ECE 3040
Dr. Doolittle
Homework 3
Unless otherwise specified, assume room temperature (T = 300K).
The goal of this homework is to design a light-dependent resistor (LDR), also known as a
photoresistor or photocell. An LDR is simply a segment of semiconductor material used as a
resistor with a resistance that is a function of illumination. CdS and CdSe are the most common
semiconductors used for this application, due to bandgaps that are well-matched to visible and
near-infrared wavelengths, respectively. The semiconductor is typically arranged in a long,
winding pattern, but for this assignment, simply design the LDR to be one rectangular block.
The circuit schematic of a lightdependent resistor
Two light-dependent resistors
Given:
The LDR is made from CdSe, designed to respond to near-infrared radiation.
The bandgap of CdSe is 1.74 eV.
The intrinsic carrier concentration of CdSe is 8000 cm-3.
The semiconductor is n-type with an electron concentration of 1013 cm-3.
The electron mobility is 500 cm2/V-s, and the hole mobility is 10 cm2/V-s.
For this assignment it is okay to assume low-level injection for all parts, thus it is valid to use the
following formula (Equation 3.34a in Section 3.3.3 of Pierret):
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Although, in reality, we would be encountering high-level injection to achieve the amount of
variation in resistance we are aiming for, and the use of the more general form would be
necessary (Eq. 3.35):
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1) Purpose: Understanding the relationships between resistance, resistivity, and geometric
dimensions of a semiconductor.
It is determined that the CdSe film has a width, w, of 0.3 mm and a thickness, t, of 5 µm.
What is the needed length, S, to achieve 100 MΩ resistance in the dark? See figure below
for reference.
Note: Assume ohmic contacts are applied to both small faces of the bar (the t x w faces),
so that carriers will flow along length S.
From lecture, we know that the resistance of a material is defined as
where ρ is the resistivity of the material, l is the length of the block, and A is the
cross-sectional area of the plane through which carriers are flowing. Therefore, in
this case,
From Section 3.1.4 in Pierret, we know the resistivity of a materially is generally
defined as (Eq. 3.7):
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For our n-type sample of CdSe, we can simplify the above as (Eq. 3.8a)
because the concentration of electrons, ND, will be much greater than the
concentration of holes in the sample. Therefore,
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2) Purpose: Understanding generation of electron-hole pairs due to light.
Assume that the only recombination mechanism possible in the semiconductor bar is
band-to-band generation. In addition, assume there is uniform absorption/generation
throughout the entire thickness of the film, and the absorption efficiency of the bar is
100% (all eligible incident photons are converted into electron-hole pairs in the
semiconductor).
a. If the intensity of the incident light is 1 µW/cm2 and the wavelengths of the incident
photons are 850 nm, calculate the generation rate of electron-hole pairs in the
semiconductor, GL.
First, calculate the energy of one photon with a wavelength of 850 nm.
where h is Planck’s constant, c is the speed of light, and λ is the photon
wavelength.
In this simplified example, band-to-band generation is the only mechanism
contributing to electron-hole pairs. The energy of an incident photon must be at
least as large as the bandgap of the semiconductor to promote an electron from
the valence band to the conduction band.
The intensity of the incident light is does not matter in this case.
b. If the intensity of the incident light is 1 µW/cm2 and the wavelengths of the incident
photons are 712 nm, calculate the generation rate of electron-hole pairs in the
semiconductor, GL.
Since the energy of an incident photon is about the same as the bandgap of
CdSe, it is possible for a photon to create an electron-hole pair. Now it makes
sense for us to proceed to determine the photon flux from the intensity of the
incident light.
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The photon flux is defined as the number of photons incident per unit area,
where the area in question is the top face of the semiconductor bar (because that
is where the photons are striking). Since photons are uniformly absorbed
throughout the volume of the bar, the photon flux divided by the thickness of the
bar will yield the generation rate of electron-hole pairs due to incident light.
c. If the intensity of the incident light is 1.36 W/cm2 and the wavelengths of the incident
photons are 712 nm, calculate the generation rate of electron-hole pairs in the
semiconductor, GL.
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3) Purpose: Getting familiar with various levels of carrier injection.
If the minority carrier lifetime in the semiconductor is 5x10-6 s, what is the resistance of
the semiconductor when it is illuminated with the following intensities? Assume the
incident photons have a wavelength of 712 nm
a. Intensity is 1 µW/cm2.
Recall from Question 2(b) that
Since we are assuming steady state conditions (the light source is continuously
shining on the semiconductor sample) and the semiconductor is experiencing lowlevel injection:
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Recall,
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The carrier concentrations are
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And resistivity can be written as
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Recall, resistance is defined as
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There is virtually no change in resistance from the low-level injection of minority
carriers.
b. Intensity is 1.36 W/cm2.
Recall from Question 2(c) that
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The carrier concentrations are
Resistivity can be written as
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And resistance is
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Resistance is drastically changed in the high-level injection scenario.
4) Purpose: Understanding Quasi-Fermi levels.
Assume that the intrinsic Fermi level lies exactly at midgap and the minority carrier
lifetime is 5x10-6 s. Calculate and sketch the Fermi and/or Quasi-Fermi levels of the
CdSe in the following conditions:
a. In the dark.
Since the semiconductor is in equilibrium, it has a Fermi level but does not have
Quasi-Fermi levels, and we use the typical equations from Chapter 2 (Eq. 2.19a):
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b. Illuminated with 1 µW/cm2 and photons with wavelength of 712 nm.
Quasi-Fermi levels are present because the semiconductor is in non-equilibrium,
and we follow the procedures outlined in Section 3.5.3. Using Equations 3.72a
and 3.72b,
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Note that the quasi-Fermi level for electrons is at the same energy as the real
Fermi level for electrons, because the electron concentration level is unchanged
in low-level injection.
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⁄
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c. Illuminated with 1.36 W/cm2 and photons with wavelength of 712 nm.
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⁄
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Note that both the quasi-Fermi level for electrons and holes move drastically in the
case of high-level injection.
FN = 1.632 eV
5) Purpose: Understanding electron and hole drift current.
If 60 V DC is applied across the length of the LDR in various stages of its illumination,
what will the electron and hole currents be? Continue to assume the minority carrier
lifetime is 5x10-6 cm2/V-s.
a. In the dark.
We know that (Equation 3.19) the total current density in a semiconductor is the
superposition of the electron current density and the hole current density.
And electron and hole current densities are comprised of both drift and diffusion
components.
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In this case, since there is no light shining on the sample, there is no
concentration gradient of carriers (assuming that the semiconductor is
uniformly doped, of course). The bar is in steady-state darkness. Therefore, we
only have the drift components of current. From Equations 3.4a and 3.4b:
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The electric field in the semiconductor bar is:
Recall, in this unperturbed case,
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Therefore,
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The current is completely dominated by the flow electrons, and the hole current
on its own would be immeasurably small.
b. Illuminated with 1 µW/cm2 and photons with wavelength of 712 nm.
Recall, in this low-level injection case,
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The total current is still dominated by electron current. The electron current is
unchanged in this case, because the total number of free electrons is barely
affected by the relatively low number of additional free electrons created by the
light. The hole current increases dramatically, because the number of holes sees
an enormous increase due to the addition of light-generated carriers, but it is
still several orders of magnitude below the electron current.
c. Illuminated with 1.36 W/cm2 and photons with wavelength of 712 nm.
Recall, in this high-level injection case,
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In this situation, both electron and hole currents increase dramatically because
of the dramatic increase in electron and hole concentrations. The electron and
hole concentrations are approximately equal because the amount of lightgenerated carriers far surpasses the original free electron concentration supplied
by dopants. The electron current is 50x greater than the hole current because the
mobility of electrons in CdSe is about 50x greater than the mobility of holes in
CdSe.
6) Purpose: Understanding minority carrier concentration transients.
Consider the case where the CdSe is illuminated with an intensity of 1 µW/cm2 for a very
long time, and the light is suddenly turned off at time t = 0 s. Sketch and label the hole
concentration as a function of time. Denote the hole concentration at time t = 0 s, after 3
minority carrier lifetimes have passed, and after 5 minority carrier lifetimes have passed.
Assume the photons have a wavelength of 712 nm.
Recall, in the situation of a 1 µW/cm2 light striking t,
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From the minority carrier diffusion equation (3.54b), we can say the concentration
of excess holes in an n-type semiconductor when a light is turned off takes the
following form:
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Our initial conditions are:
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At all times,
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Hole concentration in CdSe LDR
4E+10
Hole concentration (cm^-3)
3.5E+10
3E+10
2.5E+10
2E+10
1.5E+10
1E+10
5E+09
0
0.00E+00
5.00E-06
1.00E-05
1.50E-05
2.00E-05
Time (s)
2.50E-05
3.00E-05
3.50E-05