I'm a newbie when it comes to electricity, and this has been bugging me since like 9th grade when I first learned about V/I/R/P combinations, and to this day I still can't wrap my head around the concept.

I understand about the mathematical formulas, but what I'm having trouble with is the meaning of V/I when it comes to wiring things. I'm thinking of starting a small project wiring up a solar panel to computer fans just to learn about this.

  • Solar panel is rated at DC 6V/1W
  • Fans are rated at DC 12V/160mA

What do these rating mean?

  • Does this mean that the fans can only operate if the source voltage is 12V?
  • What if the source is under-voltage like 6V? Will the fans not power on at all?
  • What if the source is over-voltage like 120V? Will the fans start catching fire?
  • What if the source is just slightly over/under like 13V or 4V?

If my solar panel is rated at 6V/1W (~160mA max), and I connect four fans in parallel, which means each fan will receive ~40mA, does this mean that the fan will not spin? Or will the fans spin at 1/4th of the speed?

Lastly, if my solar panel is only 6V, and fans are operatable at 12V, does this mean that I have to connect two solar panel serially? Can I put a resistor between the panel and fans to increase it to 12V?


How a device will respond to the wrong voltage will depend much on the type of device. A resistor for instance getting half the voltage will dissipate 1/4 the power, and that's it. For motors like the fan it's different. A slightly lower voltage, like 10 V, will probably not be a problem, but at 6 V it may not start. Same here: it only get 1/4 of the required power, and that may be insufficient to get it going. But it may run if you would hand-start it.

120 V will kill it. Final. It's 100 times the rated power and it will burn, possible burst open with a small explosion and some smoke.

A slightly higher voltage than rated for won't do too much harm, but I would avoid it, again for the same reason: power is proportional to voltage squared, so 10 % higher voltage will give 20 % higher power, and since much of the load is resistive it will get hotter.

If your solar panel can supply 160 mA, and you would connect four 160 mA loads in parallel to it it would get overloaded and the output voltage would sag.

enter image description here

For small load changes the graph shows that the voltage won't vary very much, but at 4\$\times\$ overload you'll be in the higher part of the graph where the voltage will decrease quickly with increasing current. The fans would get too little voltage and due to the limited current won't spin.

If you need 12 V and have 6 V solar panels then placing two of them in series is indeed the right thing to do.

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  • \$\begingroup\$ What do you mean fans get too little voltage? I thought the voltage remains the same across the loads in parallel circuit, just the current that's divided up? \$\endgroup\$ – garbagecollector Sep 13 '12 at 6:50
  • \$\begingroup\$ @Dumphole - That's the case as long as the panel can provide that current. So 4 fans at 40 mA per fan would be no problem. But 4 times 160 mA is 640 mA, and at 6 V that would require 4 W, while the panel can only supply 1 W. You can't just keep drawing always more power from it (it would be great if we could!). So to stay within the 1 W the panel compromises by lowering the voltage. I've added a pretty picture :-) to my answer. \$\endgroup\$ – stevenvh Sep 13 '12 at 7:08
  • \$\begingroup\$ If I understand this correctly, a circuit will always try to balance these numbers. The fact that I'm wiring 640mA loads from a power source that can only provide 160mA, means that the power source has to lower it's voltage down to 1.5V to compensate, which then means that each fan will only receive 1.5V, not 6V. \$\endgroup\$ – garbagecollector Sep 13 '12 at 7:33
  • \$\begingroup\$ @DumpHole - The truth will be somewhere in the middle. Suppose your loads are resistors, then 6 V and 640 mA -> 9 ohm total, or 37 ohm per fan. If the voltage would drop to 1.5 V a 37 ohm would draw only 40 mA. The current also drops when the voltage decreases, so you can't just decrease the voltage to 1/4 to get back at the 1 W you can supply. But notice that the graph doesn't follow a constant power line (that would have been a hyperbole). \$\endgroup\$ – stevenvh Sep 13 '12 at 7:48
  • \$\begingroup\$ When there is a voltage drops on the power source caused by too much amperage drawn by the loads, that will short-circuit the power source, causing it to overheat or burn? \$\endgroup\$ – garbagecollector Sep 13 '12 at 22:36

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