Which way would the current flow? A quick look at Figure \(\PageIndex{1}\) 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 \(\PageIndex{2}\) is a picture of what this would look like schematically.

Photon flux enters a photovoltaic diode, which is connected to a circuit with no other components. A current of I_photo flows out of the diode anode through this circuit, and there is no voltage difference across the diode. Figure \(\PageIndex{2}\): Schematic representation of a photovoltaic cell

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 Figure \(\PageIndex{3}\) 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?

A photovoltaic diode is connected in series with a load resistor R_L in a closed circuit. A current I emerges from the diode anode, passes through the resistor, and enters the diode cathode. There is a positive voltage difference v_out across the resistor, measured from the side closer to the diode anode relative to the side closer to the cathode. Figure \(\PageIndex{3}\): Photovoltaic 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 Figure \(\PageIndex{4}\).

A current source I_photo, pointing upwards, and a diode with the anode facing upwards are connected in parallel with each other to a resistor R_L. There is a voltage V_out across the load resistor, with the upper side defined as positive. Current I_out flows into the resistor, current I_pv flows out of the upper end of the resistor, and current I_diode flows down through the diode, from the anode to the cathode. Figure \(\PageIndex{4}\): Model of PV cell

Even though we show \(I_{\text{out}}\) coming out of the device, we know by the usual polarity convention that when we define \(V_{\text{out}}\) as being positive at the top, then we should show the current for the photovoltaic, \(I_{\text{pv}}\), as going into the diode from the top, which is what was done in Figure \(\PageIndex{4}\). Note that \(I_{\text{pv}} = I_{\text{diode}} - I_{\text{photo}}\), so all we need to do is to subtract the two currents; we do this graphically in Figure \(\PageIndex{5}\). Note that we have numbered the four quadrants in the \(I \text{-} V\) 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 \(R_{L}\), the load resistor from \(0\) (a short) to \(\infty\) (an open). With \(R_{L} = 0\), we would be at point A on Figure \(\PageIndex{5}\). As \(R_{L}\) 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 \(R_{L}\) gets bigger and bigger, we keep moving along the curve until, at point C, where \(R_{L}\) is an open and we have the maximum voltage across the device, but, of course, no current coming out!

The I-V graph of I_diode is an exponential growth curve, starting at the origin. The I-V graph of I_photo is a straight horizontal line of positive I. The graph of I_pv, which is I_diode minus I_photo, is an exponential growth curve that starts at point A, at -I_photo on the negative y-axis; increases somewhat to point B, located in the fourth quadrant at the bottom right corner of a square of dotted lines with its opposite corner at the origin; and increases further to point C, on the x-axis. Figure \(\PageIndex{5}\): Combining the diode and the current source

Power is \(V \cdot I\) 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 our of out solar cell.

Figure \(\PageIndex{6}\) 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 \((0.25 \ \mu \mathrm{m})\) 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.

Side view of a solar cell waver, with the layers from bottom to top being a back contact, a 1/2-mm thick p-silicon layer, a 1/4-micron thick n-silicon layer, an anti-reflective coating, and a top contact. Figure \(\PageIndex{6}\): A real solar cell

The solar power flux on the earth's surface is (conveniently) about \(1 \ \frac{\mathrm{kW}}{\mathrm{m}^2}\), or \(100 \ \frac{\mathrm{mW}}{\mathrm{cm}^2}\). So if we made a solar cell from a 4-inch-diameter wafer (typical) it would have an area of about \(81 \mathrm{~cm}^2\) and so would be receiving a flux of about \(8.1 \mathrm{~Watts}\). Typical cell efficiencies run from about \(10 \ \%\) to maybe \(15 \ \%\) unless special (and costly) tricks are used. This means that we will get about \(1.2 \mathrm{~Watts}\) out of a single wafer. Looking at point B on Figure \(\PageIndex{5}\) we could guess that \(V_{\text{out}}\) 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.