I have a transformer that outputs 16 volts ac. I built a 4-diode full wave rectifier and added a 4700uF 35 volt smoothing capacitor. Now my dc voltage measures 21 volts? I thought the voltage would drop because it has to pass through diodes. What explains this voltage increase?

Thanks guys for answering the question. All the answers were good (every answer gave information and the ones after the first added to the previous). Thanks again!

The AC voltage--termed "RMS voltage" (which others have helpfully addressed), is one way of finding the "average electrical potential" of an AC sine wave. The actual peak voltage is higher than the RMS average (peak V is theoretically ~1.414 times the RMS voltage). __________________________________________________________________________________________ Shifting the negative wave to positive with a rectifier doesn't alter the average voltage, or the peak voltage. However, adding a filter capacitor does "fill the valleys" between each of the wave peaks. The capacitor stores electrical energy, and this allows the output voltage to approach the "peak" voltage. ____________________________________________________________________________________________ Si diodes have a voltage drop of approx 0.7V (double that for a fullwave bridge rectifier).

To add to this: the (mean) average voltage of AC is zero, because it alternates equally from +ve/-ve. At any point in time it's +this or -that, neither of these things are much use for anything. RMS being "Root Mean Square" is a math' way of getting something meaningful out - you square the voltage so that -ve becomes +ve (-2 x -2 = +4), then average it and square-root it (root +4 = 2). As AC is "lumpy" the peak voltages at the top of the sine-wave are above the RMS average, and if you're not applying heavy load the rectified DC voltage is closer to these peaks than the RMS (math' for AC) value.

When you measure AC "voltage", what your meter reports to you is the RMS (root mean square) of the sinusoidally oscillating voltage. For a sine wave, the RMS value is 1/sqrt(2) of the peak amplitude. In the U.S., the RMS is nominally 117V, for a peak value of 165V.

A simple rectifier bridge essentially flips the negative-going half of each sine wave to become positive, which moves the RMS value much closer to the peak.

Diodes don't introduce any significant voltage drop.

120 volts ac acutally has a peak of about 170 volts. And your 16 volts has peaks of about 21 volts. Divide your ac by .707 to find the peak voltage. There is a great discussion going on right now at this question.http://www.instructables.com/answers/How-do-I-make-a-160-DC-volt-power-supply/

RMS being "Root Mean Square" is a math' way of getting something meaningful out - you square the voltage so that -ve becomes +ve (-2 x -2 = +4), then average it and square-root it (root +4 = 2).

As AC is "lumpy" the peak voltages at the top of the sine-wave are above the RMS average, and if you're not applying heavy load the rectified DC voltage is closer to these peaks than the RMS (math' for AC) value.

L

Re-design gave you the numbers, but not the math.

When you measure AC "voltage", what your meter reports to you is the RMS (root mean square) of the sinusoidally oscillating voltage. For a sine wave, the RMS value is 1/sqrt(2) of the peak amplitude. In the U.S., the RMS is nominally 117V, for a peak value of 165V.

A simple rectifier bridge essentially flips the negative-going half of each sine wave to become positive, which moves the RMS value much closer to the peak.

Diodes don't introduce any significant voltage drop.