# Easy and cheap way to sink more than 5 A from a power supply

For an experiment I need to check a current sensor but I need to parallel too many resistors to achieve that without damaging the resistors. I don't want to do that. The PSU is 12 VDC, 10 A and I want to sink at least 5 A.

Is there a quick common way engineers use, or is an expensive, bulky resistor needed?

• What power are you looking at? 5A at 12V? So 60W? Why are you dumping this heat? The easiest way to dump power is a resistor, that is the easiest way, and probably cheapest. You might be able (with a decent bit of work and expensive kit) to push the power unto a battery to allow you to use it rather than waste it. Commented Sep 26, 2022 at 12:03
• Get a proper high-power resistor. Commented Sep 26, 2022 at 13:48
• Just a comment on "expensive and bulky". I see appropriate 100W metal-cased resistors are about £10 from large suppliers, and are about 65 mm long. Car headlamps are about 60 mm long and a little tricky to mount, cost about £5. Commented Sep 26, 2022 at 16:34
• @jonathanjo metal-cased resistors need to be mounted to additional heatsinks to operate at their advertised power. For example, these Ohmite resistors ohmite.com/assets/docs/acl_hs10.pdf?r=false have a series "rated" at 100 W. However, with no heatsink, they can only handle 30 W. The heatsink is almost 1000 cm^2. Commented Sep 26, 2022 at 18:31
• A cheap (i.e about $10) hair-dryer or toaster are each capable of sinking 1000W or more. You even get nice sheets of mica in each. They might be the easiest/fastest things for you to get a hold of. (Though a halogen headlight/downlight bulb is probably considerably more convenient) Commented Sep 27, 2022 at 3:18 ## 11 Answers Does the PSU have an (adjustable) current limit function? If so, then just short the output and adjust the current limit to the value you desire. Use a 55W car headlight bulb. That will take about 5A @12V • The downside is, incandescent bulbs can take a lot of current at startup, approximately 10x their rated current. Commented Sep 26, 2022 at 12:31 • @Justme if you parallel a couple of 20W bulbs and as many 10W or even 5W to make up the difference, connecting them one at a time in that order, you'll reduce the effects of inrush current. I've been known to solder directly to the base of car bulbs to make test lamps. It's not pretty but it works. Use screw terminals on the other end of the wire. Total cost 0 if like me you've got a few spare car bulbs kicking around Commented Sep 27, 2022 at 9:06 • ...there are also tricks with connecting 2 bulbs in series then shorting one of them, to preheat the filament and reduce inrush Commented Sep 27, 2022 at 9:08 • A choke like this will be quite expensive Commented Sep 28, 2022 at 10:09 • It's worth noting that such a lamp emits a lot of heat, is very dangerous to touch and the light it emits could damage your eyes if you are in its proximity. The OP would need to mount the lamp in a safe gig which leaves the lamp in open air (to avoid overheating) and which shields the light from it's workspace. Commented Sep 28, 2022 at 20:48 Cheap and ugly hack: A resistor immersed in water can dissipate 20..50x its rated power without getting off spec. Of course, common sense is required: • stay below 30V. Higher voltages are possible if you really know what you are doing or if you are extra brave. • use pure water, distilled is best for prolonged experiments. • it is advisable to protect the resistor leads from too much contact with water using paint or plastic tubes • take care not to boil off the water and leave the resistor dry, at 10x overload it will burn in seconds. Edit: Depending on the circumstances, other liquids can be used as well: • mineral oils: much less conductive than water. • transformer oil: in fact, the proper liquid for the purpose. Good well into few kV range. Expensive and probably toxic. • ethanol and ethanol-water mixtures: a lot less conductive than the pure water, much less messy than the oils, well-known toxicity. • perchlorethylene: non-messy and non-flammable alternative of the above. Dry-cleaning-shop smell. Aggressive to some metals (Al and Zn come to mind). Manageable toxicity. • This I had not thought of! I shall do some experiments at the weekend. Presumably the 'pure' water is to avoid ionic conduction? Commented Sep 27, 2022 at 18:01 • @2e0byo ionic conduction is easier. Depending on the experiment duration, one may also want to avoid solid boiling residue that will gradually worsen the heat exchange. Commented Sep 27, 2022 at 18:12 • @2e0byo if you want to experiment, see the updated answer. Commented Sep 27, 2022 at 18:34 • thank you. ofc transformer oil is best, but that's solidly out of 'hack' territory. Commented Sep 27, 2022 at 19:09 • I'd try to seal the contacts completely - solder wires to your power resistor, then cover the joints in a thick layer of epoxy, bathroom sealant, or hot glue (in ascending order of removability) without covering the housing/heatsink. Pure water has a habit of not staying pure for very long and is also rather corrosive. Commented Sep 28, 2022 at 13:52 A slightly more expensive option is to use an electronic load (like this). You can find them online for as low as 20 bucks (got mine for free after getting a refund from a supplier in China since it took nearly 6 months to arrive and we thought it had been lost by the courier). EDIT: I am not saying that you need to wait 6 months for it to arrive. By the time I finally got my first electronic load, I had already purchased another one from a local supplier for nearly £35. It was worth paying the extra money as I needed it with a bit of urgency ;) With an electronic load, you can accurately vary the current demand and get readings of power being delivered and even set timers for the stability of your circuit. Here are two examples of what you can get at different price ranges: A 12V 55/60W automobile headlight bulb may be used as the load. A proper enclosure would be required to shield one's eyes from the bright light. However, its very low cold resistance would cause the power supply to trip. The solution would be a length of galvanised curtain wire (~ 1.5m long), in series, to preheat the filaments. The connecting wire may then be slid, along the curtain wire to its opposite end, and secured with a crocodile clip. For something really cheap - we've used this in the past at my former company - just use a container of water. Then plunge the leads from the PSU into the water, turn it on, and start adding salt until you get the current draw you're looking for. Not very repeatable, and not something I would do for a deliverable item. But worked fine for a one-off experiment. We also used carbon piles, where the resistance of the pile is determined how much force is used to compress the carbon discs. • Bare in mind that this method will produce chlorine gas as the salt (assuming NaCl) gets split in the process. Commented Sep 26, 2022 at 13:56 • Baking soda (sodium bicarbonate) works as well, and it doesn't make chlorine. Keep in mind, any time you run electricity through water you will generate oxygen and hydrogen. Keep the area well ventilated so the gasses don't accumulate and explode. All it takes is a little spark to make it go "Bang!" – JRE Commented Sep 26, 2022 at 14:33 • If you do this, do not use stainless steel electrodes. Electrolysis using stainless steel electrodes releases nasty chromium compounds into the electrolyte, turning the water into no-kidding hazardous waste. Whatever you use for electrodes, they will corrode, so use some iron or copper you don't care about. Commented Sep 26, 2022 at 15:38 • When I was a kid (45 years ago .. sigh ...), the carbon center rods of zinc/carbon batteries used to make great makeshift electrodes as they weren't metallic so didn't corrode. No idea if you can even get this kind of batteries anymore, though. Commented Sep 26, 2022 at 23:23 • Over 5A at 12V would require some really salty (or baking-soda laden) water and/or big plates on the electrodes, wouldn't it? Maybe a useful idea in general, but maybe not great for this specific question. Also, please edit your answer with a warning about chlorine gas and other hazards, for the safety of future readers that don't look at comments. Commented Sep 27, 2022 at 18:26 About 1980 I put several light bulbs (130V) into a nice case with an amp meter in series and a voltmeter, it still has the original lamps. I have switches on each lamp so I can switch whatever load I want. The lamps range from 25W to a 500W photoflood lamp. It was designed to operate upto about 30V but tolerate a 120 volt line. I also have electronic loads but this is fast to connect (has banana jacks) and good enough for most of what I do, it is AC/DC hence not polarity sensitive. This gets used a lot as a load and also in series to limit current or test a potential shorted device. It was great for inrush testing with 42V systems. At 12V the resistance changes some but not a lot as the filament does not get that warm. This is more stable then the common way of using headlamp bulbs while not getting that warm The other suggestions of a ‘proper’ power resistors are good for a robust semi-permanent experiment. If it’s just a one-off, where speed and versatility is important, then using resistance wire may be a good option: For 5A at 12V you need 2.4 ohms with a power rating of 60W. around 30cm of AWG 26 nichrome resistance wire will give you the resistance you need. Coiling the wire and submerging in a large beaker of water will provide adequate cooling for short term tests as long as you keep the water topped up. • The problem is that NiCr wire could be expensive since you usually cannot buy it in short pieces, but you need to buy a whole spool. It could cost 10$ or more, depending on type. Moreover NiCr is notoriously difficult to solder, so the OP should be connecting the wire with some other mechanism (e.g. screw terminals). It all depends on the budget of the OP. Commented Sep 28, 2022 at 21:03
• @LorenzoDonatisupportUkraine You're right, you can usually only get spools from major suppliers. But you can buy short length on ebay quite cheaply - ~2-3 GBP for a 1 m length. Commented Sep 29, 2022 at 9:06

3D printer heater cartridges designed for 30-70W at 12V or 24V are dirt cheap and should meet your need. Of course you'll need somewhere to dump all that concentrated heat. Submerging it in water is probably the easiest way.

Alternatively, nichrome wire is cheap and easy to tune to the exact resistance you want.

I don't know what type or resistors you are using but you can try with an inexpensive cement resistor, there are many different values ​​and powers.

You can get 10pcs of 20W 10ohm ceramic (cement) resistors for 8 bucks on Amazon. Put 5 in parallel. Thus the total resistance is 2 ohms which is 6 amps at 12 volts. That's 14.4 Watts per resistor; entirely within technical spec of the devices (no damage) and the user's financial spec (cheap). As long as 5 resistors is not too many, this would solve your problem.

• Someone downvoted this. Dunno why. You can get 10pcs of 20W 10ohm ceramic (cement) resistors for 8 bucks on Amazon. Put 5 in parallel: 2 ohms, so 6 amps at 12 volts. That's 14.4 Watts per resistor; entirely within technical spec of the devices (no damage) and the user's financial spec (cheap). Is 5 resistors too many? Commented Sep 27, 2022 at 15:27
• @MikeS all the should be in the answer. It's pretty sparse as is. Commented Sep 28, 2022 at 20:48
• @Passerby Definitely a mediocre answer as it stands, but also marginally useful, so not deserving a downvote in my book (but neither an upvote until the OP adds more info). Commented Sep 29, 2022 at 7:48
• Thanks @LorenzoDonatisupportUkraine. I edited the article. Let's see if it's accepted. Commented Sep 29, 2022 at 15:57

Semiconductors are better at dissipating heat than resistors, in general. Make yourself a 5A current source using any of these NPNs, costing as little as \$0.71 in single piece quantities. A 0.2 ohm emitter resistor would dissipate only a watt in such a circuit.

• The fact that those BJTs can handle 5A or more doesn't mean they can do that alone. The OP must sink 5A at 12V, so 60W. Those BJTs must dissipate that much power and that requires a lot of heat sinking, which is expensive. The heat sink will end up costing much more than the transistor! Commented Sep 28, 2022 at 20:26
• For example, take the MJE3055T on that list. It is rated for 75W at 25°C. It's J-C thermal resistance is 1.67°C. This means that with 60W dissipated power you will have 1.67*60=100 °C temperature differential between case and the internal chip, which has a max temperature limit of 150°C. This means that the case cannot rise in temperature more than 50°C. This requires, in order to dissipate 60W, an heat sink having at most 50/60= 0.83 °C/W thermal resistance (and a 0°C ambient temp max). Search digikey for such heat sinks and you'll find they cost tens of dollars! Commented Sep 28, 2022 at 20:42
• In other words, you need a power BJT or MOSFET, good heat sinking and a control circuit to control the dissipated power. At this point you have an active load. Commented Sep 29, 2022 at 7:51
• Brain fart: replace "active load" with "electronic load". Commented Sep 29, 2022 at 16:30