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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?

oligofren
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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 07 '19 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 07 '19 at 16:48
  • That said, if you really really want a constant load, the easiest approach is to just buy a bench load. You can get a small one that would suit your purposes for under a grand. Even less if you buy a used one. That takes me back to the previous question: what kind of battery is this and why do you need to test the capacity? It'd have to be a pretty pricey battery to justify that kind of work; if an actual bench load is too expensive to justify here, seems like buying a new battery would make more sense. – Peter Duniho Oct 07 '19 at 16:51
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    If the voltage becomes too low then you may be in the territory of damaging the battery. – Andrew Morton Oct 08 '19 at 08:17
  • @AndrewMorton's point is good, and may justify the use of an inverter as these normally have a low-voltage cutoff to protect the battery. However the inverter should be connected to the battery by short, fat wires to avoid voltage drop in the wires causing an early cutoff – Chris H Oct 08 '19 at 09:12
  • @PeterDuniho knowing whether a large, almost new battery is enough to power an event is one potential task for this, which I've supported in the past. In that case a battery was bought to power an event of say 1 hour, would it do 90 minutes? Or should we bank on swapping batteries at a good moment? For a charitable event with no budget for hardware. – Chris H Oct 08 '19 at 09:14
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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 08 '19 at 15:28
  • @PeterDuniho that's a bit rubbish if your PA cuts out in the middle of a speech. Much better to swap between speeches. But if you swap too soon, do you use up your reserve battery? – Chris H Oct 08 '19 at 15:31
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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 08 '19 at 16:17
  • "A bit rubbish" was referring to my own performance if I was responsible for running out of battery at the wrong moment. And you can't test event audio under real world conditions without a soundproof room or very understanding neighbours - the drain on the battery depends on how much sound you're making. In that case it's better to test on a dummy load slightly higher than your maximum expected (measured briefly) output. But I think this digression isn't getting very far. It was it meant to be an example of 1 case when testing with a known load is more suitable than buying a bigger battery – Chris H Oct 08 '19 at 16:26
  • @PeterDuniho It's indeed an expensive battery: Concorde Sun Xtender PVX 2580L runs at about 1300 USD where I live. They are about as good as you can get for non-LiFePo batteries; extremely rugged, don't freeze and survive hundreds of deep cycles. They also weigh 160 pounds, which is why I don't want to transport them to/from my remote cabin without knowing they are good performers. I have recently bought three used ones. One I know is not a good performer, whereas the two others I have high hopes for. – oligofren Oct 10 '19 at 14:20
  • @AndrewMorton Yes, when you do this test you run them to about 11 volts and immediately start recharging them to avoid permanent damage. This is basically a test that battery dealers do when they test the status of the battery. Last time I tested my Concorde 2580 it had 98% capacity at 7 years of age. Pretty good stuff :) – oligofren Oct 10 '19 at 14:22
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    It seems like you already did your test, and already got your answer. A 330W load is perfectly gentle for a battery that weighs 160 pounds. I would expect 2000+ Wh out of such a battery, but 165Wh is conceivable if the battery is near end of life. – Harper - Reinstate Monica Nov 24 '19 at 17:15

8 Answers8

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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.

Ecnerwal
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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 - Reinstate Monica Oct 07 '19 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 08 '19 at 09:04
  • Low beam H4 is typically 55W, not 35. – Hobbes Oct 08 '19 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 09 '19 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 09 '19 at 21:11
  • @Hobbes oh never an H4 or any of those bitsy things, they burn at 500F, and explode if you get finger oil on them. you'd need to build a safety cage for them or they'll set your test lab on fire. I'm talking plain old sealed beams like 70‘s cars use, 5x7x5 inches. They have lots of surface area, so they run cool enough to hold in your hand. https://www.walmart.com/ip/Halogen-Sealed-Beam-Lamp-165mm-50W-GE-LIGHTING-H4651/36903380 – Harper - Reinstate Monica Nov 24 '19 at 17:29
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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)

Chris H
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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 '19 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.

IronEagle
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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 07 '19 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.

Greg Hill
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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.

Oldfart
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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 08 '19 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! https://www.aliexpress.com/item/32821877897.html – oligofren Oct 10 '19 at 14:57
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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

Sean Houlihane
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  • Agreed -- I've been happy using one like this https://www.bkprecision.com/products/dc-electronic-loads/8600-150-w-programmable-dc-electronic-load.html, though there are probably cheaper ones that fit OP's specs better. – Justin Oct 09 '19 at 19:24
  • Perfect! I found this: https://www.banggood.com/250W-DC-12V-Discharge-Battery-Capacity-Tester-Module-With-DC-Electronic-Load-Digital-Battery-Tester-p-1383823.html – oligofren Oct 10 '19 at 14:47
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    "Perfect! I found this banggood" is not a phrase I'd hope to hear very often on this site, though this is one place it makes sense. – Harper - Reinstate Monica Nov 24 '19 at 17:17
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$8 55-watt https://www.harborfreight.com/12-volt-halogen-vehicle-work-light-93904.html

$11 130-watt https://www.amazon.com/HELLA-H3-130W-High-Wattage/dp/B00IKMCEEU

stay away from LED. I think halogen bulbs will be the most power consuming

buy many wire in parallel for higher load.

ron
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For testing, you can use normal 120V bulbs, just make sure they are the old incandescent kind. I would just connect a 60 watt bulb or two.

Randomaker
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