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4hv.org :: Forums :: High Voltage
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in a villard cascade do the diodes have to be rated of 1/2 the input current?

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803
Sat May 15 2010, 06:34PM Print
803 Registered Member #2807 Joined: Fri Apr 16 2010, 08:10PM
Location:
Posts: 191
For example 10 amp input 5 amp diodes min.

Update= lots of members have said this is not pracitcal, I know that. I just wan't to know how high current rated the diodes have to be in relaction to the input current
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hboy007
Sat May 15 2010, 09:39PM
hboy007 Registered Member #1667 Joined: Sat Aug 30 2008, 09:57PM
Location:
Posts: 374
Thinking about the cascade in a simple way, it has to shift charges between two groups of capacitors which has to happen in one half cycle, so the current rating needs to be twice the desired output current. Another problem arises: as you're charging and discharging capacitors, the current waveform is strongly dependent on your driving waveform and may be a multiple of the output current. Plus, you're looking for the pulse SOA rating here.
Why would anyone build a villard cascade / CW multiplier / Greinacher circuit (the latter invented it) for 10A anyway?
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803
Sat May 15 2010, 09:53PM
803 Registered Member #2807 Joined: Fri Apr 16 2010, 08:10PM
Location:
Posts: 191
for a vttc, of course!

So what your saying is the diodes have to be rated for 2x the output current?


All I'm saying is that is 5 amps is enough for 10 amp input, or do you need 10 amp diodes?

(5 amp is contous)
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Proud Mary
Sun May 16 2010, 12:37AM
Proud Mary Registered Member #543 Joined: Tue Feb 20 2007, 04:26PM
Location: UK
Posts: 4992
803 wrote ...

for a vttc, of course!

So what your saying is the diodes have to be rated for 2x the output current?


All I'm saying is that is 5 amps is enough for 10 amp input, or do you need 10 amp diodes?

(5 amp is contous)

Voltage multipliers are never a sound means for sourcing high currents. Multiplying the voltage means dividing the current.
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hboy007
Sun May 16 2010, 11:06AM
hboy007 Registered Member #1667 Joined: Sat Aug 30 2008, 09:57PM
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Posts: 374
>> for a vttc, of course!

in this case you should start with half or full bridge converter that is fed by 350..400V generated by a PFC boost converter which will involve a serious number of well-designed magnetic components such as storage chokes, a transformer and several filter chokes. You could as well buy a neon sign transformer but that wouldn't be the same, would it?
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Steve Conner
Sun May 16 2010, 11:11AM
Steve Conner Registered Member #30 Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706
That's silly! :P If you knew how to use all that converter stuff, you could forget the vacuum tube and build a SSTC.

The time-honoured way of powering a VTTC is with the transformer, capacitor and diode salvaged from a microwave oven. You also need a RF bypass capacitor, to stop the RF from the tube oscillator going back into the power supply and destroying it.

Multiplying up 240V line voltage with a cascade is only practical for smaller, lower powered VTTCs using TV sweep tubes. (120V line is even worse.)
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hboy007
Sun May 16 2010, 11:46AM
hboy007 Registered Member #1667 Joined: Sat Aug 30 2008, 09:57PM
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Posts: 374
Of course it is, sorry for that wink Building a regulated high voltage supply with more silicon than is in the glass envelop of a tube just doesn't make sense but I can think of no better way to get rid of the line frequency/harmonics. I tend to forget what 803 wanted. So how about, let's say, 10-15 toroid core transformers with their secondaries in parallel and their 240V primaries in a diode-split assembly smile
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803
Sun May 16 2010, 03:13PM
803 Registered Member #2807 Joined: Fri Apr 16 2010, 08:10PM
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Posts: 191
No when I ment 10 amps, Iment input. I want around 600 milamps in. THis is just making things simpler, when you double the voltage , thats a voltage multiplier to. So why can't you use multipliers all thw way? It wouldn't make things worse except higher voltage drop off and lower efciency. Any way, it makes things lighter and cheaper

(I am realy bad at typing) :)
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