Electronics:Tips And Techniques

From: gingertube

This was the first time I've ever seen anyone doing it by the numbers
and I thought it important for many of us.

Given:
Aerovox BHC ALC40 Series caps 470uF and 450 V rated,
in series with 530V rails.

Leakage Current for ALC40 Series is quoted as:
0.003 Cr Vr or 10mA which ever is the smaller.
Cr is rated capacitance and Vr is rated voltage.

So do the calc Ileak = 0.003 x 470 x 10^-6 x 450
I get 0.6345 mA leakage current.

For the capacitors to share voltage properly
you want 3 to 5 times this current through
bleed/voltage equalising resistors.

So assume we want 2mA through those resistors.

With a 530V rail you have 265V across each cap.
Resistors calcuated as 265V/2mA = 132KOhms (maximum value)

I would strongly suggest you revert to 120K resistors across the caps.

Power Dissipated in each resistor = V squared on R:
R
= (265)^2 / 120K
= 0.585 Watts

Use 2W metal films, 3W metal oxide, or 5W wire wound.

Submitted and edited by SyncTroniX for use here.