1. (3 points) Unlike doped silicon, organic semiconductors have incredibly high molar absorbtivities and
therefore, much of the incident light can be absorbed over very short path lengths (100s of nm instead of
microns). But, these materials typically have a much lower free carrier mobility, which limits the overall
Part A: For a 100 nm thick "plate" of organic semiconductor with an electron mobility of 10-3
calculate how long it will take for the system to reach steady state, assuming only diffusive processes (i.e.
no applied electric field).
Part B: Holes typically have mobilities that are on the order of 0.01 that of electrons (i.e. pelectron = 100 -
How much longer will it take for the same device to reach steady state if we consider hole
Part C: Perfectly chosen contacts allow electrons to be collected efficiently (much faster than drift and
diffusion) at one side of the device and holes to be collected equally efficiently at the other. In other
words, one contact is selective for electrons, the other for holes and the device is considered to be
transport limited. Determine the steady state concentration profiles of electrons (n) and holes (p) across
the 100 nm device, assuming that the semiconductor has an intrinsic carrier density such that at x = 0, n
and p = 10 cm³ and at x = 100 nm. n = 10 cm³ and p = 0, again assuming only diffusion. Note, the
device is in the dark and hence, generation and accumulation terms are equal to zero. You should use the
diffusion coefficients for holes and electrons from above.
Part D: Assume light is then shown on the device and both holes and electrons are generated uniformly at
a rate of 1020 carriers/cm³s. Using the same boundary conditions as in Part C and the diffusion
coefficients from parts A and B, determine the carrier density profile across the device for holes and
electrons. Create a plot showing the spatial distribution of each. (You can still assume steady state).
Part E: The rate of free carrier recombination is dependent on the steady-state profile of holes and
electrons. For a bimolecular recombination mechanism, a hole and an electron must be in the same
spatial location. From your plots in Part D, how does having a mobility imbalance alter the
2. (1 point) A long copper wire 1/4 inch in diameter is held in an air stream with a temperature of Too =
100 F. After 30 seconds, the average temperature of the wire increased from 50 F to 80 F. Modeling the
wire as a cylinder, estimate the average unit surface conductance, h. by completing the following tasks:
A. Determine the Biot number in terms of h. You can assume a thermal conductivity k of 223 Btw/hour ft
B. Solve for h, assuming a Bi < 0.1, can be approximated by lump sum analysis.
C. Since we are assuming a lumped sum, we can use the following expression:
Using an expression of density of copper of 555 lb/ft³ and a heat capacity of 0.092 Btu/lb F, calculate h in
terms of Btu/hr nt2 F.
3. (1 point) Potassium ferricyanide is one of the most common aqueous redox probes used in standard
redox reactions. The ability to measure a transient redox current is strongly dependent on the magnitude
of the background current. In class on April 10th, we talked about the current measured at an electrode (x
= 0) associated with a redox process defined as:
iox = nFAD1/2 Cbulk
For the data below, calculate the current associated with the redox reaction (ica) and the capacitive current
for the following times:
12 500 us
4= 500 ms
Detection limits require a signal of interest to be at least three times that of the background. Determine
the time T when the redox current is 3x that capacitive current.
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