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I'm designing a 100W dummy load for a guitar amplifier using Arcol HS50 and HS100 resistors. Here's a link to the datasheet:

http://www.arcolresistors.com/wp-content/uploads/2014/03/HS-Datasheet.pdf

16 ohm dummy load

This is my 16 ohm dummy load. Worst case it will see 100W at its input. R1 is the resistor I'm concerned about.

I have never used these resistors before and I'm wondering if they might end up overheating or warping my board in my design.

Current board layout

You can see my first attempt at creating heat sinks here. R1 is a 16 ohm HS100 (100W) resistor, while RZ8 is 16 ohm HS50 (50W). There will be a switch to connect RZ8 in parallel with the 16 ohm load to present an 8 ohm load to the source.

Will this be enough to keep these resistors in a safe temperature range when the input is pushing 100W?

My thermal vias have a drill diameter of .1mm and they are spaced 1.2 mm apart connecting to the ground plane on both top and bottom layers. I've placed a stop mask to expose the ground plane under those resistors on both the top and bottom layers of the board.

I know this is a deep subject and I'm not asking anyone to go through and do all the math for me. I'm looking for someone with experience with these resistors to let me know if it's clearly going to end in catastrophic failure or if my heat sinks have a chance of doing the job.

I'm also curious how other people use these in their designs. Can anyone show me how they've successfully mounted these resistors to a PCB and how they were able to manage the heat?

Please let me know if I've left out important information. Thank you.

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    \$\begingroup\$ Just show the worst case configuration of resistors and add links to your question to data sheets. All the other circuitry is making it hard figuring out what your circuit really is so simplify it. \$\endgroup\$ – Andy aka May 30 '18 at 20:35
  • \$\begingroup\$ @Andyaka thanks for the quick response. I've gone ahead and edited my question, hopefully it's more clear now. There are two parts to this circuit: dummy load and pad/line out. My question only has to do with heat within the dummy load so we can ignore the line out portion for now. Sorry about the confusion. \$\endgroup\$ – Avid Pro Tool May 30 '18 at 21:04
  • \$\begingroup\$ If anyone else is browsing and looking for similar info, this is a useful document \$\endgroup\$ – Avid Pro Tool May 30 '18 at 23:01
  • \$\begingroup\$ A random comment- I wonder if the demise of ordinary incandescent lamps, clearly marked with wattage in SI units and exposed so you can burn your hand on them, is resulting in less intuitive understanding of heat dissipation in 'younger players'. \$\endgroup\$ – Spehro Pefhany May 31 '18 at 13:36
  • \$\begingroup\$ @SpehroPefhany you do have a point there, I think many younger people like myself have mostly only use LEDs in their designs instead of lamps and normally don't have to worry much about heat unless dealing with voltage regulators or power transistors, etc...In my case so far I've been able to get away with adding TO-220 heat sinks here and there for many things, but this was the first time I tried designing something that dissipates 100W. It's not like I've never felt the heat coming off a lamp and seen its power rating, I just never had to design for that or control the heat that precisely. \$\endgroup\$ – Avid Pro Tool May 31 '18 at 16:26
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Well, here is the basis for the rating from Vishay-Dale's datasheet:

enter image description here

30cm x 30cm x 3mm thick is probably considerably more than your PCB area, and the thermal conductivity of the PCB is not as good as solid aluminum so your heat spreading will not be as good. Even if it was (which it isn't), the temperature rise acceptable for an aluminium and ceramic resistor may be excessive for your PCB material.

You might be able to get 10W continuous out of it. An alternative would be to buy a finned heatsink with specified °C/W rise and design for a reasonable rise in temperature at whatever you consider to be worst-case normal operating conditions. You could consider adding a fan, but that would require a power source.

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    \$\begingroup\$ Thank you this is very helpful! So I'm guessing most people usually mount these resistors directly to heatsinks somewhere in their chassis and then run wires to the board? I'm curious if there's a way for me to keep these mounted to the board while still managing heat appropriately, but it sounds like any suitable heat sink will be too large. I'm just trying to understand how someone could get away with mounting these directly to a PCB since Eagle even makes a package for them. \$\endgroup\$ – Avid Pro Tool May 30 '18 at 21:55
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    \$\begingroup\$ Right - chassis mount or heat sink and wires. There are plenty of reasons to over-rate a resistor- to take surges, to be conservative etc. If you are not running a function generator into your amplifier the actual average output power may be considerably less than the peak, but I assume you are taking that into account. \$\endgroup\$ – Spehro Pefhany May 30 '18 at 22:22
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    \$\begingroup\$ yes when this is all finished, the source will be a guitar, not a function generator. But I wanted to be conservative and make sure this dummy load would never be damaged by any signal that could come out of a 100W amplifier. Even though most of the time I will only be using my own 22W or 50W amplifiers, probably not even at full volume. Thank you for the clarification about over-rating the resistors, I suspected that would be the only time they could be mounted to the board without heat issues. I really appreciate your thorough response, this has cleared a lot up for me! \$\endgroup\$ – Avid Pro Tool May 30 '18 at 22:44
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Standard PCB foil (the default thickness) has Rthermal of 70 degree Cent per watt PER SQUARE of foil, for any size square of foil.

In trying to move heat vertically up the PCB sketch, there are about 2 squares of foil up to the other (smaller) resistor, thus Rtherm is 70/2 = 35 degree Cent per watt.

In trying to move heat to the right and then up, you have about 3 squares of foil, thus that path has Rtherm of 70/3 ~~ 25 degree Cent per watt.

Combining these 2 paths, use product/sum = 35*25/(35 + 25) or 900 / 60 = 15 degree Cent per watt. And you have 100 watts.

Thus the temperature of the resistor will be AT LEAST 15 * 100 = 1,500 degree Centigrade, assume the top edge of the PCB are held at room temperature.

schematic

simulate this circuit – Schematic created using CircuitLab

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