I am planning to build a swimming pool heater, using a solar panel to drive a bilge pump which circulates pool water through 1/2" x (2 x 60) m poly irrigation tube to heat swimming pool.

Solar panel (see link) with following spec:

  • Solar panel = 300 W
  • Power under normal conditions: 20 - 100 W
  • Waterproof Rate: IP65
  • Size: 33.4 * 20.3 * 1.6 cm
  • Tolerance: ±5%
  • SLA battery voltage: 12 V
  • Type of solar cell: polycrystalline
  • Open circuit voltage (Voc): 12 V
  • Maximum System voltage: 12 V

BILGE PUMP (see link):

  • 750 GPH (2840 LPH) (built-in float switch)
  • 12 V 3 A, 5 A Fuse
  • Nozzle = 19 mm, ID = 16 mm
  • Length of wire = 1 m, cable= 3 mm

To test my setup I did the following:

  1. I placed the solar panel under the sun, and measured an open circuit voltage of 19.6V DC on multimeter.
  2. I connected the bilge pump to the panel, the voltage drops to 0.19V DC, and the pump doesn't start.

Can someone please explain what I am doing wrong?

Solar Panel by itself Output=19V

Solar-Pump-Tubing Connection

Solar-Pump Connected Output=0.19V

  • 6
    \$\begingroup\$ Show us a diagram of how you have everything hooked up. \$\endgroup\$
    – vir
    Nov 12, 2023 at 8:06
  • 1
    \$\begingroup\$ what voltage do you measure when the pump is connected? \$\endgroup\$
    – jsotola
    Nov 12, 2023 at 17:24
  • 1
    \$\begingroup\$ @colintd It's there, alright. \$\endgroup\$
    – MiNiMe
    Nov 14, 2023 at 10:55
  • 1
    \$\begingroup\$ @MiNiMe My fault. When I copied over the contents from the duplicate question I mangled the links. Fixed now. \$\endgroup\$
    – colintd
    Nov 14, 2023 at 11:04
  • 1
    \$\begingroup\$ @ StarCat - solar panel dimension = Size: 33.4 * 20.3 * 1.6 cm. Specifications for the solar panel and bilge pump - please click the hyperlink in Question message. \$\endgroup\$
    – Leon
    Nov 14, 2023 at 11:38

2 Answers 2


Simplistically a solar panel can be viewed like a voltage source and a resistor in series, with the value of the resistor related to the brightness of the light the panel is exposed to - the brighter the light, the lower the resistor and the higher the short circuit current from the panel.

If you just measure open circuit voltage, you will get a pretty high reading under most amounts of light. However, unless the output power of the panel exceeds the power requirements of your motor, when the load is connected the panel voltage collapses, yielding very little useable power.

(A more detailed model for PV panel output can be found in this answer to a similar question, and this one.)

The normal way to deal with this, is to add a battery to the panel. Correctly sized, the panel will keep the battery charged, whilst the battery can supply startup surges and average current draw even if the sunlight is temporarily too dim for the required motor current. Having said that, the average power of the motor (allowing for duty cycle) still has to be lower than the average power from the panel (allowing for variation in sunlight).

Ideally the battery is connected via a solar charge controller, which maximizes effective charging power, and prevents overcharge. Solar charge controllers adjust the effective load resistance to match the instantaneous equivalent source resistance of the panel based on current illumination. This is called "tracking". However, some people will just connect the panel directly to the battery. This is probably why the specs reference a 12V SLA battery.

To understand why your panel is undersized (and the 300W advert is fantasy!), observe that midday sun equates to around 1kW/m2, and solar cell efficiency of ~20% means 1m2 of panel won't give more than about 200W.

At 33cmx20cm, your panel has an area of just 0.066m2, which would give you at most 15W in bright sunlight. This is not enough to run your pump (which looks like it needs ~35W), and is entirely consistent with the panel output voltage dropping to just 0.19V under load.

  • \$\begingroup\$ thanks for a clear explanation. I did not opt for a battery because I only want the pump to run when the sun is out (and when the black PE tubing gets heated). So one of the solution is to buy 2 more solar panels 33x20cm (or 1x25W panel?) and connect in series? \$\endgroup\$
    – Leon
    Nov 14, 2023 at 23:59
  • \$\begingroup\$ In parallel might well give you enough power, but it will depend on the startup current of the motor. \$\endgroup\$
    – colintd
    Nov 15, 2023 at 0:03

You've gotten into unfriendly territory. You need a lot of power, and a puny little solar cell won't do what you need.

Your pump is rated at 750 GPH, which is about 12.5 GPM. If you look for water pumps in that range, you can find https://www.homedepot.com/p/FILL-RITE-12-Volt-1-5-HP-10-GPM-Portable-Fuel-Transfer-Pump-with-Standard-Accessories-FR1614/310528825 , which uses 1/5 horsepower for 10 GPM, or 15 - 20 GPM pumps which need 1/4 HP. So let's assume you need about 1/5 HP.

A horsepower is about 750 watts, so you need 150 watts. Worse, without a battery to provide peak (startup) power, you need about 5 times the run load for startup, so you'll need about 1 horsepower, or 750 watts from a pure array setup.

If you have your array aligned so it is pointing directly at the sun, your solar cell will put out something like 60 watts. That the advertising on your cell says 300, or even 20 to 100 means nothing.

So, since 750/60 equals 12.5, you need about a dozen more solar cells. 2 or 3 will not cut it.

Also, keep in mind that, for a minimal system, you CANNOT leave the array in a fixed position. If you have an array pointed straight up, when the sun is 30 degrees above the horizon you'll only get half the power you would when the sun is directly above. When the sun is at 45 degrees from perfect alignment you'll get about 70% of nominal power. Note that seasonal changes in sun elevation will have the same effect as time-of-day movement.

Another way to look at it is to pay attention to your startup condition. Notice that with one cell you get 0.19 volts. A solar cell which is heavily loaded behaves a lot like a current source. So, if you add a cell, you should get a doubling of voltage for this load. Under this model, you need about 60 cells to get 12 volts. Fortunately, this isn't quite true, since once the pump rotor starts to move the effective resistance starts to rise, but it does give you some idea of how underpowered a single cell is for this pump.

Fortunately, you only need about 25% of your solar cell power to keep the pump running, and you can provide the startup surge with a fairly small battery, since presumably you only start the pump once or twice a day. Not so fortunately, keeping batteries properly charged isn't always easy, so you'll need to do some homework there. In principle, you can get away with 3 or 4 extra solar cells, but if you're not willing to build a tracking mount, 5 or 6 would be better. Since you're buying from Alibaba, caveat emptor applies, and you should be aware that polycrystalline solar cells don't do well in terms of efficency for low light levels, which includes off-axis operation.

  • \$\begingroup\$ Thanks for the explanation. I thought my system will work after see these two YouTube connecting 20W solar panel to a bilge pump (1) youtube.com/watch?v=B54s2TAfsPY (2) youtube.com/watch?v=1KC0iWALT_s Any comments? \$\endgroup\$
    – Leon
    Nov 22, 2023 at 10:33
  • \$\begingroup\$ @Leon - Oh yeah. That solar panel is never, ever going to put out 300 watts. Consider a real panel from a reputable source:amazon.com/Renogy-Monocrystalline-Solar-Compact-Design/dp/… Notice that this 100-watt panel is 21 x 42 inches. \$\endgroup\$ Nov 22, 2023 at 15:54
  • \$\begingroup\$ @Leon - Plus, that pump is never a 1/4 HP unit. For that matter, I'm very suspicious about the one tyou have. Double the capacity for an extra 30 cents? Right. Pull the other one - it has a bell on it. Get a 12-volt source and see how long it takes to fill a 5-gallon bucket. (And then run the test where the bucket is significantly uphill of the pump. Prepare to be disillusioned.) \$\endgroup\$ Nov 22, 2023 at 16:03

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