Article

Reducing Power Dissipation With Dropping Capacitors

A Clever Use of Capacitive Reactance

Using dropping resistors connected in series to resistive loads often results in the dissipation of a large amount of energy. If you are working on an AC circuit, the use of dropping capacitors can reduce this to a minimum, and, for a low-power supply, they could be a valid alternative to transformers.

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Materials

Film Capacitors (Wikipedia)

Motor capacitors (Wikipedia)

Datasheet — Kemet Radial Solid Polymer Aluminum Capacitors [PDF]

Component list

Component List

Please note: the components values of this list are only intended for the circuit and the load specifications indicated in Figure 5. For different purposes, their values will have to be re-calculated accordingly.

Capacitors

C1…C6: polymer-aluminum capacitor 220 µF (Kemet A759MY227M1JAAE045) [3]

Semiconductors

D1, D2: schottky diode SB150

Miscellaneous

X1, X2: screw terminal block, 5.08 mm pitch, 2 pole

F1: 5x20 fuse holder with fuse 2.5 A

Discussion (0 comments)

vlad.shcherbakov 2 months ago

Figure 4, doesn't add up by KVL: 200V + 113.4V = 313.4V. This should be 230V. For this, I wouldn't even touch phasor math. Just use KVL/KCL. You know you want 113.4V across with 0.3A current. 230V-113.4V=116.6V. Find a resistor/impedance value that would drop 116.6V at 0.3A current. Z = 116.6V/0.3A = 388.7 Ohm. Find the capacitance value at 50Hz so that Zc = 388.7 Ohm. 388.7 Ohm = 1/(2*pi*50*C), solve for C. C = 1/(388.7*2*pi*50) = 8.19 uF.

I think you just got the wrong voltage 200V should be 116.6V, but I'm not sure where the error in reasoning is. To me, both ways look "correct".