Solar Cells2.132000/08/042007/08/14 12:08:08.624 GMT-5BillWilsonwlw@madriver.netBillWilsonwlw@madriver.netElizabethGregoryelizabeth.gregory@gmail.comJeffreyMSilvermanJSilverman@astro.berkeley.eduGerardWysockigerardw@rice.educellssolarSolar Cells
Now let us look at the opposite process of light generation for
a moment. Consider the following situation.
P-N diode under illlumination
We have just a plain old normal p-n junction, only now, instead
of applying an external voltage, we imagine that the junction is
being illuminated with light whose photon energy is greater than
the band-gap. In this situation, instead of recombination, we
will get photo-generation of electron hole pairs. The photons
simply excite electrons from the full states in the valence
band, and "kick" them up into the conduction band, leaving a
hole behindThis is similiar to the thermal
excitation process we talked about earlier.. As you can
see from , this creates excess
electrons in the conduction band in the p-side of the diode, and
excess holes in the valence band of the n-side. These carriers
can diffuse over to the junction, where they will be swept
across by the built-in electric field in the depletion region.
If we were to connect the two sides of the diode together with a
wire, a current would flow through that wire as a result of the
electrons and holes which move across the junction.
Which way would the current flow? A quick look at shows that holes (positive charge
carriers) are generated on the n-side and they float up to the
p-side as they go across the junction. Hence positive current
must be coming out of the anode, or p-side of the junction.
Likewise, electrons generated on the p-side fall down the
junction potential, and come out the n-side, but since they have
negative charge, this flow represents current going
into the cathode. We have constructed a
photovoltaic diode, or solar cell!
is a picture of what this would look like
schematically. We might like to consider the possibility of
using this device as a source of energy, but the way we have
things set up now, since the voltage across the diode is zero,
and since power equals current times voltage, we see that we are
getting nada from the cell. What we need, obviously, is a load
resistor, so let's put one in. It should be clear from that the photo current flowing through
the load resistor will develop a voltage which it biases the
diode in the forward direction, which, of
course will cause current to flow back into the anode. This
complicates things, it seems we have current coming
out of the diode and current going
into the diode all at the same time! How
are we going to figure out what is going on?
Schematic representation of a photovoltaic cellPhotovoltaic cell with a load resistor
The answer is to make a model. The current which arises due to
the photon flux can be conveniently represented as a current
source. We can leave the diode as a diode, and we have the
circuit shown in . Even though we show
Iout coming out of the device, we know by
the usual polarity convention that when we define
Vout as being positive at the top, then we
should show the current for the photovoltaic,
Ipv as current going into the top, which
is what was done in . Note that
IpvIdiodeIphoto, so all we need to do is to subtract the two currents;
we do this graphically in . Note
that we have numbered the four quadrants in the I-V plot of the
total PV current. In quadrant I and III, the product of
I and
V is a positive number, meaning
that power is being dissipated in the cell.
For quadrant II and IV, the product of
I and
V is negative, and so we are
getting power from the device. Clearly we
want to operate in quadrant IV. In fact, without the addition
of an external battery or current source, the circuit, will
only run in the IV'th quadrant. Consider
adjusting RL, the load resistor from 0 (a short) to
(an open). With
RL0, we would be at point A on . As
RL starts to increase from zero, the
voltage across both the diode and the resistor will start to
increase also, and we will move to point B, say. As
RL gets bigger and bigger, we keep moving
along the curve until, at point C, where
RL is an open and we have the maximum
voltage across the device, but, of course, no current coming
out!
Model of PV cellCombining the diode and the current source
Power is
VI so at B for instance, the power coming out would be
represented by the area enclosed by the two dotted lines and the
coordinate axes. Someplace about where I have point B would be
where we would be getting the most power out of out solar cell.
shows you what a real solar cell
would look like. They are usually made from a complete wafer of
silicon, to maximize the usable area. A shallow (0.25 μm)
junction is made on the top, and top contacts are applied as
stripes of metal conductor as shown. An anti-reflection (AR)
coating is applied on top of that, which accounts for the bluish
color which a typical solar cell has.
A real solar cell
The solar power flux on the earth's surface is (conveniently)
about
1kWm2 or
100mWcm2. So if we made a solar cell from a 4 inch diameter
wafer (typical) it would have an area of about
81cm2 and so would be receiving a flux of about 8.1 Watts.
Typical cell efficiencies run from about 10% to maybe 15% unless
special (and costly) tricks are made. This means that we will
get about 1.2 Watts out from a single wafer. Looking at B on
2.59 we could guess that
Vout will be about 0.5 to 0.6 volts, thus
we could expect to get maybe around 2.5 amps from a 4 inch wafer
at 0.5 volts with 15% efficiency under the illumination of one
sun.