# Load Resistor for high amp low voltage testing

I want to simulate a signal of 15V / 30A through a device.

I will use it to ensure that a series of prototypes are capable of switching at this current.

$$P = 15 \times 30 = 450W$$

$$R = \frac{15}{30} = 0.5 \Omega$$

To be on the safe side, (wire, etc., resistance), I intend to use three 1 ohm 225W resistors in parallel. I have a power supply where the current can be limited, so will set the limit to 30A.

simulate this circuit – Schematic created using CircuitLab

It seems really straight forward as above but I wanted to make sure I hadn't overlooked something as I have not done this before.

Can anyone tell me if the above looks ok?

• Well Three one ohm resistors in parallel is 1/3 of an ohm. Jun 17, 2015 at 13:40
• Oops, will edit the question, I have a power supply where I can limit the current. Jun 17, 2015 at 13:42
• If you limit the current at the power supply to 30 A, then the load voltage won't reach 15 V. Depending on your DUT, that might or might not matter. Jun 17, 2015 at 15:58
• Good point, didn't think of that. Jun 18, 2015 at 8:13

Assuming your Device Under Test (or DUT) and total wiring are sufficiently lossless this might work well enough.

Your basic maths steps are all in order, the power dissipation in the resistors should be fine if they are all within 25% of each other in value, which you can usually assume if they are the same type. If it says a margin on it, like 5%, you can use that to check for certainty.

If your DUT, however is very lossless when "turned on", limiting the current through the supply at 30A will not prevent a higher peak from going through. Your supply has capacitors on its output, so if at 30A your device is supposed to waste near enough to 0V (supposed to, versus what it ends up being is of course a factor in this), it will very shortly dump the full 15V with maximum current through your DUT.

Whether that is a microsecond or a milisecond is down to your lab supply, but you should then assume a peak current of:

$I = \frac{V}{R} = \frac{15V}{0.3333...Ohm} =~ 45A$

during that time, until the capacitors are empty (EDIT: Not empty, of course, but settled to the new voltage of 10V needed for 30A) and the regulation of the power supply balances out at 30A.

The resistors will not mind this at all, firstly because they are very large, bulky, 225W things, so even 45A continuous would be fine (if they are within 1% in value, at 225W things can quickly escalate into poof). But also because of their mass they can easily handle very short spikes a bit above their handling capability, as long as you then settle below their maximum power.

Whether your device will like it: That only you know.

• Good point about 10V and inrush. Thanks a lot. Jun 18, 2015 at 8:14
• @Quantum4 It's important to consider what might and might not fuse a contact. It may be a good test, or it may be the wrong thing to have, but either way you need to be aware of it, in case something "mysteriously" fails or fuses Jun 18, 2015 at 8:40

Looks OK once you get R correct.

At this resistance level you will need to check actual values as wiring drop and terminations can make a major difference.

Check temperature coefficient of resistors - it may make a difference depending on other factors.

A useful solution is Nichrome or Constantan wire of appropriate current rating. This has the advantages of close to zero temperature coefficient of resistance and the ability to make air cooled resistors of values of you choice. Self supporting coils can be wound as required.

Industrial wire suppliers have a range of sizes with upper ratings being "approximately awesome". You can use several wires in parallel if needs be but a single wire of that rating should be available. You can get the resistance you want by sliding a clamp along the wire and clamping the leads at the length desired.