Average Voltage

by Harold Hallikainen

San Luis Obispo, Calif. In the last article in this series, we noticed that the instantaneous voltage for an AC voltage is changing continuously as a function of time. The most common AC waveform is a sine wave. The voltage at any instant in time for a sine wave can be determined as:

where V_P is the peak voltage, w (omega) is the frequency in radians per second (equivalent to 2*pi*f where f is the frequency in Hz), and t is the time since the waveform started. This equation assumes the sine function is based on an angle in radians (instead of degrees). We'll be switching back and forth between using degrees and radians, depending upon which gives us the simplest math. Find that radian/degree switch on your calculator! We can also add a constant for phase shift, but we'll leave that for later.

Quite often we are interested in where we are in a waveform in terms of how much of the cycle we have completed instead of how many seconds, microseconds, or whatever, since the waveform began. Again, we can use either degrees or radians. If we want to know the instantaneous voltage of a sine waveform at any point in a cycle, independent of frequency and time (since we'll use phase instead), the above equation becomes:

where V_P is the peak voltage and a is how far we are in to the waveform in degrees or radians (depending upon which sine function we're using). Table 1 shows the voltage at various points through a single cycle of a 1 volt peak sine wave.

Average Voltage

Now that we can determine the instantaneous voltage at any point in the waveform, let's determine the average voltage through one cycle. The average is also known as the arithmetic mean. We can simply add up the instaneous voltages and divide by the number of voltages we've added. Notice that the sum is zero? For every positive voltage during the first half cycle, we have an equal negative voltage during the second half cycle. If you connect a DC meter (which mechanically averages the instantaneous voltages) to an AC source, it will read 0 volts, since the average is indeed 0 volts. However, the average for a half cycle is not zero volts. If we average the first half cycle (the first 8 voltages in the table above), we get about 0.615 volts. This is approximately the average voltage during a half cycle.

For a more exact determination of the average voltage of a half cycle, let's reconsider how we determine the average. If, for example, you travel 10 miles per hour for 1 hour and 20 miles per hour for two hours, you might be tempted to say the average speed is 15 miles per hour (add the speeds and divide by the number of speeds). However, we know that what we must really do is determine the total distance traveled (10 miles + 40 miles), then divide by 3 hours, yielding 16.667 miles per hour. Our new definition of average is the sum of the products of the value times the amount of time we spent at that value, divided by the total time. Applying this to the voltages in a sine wave, we can approximate the average voltage as shown in figure 1. In equation 1, sigma indicates we're taking the sum of several terms, V(t) is the instantaneous voltage, and delta t is the change in time from the this time to the next time we determine the voltage. Since the voltage is changing continuously, it really doesn't spend any time at a particular voltage. Therefore, we keep increasing the number of samples and multiplying them by a smaller and smaller delta t. The limit of this smaller and smaller approach turns the summation into an integral and turns delta t into dt. So, if we say the average voltage can be computed using the integral in equation 2, you know where it comes from!

Equation 2 is a definite integral. We are to do this infinite number of multiplications and additions for t values between 0 and pi (radians). Substituting the function for a 1 volt peak sine wave, we get the equation 3.

This value of this definite integral is determined by taking the definite integral of sin(t) dt, which is -cos(t) (we'll leave the calculus details for another time!), and subtracting the integral evaluated at the lower limit from that evaluated at the upper limit. This is shown in equation 4 where the evaluation limits are shown to the right of the -cos term. These evaluation limits are inserted in the equation in equation 5. Finally, the values of cos(pi) and cos(0) are substituted in equation 6. This yields an exact value for the average of a half cycle of a sine wave as 2/pi, or about 0.636 times the peak voltage.

Next time, we'll evaluate the Root Mean Square of a sine wave. To start you thinking, we're going to take the square root of the average of the squares of the voltages. We'll also discuss why we do that.

Harold Hallikainen designs transmitter control and lighting equipment for Dove Systems, a manufacturer serving the broadcast and entertainment industries. He also teaches electronics at Cuesta College and is an avid contra dancer. He can be reached at +1 805 541 0200 (voice), +1 805 541 0201 (fax), harold@hallikainen.com (email), and http://hallikainen.com (World Wide Web, where an archive of these articles is maintained).