RMS or effective ac voltage component of a full wave rectified sine (not filtered)

Question

Ignoring the voltage drop on the diodes, for a full wave rectified DC,can I say that:

1) Vrms of the original AC = Vrms of the rectified DC. ( I have just flipped one polarity. Power is same)

2) Vdc = V average = Vrms / 1.11072 ( the ratio between Vrms and V average)

( 0.7071 / 0.6366 = 1.11072)

(0.6366 / 0.7071 = 0.9003)

If I view the rectified DC as a mixture of pure DC, and AC component, the V average is the pure DC component.

3) Since I know the Vrms and the Vdc, can I calculate the AC component voltage this way:

The Vrms of the AC component would then be Vrms of the original AC sine wave / 0.435.

If this is all wrong, what is the correct ratio? I am looking for an answer without using functions. Should the AC component be referred to as Vrms or V effective or both are ok?

Answer 1

The ac component is usually referred by \$V_{ac}\$. And it is given by

$$V_{ac} = \sqrt{V_{rms}^2 - V_{dc}^2}\tag1$$ $$(1)\Rightarrow V_{ac} = V_{dc}\sqrt{1.11072^2 - 1} = 0.483V_{dc}$$ $$(1)\Rightarrow V_{ac} = V_{rms}\sqrt{1^2 - 0.9003^2} = 0.435V_{rms}$$ $$V_{ac} = 0.483V_{dc} = 0.435V_{rms}$$

The ripple factor, $$\gamma = \frac{V_{ac}}{V_{dc}} = 0.483$$ where The actual value is \$0.482\$. The variation in answer is because of the round off errors.

Answer 2

The wave is asymmetric, so you need to clarify what you mean by Vac. The Vpp of the rectified wave will obviously equal Vac (ignoring the voltage drop) of the input, or Vpp-in/2.

In the conventional sense, the Vac of the rectified signal will be the peak amplitude, i.e. the point furthest from the average. In this case this peak is at zero DC. So (again in this sense) the amplitude will be 2/pi = 0.6366.

In any case (i.e. if we choose to define Vac in some other way), the formula in 3) doesn't seem to make sense.