How is the ripple factor affected when a variable load is connected to a series inductor filter? My text book doesn't explain it in terms I understand.
A power supply first rectifies AC, changing it to pulsating DC. A .filter is used to smooth the pulsations to produce DC with no pulsations
The series inductors in power supplies are ineffective in smoothing out the current into an open circuit because there isn't any current to smooth, at least during part of the ripple cycle. This results in a large AC ripple component compared to the DC component in such a circuit. On the other hand, an ideal inductor that doesn't change its inductance with current will let only a specific amount of ripple current through, as long as the line and load voltage waveforms remain relatively constant with load, regardless of the DC component of the load current. Because the AC component of the ripple current remains the same, it will be a smaller percentage of the total load current if the DC component is large. Real inductors also have the resistance of their windings that will result in a drop in the load voltage at higher load currents but will still give the same AC ripple at high load currents as at the low ones. Power inductors can be made smaller, lighter, and cheaper if they are designed in such a way that their inductance drops inversely with their load current. Inductors designed in this way will give the same percentage of AC ripple over a large range of DC current. In practice, such an inductor will not follow an inverse inductance curve very exactly but will still be close enough to be a useful filtering component.
