Power Types (DC, RMS) • Brady Salz

There's lot of different types of power measurements to use and distinguish between, and I've noticed that lots of people have trouble keeping them straight. Terms like DC power, AC power, RMS power, available power, and plenty more all just kind of get jumbled together in people's mind. Offsets of $\sqrt{2}$ just get thrown around everywhere. Luckily, it's actually just separate and simple cases to keep track of, and will hopefully make sense after this.

Let's first define power. In electronic circuits, we define power as the real portion of the instantaneous voltage and current across a device. A positive number is said to “sink”, or use power, whereas a negative number is said to “source”, or provide power. For a simple resistor, we know $V=IR$, which means we can write our power definition in a variety of ways:

$$ P = V \cdot I = V^2/R = I^2 \cdot R $$

Now, all of these are instantaneous, and so technically the P, V, and I terms are functions of time. So when we have a static (constant) level in, we have a constant power out. That's pretty simple for most people to grasp, so let's put in a moving waveform. The input here is going to be a square wave, and the output just a simple load resistor.

power-types power-types-square

There's really only two places to calculate the power here - the “off” phase, and the “on” phase. The off phase is easy - the instantaneous power is zero, no voltage no problem. The on phase is just the applied voltage squared divided by the load resistance - we'll just say it's 1$\Omega$ to make things easy. This means we have 25 W (Watts!) of power being provided by the source, and used by the resistor. The average power here is pretty simply - we just take a weighted average of each phase. To put it in math terms:

$$ P_{tot} = \dfrac{1}{T} \sum_{i=1} t_{i} P_{i} $$

Where $P_i$ is the power during that phase, $t_i$ is the length of that phase, and $T$ is the total period. For our simple example above, we get:

$$ P_{tot} = 1/4 \cdot (2\cdot 0 + 25\cdot 2) = 12.5$$

Or correspondingly, half the power. For a smoother signal, we would use an integral instead of a summation. The problem occurs when we deliver negative voltage to our load. If I have a zero-centered sine wave, we have 0W delivered right? Every time we go up is exactly canceled by the time spent going down, per definition of sinusoids. But that's not true, as we clearly have some power delivered to the load. So what do we do?

The answer is to use RMS power. RMS is an abbreviation for root-mean-square, and that's exactly how we calculate it. We use it whenever we have a signal with a negative component. You just follow the process that the name describes in reverse order:

First you square your entire signal
Average it
Take the square root

That number is your RMS value.

Let's see what happens with a simple sine wave with an amplitude of one. First we square it, so we get a positive-only sine wave that goes from 0 to 1 over $2\pi$. After that, we go and average the signal. Since's it's a periodic signal, we can get away with only averaging one period here - 0 to $\pi$:

$$V_{avg} = \dfrac{\int_0^\pi \sin^2 (t) dt}{\pi} = \dfrac{\pi/2}{pi}$$

We then take the square root, and our final answer is $V_{rms} = 1/\sqrt{2}$. The factor, it turns out, is the relationship between the amplitude of a zero-centered sine wave and it's RMS value. To put it into context, in the United States, the mains voltage from an outlet is 120V RMS, which means me multiply by that factor to get an 170V (approx) amplitude sine wave that circuits can handle! Definitely a huge difference in voltage!