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Working on a complex power supply, but to test its limits I would like to create a set of test loads of specific voltage/current.

There are several devices for creating a set load, but they seem to start in the $300+ level.

I've found a circuit for creating a "dynamic load":

  1. http://www.ko4bb.com/Test_Equipment/DynamicLoad

...but both would only act as a single load at-a-time.

Question 1:

If have a specific Voltage/Amp load in mind, can I just use a power resistor AS that load for testing purposes?

I would like to create a set of 5V/2.1A loads.

Using Ohm's Law, that would work out to a: 2.38095Ω 10.5W resistor. Am I understanding that correctly?

I'm not finding a resistor with those specific stats, but I found:

The second gets me closest to my "target", but when I look at these two different resistors, it seems that the first is a lot "beefier" (heat sink case, etc.), but I lack enough experience to understand which would be best for my purposes. Instinct is telling me that if I'm using a resistor as a load, and I want to do any extended length of testing (several hours at least) that heat dissipation will be something I need to plan for?

Question 2:

Can I place a cheaper ceramic resistor against a heatsink to create a test load for several hours, or would the military spec resistor be a better option for my purpose?

ZacWolf
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    How accurate do you want the resistance to be? Your 10W resistors are going to get very hot and their resistance value will shift. You should be looking for a 25W resistor, at least, if you don't intend to boil water with it. – Joe Hass Jan 29 '14 at 18:05
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    using a combination of resistors is also an option, for example you can connect in parallel four 10 Ohm 10W resistors and this results to 2.5 Ohm 40W. You can also mix values to get closer to the value you are after. – alexan_e Jan 29 '14 at 18:07
  • Thanks Joe and alexan, I worked through it with JYelton, and believe I understand now. – ZacWolf Jan 30 '14 at 00:08

2 Answers2

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Your calculations are correct, you would need a 2.38Ω resistor that can handle 10.5W of power.

Both resistors you found, however, only have a 10W rating. With the 2.5Ω at 5V, you would be producing exactly 10W; I don't recommend operating a power resistor at its maximum rating for any duration of time. Components likely exceed their tolerance to some degree, but there's always a margin of error, so you should avoid operating things at the max rating unless you have good reason to do so. In the case of the 2.4Ω resistor, you would actually be running it at 10.4W (\$2.08A * 5V\$), which exceeds its rating. Definitely don't do that.

Going back to the desired load: 5V at 2.1A, 10.5W. There aren't precisely any 2.38Ω power resistors, but if you double it you get 4.76Ω and 4.7 is a common value among resistors. If you put two 10W 4.7Ω resistors in parallel, you'll get 2.35Ω total resistance, and have 20W total power handling capability. Note that they are 5% tolerance, so actual resistance will vary.

Edit, per comments:

Note that power rating on resistors is the max amount of power that they can dissipate. How much power they need to dissipate depends on how much current they draw. Remember, current (measured in Amperes) is drawn (and determined by) the load, not the power supply*.

For two equal value resistors in parallel, resistance is halved but power handling is additive. So two 10W, 4.7Ω resistors in parallel have a resistance of 2.35Ω but a total power handling capability of 20W.

JYelton
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  • If you use a simple resistor, your load current will only be as accurate as your power supply voltage. A real device that draws 2.1A might be a charger that draws a constant power load (adjusting the current upward if the voltage is lower than 5.2V). – Spehro Pefhany Jan 29 '14 at 20:59
  • So would 2 - 4.7Ω resistors, in parallel, draw only the desired 10 watts from the power supply? Is what you're saying, that because the resistors are in parallel, their current _draw_ is limited by Ohm's law to the desired ~10 watts, but they are each **sharing** that total power which means they aren't individually stressed to the edge of their rating? – ZacWolf Jan 29 '14 at 21:26
  • Current (measured in Amps) is *drawn* by a load, and power then dissipated is measured in Watts. The two resistors in my example would draw approximately 2.1A (varying based on accuracy of the supply voltage and resistor variances). In a perfect world, each resistor would draw 1.05A and dissipate 5.25W (well below their rated value of 10W). That figure will vary again because of tolerances, but you would effectively be able to load your power supply with something that will draw 2.1A, dissipate 10.5W, but *be able to* dissipate 20W max. – JYelton Jan 29 '14 at 21:32
  • OK, I think I get it now. Would you mind editing your answer explaining the _drawn_ vs _rating_ distinction as you did in the comment above? Then I can mark it the answer. THANKS! – ZacWolf Jan 29 '14 at 21:39
  • Edited as requested, feel free to ask for any other clarifications. – JYelton Jan 29 '14 at 21:48
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5 volts at 2.1 Amps is 2.3809 ohms so if you really want that value use: -

42x 100 ohms in parallel - this gives you exactly the right resistance and the power dissipated in each resistor is 0.25 watts - use 0.5 (or maybe 0.6) watt resistors and mount them on some strip/vero board.

If you can't afford 42x 0.1% resistors (probably $0.50 each making a total of $21) then use higher tolerance types like 1% providing you can live with the error.

It's a lot cheaper than $300 but also bear in mind that you'll need to space the resistors apart so that they can dissipate their heat sufficiently without exceeding their individual 0.5+ watt ratings

Andy aka
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