I have a homework in which in need to calculate the peak-to-peak voltage through a half-wave rectifier, center tapped full wave rectifier, and bridge full wave rectifier. What is confusing is that when we rectify a voltage we basically cut off the negative or the positive cycle of a signal, so basically there is just one peak either in the positive or the negative side, right? So could we say that the peak-to-peak voltage is the same as the peak voltage? Also what about Vrms do we use the formula:

$$V_{\text{rms}}=V_{\text{peak}} / \sqrt{2}$$


$$V_{\text{rms}} = \frac{V_{\text{peak-to-peak}}}{2\sqrt{2}}$$

  • \$\begingroup\$ There's still two peaks, it's just that one of them is at zero volts. \$\endgroup\$
    – Hearth
    Feb 24, 2021 at 13:26
  • \$\begingroup\$ You're overthinking peak-to-peak voltage. peak-to-peak voltage is simply the highest voltage - lowest voltage. It does not matter if that peak is really a peak or a flat line. \$\endgroup\$ Feb 24, 2021 at 13:42
  • \$\begingroup\$ Vrms is more complicated, look up the definition here: en.wikipedia.org/wiki/Root_mean_square Only for a sinewave Vrms = Vpeak/sqrt(2) is true. If the signal is not a sinewave then don't use that formula. \$\endgroup\$ Feb 24, 2021 at 13:44
  • \$\begingroup\$ As long as you DC couple after the rectifier, yes. \$\endgroup\$
    – user16324
    Feb 24, 2021 at 15:39

2 Answers 2


You can use the formula:

$$V_{\text{rms}}=V_{\text{peak}} / \sqrt{2}$$


  1. pure sine waves (with zero mean)

  2. fully rectified 1., with ideal diodes

enter image description here

The other formula is just a consequence of:

$$V_{\text{peak}}= V_{\text{peak-to-peak}} / 2$$

for any waveform with symmetrical min. and max. values.


You are correct in that the peak voltage and peak-to-peak voltages are the same for half-wave and full-wave rectified signals. However the relationships you cite between RMS and peak voltage are only valid for sine waves.


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