AC Waveforms – Why Is Sine Wave Preferred Over Other Waveforms?

Let's clarify what "effective" voltage means. First of all let's note that this is only really used to describe time-varying periodic voltages. We can imagine taking that voltage and driving across some resistor, R. Since this is a periodic wave it will on average consume some set amount of power.

If we take the same resistor and drive a particular DC voltage, we will consume the same amount of power. This voltage is the Effective Voltage. It is independent of the resistor used. Different resistors will give different power consumption, but for a given periodic voltage input, the effective voltage is the same.

If we go to work the problem out for determining what this effective voltage is for any wave, it effectively comes down to performing the following steps. I recommend working this out from V^2/R to understand why these steps are correct.

1. Divide the period up into a bunch of small samples.
2. Square the voltage value at each sample.
3. Average up all these squared values.
4. Take the square root of this average to get an approximation of the effective value.

As you do this with more and more samples (i.e. smaller time between each sample), you approach the effective value. This sequence is called the root-mean-square, or RMS. In other words, RMS Voltage is Effective Voltage.

Now, here is where people get confused. If you go and do the math, you can calculate analytically that for a sine wave $$V_{RMS} = \frac{V_{peak}}{\sqrt 2}$$ This is only true for a sine wave. For most other waves, you simply have to go do the math.

However, for some waves, like a full- and half-wave rectified sine waves, you can use the definition to determine the RMS voltage more simply.

For a full wave rectified wave driven across a resistor, it can be realized that a resistor doesn't care if a voltage is positive or negative, just what the magnitude is. The power is the same as a regular sine wave, therefore the RMS voltage is the same.

For a half wave rectified wave driven across a resistor, again, it can be worked out that the power consumed is 1/2 of the typical sine wave. The power is 1/2 the regular sine wave, but voltages are squared to get power, so:

$$\frac{P_{sine}}{P_{half}} = 2 = \frac{{V_{{RMS}_{sine}}}^2}{{V_{{RMS}_{half}}}^2}$$ which gives $$\frac{V_{{RMS}_{sine}}}{V_{{RMS}_{half}}} = \sqrt 2 => {V_{{RMS}_{half}}} = \frac{V_{{RMS}_{sine}}}{\sqrt 2} = \frac{V_{peak}}{2}$$

A bit more complicated, but it's easier than calculus.

I would like to suggest that the comparator circuit shown by @hwengmgr could use some small improvements over what was posted.

Improvements include:

1. Bigger capacitor value to reduce filter effect of the capacitor at frequency.
2. Larger value resistors in dividers to reduce current load on source supply.
3. Added hysteresis at the comparator to make the circuit less sensitive to noise that may be on the input waveform.
4. Changed pullup resistor on output of comparator to a more sane value for 1mA load instead of 33mA.

This was simulated with LT-Spice. I used a model for a comparator that was built into the package library. You can substitute a comparator of choice as long as it is specified for operation at a Vdd of 3.3V.

The addition of the hysteresis feature necessitated the swapping of the '+' and '-' inputs of the comparator which causes an inversion of the output waveform. If this is a problem it can be easily flipped with a logic element. If you select a dual comparator the other half of the package could be used to invert the signal.