0
\$\begingroup\$

Shouldn't the conduction angle for a full wave rectifier with filter capacitor be twice that of a half wave rectifier with filter capacitor? The conduction angle for a half wave rectifier with filter capacitor is:

$$\sqrt{2V_r/V_p}$$

Since the frequency of the output wave in case of a full wave rectifier is twice as that of a half wave one the conduction angle should be twice as well, but we've already taken the above into account while calculating it's Vr, so should we take it into account again?

\$\endgroup\$
  • \$\begingroup\$ Don't you imagine that the load might have something to say about the subject? \$\endgroup\$ – jonk Apr 18 at 5:21
  • \$\begingroup\$ I know load matters but here assuming loads to be same for both the cases, take load to be R. \$\endgroup\$ – Adarsh Kumar Apr 18 at 6:26
2
\$\begingroup\$

First, that equation is an approximation, based on x^2/2 being approximately equal to 1 - cos(x). It also assumes that the conduction angle is small compared to a cycle. For any reasonable conduction angle, these approximations are fairly accurate. Or, for ripple of about 15% or less it is reasonably accurate.

It is approximately valid for either full or half wave rectifiers. But, of course, everything else being equal, when you go from a full to half wave the ripple will double therefore the conduction angle will be the square root of 2 greater.

enter image description here

enter image description here

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.