EE here.

First of all, it is important to note that PV cells/arrays are *not* modeled as voltage sources; over most of their operating range, they are considered **current sources**. So what does this mean?

a) When there is no load, the current is zero, the power is zero, and the voltage is at a maximum. This voltage is called Voc. This is also what you would measure if you hooked the array to a digital voltmeter.

b) When the array is short-circuited, the current is at a maximum, the power is zero, and the voltage is zero. This current is called Isc.

c) When there is a resistive **load** on the array, you’ll have non-zero power, voltage, and current. In other words, you’ll be somewhere between the two (extreme) conditions described above.

d) There is a voltage at which power is at a maximum. This voltage is called the Maximum Power Point (MPP).

e) When the voltage is below the MPP, the array behaves like a current source.

f) When the voltage is above the MPP, the current takes a nosedive. As voltage continues to increase the current goes to zero.

g) For a typical PV array, the MPP voltage is around 80% of Voc.

Second, PV array manufacturers are not conservative when quoting specs. When they quote a voltage, it is usually Voc. This is very deceptive because you can’t operate the PV array at Voc. And when they quote a power rating it is usually at the MPP voltage. This is just about as deceptive, because you can only get that power when the load is at one particular value of resistance.

Now to the analysis…

According to your first post, you said the panel is “12V / 2 watt / 100 mA.” This really doesn’t “add up,” if you know what I mean. If we assume 100 mA = Isc, then the MPP voltage would be 20V, and Voc = 25V. Did you measure 25V with your voltmeter? Probably not. I would assume, then, that Voc = 12V (is *this* what you measured with your voltmeter?), and the MPP would be at approx. 9.6V. Therefore, the current is approx. 210 mA at the MPP and, according to e) above, the current is approx. 210 mA when the voltage is between 0 and 9.6V.

I = 210 mA when V < 9.6 volts.

With me so far? Now what we need to know the resistance of the load . Once we know this, we can determine the operating point.

Let’s assume your motor draws 100 mA at 6V. This means the resistance is 60 ohms. Now hook the motor up to your PV array. Now remember, the PV array is a current source, so it tries to push 210 mA through a 60 ohm load. This requires it to supply 12.6V. And guess what? It can’t do it. Because at 12V (assume Voc = 12V), the current goes to zero. (And in a way, be *glad* it can’t do it, else your motor would smoke.)

I made a lot of assumptions, but I believe my analysis is correct. If you really want to know the answer, do the following:

- In full sunlight, measure the PV array with a digital voltmeter. This value is Voc.
- Hook the vacuum cleaner up to a 6V power supply and measure the current (I_motor). Calculate the resistance using R_motor = 6/I_motor, where I_motor is in amps.
- Calculate the MPP voltage of your array. The MPP voltage is approx. 0.8*Voc.
- Calculate the current of the array: I_array = 2/MPP voltage.
- If I_motor > I_array, then you’re automatically out of luck.
- If I_motor < I_array, then you’re still not out of the woods. You must now determine where you’re at on the curve. To do this, calculate the voltage the array would generate if you hooked the motor up to the array: V_array = I_array*R_motor. What is this voltage?? If it’s around 6V, then it would work. If it’s much above 6V (like above 9V), you might burn up the motor! But if it is much below 6V (like below 4V) then the motor won’t run.

Hope this helps.