
First, we look at
the junction in equilibrium, i.e. there is no net current and the Fermi energy
is the same everywhere (we have the same situation as
shown before; but for diversities
sake, the p and nside reversed). 




In what follows, there will be a lot
of shuffling formulas around  and somehow, like by magic, the
IVcharacteristics of a pnjunction will emerge. So let's
be clear about what we want to do in the major steps (highlighted in blue) 

The first basic goal is to
find an expression for the carrier density at the edge
of the space charge region. 


We know in a qualitative way
from the consideration of pure
diffusion currents that the minority carrier concentration around the edge
of the space charge region is somewhat larger then in the bulk under
equilibrium conditions  there is a Dn_{min} given by Dn_{min} = n_{min}(x)
 n_{min}(bulk) and Dn_{min} will increase for forward
bias, i.e. for nonequilibrium conditions. 


We
also know that Dn_{min} induces a diffusion current
and that we therefore need a "real" current to maintain a constant
Dn_{min}. Finding Dn_{min} and, if necessary, Dn_{max}, thus will automatically give us the necessary currents
belonging to the nonequilibrium as defined by the voltage. 

We first look at the various energies
involved: 


The energy difference between the left side (=
nside; index "n") and
the right side (= pside; index
"p"; these indizes will
appear "up" or "down" as dictated by the limitations of
HTML) of the junction is given by 


E _{C}^{n} –
E_{C}^{p} 
= 
E_{V}^{n} –
E_{V}^{p} 
= 
e · (V ^{n} – V
^{p}) 
= 
e · DV 




With V ^{n}, V
^{p} = constant potentials at
the n or pside and far away from the junction, respectively.


The band energy levels
E_{C}^{n, p} = E_{C}^{n,
p}(x) and the potential V(x) are functions
of x, which makes all concentrations functions of
x, too. We will, however, write all
these quantities without the "(x)" from now
on. 


As long as we discuss
equilibrium, the Fermi energy is constant and the carrier concentrations are
given by their usual expression. We
consider them separately for the left and right hand side of the junction i.e.
for the n and ppart. 


n^{p}_{e} 
= 
N ^{e}_{eff} · exp

E_{F} –
E_{C}^{p}
kT^{ } 
= 

concentration of
electrons on the pside 







n^{p}_{h} 
= 
N ^{p}_{eff} · exp

E_{V}^{p} –
E_{F}
kT^{ } 
= 

concentration of
holes on the pside 







n^{n}_{e} 
= 
N ^{e}_{eff} · exp

E_{F} –
E_{C}^{n}
kT^{ } 
= 

concentration of
electrons on the nside 







n^{n}_{h} 
= 
N ^{p}_{eff} · exp

E_{V}^{n} –
E_{F}
kT^{ } 
= 

concentration of
holes on the nside 




We also have the
mass action law
applicable everywhere in equilibrium 


n^{p}_{e} ·
n^{p}_{h} 
= 
n^{n}_{h} ·
n^{n}_{e} 
= 
n_{i}^{2} 
= 
N ^{e}_{eff} · N
^{p}_{eff} · exp – 
E_{g}
kT 



Algebraic manipulation gives us any
number of relations, some of which may be useful later on. Here some
examples: 


E_{C}^{p} –
E_{V}^{n} 
= 
– kT · ln 
n^{n}_{e}
N ^{e}_{eff} 
– kT · ln 
n^{p}_{h}
N ^{h}_{eff} 
e · (V ^{n} – V
^{p}) 
= 
E_{C}^{p} –
E_{C}^{n} 
= 
E_{g} + kT · 
n^{n}_{e} ·
n^{p}_{h}
N ^{e}_{eff} · N
^{h}_{eff} 



This is about all
we can do in full generality. 


What we need to know to get on is the xdependence of the energies or of the
potential V(x)  this simply measn we need the
(quantitative) banddiagram that so far we always just drew "by
feeling". 


For this we need to solve the
Poisson equation and
this demands to specify the total charge r(x) so we can write dow the charge as a
function of x. This is easy in
principle: 

The total (space) charge r(x) at any point along the junction is the sum
of all charges: Electrons, holes, ionized donors
(N^{+}_{D}(x)), and ionized aceptors
(N^{–}_{A}(x)). We have
as before (with the necessary
"minus" sign being part of the elementary charge e) 


r(x)

= e · 
æ
è 
n^{h}(x) –
n^{e}(x) +
N_{D}^{+}(x ) +
N_{A}^{–}(x) 
ö
ø 




Inserting r in
the Poisson equation gives 


e e_{0}
e_{ } 
· 
d^{2}V(x)
dx^{2} 
= 
n^{h}(x) –
n^{e}(x) +
N_{D}^{+}(x) +
N_{A}^{–}(x) 




Solving this equation with the proper boundary
conditions will yield V(x) and everything else  but not so easily because
n^{h}(x) and n^{e}(x)
are complicated functions of V(x) via the Fermi
distribution. 


It is, however, not too difficult to find
good approximations for
"normal", i.e. highly idealized junctions; this is shown in an
advanced module accessible through the link. 

For our final goal, which is to
describe the currentvoltage characteristics of a pnjunction, we use
the same approximations and conventions,
namely 

1. The zero point of the electrostatic potential is
identical to the valence band edge in the pside of the junction , i.e.
e · V ^{p} = E_{V}^{p} = 0 as
shown in the complete
illustration to the situation shown in the picture above. This is a simple
convention without any physical
meaning. 

2.
All dopants are ionized, their
concentration is constant up to the junction, and there is only one kind on
both sides of the junction, i.e. 


n^{h}_{p}(bulk) 
= 
N_{A} 
= 
N ^{–}_{A} 





n^{e}_{n}(bulk) 
= 
N_{D} 
= 
N ^{+}_{D} 




This is a crucial
assumption. Note that while n^{h, e}_{p,
n}(bulk) are constant, this is not required for n^{h,
e}_{p, n}(x) around the junction. 

3. We also
assume that the Si extends into infinity (or at least to a distance much
larger than a diffusion length) to both sides of the junction  in total we use
the "abrupt" "large" junction approach 


This gives us for the carrier concentrations in
equilibrium anywhere in the junction: 


n^{h}(x) 
= 
N_{A} · exp – 
e · V(x)
kT 





n^{e}(x) 
= 
N_{D} · exp – 
e · [V(x) –
V ^{n}]
kT 




With V ^{n} = constant
value of the potential deep in the ntype region. 


These equations are already a special case of the more general equations above.
What they mean is that the carrier concentration is whatever you have in the
undisturbed p or npart (i.e the dopant concentration) times the
Boltzmann factor of the energy shifts relative to this situation. 

V ^{n} is the
difference of the builtin potentials for equilibrium conditions, it is
thus determined by the difference
in the Fermi energies of the two semiconductors before contact  our
simple view of a junction is totally
correct on this point 


With and without out an external voltage
U we have 


V ^{n}(U = 0) 
= 
1
e 
· 
(E ^{n}_{F}
– E ^{p}_{F}) 





V ^{n}(U) 
= 
1
e 
· 
(E ^{n}_{F}
– E ^{p}_{F} + e ·
U) 




The maximum potential at the nside,
V ^{n}(bulk) becomes 


V ^{n}(bulk) 
= 
V ^{n}(U = 0) + U 
= 
1
e 
· 
DE_{F} +
U 
:= V
^{n} 
for brevity 




Looking at the
proper solution of the
Poisson equation for our case, we realize that the space charge region was
defined as the part of the Si where the potential was not yet constant.
This means that V ^{n}(bulk) = V ^{n} =
V ^{n}_{edge SCR} on the nside, and
V ^{p}_{edge SCR} = 0. This is an essential
point, even so it is matterofcourse. 

We know can move
towards our primary goal and find an
expression for the carrier density at the edge of the SCR by inserting
the proper potential in the n ^{p}_{e} and
n ^{n}_{e} equation from above. We obtain for the edge of
the SCR: 


n^{p}_{e}
n^{n}_{e} 
÷
÷
÷ 
edge
SCR 
= 
n^{n}_{h}
n^{p}_{h} 
÷
÷
÷ 
edge
SCR 
= 
exp – 
e · (V ^{n} + U)
kT 




The minority carrier concentrations (always at the edge of the SCR without indicating it
anymore) can now be written as 


n^{p}_{e}(U) 
= 
n^{n}_{e}(U) · exp
– 
e · (V ^{n} + U)
kT 

electrons on
the pside 






n^{n}_{h}(U) 
= 
n ^{p}_{h}(U) · exp
– 
e · (V ^{n} + U)
kT 

holes on
the nside 




These equations are nothing but the Boltzmann
distribution giving the number of particles (n_{min})
that make it to the energy e(V^{n} + U) out of a
total number n_{maj}  in thermal equilibrium.
We used essentially the same equation
before, but now we know the kind of approximations that were necessary and
that means we also know what we would have to do for "better"
solutions of the problem. 

Since this is important, let's
review the approximations we made: 


Besides the
"abrupt" "large" junction, we used the approximations
from the simple solution
to the Poisson equation which implies that the potential stays constant right
up to the edge of the SCR and than changes monotously. 


This means that for equilibrium we must obtain
the same equations by computing the minority carrier concentration from the
mass action law, i.e.



n^{p}_{e}(U = 0) 
= 
n_{i}^{2}
n ^{p}_{h}(U = 0) 




We will see if this is true in a little
exercise 


Exercise 2.3.51 
Show the equality of the two
equations for the minority carrier density 
