Diode Current in CW multiplier

Thread Starter

thingmaker3

Joined May 16, 2005
5,084
I know how to calculate the voltage drop of a Cockroft Walton multiplier, and how to estimate output voltage. I know that voltage ratings of the capacitors and diodes must be a minimum of 2x the input peak voltage.

What I have trouble finding is the ammount of current that goes through the diodes. Does the same ammount of current go through each diode? Does more current go through the diodes in the lower stages?

I am interested in both the half wave and full wave CWs.
 

beenthere

Joined Apr 20, 2004
15,819
Hi,

Don't know for sure, but the nature of the beast suggests that all diodes conduct similarly. To pass enough charge to float up each capacitor means that equal charges has to pass each diode.
 

n9352527

Joined Oct 14, 2005
1,198
The difference would be minimal and mainly associated with stage losses. In the case of perfect components, where there is no loss whatsoever, the currents would be equal.
 

Thread Starter

thingmaker3

Joined May 16, 2005
5,084
Thank you, HVFan. I am familiar with the Blaze Labs site and have already run the math on a spreadsheet. I had somehow missed the java calculator though.

I am curios about diode current. What is the value for Id1? For Id2? Is it equal to load current? Blaze Labs text says "equal to the diode's rated current" but I can't make sense of that...
 
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