I have a pair of solar panels mounted on my roof. They are rated 90W each, but I want to know, given my latitude and installation, what amount of power they are actually able to generate. I have a test setup that includes two 2-ohm, 100W resistors, a ground, a voltmeter, and ammeter. I can attach one resistor to make a 2-ohm drain, or I can attach them in parallel or series to make the drain 1 or 4 ohms, respectively. If I put this drain circuit on my solar panels and I read 15 volts at 7 amps, then I know that my nominally-rated 180W panels are realistically capable of producing 105W on that date and time at my latitude and under my installation conditions.

I have made 36 sets of measurements at various dates and times, always in full sunlight (no clouds). When I make them, I always take readings with the resistors in the 1, 2, and 4-ohm configurations. I'm puzzled by the fact that their power draws are rarely equal. For example, here are the readings I took yesterday:

No drain: 19V 0A

1-ohm drain: 8V 7.2A = 58W

2-ohm drain: 14V 6.8A = 95W

4-ohm drain: 17V 4.4A = 75W

This is pretty typical in that the 2-ohm drain produced the highest power draw. In fact, out of 36 measurements, that has been the result 23 times. 10 times, the 4-ohm drain produced the highest draw; 2 times, the 1-ohm drain did, and once, all three readings were equal. I haven't been able to detect any pattern as to under what sunlight conditions a given resistor configuration will produce the highest draw.

First question: why does the power produced by the solar panels in my test setup differ depending on whether the drain circuit is 1, 2, or 4 ohms? Second, any ideas why the 2-ohm drain usually does a better job of maxing out the solar panels than the 1 or 4-ohm drain?

Incidentally, in case anyone is wondering, the highest power output I've been able to get out of this 180W solar panel setup is 112W. That was 14.0V, 8.0A, with a 1-ohm drain.

1 Answer 1


A solar panel is a diode. It's just a diode where light generates charge carrier pairs. So it obeys the ideal diode law.

An ideal diode law with charge carriers created by light is as follows:

I = I_l - I_s*(exp(V/(n*V_t)) - 1)

...where I is the external current, I_l is the photogenerated current, I_s is the saturation current (a diode model parameter), V is the voltage at the diode, n is the ideality factor (usually between 1 and 2), and V_t is the thermal voltage depending on the temperature of the diode (26 mV at room temperature).

The power created for an external load is:

P = V*I = V*(I_l - I_s*(exp(V/(n*V_t)) - 1))

You see that in this expression, the V inside the diode model isn't cancelled by the V by which the current is multiplied.

You may consider the following thought experiment:

  1. You leave the solar panel at an open circuit, with infinite resistance across its terminals. Will the power still be the nominal power of the solar panel? One way that would allow open circuit to create power is an infinite voltage, but can sunlight really create an infinite voltage?
  2. You short-circuit the solar panel, so that resistance is zero across its terminals. Will it still generate power at the short circuit? P = R*I^2 so if R is zero, I would have to be infinite to have something else than zero. Can finite sunlight generate infinite charge carrier pairs?

These thought experiments demonstrate something: you can't expect a solar panel to produce constant power all the time. In particular, if you short circuit a solar panel, power has to be zero since voltage is zero. If you leave it with an open circuit, power has to be zero since current is zero.

Why would you therefore expect that if the resistance is something else than zero or infinite, that the power would not depend on the resistance?

Another way to think about this is as follows: light creates a current I_l. Part of that current is recombined already at the diode according to the Shockley diode equation (I_s*(exp(V/(n*V_t)) - 1)) and only part ends at the external circuit where it can do useful work. How much work it can do depends on the voltage (because voltage affects power, and also because voltage affects how much current is recombined already at the diode), which depends on the resistance. So actually the solar panel may be thought of as a device with constant power, only some of that power gets lost in the panel itself as heat, and how much gets lost as heat depends on the operating conditions.

This is exactly the reason why we have maximum power point tracking charge controllers (to allow better efficiency than PWM charge controllers). This is exactly the reason why any decent grid-tie solar inverter already has a maximum power point tracking device built in.

If you want maximal power, there is only one (voltage,current) operating point that generates maximal power, although with partially shaded complex solar systems, there may be local false optima in addition to the true global optimum.

If you want to do something useful with your panels, other than testing it with resistors, you have two ways:

  1. Attach a battery into it, with either:
    • PWM charge controller (cheaper)
    • MPPT charge controller (more efficient)
  2. Attach a grid-tie inverter into it and sell the electricity to grid
  • "Why would you therefore expect that if the resistance is something else than zero or infinite, that the power would not depend on the resistance?" This verges on calling the question a stupid one. I expected what I did because I didn't understand how it worked. "If you want to do something useful with your panels, other than testing it with resistors," I already am, but be that as it may, I consider doing tests, writing down measurements, and studying the results to be an excellent use of my time and equipment, regardless of whatever else I use my panels for.
    – davo
    Aug 9, 2023 at 15:28

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