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Circuit Design Help Answered

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?



cap sizing anyone?

The starting current of the motor could be more than the current provided by the solar panel. Measure the winding resistance of the motor. Divide the operating voltage of the motor by the winding resistance. This will give you the starting current of the motor. You could add one or two large capacitors across the supply. Lets the capacitors charge then use a switch in series to turn the motor.

Remember: There is a difference between starting current & nominal/continuous current which should be supplied to a motor. The motor cannot be supplied starting current at all times. Starting current should only be provided during initial conditions & the nominal voltage & current should be supplied. If starting current is supplied after motor starts to turn, the motor will get damaged. Hence, I suggest using a starter.

I love curves !!! thanks!

The Impp is only 7.43A at NOCT. I hooked two panels together in parallel to try and improve performance. Still no turn of the simple water pump using solar. (The car battery does turn the pump btw). I think we are back to needing a start capacity. Any idea on how to size it? The DC to SDC converter works great ! Thanks!

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

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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.

The DC to DC Converter seems like the ticket. BTW does anyone know what "Buck" refers to? I've attached a simple diagram , does anyone see any gotchas?

Thanks! (I'll take pictures once I get the system up and running. The ultimate purpose of this system is to run water through pool thermal panels under a green house to keep it warmer at night. It is a closed thermal system. Picture two black thermal panels in the sun on the ground, and 4 black thermal panels located under the green house. Towards the end of the daylight, I turn the pump on and the water is warmed , and the pump is driven , both by the sun.)

A buck converter is a step down DC to DC converter that uses a (SMPS) convert the power output to the desired levels. (Switch Mode Power Supply)

Basically it turns the power on and off on one side of a transformer so the other side puts out the desired value.

Very cool ! Never heard of a SMPS, and I'm guessing a guy named Buck invented it. It is only been about 30 years literally since I took electronics in high school (we had vocational schools back then).

I went to a vocational school; South Peel Secondary, after 4 years I was to graduate as a 3 year electrician apprentice. On my third year they abandoned vocational schools and I was not going to get my apprenticeship. When I had my 27 credits for grade 12 I quit school six months early and got a job as a truck driver.

I don't know if the inventors name was Buck but SMPS's are handy just about every DIY person makes a bench supply out of one.

You can salvage one out of any DVD player or some other home entertainment system check out this bench supply made by one of the Instructable members.


I was an International Truck Driver, Canada, USA, and Mexico. Lost my licence 7 years ago when I woke up blind one morning.

I wonder if I still need the capacitor if I have the Buck?

Thanks ! I'll consider these ideas. God Bless.

Ok so you have 500 watts in solar panels and you can't run a 110 watt pump, about right?

Each of those panels should be 250 watts and that motor should be 110 watts.

I know you might not want to here this but you are doing everything wrong.

You need to charge at least a capacitor up to full voltage before you try to start a motor.

Next a full power switch, the full power switch will not turn the motor on until you have enough power to run the motor on the pump. For this you will need to make simple circuit, zener diode, resistor, and a transistor you can use a mosfet, triac, or SCR also.

You could use a photo cell switch works the same only it measures sun and not power in the circuit.

One panel should be plenty to run that motor, in fact it should be twice what you need in full sun.

So once you have full power from your cell about 24 volts run it through a DC to DC converter 24 volts to 12 volts. The DC to DC converter should half the voltage and double the current bringing you well within the pumps needs. DC to DC converters are easy to buy. I would go with 24 volts 5 amps in 12 volts 10 amps out minimum.

1. Mistake: Taking a pump that is already rated higher in current than what the panel can supply in perfect conditions.
2. Mistake: Trying to limit the current - A pump won't start unless there is sufficient current available.
You can expect that your pump, with water in it and a bit of pumping distance needs at least 15-18A to get to normal speed, after this the current will drop down.

Best option is to add proper batteries, charge controller and to either add more panels or to use a pump that is less power hungry.