Trying to build a simple circuit to power a small pump while the sun is out. Here are the specifics:

Solar Panel (60 cell mono panel) Open circuit voltage 36.4V DC, Short circuit current 7.96A. Pump 12 V DC, 9A max

Using E=IR , R=E/I, (36.4-12)/7.96 = 3 ohms Power = V(sq)/R = 24.4/3 =198 watts

I purchased a resistor that meets these requirements and hooked everything in series. (Solar panel lead to resistor, resistor to pump, and pump to other solar panel lead.

The pump does not turn on when the panel is in the sunlight. I measure the voltage coming out of the panel and it is within specification. I also tested the pump by hooking it up to a car battery and it works fine.

Any ideas what I am doing wrong?

  • 1
    \$\begingroup\$ What you are doing wrong is thinking that you can have maximum voltage and maximum current at the same time. You need to look at the voltage - current curve for your panel and figure out the relationship. \$\endgroup\$
    – Transistor
    Commented Mar 26, 2016 at 9:20
  • \$\begingroup\$ 2 things: 1) By connecting the panel to your motor, you won't be operating at Voc(open circuit voltage), nor you can expect Isc(short circuit current). - solution: check the data sheet of the panel, you will find some graphs called as VI characteristics, it will give you some idea. 2) you are using a solar panel, which clearly is not a battery, it's output power will vary based on irradiance and temperature, so you won't be getting a constant voltage or current, make sure you are operating within the absolute maximum ratings of your motor. \$\endgroup\$ Commented Mar 26, 2016 at 10:49
  • 1
    \$\begingroup\$ And you shouldn't need a resistor... that just burns power. \$\endgroup\$ Commented Mar 26, 2016 at 14:45

2 Answers 2


You can estimate the V - I characteristics of your solar panel to get a general idea of the performance. Here is an estimate (corrected):

enter image description here

I am assuming that the 9 amps maximum motor current is the maximum continuous operating current, the current for rated motor power output. The input power to the motor for full load operation would then be 108 watts. I am also assuming that this is a permanent-magnet motor with a commutator, not a brushless motor with an electronic control or commutating circuit. The full-load losses due to the resistance of the armature winding would probably be something like 10 watts. If we assume that the resistance of the armature and brushes is (correction) 0.125 ohms, the load line of the motor plus series resistor intersects the solar-panel curve at about 24 volts and 7.7 amps. When the motor is not turning, there is no back EMF, so the solar-panel current is determined by the series resistance plus the resistance of the motor.

(correction) That would indicate that the motor should be able to overcome friction and accelerate the pump to the point at which the load increases to the point that the power required is less than the power available. If the solar panel is receiving maximum sun, the motor should operate al least slowly.

You need a solar panel with a maximum-power-point (MPP) voltage that is equal or slightly above the rated voltage of the motor. The MPP current can exceed the motor current rating by as much as you want, but the motor will need to be protected. It could draw too much current and overheat if there is a mechanical problem that causes an overload. The You can then connect the motor directly to the solar panel without a series resistor.

When there is less than the maximum sunlight, including the effect of the sunlight hitting the panel at an angle, the V-I curve will have a similar shape, but lower current values. At lower sun brightness, the load line will intersect the V - I curve at a lower voltage point and the motor will slow down. If it slows down too much, it could stall, so it should be turned off at some minimum speed. It could overheat even if the current is not too high because it will not cool properly if it is running too slow.

The nature of a centrifugal pump load will help with the performance.

Note that the load line with an operating motor and load will not be a straight line and will change as the motor speed changes.

  • \$\begingroup\$ If you have a brushless motor, the initially applied, high open-circuit voltage may damage the controller or activate some protection that shuts it off. \$\endgroup\$
    – user80875
    Commented Mar 27, 2016 at 16:44

The battery runs the motor because it can provide startup current to the motor. The start up current is several times larger then running current. Solar panels may keep the motor running but wont start the motor. Either increase the solar panels until there is enough in rush current (startup current) or charge the battery through the solar panels and have the battery run the motor. There are some circuits that incorporate capacitors that provide the startup current. The startup current lasts for about 0.2 seconds.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.