Current and Voltage from photovoltic cells

[HyperSniper]

I'm pretty confused as to how circuit analysis works when using solar cells. They all have ratings for voltage AND for current, but I'm not sure exactly what that means as far as using a cell in a circuit goes.

Let's say I have a solar cell rated at 0.5V and 3.5A wired in series with a 1 ohm resistor and nothing else (so it forms a complete circuit). How would I determine the current and voltage across the resistor? Do I treat the cell like a current source or a voltage source, or something else entirely?

 

[Bob S]

Every solar cell has a V-I curve output which depends on the incident sunlight and the output load. This is (should be) available from the manufacturer. The resistor V-I curve is V = IR. You have two equations in two unknowns. Draw the V-I curve for the solar cell on a V vs. I plot. Draw resistor load lines as diagonal lines on the same plot. Solve it graphically. You already have the solution for a 0.143 ohm resistor (0.5 volts, 3.5 amps) for (I presume) full sunlight. For a 1-ohm resistor, the solar cell voltage will be higher, and the current lower. I show typical solar cell V-I curves for different insolations (solar irradiance) and a resistor loadline in the attachment.

Bob S

 

[gnurf]

You can think of it as a current-limited voltage source. Stare at the solar cell's equivalent circuit for a while:

I_L is a function of the solar radiation and is proportional to the cosine of the angle between the cell normal vector and the Sun vector. If your cell is rated 3.5A, that means at full Sun (AM1.5 if they're terrestrial solar cells) and with a maximum effective area (i.e., cos(0) = 1), I_L = 3.5A. If you shorted the output with an ampere-meter, this is the current that you would measure. It also known as the cell's short-circuit current, I_SC. Of course, a short-circuit also means zero voltage and thus zero power.

If we ignore leakage current caused by the large shunt resistor R_SH and assume an open-circuit as shown in the figure above, all of I_L will flow through the diode. The voltage drop over the diode will follow Shockley's diode equation as usual (notice the temperature dependency), and if we ignore the small series resistor (it represents contact resistance and resistance in the electrodes, etc), you have the other commonly rated parameter; the open-circuit voltage V_OC which in your case is 0.5V.

With that out of the way, you can think of the open-circuit as a very large resistor. As this imaginary resistor gradually becomes smaller and smaller, more and more current will be diverted away from the diode, reducing I_D to the point at which the diode no longer can maintain the voltage across it---the output voltage collapses.

Hope that helps you along.