UCSD Department of Electrical and Computer Engineering
ECE 183 Winter 2005
Homework No. 5: Solutions
Assigned: March 7, 2006. Due: March 16, 2006 in class
1 Photodetection
A photodetector consists of two p−i−njunctions in series (back-to-back). Light that
reaches the second junction must travel through the first junction. Since the junctions are in
series, the same photo-current should be generated in each section, otherwise one of the two
sections limits the performance of the device, i.e., is the limiting factor in the short-circuit
current (density).
The total thickness is 1 µm. Assume that the pand the nregions are of negligible thickness
and that there are no R-G processes, or radiative recombination in the iregion. (Assume
that the semiconductor has a direct band-gap and is characterized by parabolic bands as
discussed in this class.)
(a) Assume that the absorption of the material is α= 3 ×106m−1at the wavelength of
interest. What should the thickness of each of the iregions be?
Since the two diodes are in series, light that is incident on the second section must have
traveled through the first section. In each section, the photocurrent that is generated
is proportional to the absorbed flux. Therefore, in order to balance the photocurrents,
I0Zd1
0
e−αxdx =I0e−αd1Zd2
0
e−αxdx (1)
where d1+d2= 1 µm. Solving the above equation numerically, d1= 0.215 µm and
d2= 0.785 µm. P.S. Under the substitution α→2α, the values are changed to
d1= 0.115 µm and d2= 0.885 µm
(b) Explain in words what would happen to the short circuit current (density) if the
wavelength of light is increased or decreased. (Hint: what is the relation of αto
E=hc/λ for a semiconductor with parabolic bands?)
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