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# What is the best form of charge a capacitor bank? Answered

I want to charge a capacitor bank of 450 volts, 4000-6000 uf. With dc.
I have these ideas to charge it:
Cockcroft walton
Flyback

What is the best way to charge it, i want that the charge time be like 2-10seconds
Thanks

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## 7 Replies

steveastrouk (author)2011-10-16

Chop the DC into the parallel secondaries of two small transformers, then tie the primaries in such a way that you get 450 V out, then rectify, smooth etc.

Steve

iceng (author)2011-10-16

+1

sshuggi (author)2011-10-15

It's pretty simple: transformer up to 450V and a rectifier to make it +DC. If you have no resistance in the in the wires it charges pretty much instantaneously. You could have like 1kOhm of stray resistance in your wires and rectifier and still have it charge in 10s. But, you'll get nowhere near that much resistance in wires and diodes. (It'll be something in the single digits.) Anyway, here's an instructible that shows the setup. The video posted just shows where he got all his parts, but if you go to the page of that guys account on YouTube. It shows him discharging it on various bits of metal. (Then, it charges back immediately.)

Kurt Gerhard (author)2011-10-15

Thanks, but I dont want to use ac i want to use dc

sshuggi (author)2011-10-16

You can always convert it to AC and then back. You'd need an oscillator that can handle the current you'll need to charge it in 10 seconds though.

Things to consider whichever way you go: The current into a charging capacitor (and voltage) are non-linear and modeled by:
I(t)=V0/R * e^[-t/(RC)]          RC=Tau (time constant)

This has a few implications to your problem:
1. At t=0, the current is at it's maximum of V0/R. If you have no resistance, this current is very high. You're going to want some kind of resistance.
2. This resistance is what determines how long it takes to charge. A capacitor is considered fully charged after 5 time constants. If you want this to be 10s, then:
10s=R*4000uF  → R=500 Ohms
3. Going back to 1, you get your max current at 450V to be 900mA.

Therefore, your components are all going to have to handle 900mA (for a split second.) That's pretty insane... Of course, this current will drop by 64% to 424mA after one time constant (2s.) But, that's still a lot. That's why the transformer and rectifier in that guys video were huge and industrial quality.

My suggestion, go for a longer wait and/or use a current regulator.

sshuggi (author)2011-10-16

Correction: 10s=5*R*4000uF → R=500 Ohms

iceng (author)2011-10-15

Cockcroft walton is way too slow