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Registered Member #124
Joined: Fri Feb 10 2006, 01:30PM
Location: Portland Oregon
Posts: 35
I want to build a little bitty CW: 2Kv at .5A load current. I know my caps are good (47% more than 2x input peak E) but I want to insure that my diodes won't burn up.
My Google and book searches have failed. How does one calculate current through the diodes of a CW multiplier? Is it simply load current? Or do the diodes in the earlier stages carry more current than in the later stages?
Registered Member #87
Joined: Thu Feb 09 2006, 01:36PM
Location: San Jose
Posts: 191
I'm sure 1n5408's will do fine, .5A load wont bother them much. I beleive the current passed through the diode is mostly load current, and some shared current drawn by the rig itself.
Registered Member #75
Joined: Thu Feb 09 2006, 09:30AM
Location: Montana, USA
Posts: 711
Conservation of energy: If you put out 10x times the voltage you put in, you have to put in 10x the current of what you get out, just like with a transformer. Since all the diodes are connected in series, there is no load sharing either. You need to size you diodes for 2n x .5A, where n is the number of stages.
Registered Member #15
Joined: Thu Feb 02 2006, 01:11PM
Location:
Posts: 3068
joe doh wrote ...
Conservation of energy: If you put out 10x times the voltage, you have to put in 10x the current, just like with a transformer. Since all the diodes are connected in series, there is no load sharing either. You need to size you diodes for 2n x .5A, where n is the number of stages.
Not quite. You can use the following equation to determine output voltage vs. load current. From this relationship you can determine output voltage and then figure out current through the diodes. As you can see, output voltage is inversely proportional with load current so the more current you have, the less voltage.
Voltage drop under load = I1/ (f*C) * (2 /3*n^3 + n^2/2- n/6)
Registered Member #162
Joined: Mon Feb 13 2006, 10:25AM
Location: United Kingdom
Posts: 3141
The diode current peak can be relatively high if you drive a CW-Multiplier with a flyback for average rating the conservation of energy approach seems easiest but unless you protect the output with a fun-killing resistor the largest diode currents will be due to pulse discharges so over-rating for current is a good idea. (I've killed 1N4004's with a 50 Hz 240 Vac 15-stage multiplier using cheapo 10 nF disk ceramics)
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Caclulating diod current in CW?
