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I need to test a battery for its remaining capacity and to do that, I need to generate a constant load to be able to measure how long it can sustain that load.

I tried using a 330W load (290W shoe heater + 85%-ish efficient inverter), but I found out that it couldn't sustain that load for more than half an hour before the voltage was too low. Lower loads seemed to work, though, so I'll need to find a way of creating something that has constant load, but at a lower wattage.

Now, this seems silly-stupid simple, right? Just find something that draws, say 100W and see if that works. Or 50W if that should fail. The problem is, I have a hard time finding anything with a constant power draw in the range 15W-300W. It's impossible to find light bulbs that are not LEDs anymore (1-2W instead of 50W), and my vacuum starts at 350W at the minimum setting. Nothing inbetween.

So I guess I could hack my own? I'm not that great at electro-physics, but I understand it's possible to make some kind of super-basic heater: something with a high resistance causing a lot of thermal heat. I have no idea, though, how to make it work at 12V. Is this something I could hack together?

Alternatively, are there common household appliances that can generate a constant load at 50-150W for hours without inducing mechanical failure of some sort?

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    You can buy Nichrome wire and make your own heater. Different lengths of wire will make different amounts of heat (and use different wattages). Does that sound like something you would want to do? It's easy to burn yourself going this route, so maybe there is something safer, but turning power into heat is probably the simplest. – JPhi1618 Oct 7 at 14:28
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    "I need to test a battery for its remaining capacity" -- could you elaborate? At some point, if the capacity of the battery is in question, it probably makes more economic sense to just replace it. Note also that for most battery chemistries, draining a battery all the way to empty will reduce the battery's lifespan. Maybe from a practical point of view, the tests you've already done suffice. But if not, I don't think you're going to get constant load from any random resistive load (like an incandescent light bulb). As V drops, so will I (current). – Peter Duniho Oct 7 at 16:48
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    If the voltage becomes too low then you may be in the territory of damaging the battery. – Andrew Morton Oct 8 at 8:17
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    @ChrisH: "knowing whether a large, almost new battery is enough to power an event is one potential task for this" -- for that task, I would just operate whatever equipment the battery was intended to operate, until the equipment didn't operate any longer. The theoretical max capacity of the battery, computed based on a simulated load, isn't actually going to give you that answer anyway. – Peter Duniho Oct 8 at 15:28
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    @ChrisH: I have no idea what you're trying to say. No one, least of all me, is saying you should run your "test to depletion" during an actual live performance. You do the test before you need the equipment for real. The point is that you do a test that is identical to the real-world conditions. There's nothing at all about that that's "a bit of rubbish", as you so snidely put it. – Peter Duniho Oct 8 at 16:17
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There are 'off the shelf' battery tester/electronic loads which handle up to 20 A/150 W at a fairly nominal price (considering that they have some degree of monitoring built in) - assuming that you are not really interested in designing your own here.

These use some sort of power FET coupled with a fan/heatsink and constant current control.

For clarity, this is exactly the sort of device where I would buy a low cost, bare PCB style product and import directly (for ~€20) rather than look for a traditional lab instrument.

Electronic Load

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Automotive headlamps are the obvious, easy 12V choice, being typically in the 45-55W range (don't get LED headlamps.)

If you ask the right folks you can often get dual-filament bulbs (high/low beam) with one filament broken for free.

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    This takes the medal. The low beam is typically 35W and high beam 55w, and the low beam usually burns out first. DON'T run both hi and low at once, they're not made for that. – Harper Oct 7 at 20:40
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    This is the obvious answer, but be careful with the temperatures involved and possible fire risks. We once built a test rig using a couple of 12V 55W headlight bulbs, that would heat up a small target (1mm square) to about 1000C in a few minutes!! – alephzero Oct 8 at 9:04
  • Low beam H4 is typically 55W, not 35. – Hobbes Oct 8 at 18:38
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    You can put the bulb & associated wires into a bucket of mineral oil. That's a standard technique for cooling DIY dummy loads. – Paul Uszak Oct 9 at 16:18
  • I've used incadescent loads to test power sources many times. DO care for the heat dissipation. I've burnt insulation by using 4 kW loads... – Crowley Oct 9 at 21:11
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You don't need to step up to mains voltage, in fact for experimenting it's best not to.

There are such things as 12V car fan heaters that are typically 120-150W.

Or you can buy wire-wound load resistors. A 3 Ω resistor would dissipate 48 W at 12 V, so wiring 3 in parallel would work and allow you to adjust the load. I'd buy 100 W-rated models rather than 50 W, which would get rather hot. Either way they should be screwed to a big heatsink or lump of metal, and connected using cables/connectors/switches rated for the current.

If you really want to use an inverter, halogen floodlights are still available. The one I've linked is 120W 230V. Another mains option is one or more low power tube heaters. They range from 40 W up to about 150 W and are typically used to provide background heat against freezing in outbuildings etc. Halogen lamps would be cheaper though.

(UK links because that's what Google assumes I want)

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    The inverter was just a necessary evil: I use 12V at my cabin, but the battery bought in an urban area where I had no 12V consumers. Buying a small, used inverter to use some of my existing gear was the quickest route. – oligofren Oct 10 at 14:42
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For 12V DC, you are probably looking for halogen bulbs, not incandescent. You should be able to get a few hundred watts of 12v halogen bulbs for about $10 online, and you could even get them in units of 20W or so if you want to refine your testing.

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    Halogen bulbs are a subtype of incandescent bulbs, and 20W 12V bulbs are rarely halogen anyway (45W+ headlamp bulbs are). But +1 for a good idea – Chris H Oct 7 at 15:42
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As you know the battery voltage will fall as its charge depletes. Depending on how constant you want the load to be it may need to self-adjust to hold a steady load (wattage).

Ham radio enthusiasts routinely require a "dummy load" for testing radio transmitters without actually transmitting anything. These are basically high-power 50 ohm resistors.

On its own a 50 ohm resistor won't do much for you. Connected directly to the 12 V supply it'll consume only V^2/R=2.9 watts. However, if you put a dc-dc boost converter between the battery and the dummy load, then you can push some serious power into it.

A boost converter that can reach 100 V output from 12 V input will do two things for you: it'll pump 100^2/50=200 watts into the load, and it'll maintain that power as the battery voltage falls. The power level can be adjusted indirectly by adjusting the output voltage of the boost converter. A search for '100 v dc boost converter' on eBay or Amazon yields many options.

5

Officially what you are actually asking for is a constant current sink: something which uses the same current even if the voltage drops.

A light-bulb does that somewhat as it has a PTC thermistor characteristic (The resistance increases when it gets hot).

If you really want constant current you would need to build something electronically. That is probably beyond what you want (too complex and too expensive) but I am mentioning it anyway as there maybe others who want/need a top-notch solution.

Two solutions spring to mind:

1/ Build a real constant current circuit. This is the most accurate and most complex. You can find electronic diagrams using the search term "constant current load" and select images. You probably have to build your own as complete ones are only uses in electronic labs and thus will cost waaaay too much. Also it will have to burn away 150Watts* so think big!

2/ Build/buy a linear 150W 5Volt (which is about 150/5=30Amps) supply. They are a lot more common. To draw a current of X ampere you connect a resistor of 5/X Ohm. Thus for 5V and 1A you use a 5 ohm 5 Watt resistor. The trick is that the supply will use up the rest of the wattage* and adapt if the input voltage drops. It behaves like a constant current load until the input voltage gets too low, probably around 6V at which point your battery voltage is too low to be any use.

*and will get very, very hot. You will need a fan and/or a big heat-sink.

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    'Current sink' is normally used for a negative current source, so supplying power rather than a load. 'Constant current load' is the more usual term, 150W ones specifically for battery testing cost a few tens of pounds on ebay, search for something like '150W 20A Constant Current Load' depending on current requirements. – Pete Kirkham Oct 8 at 12:16
  • @PeteKirkham Great tip! I found exactly one that was just 25$ and could do constant loads from 0-150W and from 1-200V! aliexpress.com/item/32821877897.html – oligofren Oct 10 at 14:57

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