Question

Given that ripple factor can be defined as:
RF = 1/(2*sqrt(3)*fCR)
Does that mean that (in an ideal circuit rectifying circuit) without a capacitor for voltage smoothing, the ripple factor approaches infinity? Any clarification would be appreciated.

Answer 1

While rectifing an AC signal, the output obtained would be a dc signal with frequecy twice(full wave rectifier) as that of the input AC.
Ripples are defined as the AC components alone in the rectified output if the DC is filtered out from it.
Any rectifier output would have a very high value of ripples in it and thus inorder to make the ripples negligible we use a smoothing circuit.When a capacitor is introduced after with the rectifier section, the ripples are considerably reduced and the output obtained is similar to a sawtooth waveform. The rms value of a sawtooth waveform is (1/(sqrt(3)))peak value.
Thus while computing the ripple factor we are using the formula you mentioned. The (1/2) in the formula comes from half of the peak voltage(since rectified output does not have a negative peak)

Ripple factor is also defined as Vrms/Vdc... when the capacitor is removed or that path is opened, the ripple factor should go to infinity by the formula..but by looking at the circuit it is clear that this wont happen because when the cap is removed it comes back to the normal full wave with ripple factor of .482.. if the rectifier is also removed the ripple factor of the sine wave goes to infinity since Vdc=0 and Vrms= 1/sqrt(vp)
dividing both we get infinity..

I guess that formula applies only for the combination of a rectifier plus smoothing circuit.. as soon as the smoothing circuit is removed, the ripple factor comes back to the RF of the prior circuit