# Is the RMS value the same after you pass it through a full wave bridge rectifer?

Is the RMS value the same after you pass it through a full wave bridge rectifier?

That is if the value was 230VAC (RMS) before the rectifier will it still be so after the rectifier? (ignoring any loss of the diodes)

I tried to google this but found different pages saying different things, so I thought I would try here.

• Unless you smooth it... Mar 31, 2023 at 10:40
• Don't forget the voltage drop of two rectifier diodes, one on the plusitive and one on the minusitive. Mar 31, 2023 at 18:45

Is the RMS value the same after you pass it through a full wave bridge [ideal] rectifier?

Yes. Regardless if you take the discrete or continuous definition:

$$\ x_\text{RMS} = \sqrt{ \frac{1}{n} \left( x_1^2 + x_2^2 + \cdots + x_n^2 \right) }. \$$

$$\ f_\text{RMS} = \lim_{T\rightarrow \infty} \sqrt {{1 \over {2T}} {\int_{-T}^{T} {[f(t)]}^2\, {\rm d}t}}. \$$

The input is squared, so the polarity is taken out of the calculation.

If you think about the meaning of RMS (the power dissipated by a resistor subjected to an equivalent DC voltage) it also makes sense, since a resistor will dissipate the same instantaneous power regardless of the current direction.

From this diagram on Wikipedia, you can see that it must be. You can think of it like this: the RMS value of the positive half-cycle is calculated up to the first zero crossing. A wave which repeats all the positive half-cycles will thus have the same RMS value.

Alternatively, from the equation, you can see we're doing something with the square of the values, which are obviously the same for a rectified wave as the unrectified negative cycle.

RMS means taking the square root of the mean value of the square of a signal. So, if you square a sinewave, the waveform becomes exclusively positive and, is exactly the same shape and size as a squared rectified sinewave.

It works with any waveform shape also (not just sinewaves).

RMS voltage through an ideal rectifier doesn't change because RMS is independent of sign of the voltage.

And, if one connects a resistor to the output of an ideal rectifier, the power dissipated will be indeed the same as if it was connected to the rectifier input.

But. If the load impedance is not purely resistive, and that includes adding capacitors to the output - the RMS power dissipated in the load does not need to be the same as when the load is connected to the un-rectified AC.