I'm looking for the simplest/cheapest (but safe!) way to build a 4.5A test-load for an LMR14050-based 24Vdc to 12Vdc/5A step-down voltage regulator circuit.

LMR14050 Datasheet

Circuit Schematics


Edit (30.7.18): I commented on the responses below, but it seems my follow-up questions may be rather hidden in all the text, so I figured I'd enter them here to:

As it's a bit hard for me to find the suggested bulbs, I thought of hooking-up 9 of these alternative 5W bulbs from eBay in series to give me more or less the power draw I'm after - Would these bulbs work for the intended purpose?

Also, what would be the recommended capacitor value for each bulb?

Edit (31.7.18):

So apparently the bulbs I found aren't suitable - does anyone know where I can order the correct bulbs (i.e. each drawing 500mA more or less at 12Vdc) online?

  • \$\begingroup\$ What do you mean "safe"? Are you afraid to burn your fingers? What resistor value do you expect from your load? \$\endgroup\$ Jul 29, 2018 at 0:58
  • \$\begingroup\$ You'll want at least a 100 W dissipation-capable resistance (about twice the actual dissipation.) One dozen, 33 Ohm, 10 Watt resistors in parallel would get close. Or you could make your own out of surplus stuff laying about the home (toasters and space heaters have nichrome wire, ceramics and mica in them, for example.) If you make it yourself, getting burned is the safety issue so protect/block the area around the dissipating device (or else add a lot of dissipating surface area -- though that is probably harder and more expensive than just blocking the area.) \$\endgroup\$
    – jonk
    Jul 29, 2018 at 1:44
  • 2
    \$\begingroup\$ I've seen a guy (MikesElectricStuff on yt I think) use a spool of very thin enameled wire in a bucket of water. \$\endgroup\$
    – Wesley Lee
    Jul 29, 2018 at 3:03
  • \$\begingroup\$ Thanks for the suggestion, jonk. Could you please explain how you arrived at the 33ohm value? \$\endgroup\$
    – SnowCrash
    Jul 29, 2018 at 18:41
  • \$\begingroup\$ Those bulbs you suggest are LEDs intended to replace Halogen bulbs - they would draw much less than 5 watts. \$\endgroup\$ Jul 30, 2018 at 20:41

3 Answers 3


Automotive incandescent ight bulbs are what most people use for 12V load testing.

These are available in power ratings starting at about 5 Watts all the way up 55 Watts or more. The ones that I use in my shop draw about 1A each and I simply use as many as required to get the desired load current.

Incandescent lamps have the advantages of being low cost and readily available. enter image description here

These are 1141 lamps

Lamps similar to these and their sockets can be had almost for free if you visit a local automobile wrecking (salvage) yard.

  • \$\begingroup\$ Though I also use incandescent lamps, one possible issue is that the cold resistance is far lower than when warm. Initial current inrush can be 10 times the rated load. If that is a problem, put a large capacitor across the output before connecting the lamp load to accomodate the surge needed. \$\endgroup\$ Jul 29, 2018 at 14:00
  • \$\begingroup\$ Assuming you are using multiple lamps, connect them one at a time. \$\endgroup\$ Jul 29, 2018 at 15:28
  • \$\begingroup\$ This is probably the simplest & cheapest way to do it, thanks Dwayne. Not sure where I might find these (not many automobile wrecking yards where I live). However, I did a quick search and came up with these bulbs on eBay: ebay.co.uk/itm/132662063001 Do you think they'd work? Also, no need for a current limiting resistor in this setup right? I thought I'd hook-up 9 of these in series to give me more or less the power draw I'm after. \$\endgroup\$
    – SnowCrash
    Jul 29, 2018 at 17:52
  • \$\begingroup\$ Thanks for the heads-up about the capacitors, DrMoishe. Assuming the bulbs in the previous comment are adequate, could you tell me what would be a good value for the capacitors? \$\endgroup\$
    – SnowCrash
    Jul 29, 2018 at 17:56
  • \$\begingroup\$ sorry, just to note that in the eBay listing above, I was referring to the 5W version of the bulbs. \$\endgroup\$
    – SnowCrash
    Jul 29, 2018 at 18:02

I would just use a couple of power resistors. Just monitor the voltage under load and make sure it stays at 12V.


For fun, get a single 2 ohm resistor, rated at 10 watts (yes, you are dissipating 40 watts). If possible, get a power resistor with hollow core.

Place this in a beaker or flask 2/3 filled with WATER.

Connect this water-cooled load to your 12 volt supply.

Enjoy the bubbles.

  • \$\begingroup\$ Thanks for the interesting suggestion, analogsystemsrf. Could you please elaborate on the math a bit? How did you arrive to the specific values of 2ohm and 10W? Also doesn't the power dissipation come to 12V x 4.5A = 54W? \$\endgroup\$
    – SnowCrash
    Jul 29, 2018 at 17:13
  • \$\begingroup\$ 12volt/5amp = 2.4 ohms; the 10 watt resistor will become quite hot given the 40 watt dissipation, and the water will boil. Be careful. \$\endgroup\$ Jul 30, 2018 at 5:54
  • \$\begingroup\$ Unless you're using distilled water this is a bad idea...the contaminants in water will cause a short unless you do all sorts of fun stuff with your ends. \$\endgroup\$
    – biggi_
    Jul 30, 2018 at 20:51
  • \$\begingroup\$ Makes sense, biggi, thanks. I'm leaning heavily toward the bulb solution as it seems much less 'sparky', but having trouble finding suitable bulbs for this purpose (see comment above). \$\endgroup\$
    – SnowCrash
    Jul 30, 2018 at 23:23
  • \$\begingroup\$ @biggi The water will probably be more than 2 ohms, especially after the wire forms an oxide from electrolysis \$\endgroup\$
    – C_Elegans
    Jul 31, 2018 at 0:44

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