The "RMS" value of an alternating voltage is equivalent to a DC voltage that would generate the same amount of heat in a resistor as the AC voltage would. And RMS = 0.707 * Peak value [N.B. Sine wave]

But the DC equivalent of the same wave is 0.636 * Peak value [N.B. Full wave]

I've seen the mathematics related to it for many years. But what does this mean? If the RMS is the heat equivalent, then what is DC? How are they different?

AC - RMS Reference
AC - DC Reference

I've included a full-wave rectifier and it's filtered output below.

1.Full bridge rectifier circuit.

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2.Input of rectifier.[RMS]
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  1. The output of the rectifier.
    enter image description here
  • \$\begingroup\$ Thanks. I've edited the errors. \$\endgroup\$
    – Sadat Rafi
    Apr 13, 2020 at 17:43

1 Answer 1


I will assume we're only talking about pure sine waves here.

"DC equivalent" is wrong. You mean 1/2 cycle average (average over a whole cycle is obviously zero.

If you have a precision full wave rectifier (such circuits exist) and then filter the output with a low-pass filter you will get the average of about 0.636 times the peak.

Because power is proportional to voltage squared when applied across a fixed resistance we can calculate the DC voltage that would result in equivalent heating as the RMS voltage. For a sine wave that's about 0.707 times the peak.

The RMS voltage is about 1.11 times higher than the average for a sine wave. A cheap multimeter on AC current and voltage ranges will full-wave rectify the input signal and then display a number 1.11 times higher than the actual voltage to give the user an approximately correct number for a sine wave input. For example on a North American mains it will show about 120VAC.

If the waveform is not sinusoidal then the numbers will not be 0.707 or 0.636 but will generally be something else. Of course in the degenerate case of DC the average is exactly the same as the RMS. For a square wave, the peak is the same as the average of the absolute value is the same as the RMS value.

You can see enormous divergence between the two if the waveform is 'spikey', for example, a waveform consisting of +63V pulses at a 1% duty cycle (say 10us on, 0.99 ms off) has an average of only 630mV but an RMS value of 6.3V so a 6.3V pilot lamp will illuminate normally.

With regarded to your added info, the diode + capacitor acts more like a peak detector than a low-pass filter.

With no load (ignoring the diode drop) the output voltage will be equal to the peak input voltage.

When you add a load, the voltage drops between the peaks, the drop is proportional to the current and inversely proportional to the capacitance and frequency. In a real circuit you'll also see the transformer voltage drop and the diode drop increase a bit as the current increases.

If you want to see something approaching the 0.636 (ignoring the diode drop), remove the capacitors and measure the voltage on a DC range.

  • \$\begingroup\$ I've included some practical data in my question description. I've never got the filtered output 0.636 times of peak. Even under the loaded condition, it stays near the peak. According to the rule, it should be 16 volts. But it's 24. Even when it's driving a load. \$\endgroup\$
    – Sadat Rafi
    Apr 13, 2020 at 18:16
  • \$\begingroup\$ Thanks a lot. :) \$\endgroup\$
    – Sadat Rafi
    Apr 13, 2020 at 18:23

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