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What is the cheapest and simplest way to dummy load 450w, 15V, 30A, only for a few seconds?

I would like to measure the max output capability of an MPPT charger controller, basically as a "all losses included" way of measuring actual solar panel maximum power capability.

The theoretical maximum output is 440w. In reality, the solar panels probably won't output more than 350-400w on the sunniest day.

The controller has a load-side shunt, allowing me to measure current easily.

I'm planning to use a MOSFET to short-circuit the load-side through a dummy load. The resistance needs to be very low to allow 30A to flow at 14.4V. I've considered using a ceramic heater, but they are typically dimensioned to have a resistance that restricts current to their dissipation-capability under continuous load, to prevent overheating. I could buy a heater dimensioned at 500w, but it would probably take up a lot of space. Since it only needs to operate for a few seconds, the concern is different.

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  • \$\begingroup\$ You really need something rated for the output at multi-second time periods. If this were milliseconds, I'd say differently. But this is a few seconds... so I think something rated for perhaps 100 watts is the most you should push this. (2000 Joules.) 25 feet of #24 gauge copper wire -- wound around something ceramic would be more than enough. But you could also use a short bit of nichrome from an old toaster, as well, measured to the required resistance. You might spend money, too: mouser.com/ProductDetail/Ohmite/HS200-R47-J/… \$\endgroup\$
    – jonk
    Commented Dec 23, 2016 at 9:54
  • \$\begingroup\$ A suitable light bulb! Or several. \$\endgroup\$
    – winny
    Commented Dec 23, 2016 at 10:00
  • \$\begingroup\$ @winny Ah!! That caused me to think of automotive headlamps, which then caused me to think of pool lamps. Try this: replacementlightbulbs.com/lamp300par56wfl12v.html That is very close to the right value, I think. \$\endgroup\$
    – jonk
    Commented Dec 23, 2016 at 10:03
  • \$\begingroup\$ Any 12 V bulb will do the job at 15 V. Keep in mind that the resistance is about 12 times lower for a cold bulb and the tie constant is around 100 ms. You can't beet the price/W and ease of cooling though. \$\endgroup\$
    – winny
    Commented Dec 23, 2016 at 11:33
  • \$\begingroup\$ I think that the benefit of the single, but expensive, power resistor, is that it is in an aluminum housing which also functions as a heat sink. But if I use 9x 10w cheap resistors, they will also have a larger surface area, so perhaps it is roughly equal? \$\endgroup\$
    – user95482301
    Commented Dec 23, 2016 at 17:05

4 Answers 4

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Why not use the appropriate resistor? Some heavy duty wirewound like this:

enter image description here

You will certainly need to put several in parallel/series to handle all the power, but note that they can be overloaded for a short time without damage. See datasheet. For example, for Tyco THS series (which is available in a wide range of values, from 10 to 75W):

enter image description here

So, if, for example, the pulse lasts for 5 second, you can handle 5 times the power. So you just need a 90W resistor. Take two 1ohm 50W resistor in parallel (THS501R0J), you're done. 3,86 € each at mouser. Can't be cheaper.

However, you'll need a hell of a MOSFET to switch that much current. And the MOSFET will itself dissipate a lot and probably need a heatsink. I would rather use a relay, as RoyC suggested.

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  • \$\begingroup\$ I found some 10w 0.1 Ohm resistors at 0.15 USD per piece, so for 10 pieces that's 1.5 USD, and in parallel the resistance becomes 0.01 Ohm (+ wiring resistance). The MPPT charger naturally limits to max 30A, so this is perfect. Thanks! \$\endgroup\$
    – user95482301
    Commented Dec 23, 2016 at 10:27
  • \$\begingroup\$ What about IRF3205? 110A at 8mOhm when fully switched on? Am I missing something? \$\endgroup\$
    – user95482301
    Commented Dec 23, 2016 at 10:29
  • \$\begingroup\$ IRF3205? Yes, in the worst case, with a 10V gate voltage and at 30A, it will have to dissipate about 6W. You still need to heatsink it, but this is handlable, indeed. \$\endgroup\$
    – dim
    Commented Dec 23, 2016 at 10:31
  • \$\begingroup\$ If the charger limits the current you don't even need the resistors, just use a relay. \$\endgroup\$
    – Vladimir Cravero
    Commented Dec 23, 2016 at 11:00
  • \$\begingroup\$ @user95482301 Vladimir is actually right, I was just thinking about that. It took some time to reach my neurons, but yes, you don't need any actual load, then. Voltage output of MPPT will be much lower than 15V, and you won't dissipate the whole 450W. But then, you're not testing the MPPT in the nominal case. \$\endgroup\$
    – dim
    Commented Dec 23, 2016 at 11:04
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Use an inverter and a few old filament 240V or 120V light bulbs. You may need to replace your Mosfet with a relay. This has the advantage of being easily adjustable (more or different bulb ratings). I use this for battery capacity testing on boats.

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  • \$\begingroup\$ This is a good trick .Better still it does not matter if you have a square wave or sine or even modified sinewave unit. \$\endgroup\$
    – Autistic
    Commented Dec 23, 2016 at 19:38
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Automotive headlamp. That will have no problem with 15V since the electrical system is typically at 14.4V with the motor running.

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The obvious thing that springs to mind is filament lights. Unfortunately, they have this nasty feature whereby the running resistance is often an order of magnitude more than the cold resistance. If you plan to use a nominal 500W of filament lights for a 500W load, then they will tend to overload the output of whatever you connect them to.

This may not matter if the source is a solar panel (which can tolerate a short circuit), or a current-limited output inverter, but it does matter if it's a voltage-output supply with a current trip, try testing a PC power supply by plonking cold 12v halogens onto it!

If you plan to use you bulbs at perhaps 30% of their rated current, they won't change resistance by more than a factor of 2, but you'll need a lot more of them.

As your load only needs to dissipate for a few seconds, you could use something much dirtier. If you have, or can borrow, a 50m length of 3 core cable of 1mm2 cross section, that has a resistance of about 0.8ohms per core cold, increasing to 1ohm for 'not too hot to touch'. Two or three of those cores in series would give you an appropriate sized resistor, for a few seconds. Don't forget to bypass any fuse that's included in plugs if you borrow an extension lead.

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