I am going to build an Audio Amplifier Dummy Load. I found this article and plan to build mine like it. Attached is the schematic. And here is the link to the build:
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Audio Dummy Load Project Kit

He says the power handling capabilities are for each channel is 200W. From what I remember about the formula for adding resister wattage is:

• When wire in parallel, add the values.
• When wire in series, the value is the lowest wattage of any in the series.

If this is correct, looking at his schematic I think the power handling capability of each channel should be only 100Watts instead of 200Watts.

What am I missing?

I want to build mine using (8) 8ohm resisters rated at 100Watts each (4 per channel).

Using is schematic will my dummy load be 200 or 400 watts in each channel.

I am asking to learn, not just get to correct answer. I am looking for the math formula so that I can also do this in the future if need presents itself again.

Thanks

Dennis

 

It does not matter how you wire your resistors, whether in series or parallel, assuming the resistance value is the same for all resistors.

Think of it as each resistor is sharing the load equally.

If each channel is rated at 200W and you use 10 resistors, then each resistor will have to dissipate 200W/10 = 20W.

Use a wattage rating at least twice that, i.e. use a resistor that is rated for at least 40W.

A lot depends on the physical construction and mounting of the resistors. Power resistors come in a metal housing that is meant to be bolted to a chassis or heat sink.

 

As MrChips says, when all of the resistors are the same value, it doesn't matter as long as they are connected in such a way that they will share the load equally.

But, since you indicated you are interested in learning the concepts, in general there is no simple way to say how wattage ratings add when you combine resistors.

For instance, consider taking ten 8Ω resistors and making a dummy load for one of the channels as follows:

Make R1 by using two resistors in parallel. (R1 = 4Ω)
Make R2 by using four resistors in parallel. (R2 = 2Ω)
Make R3 by using four resistors in parallel. (R3 = 2Ω)

Make the dummy load by putting R1, R2, and R3 in series.

Now let's put enough voltage across the dummy load to put 160W into it (to make the math easier). Because the three effective resistors are in series, they will have the same current. Since power in a resistor is \(I^2R\), the 4Ω resistor will have twice the power dissipation as each of the other two, meaning 80W in R1 and 40W each in R2 and R2. By symmetry, this means that each of the two 8Ω resistors that make up R1 will sink 40W while each of the four resistors that make up R2 and R3 will sink 10W each.

The problem becomes even more pronounced if the individual resistors are not the same size (which is actually show above with R1, R2, and R3). So, in general, unless you can exploit symmetry arguments, you have to just work out the analysis to see how much power each resistor is going to have to handle and sizing each one accordingly.

 

Thanks MrChips and WBahn.

I understand now.

As long as the dissipated power of each individual resister used to construct the dummy load does not exceed its manufactured rated value then my dummy load will not burn up.

I just wanted to say this site has a lot of great information. I look forward to learning more.

 

As long as the dissipated power of each individual resister used to construct the dummy load does not exceed its manufactured rated value then my dummy load will not burn up.

You are more than welcome.

In theory, you are correct. However, there are some real-world practical considerations as well. As MrChips said, don't push the rated power limits too hard. Try to use a power rating at least twice what you expect to actually dissipate. Furthermore, the power ratings make some assumptions, the most notable probably being that the resistor is in free air at room temperature. If the ambient temperature is higher, then the power rating must be "derated" for the actual expected conditions. If all of these resistors are in a box, then you have to have some way to get that 200W of heat out of that box efficiently. Otherwise, the tempurature will quickly rise to a point where the resistors are operating beyond their properly derated limits.

Think of putting your hand on a 60W light bulb. Now consider that the surface temperature of that bulb, dissipating only 60W, is probably much higher than you dare let those resistors get.

Many power resistors are in metal cases where one side that is flat and has a couple of mounting holes. One way to get the heat out quickly is to get some heat sink compound and mount them onto a metal plate and mount finned heat sinks directly on the other side of the plate. If possible, blow air over both the resistors and the heat sink fins. You may not need to go to those extremes, but be prepared to.

If I were doing this project, I would be tempted to make each dummy load out of a 4x4 array of 8Ω power transistors, each rated at 25W or, even better, a 5x5 array of 15W resistors. If you step up even further to a 6x6 array, then you can use 10W resistors, which frequently are more common and better priced than higher rated resistors. The more you spread out the heat (meaning both the more resistors involved and the greater the spacing between resistors), the more likely you are to avoid having to use additional heat sinks for forced air flow. How far you go depends on how much one-time money and effort you are willing to invest in this. If it is something you expect to use quite a bit, then it may well be worth doing it well once so that you lessen the risk of doing it again later.

Now, one thing that hasn't been discussed is what type of resistor to use. I don't know the best answer for that, but it is very possible that you want to avoid certain types of resistors, most probably wire-wound resistors, because the inductance may be undesireable. I don't know whether this is an issue at audio frequencies or not. Perhaps not since, at least in the old days, you were driving a speaker coil. Hopefully someone else can shed more knowlegable light on this point.

 

Here is a simple trick that comes in usefull in the lab.

If you need a 100 ohm, 1/2W resistor and all you have is 100 ohm 1/4W resistors ... put 2 in parallel, then 2 of these parallel combinations in series. Very convenient when you need a load quickly and don't want to fool around calculating values to get the right resistance and power.

If you instead need a 1W resistor, then the solution is 4 series combinations of 4 parallel 1/4W resistors.