 0.
Our previous notation
was
E

x
(    const .)
In the depletion layer –dp < x <dp
 n
J
 e 0
x
e n n
 p
x

Jp
e p p
0
since n and p are sharply decreasing functions in the
depletion layer. Also, we assume that passage of carriers
across the junction is very fast so that generation and
recombination currents in the depletion region are negligible.
This leads to assumption that Je and Jp are constant in the
depletion region –dp < x <dp
 eV 
 eV 
ni2
  n po exp 
 (1)
exp 
Consider n at x = -dp (on the p-side): n p | x  d p 
pp
 k BT 
 k BT 
Note that npo is the electron density on the p-side at x = -.
 eV 
 (2)
pn | x  d n  pno exp 
 k BT 
Similarly, at x = +dn
And pno is the hole density
on the n-side at x = +
These equations serve as boundary conditions for the I-V equation of the ideal
p-n junction.
Recall that just outside of the depletion region (i.e. |x| > dn, dp the region is
neutral, E0, and this is called the diffusion region. In these regions, the
following diffusion equations apply for the respective minority carrier densities:
Dn
d 2n p
dx 2

n p  n po
n
;
d 2 pn pn  pno
Dp

dx 2
p
Using boundary conditions (1) and (2), the solutions are:
 x  dn 
  eV  



pn  pno  pno exp 
 1 exp  


L p 
  k BT  

  eV  
 x  dp 

  1 exp 
n p  n p 0  n po exp 
 Ln 
  k BT  
x  dn
x  d p
Note that the expected
limits are observed: np 
npo at x = - while pn 
pno at x = +
We can now calculate the electron and hole currents entering the diffusion region
where E0.
At x = dn
  eV  
  1
exp 
k
T
  B  
n p
eDn n po   eV  
  1
J e  eDn
|x d p 
exp 
x
Ln   k BT  
eD p pno
pn
|xdn 
x
Lp
J p  eD p
for the n-side.
for the p-side.
  eV  
  1
The total current is given by J = Je + Jp  J S exp 
  k BT  
eD p pno eDn n po
where J S 

Lp
Ln
This result is the well known Shockley equation. Previously we wrote the
current in terms of electron and hole generation currents:

j  e J hgen  J egen
J
gen
e

Dn n po
Ln
  eV  
  1
exp 
  k BT  

Note that
p po  N A
nno  N D
2
Dn ni2 (on p-side) and gen D p pno D p ni (on n-side)

Jh 

Ln N A
Lp
Lp N D
 Dn
D p    eV   Contains fundamental parameters

 exp 
  1 describing the rectifying behavior.
 J  en


L N

 n A L p N D    k BT  
2
i
Examine the minority carrier densities and current densities for
forward and reverse bias:
p-side
n-side
p-side
n,p
n-side
n,p
npo
pn
np
pno
pno
npo
np
-dp 0 dn
J
x
pn
-dp 0
dn
x
J=Jn+Jp
J
Jp
J=Jn+Jp
Jn
Jp
Jn
Note that (dp+dn)
for reverse bias is
greater than (dp+dn)
for forward bias.
-dp 0 dn
x
Forward Bias (V >0)
-dp 0 dn
Reverse Bias (V<0)
x
Possible to examine details of diffusion, using the Haynes-Shockley Experiment
Consider changes in minority carrier density with time:
pn
J
p  pno
 h  n
;
t
x
p
J h   p pn E  D p
pn
x
pn
pn
 2 pn pn  pno

  p E
 Dp

2
t
x
x
p
For E = 0, the solution follows the solution of the diffusion equation:
2

N
x
t 
pn ( x, t ) 
exp  
   pno
 4D t  
4D pt
p
p 

Where N= number of holes/Area generated (laser light pulse).
V t
1
(E = 0)
t2
t3
If E 0, x  x - pEt
h
Sample
V
t1
t
L
0
Et1
Oscilloscope
Et2
t2
(E > 0)