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The electrostatics of pn-junctions, discussing poisson's equation, built-in potential, depletion approximation, and step-junction solution. It covers the concepts of charge density, electric field, and electrostatic potential inside the diode under equilibrium and steady state conditions. The document also explains the fabrication processes of pn-junctions, including diffusion, ion-implantation, and epitaxial deposition, and their resulting doping profiles.
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You will also learn about:Poisson’s EquationBuilt-In PotentialDepletion ApproximationStep-Junction Solution
In this chapter you will learn about pn junction electrostatics:Charge density, electric field and electrostatic potential existinginside the diode under equilibrium and steady state conditions.
PN-junctions are created byseveral processes including:1. Diffusion2. Ion-implantation3. Epitaxial depositionEach process results indifferent doping profiles
kT
n
p
F
i
i
exp
F
= same everywhere
under equilibrium
Join the two sides of theband by a smooth curve.
kT
n
n
i
F
i
exp
x E q x E q
d d 1 d d 1
i
C
Potential,
q
C
ref
). So,
potential difference between thetwo sides (also called built-involtage,
bi
) is equal to
q
C
ρ^ ε
x
d d
ref
C
q
ρ
= charge density
ε
=
K
s
ε
o
bi
When the junction is formed, electrons from the n-side and holesfrom the p-side will diffuse leaving behind charged dopant atoms.Remember that the dopant atoms cannot move! Electrons willleave behind positively charged donor atoms and holes will leavebehind negatively charged acceptor atoms.The net result is the build up of an electric field from the positivelycharged atoms to the negatively charged atoms, i.e., from the n-side to p-side. When steady state condition is reached after theformation of junction (how long this takes?) the net electric field(or the built in potential) will prevent further diffusion of electronsand holes. In other words, there will be drift and diffusion currentssuch that net electron and hole currents will be zero.
Under equilibrium conditions, the net electron current and holecurrent will be zero. electron drift currentopposite to electron flux
A
17
cm
−
3
D
16
cm
−
3
electron diffusion currentopposite to electron flux hole diffusion current
hole drift current
net = 0
net = 0
-field
bi
The built-in potential,
bi
, measured in Volts, is numerically
equal to the “shift” in the bands expressed in eV.
bi
q
i
F
p-side
F
i
n-side
side
n
on
ion
concentrat
electron
and
side
p
on
ion
concentrat
hole
where
ln
ln
ln
n
p
(^2) i
n
p
i
i
n
p
n
n
p
q kT
n n
q kT
n
p
q kT
kT
bi
V
q
n n
p p
exp
n p
p n
p
p
n
n
p
n
p-side
A
D
n-side
else
everywhere
0
for
)
(
0 0
d d
n p A D s s =
≤
≤
−
−
⎧⎪ ⎨⎪⎩
ε
=
ε ρ
=
x x x N N K
q
K
E x
We
assume
that the free
carrier concentration insidethe depletion region is zero.
A p-n junction is formed in Si with the following parameters.Calculate the built-in voltage,
bi
D
16
cm
A
17
cm
Calculate majority carrier concentration in n-side and p-side.Assume
n
n
D
16
cm
−
3
and
p
p
A
17
cm
−
3
(^2) i
D
A
(^2) i
n
p
bi
ln
ln
n
q kT
n
n
p
q kT
Plug in the numerical values to calculate
bi
