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?


1 Answer 1


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


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