Charging time needed for 10-stage Cockcroft-Walton voltage multiplier

[woelen]

I plan to build a 10-stage dual-branch Cockcroft-Walton voltage multiplier. The circuit looks as follows, but has more stages:

 

 

My circuit will use 30 capacitors and 40 diodes.

 

I want to connect it to a center-tapped transformer, each half giving 120 V AC. So, per capacitor, I can obtain appr. 350 V DC and the total output voltage can be 3500 V. I want to use 220 uF/400 V capacitors. This allows for an energy storage of 400 J in the total circuit!

 

I have one problem. If I connect the device to my output transformer, then I expect it to blow out my fuse, due to the enormous initial charge currents. I already built a 3-stage device with 100 uF/400 capacitors and when this is powered up, then sometimes the fuse is blown out, so with the more than 3 times as large circuit and double capacitance I expect major power up surges.

 

I do not need high power output, I only want to charge the device and then use the fully charged device. So, I have the idea to place a series resistor of e.g. 1 kOhm in series with the left branch and right branch, immediately in series with the trasnsformer output. This limits the charging current, but also the time, needed to fully charge the device increases. THen I wait, until the voltage of 3500 V is reached and then start the experiments. My question, however, is how long it will take before the circuit is charged to e.g. 95% or so.

 

I've no idea how long it will take to charge the device with 1K resistors, when a 10-stage circuit is used. My input is 50 Hz AC. If someone did simulations of this circuit then that would be nice. Any suggestions? I did some experiments with my current 3-stage multiplier circuit and 2.2 kOhm resistors in series with the transformer AC-leads (GND was connected directly). That already gives very long charge times.

Are there other safe ways of limiting the initial charge currents?

[Externet]

I would say 10 stages x resistance in Ohms x capacity in Farads = time in seconds to reach 66% of final tension.

Miguel

[woelen]

I would say 10 stages x resistance in Ohms x capacity in Farads = time in seconds to reach 66% of final tension.

Miguel

That is correct for charging a capacitor through a resistor from a DC source, it is the simple exponential law Vout = Vin*(1 - exp(t/RC)), which you express.

But... here, we are talking about AC-charging, so the actual charging only is at part of the sine wave of the applied input voltage and also, the voltage is not constant. Added complexity is the staging of charges. Do you still think that this simple RC-combination holds?

[Externet]

I believe it does; Half cycle is handled by one branch of the circuit, the other half on the other branch, equivalent to a continuos voltage applied in the amount of the RMS value of the AC.

The actual charge voltage in the capacitors will be more like the peak voltage of the AC, but the charge time is independent from voltage and will not be affected.

Miguel

[woelen]

Half cycle is handled by one branch of the circuit, the other half on the other branch

That is a good point. So, I only need to use one of the series resistors in the time constant computation? That is a good thing, because I don't want too large time constants.

[Externet]

What am unsure of is what you mean by 10 stages... 2 branches of 5 or 2 branches of 10.

A similar circuit without one branch would mean 10 stages using 20 diodes; i think.

 

On this one, a single resistor should be at the centre 'in' to function equally on the entire cycle but the 'out' terminal could be tapped without the resistor.

 

For low energy requierements, just use much, much smaller capacitors; still will yield same tension with barely inrush current.

Miguel

[YT2095]

10 stages means 10 charge caps and the accompanyied 10 coupling caps (10 for a regular CW and 20 for the double bridged type)

you need only calculate the Charge caps.

although as fullwave on a double bridge, and half wave on a single bridge.

it`s the charge caps that count.

 

personaly I`de use an MOT and a single stage Vmux to get that sort of voltage and current though, I wouldn`t mess about with a CW for such low voltages.

[woelen]

The reason why I do this is that I can obtain a LOT of 220 uF / 400V capacitors for a bargain. Usually these things cost $5 or even more, I can have them for $1 or even less per piece. Once, I have this circuit, I have a very stout DC power supply for 3.5 kV, with 400J of energy stored in the circuit. This circuit easily can give several tens of Watt of power with acceptable ripple. It also saves on transformer costs. The center-tapped transformer I already have anyways. It is just a matter of economics .

 

I intend to use the circuit for research on relaxation oscillators, based on just an RC-combo and a spark gap of a mm or so. Of course, the thing also is fun for other things, such as gas discharges and so on.

 

you need only calculate the Charge caps.

What do you mean with this? The central row of capacitors determine the total voltage, but the side capacitors also need to be of sufficient size, as I understood from many Internet pages. Usually a CW-cascade is made of all capacitors of equal size, and all of them are fully charged to 2*Uin, except the first one of the side branches, which is charged to Uin.