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Registered Member #1829
Joined: Sun Nov 30 2008, 01:06AM
Location: Raleigh N.C.
Posts: 74
Yes, Dr. 2n3055's circuit is similar to mine, a flash inverter with a Cocroft Walton cascade after it, mine just has a fullwave cascade and two inverters. I'd suggest building his simply because it's made entirely from parts from the cameras. If you can beg a bag of these from a photo shop you'll be set. Output will depend on the size of the resistor at the end of the cascade and the shape of the discharge terminals as well, (pointy vs. rounded). Here's you a video of my circuit without any resistor and with a 1k resistor. The arc starts at about 4 mm with pointy pieces of wire.
Yes, Dr. 2n3055's circuit is similar to mine, a flash inverter with a Cocroft Walton cascade after it, mine just has a fullwave cascade and two inverters. I'd suggest building his simply because it's made entirely from parts from the cameras. If you can beg a bag of these from a photo shop you'll be set. Output will depend on the size of the resistor at the end of the cascade and the shape of the discharge terminals as well, (pointy vs. rounded). Here's you a video of my circuit without any resistor and with a 1k resistor. The arc starts at about 4 mm with pointy pieces of wire.
Wow I think I need exactly this thing! I am going to build yours, not 2n3055's, after viewing the video you just gave me!
But I still got 3 questions about it... 1) What do you mean with "without any resistor and with a 1k resistor." (your setup for the video) 2) How bad would it be if I would only do a "regular" multiplier instead of the fullwave type? Where is the difference anyway? 3) Where is the ground of your multiplier? I've looked at the picture and identified it as the battery's minus. but why?
Registered Member #1911
Joined: Mon Jan 05 2009, 06:30PM
Location: Salem, Oregon, USA
Posts: 165
A full-wave CW multiplier is superior to a half-wave multiplier because it puts all of the power available (for the most part) to work. If you were using half-wave, you would end up with half of the wave-form missing and, therefore, half the power.
If you use full-wave, you will get twice as many discharges in the same time frame.
Registered Member #1829
Joined: Sun Nov 30 2008, 01:06AM
Location: Raleigh N.C.
Posts: 74
The fullwave CW is designed to multiply voltage from two 180 degree sine waves instead of one. Say, from both bushings of an NST, or any center tapped transformer for that matter. With this setup it's a bit different however. The inverter transformers don't generate a nice sine wave. They make a spike on one side of the ground and then ring down on the other. Still, there is oscillation, albeit not sinusoidal, so the CW will work. I used a fullwave CW here because it seemed the most elegant way to combine the outputs of two inverter transformers. As to the resistors. The output of a CW should have some kind of resistive load. Otherwise you'll have very large currents which your caps won't like for long. When creating arcs, you need a resistor between the electrodes and the output of the CW. I was showing an arc with a 1k resistor and one with no resistor at all (because I was only doing it for a few seconds). You could probably get away with a resistor of only a few hundred ohms for a better arc. Finally, I tend to think of - as ground. If only Franklin had got it right we could avoid a lot of confusion about current and polarity in physics 101. Anyway, you can reverse the polarity by interchanging the transistors' C and E and reversing the battery but doing this will only interchange the +HV transformer and the -HV one. If you want to change the polarity of the output you simply need to reverse all the diodes in the CW stack.
ah I see. So with the way you combine 2 transformers, I have to use a fullwave CW grounded on battery -
But what if I drive the 2 transformers like I would drive only one, with just putting the transformers' primaries parallel and the secondaries in series? I would have 660V as well and could drive a regular CW.
Would there be a drop in output, spark frequency or current?
Thanks so far, I feel really close to my goal! The bright arc from your video (no resistor) is looking gooooood!
Registered Member #1829
Joined: Sun Nov 30 2008, 01:06AM
Location: Raleigh N.C.
Posts: 74
Yes, this way you do need a fullwave CW. However, the only reason the HV ground is the battery - is because, with these transformers, the HV out coil is tied to the drive coil. If you can get some that are separate (6 pin transformers) you can have the HV ground be anything you want. If you have a number of these transformers, you can drive each of them with their own transistor and synch them up by tying all the transistor bases together. This means that all the transistors will switch simultaneously and will only switch when the last of the transformers saturates. If your transformers are different, then some may saturate before others and stay in saturation for awhile before the transistors switch. Alternately you could parallel several transformers and drive them all with one transistor. The transistor will be handling the current for all of the transformers so it must be sized appropriately. However, I don't think there's any way to series the outputs of these transformers (5 pin type). You can parallel them and get more current. Then, if you want more voltage, just add more stages to the CW.
First, can I use regular CW multiplier without loss, for - multiple transformers in parallel, outputs parallel, too, 1 transistor - driving each one with 1 transistor and sync by connecting all bases, outputs parallel, ?
And, if I drive two transformers in series, with double input voltage, like in the attachment, do I need a regular CW or a fullwave CW?
taken from , creates ~600V with 76% efficiency, ignore the diode!
Of course I could just build a fullwave CW but it would be better to achieve the same results with a regular CW
So that would be pretty much everything I would like to know now
Registered Member #1829
Joined: Sun Nov 30 2008, 01:06AM
Location: Raleigh N.C.
Posts: 74
Question one: Yes for both. As long as you only have one oscillating HV output (transformers outputs tied together) you only need a halfwave. There's no matter of loss between HW and FW configurations. Say you have 2 waveforms at x voltage capable of delivering y current. You can feed these into a fullwave CW. It will generate x' voltage at y' current and each leg of the CW will only need to be rated for y current. If you combine the two waveforms and feed them to a halfwave CW then you'll have x voltage and 2y current going into the CW so the components must be rated accordingly. My circuit generates two opposing waveforms and with a fullwave CW can generate twice the output voltage with half as many CW stages. If you were to build a halfwave CW capable of the same voltages from two paralleled transformers I think you would wind up using more components which is what I assume you mean by better/worse. So it's not a matter of loss so much as effectively converting current to voltage in as little space as possible. As for the second question, again one HV output = halfwave. However, I've never seen a photo inverter made like this. It seems he's found some where the feedback is tied to the output coil and the drive coil is separate. Or perhaps he just switched the feedback and drive in the circuit and it still worked. I'll just say, you've seen what my circuit can do, and that's the only one I have personal experience with. If you want to try something else you should experiment and see what happens. You can get a bag of these cameras free after all. And if you do try any other circuits please keep us posted on the results. Good luck.
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Difficult: Miniarc
