Resistor Power Rating and headroom for AC circuit

Question

Background

I'm going to put together a resistive load, to load a 100W source with an \$8\Omega\$ or \$16\Omega\$ resistor. I'm finding that when to use peak, peak-to-peak or RMS values of voltage, current or power is a subject still shrouded in mystery for those new to the subject or not working in the industry.

Understanding which power to use (RMS, Peak or Peak-to-peak)

When selecting a power rating for the resistor

• do you assume that the power rating on the component website is for a DC supply
• should you calculate the power rating of a component from the RMS power, Peak Power or Peak-to-peak power?

Example: Without taking into account the extra headroom needed (let's run this resistor at maximum permissible power), $$100W_{RMS}=200W_{Peak}=400W_{Peak-to-Peak}$$ therefore should you select a 400W resistor?

Headroom

What is the convention when it comes to selecting headroom on the component's power rating. Is, 1.2 - 1.5 a reasonable factor to multiply by. Let's for the sake of safety put a fuse in line with the resistor to protect it.

My intuition

My intuition tells me to place a fuse in series and either:

• work with the Peak power value in mind because the AC signals will only ever reach a maximum of the peak value, not the peak-to-peak value, and to select headroom in the region of a factor of 1.25 - 1.50. From the example, this would mean selecting a resistor with power rating of \$200W_{Peak} \cdot 1.2 = 250W\$ or \$200W_{Peak} \cdot 1.50 = 300W\$.

or

• work with the peak-to-peak power value which according to the logic in the previous bullet point will surpass the peak power value by a factor of 2 and not chose to use additional headroom. From the example, this would mean selecting a resistor with power rating of \$400W_{Peak-to-peak} = 400W\$

Of these two estimations, the second seems to be sensible due to the apparent large headroom, and therefore component lifespan, but perhaps at an extra financial cost.

Simplification of finding a component

To make it easier to find components within a restricted budged (in the tens of pounds rather than hundreds of pounds) I intend to connect a number of resistors in parallel to distribute the current and therefore power/ component heating. i.e. \$5 \times 40\Omega\$ resistors in parallel \$=8\Omega\$ and \$5 \times 80\Omega\$ resistors in parallel \$=16\Omega\$.

What say you? Thanks in advance

Daniel

Answer 1

For determining the power a voltage will dissipate when applied to a resistance, use the RMS voltage.

One way to look at the RMS voltage is that's the equivalent AC to the DC voltage that would dump the same power into the same resistance.

For example, 5 V across a 2 Ω resistor will dump (5 V)²/(2 Ω) = 12.5 W into the resistor. That is true whether the 5 V is DC, or 5 V AC RMS. Actually, the voltage in the equation is always RMS. For DC, the RMS is the DC voltage.

Answer 2

Using the formula E=sqrt(P*R) where E = rms volts, P = power in watts, R = resistance in ohms, you can figure out the maximum rms voltage your resistor can tolerate. So for 8 ohms and 100 watts, you can put 28.28 volts rms on the resistor.

Depending on the type of resistor, there will be some dynamic headroom there. A 100 W wirewound resistor can usually handle 150 or maybe 200 W for at least a few seconds, maybe more.

If you run a resistor at full power on a sustained basis, you have to be careful with its temperature. A fan can be helpful to keep it from overheating. The data sheet should indicate a maximum temperature.

Also if you run a resistor at full power on a sustained basis, you might be pushing your luck. I would suggest allowing a margin; 200 W resistors are commonly available and if you want to run 100 W for hours then that would be a better choice; also they won't run as hot. Note that the full power rating will be given at a certain maximum temperature; sometimes there is a derating curve you can use to find what it can do when it heats up.

Fuses add resistance, typically of uncertain value, and depending on how critical your application is, that might not be a good idea. For testing amplifiers, I would never use a fuse; instead just be careful about how much voltage you put on the load.

A good true-rms voltmeter will show you (with E^2/R) how much power you are delivering. Such a meter should be sufficient for whatever crest factor you would find with an audio (music or voice) signal. If you are using a pulse generator or something like that then you should check the crest factor of the signal and the meter spec in order to accurately show rms voltage which will yield an accurate power measurement. Hope that helps!

Answer 3

In detail, you will tend to care about average power in circuit components. There are times when instantaneous power may matter more (the "current action integral") when explosive situations are being examined (fuses, for example.) But usually, it's average power.

In general, the average power over some time from \$t_0\$ to \$t_1\$ is:

$$\begin{align*} \overline{P}=\frac{1}{t_1 - t_0}\int_{t_0}^{t_1} V_t\: I_t\:\:\textrm{d}t&=\frac{1}{t_1 - t_0}\int_{t_0}^{t_1} \frac{V_t^2}{R}\:\:\textrm{d}t=\frac{1}{t_1 - t_0}\int_{t_0}^{t_1} \frac{V_{_\text{PEAK}}^{^2}\cos^2\left(\omega\,t\right)}{R}\:\:\textrm{d}t \\\\ \text{Set }x=\omega\, t\:\therefore \text{d}x&=\omega\,\text{d}t\:\text{ and pick convenient }x_0=0 \\\\ \text{ and } x_1&=\pi\text{ so that }t_0=0 \text{ and } t_1=\frac1{2\,f} \\\\ &=\frac{1}{\frac1{2\,f} - 0}\int_{0}^{\frac1{2\,f}} \frac{V_{_\text{PEAK}}^{^2}\cos^2\left(\omega\,t\right)}{R}\:\:\textrm{d}t \\\\ \text{substituting } &x \text{ into the integral and moving things around}, \\\\ &=\frac{2\,f\,V_{_\text{PEAK}}^{^2}}{R}\int_{0}^{\pi} \cos^2\left(x\right)\:\frac1{\omega}\:\textrm{d}x \\\\ &=\frac{V_{_\text{PEAK}}^{^2}}{\pi\,R}\int_{0}^{\pi} \cos^2\left(x\right)\:\textrm{d}x \\\\ &=\frac{V_{_\text{PEAK}}^{^2}}{\pi R}\left[\frac{1}{2}\left(x+\operatorname{sin} x\cdot \operatorname{cos} x\right)\right]\bigg\rvert_0^\pi \\\\ \overline{P}&=\frac{V_{_\text{PEAK}}^{^2}}{2 R} \end{align*}$$

Given \$V_{_\text{PEAK}}=\sqrt{2} \:V_{_\text{RMS}}\$ and \$V_{_\text{PEAK}}=\frac{V_{_\text{PP}}}{2}\$ you can easily find:

$$\begin{align*} \overline{P} &= \frac{1}{2}\frac{V_{_\text{PEAK}}^{^2}}{R}\\\\ &=\frac{V_{_\text{RMS}}^{^2}}{R}\\\\ &=\frac{1}{8}\frac{V_{_\text{PP}}^{^2}}{R} \end{align*}$$

In the above, note that I said "if." I said that because now it's easy to see why \$V_{_\text{RMS}}=\frac{V_{_\text{PEAK}}}{\sqrt{2}}\$ is useful. It gets rid of that preceding constant and returns the equation to a very simple formula that looks exactly like the DC equivalent and the effect (in average power) will also be the same. This is why the concept is so important.

It's probably also important to understand the development here and not just "memorize" the equivalence. The idea of RMS is meaningful with sine and cosine (had I used sine above, the sign of a term that resolves to zero regardless would have changed) curves as it "normalizes" the result so that it follows the usual Ohm's Law without extra constants. But it also helps to know the development for cases when the time series shape of the curve isn't so convenient. You can now see that, for example, an "audio" signal might not be so conveniently described, relative to its peak values for example. And I think you can also see that if there were an added DC bias to the voltage, the results would also be different. (The constant DC bias would "come out" of the integral and present an obvious addition to the final formula.)

By the way, you are right and since \$V_{_\text{PP}}=2\sqrt{2}\:V_{_\text{RMS}}\$ for sine/cosine voltages, it follows that for \$V_{_\text{RMS}}=5\:\textrm{V}\$ that \$V_{_\text{PP}}\approx 14.14\:\textrm{V}\$.

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You can find all of the above development in a lot of places on the web and I'd recommend that you get comfortable with the ideas. It was just easy to write it out for you here. (Hopefully, I didn't make a mistake.)