PHYSICS EXPERIMENTS — 133
19-1
Experiment 19
Solar Cell Investigation
INTRODUCTION: This experiment investigates
the behavior of two simple circuits; each with a
"seat of emf" connected to a (variable) resistor.
One seat of emf is a box acting as a model of a
battery; in the box is a dry cell connected to an
unknown "internal" resistance. The second seat of
emf is a photovoltaic solar cell. We will be
interested in the following: 1) the voltage across the
resistor, 2) the current through the resistor, and 3)
the power delivered to the resistor. This will be
repeated for different values of the resistance. We
will set the resistance to a known value and measure
the voltage across the resistor. With voltage V and
resistance R known, we can calculate the current I
through the resistor (V = IR) and the power it
dissipates (P = IV).
Our goal in using the model battery is to
determine its internal resistance and the value of the
external resistance that enables the maximum power
to be delivered from the battery. When using the
solar cell our goal is to determine the maximum
output power of the cell and, by comparing this to
the input power, to determine the efficiency of the
cell as it converts solar power in the form of
sunlight into electrical power. Furthermore we will
discuss some factors that determine the solar cell's
efficiency and we will try to compare the efficiency
of the cell in sunlight to its efficiency in ordinary
indoor lighting.
ORDINARY BATTERIES: An ideal battery in a
simple circuit does work on charge carriers passing
through it, maintaining a constant potential
difference in the circuit. If the circuit resistance
changes, the current changes accordingly while the
potential difference between the battery's terminals
remains constant. The current depends on the emf
(ε) and resistance according to: I = ε/R. A real
battery differs in that it is not a pure source of emf
but has some internal resistance within it. The
potential difference between the battery's terminals
is now V=ε - Ir, where r is the internal resistance.
The current in the circuit is now given by I = ε/(R +
r).
From the above discussion you now have an
expression for V in terms of I and the constants r
and ε. Assume measurements are taken on a real
system and the data is plotted as V vs. I. It should
be clear that the magnitude of the slope of the line is
r and that the y-intercept is the emf, ε.
The power delivered to the external resistor (P =
IV) depends on the values of the external resistor R
and the battery's internal resistance r. You should
be able to derive an expression for P in terms of ε,
R, and r. In this experiment we can only vary R.
We will look for the value of R that gives the
greatest output power and see how that value relates
to r (obtained from the slope of V vs. I). You
should be able to derive theoretically what the
optimum value of R is by differentiating your
expression for P with respect to R, setting equal to
zero, and solving for R.
A new battery has a very small internal
resistance; as it ages and deteriorates its internal
resistance increases. Thus a plot of V vs. I for a
new battery should be a straight line with a small
negative slope. A weak battery should show a
steeper slope since it has a larger internal resistance.
Other curves may be possible for different seats of
emf. In particular, all V vs. I graphs need not be
straight lines; if the slope varies we could interpret
this as a changing internal resistance. We may find
very non-linear behavior during the investigation of
the solar cell.
V
new battery
weak battery
variable r
I
19-2
RADIATION AND SOLAR CELLS:
It is beyond the scope of this class to go into the
details of photocells. Briefly, the photocells we are
using are made of a very thin layer of "n-type" (or
"p-type") semiconductor silicon that has been
deposited on a p-type (or n-type) silicon base. Each
of these "types" is normally electrically neutral but
the n-type contains loosely bound electrons (which
are negative and hence the name n-type). In p-type
materials there are "absences of electrons" which
can be considered to be positive "holes". The
loosely bound electrons and holes are present in the
material because about one in a million of the
silicon atoms in the silicon crystal have been
replaced by other atomic elements such as arsenic
or boron. Electrons and holes diffuse across the p-n
junction boundary and set up a positive and
negative charge separation region there. This
results in a permanent electric field across the p-n
junction; the electric field is directed from the ntype material to the p-type.
Light can strike the p-n junction. Incoming light
photons supply energy and produce free electrons
and holes in both the p- and n-type material near the
junction. The electrons and holes are then forced by
the electric field to move across the junction, giving
the n-type material a net negative charge and the ptype material a net positive charge. Output wires
connected to the p- and n-type materials will then
be charged + and - , just like the terminals of a
battery. If the wires are connected across a
resistance then charge will flow through the
external resistor delivering power to it.
The efficiency of a photocell depends on whether
or not light is absorbed near the region of the
permanent electric field. Thus the absorbing properties of the photocell itself will effect the efficiency
of the cell. The temperature of the cell can also
effect its efficiency. We will not be able to study
these aspects of a photocell's performance.
It is beyond the scope of this class to discuss in
detail the nature of light. Radiation emitted by a hot
object comes off in very small quantized bundles
called photons. A photon's energy is inversely
proportional to its wavelength. Two things happen
PHYSICS EXPERIMENTS — 133
as an object is heated: it emits more photons
(becomes brighter) and the photons emitted tend to
be at shorter wavelengths (becomes bluer in color).
These ideas help explain photocell behavior. Long
wavelength photons do not have enough energy to
create electron-hole pairs and their energy is
converted to heat in the photocell.
Short
wavelength photons have enough energy to produce
an electron-hole pair but their excess energy also
goes into heat.
The most efficient way to generate electrical
energy in a photocell is to shine photons that have
just enough energy to produce electron-hole pairs.
The wavelength of the minimum energy photon that
can create an electron-hole pair in our silicon cell is
about 1100 nm. The sun's surface has a temperature
of about 5800 K and the wavelength at which the
maximum power is emitted is about 500 nm. For a
typical incandescent light bulb the corresponding
values are 2800 K and 1000 nm. In this experiment
we will see if we can determine a difference in cell
efficiency under these two lighting conditions.
BATTERY EXPERIMENT:
You will use the model battery: it is a box with a
real battery inside along with an unknown resistor.
The real battery provides the emf ε of our model
and the unknown resistor takes the place of the
internal resistance, r.
1) Attach the decade box variable resistor R to
the model battery. Attach the DC voltmeter across
R. Take sufficient data to be able to plot useful V
vs. I and P vs. R graphs. The data consists of a
measured voltage value V for each R that you
select. On your data table you should also have
columns for the calculated values of I and P for
each point. There is no need to choose the R values
uniformly; in fact, it may be preferable to have
closely spaced R points near the maximum of the
power in the P vs. R graph and more widely spaced
R values away from the peak. Roughly determine
where the maximum P is while taking data and then
concentrate your measurements near the maximum.
You will want to be able to accurately locate the
PHYSICS EXPERIMENTS — 133
maximum and yet have enough data points far from
the maximum to give a useful plot.
2) Plot V (vertical) vs. I (horizontal) using an
appropriate scale. What do you expect the slope
and intercept to be? From the plot of your data
determine the internal resistance and the emf of
your model battery. Now open the wooden box
and using a digital multimeter measure the actual
internal resistance (resistor in box) and emf (voltage
of the battery). Compare the actual measured
values with those obtained from your graph.
3) Plot P vs. R and then determine the value of R
that gives the maximum output power. Compare
this optimum R with the value you found for the
internal resistance, r. How well do your results
compare with what is expected from theory?
OUTDOOR SOLAR EXPERIMENT:
The experiment proceeds just as was done for the
model battery, except the photocell will be used
instead of the model battery. Connect the cell to the
variable resistor with the voltmeter across it.
4) Start outside with bright sunlight incident on
the cell. Arrange the cell so that sunlight falls
perpendicular to its surface.
5) Take the necessary data to obtain useful V vs.
I and P vs. R graphs. Find the optimum resistance
that yields the maximum output power from the
cell.
6) In order to determine the cell's efficiency you
will need to know the amount of solar energy
entering the cell. At the top of the atmosphere the
accepted value for solar intensity is 1400 watts per
square meter. We will use a calibrated meter to
determine the solar intensity at our location; the
expected value is about 1000 W/m2. This measured
solar intensity multiplied by your cell's collection
area gives the input power to the cell. (A
disassembled cell is on the front table for
inspection; the actual collector area of the cell is
about 6.0 cm2.)
19-3
7) Do the necessary calculations and plot the two
required graphs. Determine the maximum output
power delivered from the cell to your resistor.
8) From part (6) you know the input power to
your cell. From part (7) you know the maximum
output power delivered by your cell. Determine the
efficiency of your cell in %.
9) The experiment using the model battery
should have shown you that maximum power
transfer occurs when R equals r (a condition usually
referred to as impedance matching). We will
briefly investigate this for the case of the photocell.
Look at your V vs. I graph for your cell. Determine
the slope of the graph (and thus the internal
resistance of the cell) at the point where the
maximum power occurs.
Compare this
experimental value of r with the R at which
maximum power is delivered. How do they
compare?
INDOOR PHOTOCELL EXPERIMENT:
10) Repeat your measurements with your cell
indoors. Use an ordinary 100 W electric light bulb
placed 20 cm above the cell. (Unfortunately, the
bulb's output is not spherically symmetric; for
repeatability of results always try to align your bulb
with its filament horizontal.) In this case we do not
ask you to make complete graphs or take extensive
data. Merely take enough data to enable you to find
the maximum power delivered to your resistor and
the value of R at which this occurs. Again, this is
determined from the maximum of your P vs. R data.
(Take large resistance steps and then go back and
fill in.)
11) Determine the efficiency of your cell. This is
slightly more difficult to obtain and below are step
by step instructions:
a) Obtain the output power from your bulb.
Nominally it is the wattage marked on it. Check
this value using the power meter on the front table.
b) You will need the power per unit area emitted by
the bulb at the location of your cell. Assume that
the power of the bulb is distributed uniformly over a
sphere whose radius equals the distance between the
bulb and the cell. To calculate the intensity of the
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PHYSICS EXPERIMENTS — 133
light at the cell's location take the bulb's power (part
a) and divide by the surface area of the sphere.
surface area = 4!r2
r = radius
photocell
c) You need the power entering the cell. This is the
product of part (b) times the area of the cell. (A
disassembled cell is on the front table for
inspection; the actual collector area of the cell is
about 6.0 cm2.)
d) You have already determined the maximum
output power of your cell in part (10), above.
e) Find the efficiency of your cell under indoor
conditions by taking the ratio of part (d) and part
(c).
How does the efficiency of your cell compare to
its previous value under sunlight conditions?
OTHER INDOOR MEASUREMENTS:
12) We ask you to do the indoor measurements
again, but under different circumstances. In this
case use a lower wattage bulb instead of the 100 W
bulb used previously. You knew the intensity of the
light at the cell when the 100 W bulb was 20 cm
above the cell (part b). Now calculate the distance
above the cell where the new light bulb should be
placed in order to have the same intensity at the
cell. Place the new bulb at that distance from the
cell and repeat the experimental work. Remember
to keep the bulb's filament horizontal. Determine
the cell's efficiency under these circumstances.
13) We ask you to do the indoor measurements
yet again using the 100 W bulb, but this time
connect the bulb to the "variac". Adjust the variac
until the bulb's output power is the same as the
lower wattage bulb that was used in part (12) and
keep the bulb in the same location as in part (12)
with its filament horizontal. What you are doing is
lowering the temperature of the 100 W bulb's
filament and changing the distribution of photon
energies being emitted. Again repeat enough
experimental work to enable you to determine the
peak power delivered to the bulb and the efficiency
of the cell under these circumstances.
rev. 8/05