4.4 The electrostatic analysis of a p-n diode
Table of Contents -
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In this Section:
- The full depletion approximation
- Calculation of the charge density
- Calculation of the electric field
- Calculation of the potential
- Calculation of the depletion layer width
- Calculation of the energy band diagram
4.4 The electrostatic analysis of a p-n diode
The electrostatic analysis of a p-n diode is of interest since it provides knowledge
about the charge and field in the depletion region. It is also required to obtain the
capacitance-voltage characteristics of the diode.
The general analysis starts by setting up Poisson's equation:
(pn5)
where the charge density is written as a function of the electron density,
the hole density and the donor and acceptor densities. The electron and hole densities can then
be expressed as a function of the potential yielding:
(pn5a)
with
(pn5b)
where the potential is taken to be zero in the n-type region.
This second order non-linear differential equation can not be solved
analytically. Instead we will make the simplifying assumption that
the depletion region is fully depleted and that the neutral regions
contain no charge. This
full depletion approximation is the topic of the next section.
4.4.1 The full-depletion approximation
The full-depletion approximation assumes that the depletion region around
the
metallurgical junction has well-defined edges with an abrupt transition
between
the fully depleted region where no carriers are present and the
quasi-neutral region, a neutral region where
the carrier density is close to the doping density.
This approximation is justified by the fact that the carrier densities
change
exponentially with the position of the fermi energy relative to the band
edges. For
example as the distance between the fermi level and the conduction band
edge is
increased by 59 meV, the electron concentration at room temperature
decreases to one tenth of its
original value. The charge in the depletion layer is then quickly
dominated by the remaining
ionized impurities, yielding a constant charge density for uniformly
doped regions.
We will therefore start our analysis using an abrupt charge density
profile, while
introducing two unknowns, namely the depletion layer width in the p-type
region,
xp, and the depletion region width
in the n-type region, xn.
The
sum of the two depletion layer widths in each region is the total depletion layer
width w, or:
(pn17)
From the charge density profile we then calculate the electric
field and the potential across the depletion region. A first relationship
between the two unknowns is obtained by setting the positive charge in the
depletion layer equal to the negative charge. This is required since the
electric field in the quasi-neutral regions must be zero. A second
relationship
between the two unknowns is obtained by relating the potential across
the depletion layer width to the applied voltage. The combination of both
relations yields a solution for xp and
xn, from which all
other parameters can be obtained.
4.4.2 Calculation of the charge density
Once the full-depletion approximation is made it is easy to find the
charge density profile: It equals the sum of the charges due to the holes,
electrons, ionized acceptors and ionized holes as given by:
(pn1)
where it is assumed that within the depletion region no free carriers
(i.e. electrons or holes) are present. An example is shown in the
figure below. The figure shows the charge density at -5 Volt bias (thick
line) and at 0 Volt bias (thin line).

pncharge.gif
Fig.4.4.1 Charge density in a p-n junction at
-5 Volt bias (thick line) and at 0 Volt bias (thin line).
The p-type region is on the left with
Na = 1016 cm-3 and
the n-type region is on the right with
Nd = 2 x 1016 cm-3
As can be seen from the figure, the charge density is
constant in each region, as dictated by the
full-depletion approximation. The depletion region width
also increases as a more
negative bias is applied.
The charge throughout the diode is given by the following equations:
(pn1a)
(pn1b)
(pn1c)
(pn1d)
4.4.3 Calculation of the electric field
The electric field is obtained from the charge density using
Gauss's law,
which states that the field gradient equals the charge density divided by
the dielectric constant or:
(pn2)
As a result one finds that the electric field changes linearly with
position. The integration constant is obtained by assuming that the
electric
field is zero at both ends of the depleton region, or at
-xp and
xn. The electric field has to be zero outside the depletion
region since any field would cause the free carriers to move so that the
associated charge eliminates that field. The fact that the electric
field is zero on both ends of the depletion region also implies that
the total positive charge per unit area in the depletion layer equals the
total negative charge.
An example which is calculated from the charge density shown
in the figure above is provide below. Again the figure is calculated for a
bias of -5 Volt (thick line) and 0 Volt (thin line).

pnfield.gif
Fig.4.4.2
Electric field in a p-n junction at
-5 Volt bias (thick line) and at 0 Volt bias (thin line)
(Na = 1016 cm-3 and
Nd = 2 x 1016 cm-3)
The maximum electric field occurs at x = 0 and can be obtained
by integrating from -xp
to 0 or by integrating from xn to
0 yielding:
(pn6)
Both calculations must yield the same result since the dielectric constant
is the same in both regions. This provides the first relationship between
the two unknowns, xp
and xn, namely:
(pn18)
This equation expresses the fact that the total positive charge
in the n-type depletion region exactly balances the total negative
charge in the p-type depletion region. We can then
combine this equation with the expression for the
total depletion layer width, w,
(pn17)
and rewrite the depletion layer width in
each region
as a function of the total depletion layer width, yielding:
(pn10)
and
(pn11)
4.4.4 Calculation of the potential
The potential in the semiconductor is obtained from the electric
field using:
(pn4)
We therefore integrate the electric field yielding a
piece-wise parabolic
potential versus position. An example is shown
in the figure below, again
for an applied bias of -5 Volt (thick line) and
0 Volt (thin line).

pnpot.gif
Fig.4.4.3 Potential in a p-n junction at
-5 Volt bias (thick line) and at 0 Volt bias (thin line)
(Na = 1016 cm-3 and
Nd = 2 x 1016 cm-3)
The total potential across the semiconductor must
equal the difference between the built-in potential and the applied voltage,
which provides a second relation between xp
and xn, namely:
(pn7)
4.4.5 Calculation of the depletion
layer width
The depletion layer width is obtained by substituting the expressions
for xp and xn into the
expression for the potential across the depletion region, yielding:
(pn9)
from which the solutions for the individual depletion layer widths,
xp and xn
are obtained:
(pn12)
(pn13)
4.4.6 Calculation of the energy band diagram

pneb.gif
Fig.4.4.4 Energy band diagram of a p-n junction at
-5 Volt bias (upper set of curves) and at 0 Volt bias (lower set of curves)
4.3
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4.5
© Bart J. Van Zeghbroeck, 1996, 1997