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Load Tester — Parallax Forums

Load Tester

LightfootLightfoot Posts: 228
edited 2006-05-30 02:56 in General Discussion
I have a power supply that supplies 12 volts. I want to build a circuit that draws 5 amps, or a 60 watt load. What is the easist way to accomplish this?

Thanks [noparse]:)[/noparse]

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Well well, I'm seeing things, three of them.

-Stanley Blystone

Post Edited (Three of Them) : 5/29/2006 6:44:48 PM GMT

Comments

  • PJAllenPJAllen Banned Posts: 5,065
    edited 2006-05-29 18:57
    · Resistors -- power resistors.· You're looking for 'vitreous enameled'.· You can probably buy a power rheostat, but that'll cost a lot.· You could have a combination of fixed resistors and a power rheostat, if you're intending to do a load curve or take such data.· Those would·be the most cost-effective "solutions."
    · Your resistors' power rating should be twice the power anticipated (60W, therefore 120W of rating.)
    · You could make an electronic load with, basically, a FET (or several in parallel), the more expensive way to go.
    · An ammeter would be a nice touch.
  • LightfootLightfoot Posts: 228
    edited 2006-05-29 19:09
    For starters I put 4 10 watt 50 ohm resistors in parallel. Boy those things get so hot I burned by finger!

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    Well well, I'm seeing things, three of them.

    -Stanley Blystone
  • PJAllenPJAllen Banned Posts: 5,065
    edited 2006-05-29 19:32
    Yep -- heatsink 'em or get a fan blowing on them.· 50W is a LOT of heat.
  • LightfootLightfoot Posts: 228
    edited 2006-05-29 20:15
    Update:

    I increased the load to 50 watts (40 before). The bridge rectifier is an 8 amp, it gets very hot too. I took a copper heatsink that came with a video card GPU cooler kit and placed two very tiny drops of quick set epoxy on top of the rectifer and put the heatsink on. Would this work well enough?

    ▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔
    Well well, I'm seeing things, three of them.

    -Stanley Blystone
  • PJAllenPJAllen Banned Posts: 5,065
    edited 2006-05-29 21:22
    If this stuff is from the kit, thermally-conductive epoxy, that's one thing.· You need to clamp the component in question to a chunk of metal for efficient heat-transfer --·mass & solid-contact are·what it is all about.
  • StarManStarMan Posts: 306
    edited 2006-05-29 22:29
    For high heat situations, epoxy is typically not a good solution.· The thermal expansion of the components is very likely greater than that of the epoxy and the component will simply pop off when the stress becomes high enough.· Better to clamp the component with some thermal grease.· The more grease you can squeeze out of the joint, the better.· It is best to have the surfaces of the component and heat sink attain as much contact as possible and the grease just fills in the very thin voids.

    BTW, thermal greases and epoxies do not have any where near the thermal conductivity of, say, aluminum or copper.· It is just a heck of a lot better than having air gaps.

    I'm assuming the heat sink came with a fan.· If not, get one and use it.

    Chris I.
  • MikeKMikeK Posts: 118
    edited 2006-05-29 22:33
    I think I missed something. 4 50ohm resistors in parallel is equivalent to a 12.5 ohm resistor. Power = V^2/R = 12*12/12.5 < 12watts. Each resistor would be dissipating 1/4 of this or about 3watts. Why should a 10W resistor have been too hot to touch, and where did the 40W number come from?
  • Robert KubichekRobert Kubichek Posts: 343
    edited 2006-05-29 23:14
    Three of Them said...
    For starters I put 4 10 watt 50 ohm resistors in parallel. Boy those things get so hot I burned by finger!

    You would need 4 10 ohm 30W power resistors in parallel to make a 60W load.

    A 60W load at 12v is 2.4 ohms at 5 amps, so 4 30W resistors in parallel will give you a capability of 120W, and run a little cooler.
    BUT mount em on a good heatsink with thermal grease " Artic Silver " is one of the best, and is used to sink CPU's to keep em cool.

    Bob N9LVU scool.gif
  • pjvpjv Posts: 1,903
    edited 2006-05-29 23:27
    Hi All;

    Yup, MikeK is correct; someone is not doing their math right.

    Stanley; just because you put a (actually four of them) 10 watt resistor accross a 12 volt source, that does not imply that the power drawn by each is 10 watts. It simply means that the resistor is specified not to burn up when it is asked to dissipate 10 watts. The power drawn is a function of the resistance.

    As MikeK stated, a 50 ohm resistor accross 12 volts dissipates 12*12/50 = 2.88 watts. Probably hot enough to burn your fingers after they've been on for a while. So the four of them in parallel will dissipate a total of 11.5 watts; nowhere near the load of 60 watts required.

    In order to accomodate that 60 Watt load, and it is quite a substantial amount of heat, a resistance of 0.42 Ohms is required. That can be made up of two 0.84 Ohms in parallel, or four 1.68 Ohms in parallel, eight 3.36 Ohms in parallel, etc.

    If this requirement is not for a permanent installation, then submersing the resistors in a small can of oil will help distribute the heat away from the resistors. This is not suitable for a permanent installation.

    Hope this helps,

    Cheers,

    Peter (pjv)
  • Robert KubichekRobert Kubichek Posts: 343
    edited 2006-05-29 23:52
    pjv said...
    Hi All;

    Yup, MikeK is correct; someone is not doing their math right.

    Stanley; just because you put a (actually four of them) 10 watt resistor accross a 12 volt source, that does not imply that the power drawn by each is 10 watts. It simply means that the resistor is specified not to burn up when it is asked to dissipate 10 watts. The power drawn is a function of the resistance.

    As MikeK stated, a 50 ohm resistor accross 12 volts dissipates 12*12/50 = 2.88 watts. Probably hot enough to burn your fingers after they've been on for a while. So the four of them in parallel will dissipate a total of 11.5 watts; nowhere near the load of 60 watts required.

    In order to accomodate that 60 Watt load, and it is quite a substantial amount of heat, a resistance of 0.42 Ohms is required. That can be made up of two 0.84 Ohms in parallel, or four 1.68 Ohms in parallel, eight 3.36 Ohms in parallel, etc.

    If this requirement is not for a permanent installation, then submersing the resistors in a small can of oil will help distribute the heat away from the resistors. This is not suitable for a permanent installation.

    Hope this helps,

    Cheers,

    Peter (pjv)

    Peter, 12V or E / .42 ohms or R = 28.57 Amps or I ( E/R=I ) at 342 W ( E*I=W)

    12V or E / 2.4 ohms or R = 5 Amps or I at 60W

    With a 12v supply to make a 60W load you need to sink 5A, hence 4 10ohm 30W resistors in parallel will handle 120W even 20W will work, but 10W
    will have a 50% overload and run very hot or burn out. Even with a 120W rating, I would mount em on a GOOD hefty heatsink...

    Bob N9LVU
  • BongoBongo Posts: 65
    edited 2006-05-30 00:22
    From what I can gather, a dummie load is being sought.

    If this is true, a light bulb is cheap and easy.

    Vehical headlight bulb is usually rated for 12 volts, 50 watts on the low beam element and 60 watts on the high.

    You can mix and match with tail lamps, 10 watts, stop lamps 20 watts.· Interior lamps , 5 watt, 10 watt, 20 watt.

    Indicators, dash lamps, boot lamps, door lamps, licence plate lamps - the range is quite good really, and none of them require a heat sink.

    bongo
  • pjvpjv Posts: 1,903
    edited 2006-05-30 02:56
    Hi All;

    Boy, do I feel like a dummy.

    I got an inversion in that calculation; my 0.42 Ohms should have been 1/.42, or 2.4 Ohms. Sorry about that.

    I'll crawl into my hole now..........

    Cheers,

    Peter (pjv)
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