EC ENGR 101A Problem Set#6 Due 2/28/23, 11:59PM
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Problem #2 (30 points) Poisson’s Equation and Dielectric relaxation
Consider a leaky parallel-plate capacitor with area A, thickness d, and a lossy dielectric
with permittivity ε and non-zero (but small) Ohmic conductivity σ. At time t = 0, there is
a free volume charge density ρ(x,t=0) =ρ0 x /d inside the dielectric (0<x<d). You may
neglect any fringing fields (i.e. assume width >> d) to use a 1D model. The two perfectly
conducting plates of the capacitor are short-circuited together.
a) At time t=0, what is the potential V(x,t=0) and the electric field E(x,t=0) in between
the conducting plates? Use Poisson’s equation for the dielectric and apply the
boundary conditions set by the plates. (You can ignore any time dependence and
conductivity for the initial condition at time t=0, because any free charge hasn’t had
time to react yet).
b) Given this initial condition, what is the charge density ρ(x,t) as a function of time?
To solve this, use the charge continuity equation that we discussed in class, and
substitute Ohm’s law to obtain a first order partial differential equation that
describes the time evolution of ρ(x,t). (It is discussed in Chapter 6.9-6.10).
c) Use your expression for ρ(x,t) and find E(x,t) as a function of time.
d) What is the surface charge density ρs at x=0 and x=d as a function of time?
e) What is the short circuit current i(t) as a function of time?
2
Problem #3 (30 points) Semiconductor pn junctions
The junction between positively (p)-doped and negatively (n)-doped silicon is at the heart
of modern semiconductor devices, including p-n diodes and transistors. We can model an
“abrupt” junction as a dielectric material (ε = 11.9ε0 for silicon) with four regions with
charge density ρ(x).
In this expression, ρ0 is a constant charge density, and x0 is a length of the depletion region
on the p-side (-x0<x<0) and n-side (0<x<x0). The problem has planar symmetry, so there
is no variation of ρ in the y or z directions. Without worrying about the physics of how
these charge distributions are created, we can find the electric field and scalar potential
within the junction region.
a) The Electric field is zero in the regions x<-x0 and x>x0. Derive a piecewise
expression for E(x) in the region between –x0<x<0 and 0<x<x0.
b) Find an expression for the scalar potential V(x) in the region –x0<x<x0. You may
assume that V(-x0) = 0.
c) The potential difference V(x0)-V(-x0) is known as the “built-in potential” of a
semiconductor junction. Using the values ρ0=160 C/m3, and x0=0.65μm what is the
built-in potential for this junction? (While it is not necessary to know this to solve
the problem, this charge distribution is created at the interface between n-doped and
p-doped semiconductor. The positive charge is created by ionized donor atom
cores, and the negative charge is created by ionized acceptor atom cores. The values
given above are typical for a Si p-n junction with doping levels NA=ND=1015 cm-3.
This material is covered in great detail in EC ENGR 2.)
d) There are other types of junctions where the charge distribution is not so abrupt.
These are called “graded” junctions. Find E(x) and V(x) for the following charge
3
distribution:
where G is a constant of units (C/m5)
Problem #4 (20 points)
Hint: Remove one of the conductor planes by reflecting a charge about it, then remove
the reflecting both the charges (original and image charge) around the other conductor
plane.
4
Problem #5 (30 points)
(c) Find the surface charge density "! ($) on the surface of the sphere.
(d) Consider the expression for an electric dipole given in Ulaby chapter 4,
"⃗⋅%&
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. We see above that we can obtain our solution for &((, $) by
considering the sum of a constant field term and a dipole term. What is the
equivalent dipole moment *⃗ required to obtain our solution in part b?
5
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