I've built a small electronic dummy load. From the EEVBlog site - turned out to be a pretty interesting project....

It is a mosfet that is driven by an op-amp keeping the voltage across a load resistor between mosfet source and ground at a set voltage. It is explained really well in the short video if anyone is interested....

My load resistor is 1Ω (10x10Ω in parallel, each metal film 1W).

I had it set to 2A - so the power dissipated is 4W in the 10W resistor bank. With a thermocouple I measured the temp of the resistors around 150°C (300°F). Seems high?? What sort of temp would the average resistor tolerate??

I'm planning a Mk 2 version with a 10x10Ω or 12x12Ω bank in 5W wire wound resistors, hopefully to handle 5A or so.... might be able to use the enclosure to bake cookies!

Lee

 

Pick a resistor from this list.

Download and study its data sheet.

Check what the maker says about how the resistor is mounted, vertical, horizontal, how much air circulation, etc. All of those effect the resistor's ability to dissipate heat....

 

When I use a resistor it dissipates exactly what I want it to. No more, no less.

A resistor can handle what it is designed to handle, and every resistor is different. You can tell what it does by looking at its spec sheet.

Don’t have a spec sheet for your parts? Then sorry to say throw them away (at least for this application) and buy some once you know what it can do.

Many power resistors I’ve seen are only rated for full power up to 80 °C, down to zero watts at 150 °C. So you’re cooking your parts like that.

I can’t get Dave’s site here but typically when a MOSFET is involved it’s the FET that dissipated the big share of the load power.

2A into 1 ohm is 2 watts so the FET is doing some of the work but not the majority.

If you can post your circuit I may be of more help to you.

 

2A into 1 ohm is 2 watts so the FET is doing some of the work but not the majority.
.........................

You forgot to square the current, making it 4W.

 

When I use a resistor it dissipates exactly what I want it to. No more, no less.

A resistor can handle what it is designed to handle, and every resistor is different. You can tell what it does by looking at its spec sheet.

Don’t have a spec sheet for your parts? Then sorry to say throw them away (at least for this application) and buy some once you know what it can do.

Many power resistors I’ve seen are only rated for full power up to 80 °C, down to zero watts at 150 °C. So you’re cooking your parts like that.

I can’t get Dave’s site here but typically when a MOSFET is involved it’s the FET that dissipated the big share of the load power.

2A into 1 ohm is 2 watts so the FET is doing some of the work but not the majority.

If you can post your circuit I may be of more help to you.

I got the resistors off ebay several years ago, no datasheet.....
I wonder about their true rating....

I think the majority is done by the FET generally - if I put 12V through the circuit at 1A, the resistors are still (trying to) dissipate 1W, but the other 11W is at the FET, which is well heatsinked, but still runs around 80°C (180°F). I'll attach the circuit in a minute....

 

Circuit.

The voltage divider at the op-amp input gives between 0-2V. The op-amp drives the MOSFET, keeping the voltage across the 1Ω resistor at the input voltage. As the resistor is 1Ω, that voltage is the current drawn from the PSU/battery.

Original circuit by Dave Jones @ EEVBlog.com, modifications suggested by others on his forum....

 

From a couple of datasheets/manufacturer info I've seen, it seems fairly routine for resistors to run at this sort of temp at 50% load approx....

I made up a large 1Ω resistor from 12x12Ω 5W wirewounds, it too runs over 150°C when dissipating 30W (50% load).... Keep fingers clear!

 

For a given power dissipation the free-air resistor temperature is basically a function of the ambient temperature and the dissipation surface area of the resistor. So for a given physical size of a resistor it must reach a given surface temperature above ambient to dissipate away the power by radiation and convection. This is unrelated to the power rating the manufacturer gives to the resistor, it's just a matter of physics.

A smaller physical resistor for a given power rating will have to operate at a higher temperature and must be made of materials that can tolerate that temperature.

 

Hi,

If you are making a load bank just for testing power supplies or other devices then the resistors do not have to run continuously and this offers a HUGE advantage in heat sink size and makeup compared to resistors that have to operate continuously as in most products.

This means you can even resort to water cooling with little or no extra effort and see your resistor power rating jump up to a much higher rating and it only costs about a dollar to do.

For one thing, water has a specific heat capacity of something like 10 times that of aluminum (the typical heat sink metal) and that means it takes 10 times longer to heat up. This in turn means that if you submerse the resistors in distilled water while doing the test, you can get a decent run time for testing before the water heats up.

Power resistors work great like this. The water gets inside the resistor hollow body and flows naturally due to convection currents, and plus the whole outside is in contact with the water too, so the resistors stay very close to the temperature of the water for a long time.
The power limit then is the power that it takes to cause a temperature gradient that is too high, such that the resistor starts to boil the water in the local vicinity of the body of the resistor while the rest of the water stays cool.
I would not doubt it if you could get at least 2 times the power rating but it is probably much better than that.

We did some simple experiments a while back with power resistors in tap water and distilled water. The tap water works too but it may vary from place to place due to the mineral content so it may be more conductive in some areas of the world than others. Around here it is not too conductive.

Ice cubes in the water will give you even longer run tme, as long as the water used to make the ice cubes is not too conductive either

It's truly amazing how good this works.

 

I think the majority is done by the FET generally - if I put 12V through the circuit at 1A, the resistors are still (trying to) dissipate 1W, but the other 11W is at the FET, which is well heatsinked, but still runs around 80°C (180°F). I'll attach the circuit in a minute....

OK, someone caught my math previous error....

80 °C is a FET is cruising, quite safe. So you have an excellent heat sink. The resistors are the issue.

One way to get the heat out of the resistors is to never put it in to them in the first place: instead of using 1 ohm use 0.1 ohm. So the power goes down to 1/10th of what it was.

Of course, you'll have to adjust either the 39K or 10K pot (or both) to give a lower input. Adding another voltage divider at the amp input also works.

 

Thanks for the advice.... water cooling would obviously help a lot, but not really convenient given the rest of the electronics....
The temp of the resistors themselves doesn't bother me too much, esp the 5W ceramic ones, just makes me revise the interior of the case, I don't want to cook the op-amp or battery.....

 

Resistors have a range of, "rated" temperatures that can make green circuit boards turn brown. I don't like that! Change your thinking to use the case for the heat sink and use transistors or resistors that mount easily to the case. 3mm or 4mm aluminum isn't all that expensive, and you'd be surprised at how much difference even a tiny video card fan will make by pushing only 1 or 2 CFM through the area. That's my opinion of the best way to get rid of heat without cooking the important parts. I promise, I have made this mistake myself.