0
\$\begingroup\$

I've come across a formula for calculating the ripple voltage from a wall adapter AC to DC power supply, which uses a step down transformer, fed to a rectifier, with the rectifier output connected to a ripple smoothing capacitor in parallel with the load.
enter image description here The formula would be: V(ripple)=V(DC)/(RL* C*f).
Going by this formula, if the load is not purely resistive, can the resistance (RL) be replaced by impedance magnitude?

\$\endgroup\$
1
\$\begingroup\$

This formula works OK as an estimate using the Vdc min and f = 2x input with a resistive load, but not with a reactive load.

Consider that if the load were pure C then there would be zero ripple ( +ve peak detector) and replacing R with L increases the ripple.

However the shape of the ripple does not "overshoot" the peak as shown but rather is a fairly straight as the Cap decays towards T=RC

\$\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.