Student Project Physics of Semiconductor
Devices
WS 2011/12
A Silicon p-n junction
Markus Polanz
0630160
Introduction:
In this student project the following exam question about a silicon p-n junction
will be solved and discussed:
a) Band diagram (no voltage applied)
The pn-junction consists of an acceptor doped area (p-zone), which borders
on a donator doped area (n-zone). In the p-zone the charge carriers are holes
and in the n-zone they are electrons. At the crossing, the two sorts can
recombine. Only immobile charge carriers are left in this area. These are the
negative and positive space charges of the impurity atoms respectively in the
so-called depletion zone. The electric field lines in this zone start at the
donator atoms and end at the acceptor atoms. Because of the electric field
there is a potential difference between the two space charge areas. There is
a thermal equilibrium between the field current and the diffusion current.
Thats why the fermi energy is on a constant level. The total potential
difference between the left and the right zone is given by the built in voltage
Vbi and results out of the diffusion of the charge carriers.
b) Band diagram in forward and reverse bias
If we apply a forward bias then charge carriers are pushed into the depletion
zone and the depletion zone gets smaller. If we apply a reverse bias electrons
are picked up on the right side and holes are picked up on the left side and
the depletion zone gets bigger. In both cases there is a energy difference
between the p-zone and the n-zone. The sign of the bias tells us if the energy
is lifted or lowered.
c) Concentration of holes
For the following calculations some physical constants and informations
about silicon are needed:
kB = 1.38E-23 J/K …Boltzmann constant
ε0 = 8.854E-12 F/m …dielectric constant
e = 1.602E-19 C …elementary charge
εr = 11.9…dielectric constant for silicon
ni = 1.5E10 cm-3…intrinsic carrier concentration for silicon
Nd = 5E15 cm-3…donator concentration at T = 300 K for silicon
Na = 1E17 cm-3… acceptor concentration at T = 300 K for silicon
On the p-side the concentration of holes np is equal to the concentration of
acceptors Na:
np = Na
On the n-side we can calculate the concentration of holes np with the law of
mass action:
n2i
np =
Nd
All in all we get for the hole concentrations np on both sides:
p-side: np = 1E17 cm-3
n-side: np = 45000 cm-3
d) Calculation of the depletion width
To calculate the depletion width W we need to determine the built-in potential
Vbi:
k B T Nd Na
Vbi =
ln( 2 )
e
ni
Now we are able to calculate the depletion width W:
2 ε0 εr (Nd + Na )Vbi
W=√
e Nd Na
Built-in potential: Vbi = 0.73 V
Depletion Width: W = 0.45 μm