Now let us look at the opposite process of light generation for a moment. Consider the following situation.

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 valance band, and "kick" them up into the conduction band, leaving a hole behind. As you can see from Figure 1, this creates excess electrons in the conduction band in the p-side of the diode, and excess holes in the valance 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 Figure 1 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 (Figure 2)! Here 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 potential 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 Figure 3 that the photo current will go through the load resistor in such a way that 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?

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 Figure 4. Even though we show Iout Iout coming out of the device, we know by the usual polarity convention that when we define Vout Vout as being positive at the top, then we should show the current for the photovoltaic, Ipv Ipv as current going into the top, which is what was done in Figure 4. Note that I pv = I diode − I photo I pv I diode I photo , so all we need to do graphically is to subtract the two currents 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 II and VV is a positive number, meaning that power is being dissipated in the cell. For quadrant II and IV, the product of II and VV 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 RL , the load resistor from 0 (a short) to ∞ (an open). With R L =0 R L 0 , we would be at point A on Figure 5. As RL RL starts to increase from zero, the voltage across the device will start to increase also, and we will move to point B, say. As RL RL gets bigger and bigger, we keep moving along the curve until, at point C, where RL RL is an open and we have the maximum voltage across the device, but, of course, no current coming out! Power, of course, is VI V I 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.

Figure 6 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 contact 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.

The solar power flux on the earth's surface is (conveniently) about 1kWm2 1 kW m 2 or 100mWcm2 100 mW cm 2 . So if we made a solar cell from a 4 inch diameter wafer (typical) it would have an area of about 81cm2 81 cm 2 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 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.

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This work is licensed by Bill Wilson under a Creative Commons Attribution License (CC-BY 1.0), and is an Open Educational Resource.

Last edited by Elizabeth Gregory on Jun 20, 2003 12:00 am GMT-5.