# Is the peak-to-peak voltage the same as the peak voltage in rectified voltage?

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:

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

or

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

• There's still two peaks, it's just that one of them is at zero volts. Feb 24, 2021 at 13:26
• 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. Feb 24, 2021 at 13:42
• 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. Feb 24, 2021 at 13:44
• As long as you DC couple after the rectifier, yes. Feb 24, 2021 at 15:39

You can use the formula:

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

for:

1. pure sine waves (with zero mean)

2. fully rectified 1., with ideal diodes 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.