Ripple Voltage from Constant Power Load

For a capacitive rectifier, the droop in the capacitor will be the same for constant power as it will be for equal average power.

The problem is determining the exact intersect between the rectified sine and the capacitor droop to get the discharge period. That sounds like integral calculus to me, but its probably just algebra and trig.

For rough estimation you can assume 2mSec charging in a 6mSec period, that leaves ~4mSec to discharge.

C/2 x ( V1^2 - V2^2) = delta j delta j = P x t j = CV^2/2

root{ 160^2 - ( 180 x 4E-3 / 1.8E-4 ) = V2 = 147V

ppk ripple = 166 - 147 = 20V

Worst case ripple occurs at low line as the initial energy stored in the caps reduces.

RL

I figured out a model to confirm.

Good math.. Your rough estimate was accurate enough for my app. :)

Thanks

D from BC British Columbia Canada.

If V is the DC capacitor voltage then the DC load current is I=P/V making the incremental negative conductance dI/dV=-P/V^2 which is pretty small. Taking the average V to be say 90% of Vpeak~170V, this gives

-4mA/V incremental or -250 ohms, which is less than 1Watt incremental, and not worth considering in more detail.