The Light Emitting Diode (LED)
Workshop #11
Physics 102
Name: _________________________
Instructor: __________
Name of Partner(s): _________________ Day/Time of Workshop: ____
Introduction
In previous lectures and workshops, experimental evidence was presented
for the following two postulates:
a) The energy of light is in the form of indivisible “lumps” known as
“photons”.
b) The energy values of electrons bound to atoms take on discrete values,
only. The values between the discrete allowed values are forbidden.
The evidence presented for a) was the photoelectric effect. The evidence for
b) is the discreteness you observed for the frequencies emitted by excited
gases in the workshop Atomic States.
Statements a) and b) are part of quantum mechanics, the modern model of
light and matter.
Today, you will reinforce your understanding of a), by performing a new
experiment that supports the photon hypothesis. To do this, you will produce
light in materials known as Light Emitting Diodes (LEDs). By doing so, you
will introduce yourself to an important and widely-used application of
quantum mechanics – the LED.
In addition, you will also give support for a basic property of materials
known as semiconductors. The property is the existence of a forbidden range
of energies. This property has similarities to property b) of electrons in
atoms.
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Part 1: The Nature of an LED
An LED is a source of light that is less fragile and more efficient than an
ordinary light bulb. Its applications include use as indicator lights, large
bright outdoor television screens, and traffic and brake lights. Solid state
lasers, such as those used to read compact disks, are based on LEDs. It
actually emits light by a process much simpler than that of a light bulb.
First, a single electron of the material of the LED is given energy by an
outside source, such as a battery. Let E be this energy. Second, that energy is
quickly released by the electron. In this release, a photon is created of energy
E, consistent with energy conservation. The process is repeated many times,
resulting in many photons. The stream is the light produced by the LED.
The diagram below summarizes the two processes.
First Step: An electron in the LED accepts energy
e.g., from a battery. Its original state is in the
valence band. Its final state is in the
conduction band.
Second Step: The electron goes
from the conduction band back
to the valence band. To conserve
energy, a photon is released.
Remarks on terminology: The lowest energy set of states has the technical name valence
band. For the unexcited material, at room temperature or below, essentially all of these
states are filled. Each state in the valence band contains exactly one electron.
The highest energy set of states has the technical name conduction band. For the
unexcited material, at room temperature, essentially all of these states are empty. That is,
they have no electron occupying the state.
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In between these two bands there are no states. This region is the forbidden region.
What is the frequency f of the emitted light in the second step in the diagram
on the preceding page? The Einstein-Planck relation E = hf answers this
question. Here, E is the photon energy. By energy conservation, E must also
be the energy accepted by the electron. This relation gives us the frequency
f, and hence the color of the light produced.
Equipment
1.
2.
3.
4.
Two or three LEDs that produce different colors (red, green or blue).
A black power supply. It is a small box powered by a 9 volt battery.
A diffraction grating mounted on a white slide (three per table).
A digital voltmeter, to measure voltages produced by the power
supply.
5. A magnifying glass (also called a loupe), for examining the LED.
If you are uncertain how to make electrical connections, consult the
instructor. Do not stare at any brightly-lit LED. Further, when asked to light
an LED, do not turn the dial on the power supply any further than is
necessary to light the LED. Otherwise, you may burn out the LED.
A. Introduction to the Voltmeter
Turn on the digital voltmeter by turning the dial to 20V (twenty volts). The
meter will then display the voltage (also called the potential difference)
between its two input wires. These are the black and the red wires.
The voltage Vbattery between the two terminals of a battery is the energy per
charge that the battery can give to a charge.
Connect each of the two probes to the voltmeter (black to black, red to red).
Touch the sharp end of one probe to a pole of the battery; touch the other
sharp probe to the other battery pole. Record the voltage across the battery.
Vbattery
=
__________ .
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B. Introduction to the LED
Select one of the LEDs at your table, and examine it. The two wires on
the LED are the “leads”. Note: one wire is shorter than the other.
Now, examine the LED, using the magnifier. Sketch below what it looks
like, particularly inside the clear plastic case. Indicate with words the
apparent shape that each piece of the LED has.
Part 2: Electrical Energy into Light Energy
a). If the battery is not connected to the power supply, insert it in. Turn the
dial on the black box clockwise as far as it will go. This will bring the output
voltage to near zero.
b) Note that the socket on the power box has two rows of small holes. Insert
any one of the LEDs into the socket, putting the short lead into a small hole
that is in the row marked by the shorter yellow line.
c) When you think that you have a good connection, slowly turn up the
voltage on the power supply. Do this by rotating the dial counter-clockwise.
Stop, once the LED lights up.
d) Record below the minimum voltage Vcritical needed to light up the LED.
Do this by measuring the voltage across the LED when part c) is completed.
To measure this voltage, put each of the sharp probes against the two leads
of the LED. Also, record the color of the light.
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e) Still holding the voltmeter probes across the LED, reduce the voltage
across the LED by rotating the dial clockwise. Does the LED light up when
the voltage across it is less than Vcritical?
f) Use your magnifier to note which part of the LED lights up. Note this on
your sketch on the preceding page.
Part 3: Differently Colored LEDs
a) Recall: A diffraction grating separates light into its pure components. Try
out your diffraction grating by looking at the white light from the ceiling
lights. To do this, hold the grating almost up to your eye. Then, look to the
sides, or else look up and down, by moving your eyes. Do this without
moving your head.
Describe the spectrum of colors you see.
b) Repeat for the light you see from the LED. Note below the major
differences in the light produced, relative to part a). Does one of your
observations include a wide range of frequencies, in contrast to the other?
c) Choose one of the other two LEDs. Repeat subparts a), b) of this Part for
this LED. Record your results below.
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d) In the two rows below, summarize the color of the LED and the minimum
voltage needed to light up the LED. Do this for each of the two LEDs.
LED :
Color
Critical Voltage
#1
#2
e) Which of the two colors has the higher frequency f for the light produced?
How do you know? (The dependence of color on wavelength is below.)
f) Which of the two LEDs requires the stronger voltage to light up?
g) Compare your answers to parts e) and f). Discuss, in a few sentences,
whether or not your results are consistent with the photon hypothesis that
connects the photon energy with its frequency.
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Part 4: The Energy Gap
You have noted from your experiments on both LEDs that:
There is a critical voltage below which the LED does not light up.
Hence: When the voltage is below this critical value, the electrons in the
LED material do not accept the energy from the battery.
(How do we know this? Reason: If they did accept energy from the battery in this case,
the electrons would go to higher energy states. Once this happens, by spontaneous
emission, they would release their energy and thereby produce light.)
In the recent lectures on atomic states, it was pointed out that atoms are
“fussy”, in that they do not always accept energy offered to it from the
outside. To accept energy, the energy offered must match an energy
difference of two allowed discrete states of the atom.
Similarly, you have today seen that the LED, a solid, is also “fussy”. It will
not accept energy lower than a critical value.
This critical value is called the gap energy EG. The reason for its name is
that EG is the size of a gap in the energy values. Values of the energy inside
the gap are forbidden.
Consider an LED, made of a particular material. Then, there is one
characteristic value for EG, for that material. The value of the gap energy EG
gives the size of the forbidden energy range.
This is illustrated by the energy level diagram on the next page. Let E0 be
the initial energy of an electron in the LED that can receive energy from the
battery. This value is the largest electron energy of the unexcited material.
Above the value of E0 lies the forbidden region. The size of the forbidden
region is the gap energy EG.
The material of the LED is a type of solid known as a semiconductor. A
semiconductor is a material which has an energy gap of this type, with the
size of the energy gap between 0.5 eV and 10 eV, roughly.
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Energy Level Diagram for a Semiconductor
.
a) Compute, on the next page, the size of the energy gap for each of the two
LEDs. Express your result in the unit of the electron volt.
Hint: Use the relation that gives the potential energy of a charge q placed at a point for
which the electric potential is V, relative to zero energy. The relation is that the potential
energy received is qV.
Also use the fact that the magnitude of the electron charge is 1.6 x 10-19 Coulombs.
Reminder about units: The Coulomb is the Standard International (SI) unit for charge. For
electric potential, the SI unit is the volt. Now, whenever you use SI units in any relation,
your result will remain in SI units. Hence, your result for the energy will be in the SI unit
for energy – the joule.
However, the joule is rarely used to express atomic energies or energy gap sizes. The
reason is that you will then wind up with clumsy factors such as 10-19. Hence, the
electron volt (eV) is almost always used to express these quantities.
Note that the definition of the electron volt is: 1 eV = 1.6 x 10-19 joules.
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Calculation for the two values of the energy gaps, for LED #1 AND #2:
b) Explain why the presence of the energy gap prevents the LED from
lighting up when the voltage across the LED is less than Vcritical..
Do this in a paragraph or two below.
Hint: Can the electron accept energy if acceptance would put its energy in the gap
region?
c) Consider two LEDs. One produces red light. The other produces blue
light. Which LED has the larger gap? Why?
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Part 5: The Conversion of Light Energy into Electrical Energy
In Parts 2, 3, and 4, you converted electrical energy into light energy. You
gave electrical energy to the LED by connecting it to a power supply. The
power supply is powered by a battery.
In this part, you will produce the opposite transformation. Light energy will
be transformed into electrical energy. This is exactly what a solar cell does.
In the case of the solar cell, the light energy is from the sun.
A solar cell can generate a voltage when sunlight strikes it. In both the LED
and the solar cell, energy in the form of light can be changed into electrical
energy. The material that receives light is called a photovoltaic. This is
because when light (photo) is received, electrical energy (voltaic) is
generated in the material. The material of the LED is a photovoltaic, similar
to that in a solar cell.
In the experiment of this part, one LED (to be called the illuminator) will be
the source of light. A second LED (the receiver) will be the photovoltaic.
a) Select two LEDS of differing colors. Tape them together with Scotch tape
“nose-to-nose”. That is, the curved ends of both LEDs should face and touch
each other.
b) Identify the LED that produces light of the higher frequency of the two
LEDs. This LED will be the illuminator LED in this part. It will produce the
light whose energy you will try to convert into electrical energy. Insert its
two leads into the power supply box.
Turn up the voltage on the power supply until the LED is brightly lit, (but
not too bright!).
c) The other LED is the receiver. Its light has a lower frequency than the
illuminator. Its face is now down. Hold each of the two voltmeter leads,
against the two leads of the receiver. It may be easier for two students to
work with the receiver LED. Pinch with your thumb and finger to hold the
leads together at the places shown by the arrows below.
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d) Another partner should now turn the dial on the power supply back and
forth. Then, the light on the illuminator will be turned on and off.
Is there any variation in the voltmeter reading, as the illuminator is
turned on and off? Remark: Stray light from lamps or fixtures in the room
may activate an LED. To avoid confusion from this, it is helpful to turn up
and down the brightness of the illuminator and then see if the voltage across
the receiver changes. It is not necessary to darken the room.
Discuss your observations. Have you transformed light energy into
electrical energy? How do you know?
e) Repeat steps b) through d), with the following change: Choose your
receiver LED to have a higher frequency than that of the illuminator. That
is, reverse the roles of illuminator and receiver.
Have you succeeded in transforming light energy into electrical energy
for this case? Discuss your observations.
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g) You may have noted a clear difference between your two sets of
observations. That is, whether or not electrical energy is imparted to the
receiver may depend on the whether the receiver has a low or a high
frequency, relative to the illuminator.
Draw, on the diagrams below, the photon energy of the illuminator, and
contrast it with the energy needed for the receiver LED to accept the photon
energy.
Case where filluminator > freceiver.
Case where filluminator < freceiver
Note: For both of these diagrams, the LED represented is the receiver LED.
Write a paragraph or two below to explain the difference between your
results for the two cases.
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