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gwalters
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[*] posted on 9-6-2004 at 20:49
very high voltage


i read that a microwave transformer takes 120 volts up to ~ 4000. 120/4000=.003 so if i connect 2 trans. together in series i get 4000/.003 = 1 333 333. will that work? will the wires in the trans. arc at that voltage? what about if i run it in oil? any other ideas?
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chemoleo
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[*] posted on 9-6-2004 at 21:05


well of course, the insulation of the transformer wiring will not withstand such high voltages. Look at power lines, and see how far they are separated, or how long the insulator are (connecting the mast to the powerline). It's not for no reason!
The only way to easily generate high voltages like that is to either build an appropriate transformer with massive insulation, or better, to work via capacitors in series. Safer, but still VERY dangerous.
You can get high voltage capacitors from TVs by the way, connect them up and charge them. If you short cut them, the power is so high that it makes a massive spark and evaporates half of the wire!


Ever heard of Leyden jars (sp?) by the way?

[Edited on 10-6-2004 by chemoleo]




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[*] posted on 9-6-2004 at 21:17


ok then....say i connnect a bunch of caps. in series:

-cap+-cap+-cap+

like that...the voltage will add up but will i have to charge it with the max voltage or could i charge it with say 4000 volts until it is charged?
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[*] posted on 9-6-2004 at 21:41


Well thats what I did a long time ago.
Connect them all in series, and charge them with an excessive voltage (doesnt matter if it is a bit more than needed ;) ). I attached a voltmeter to it, so I could monitor the voltage rising... after a few minutes they would reach nominal/calculated voltage... and then it was time to shortcut them!
Be warned though, this is very dangerous. Plus, after treating the capacitors a few times like that, they refuse to operate any further... so clearly that's not the standard way to go about this!

Also, I remember our central heating being replaced. Guessing that the oil is ignited by electric sparks, I thought, let's have a look at the old central heating system. Indeed, very soon I found a lovely transformer that would amplify 220 V to 15000 Volts! (the engineers didnt mind me scavenging the transformer :) ) How did I find that out? Well obviously I connected it to something like 20 V and measured the output on the second coil, and extrapolated the final voltage from it! With this I made many arcs, and jacobs ladders. My most impressive experiment I remember doing is to cut a piece of obsidian apart (which is a volcanic black glassy like mineral), whereby the sheer heat generated by the current flowing through the obsidian melted it so that it would conduct current.
After that I tried all sorts of stones... and they all fell apart. Neat stuff ... at least for the crazy mad scientist :D
Again, be warned. I once managed to shortcut this through my fingers (despite precautions... it went through several cm of insulation)... and it left two little black craters in my finger tips, which were painful. I was lucky my heart didnt give out!

[Edited on 10-6-2004 by chemoleo]




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[*] posted on 10-6-2004 at 01:02
Parallel, Not Series


Run them in parallel, not series, if you want to accomplish anything. Running them in series will be a waste of capcitors.
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[*] posted on 10-6-2004 at 02:44


Quote:
Originally posted by Turel
Run them in parallel, not series, if you want to accomplish anything. Running them in series will be a waste of capcitors.


You're wrong.

If you want to generate high voltages, the charged capacitors must be connected in series (just as batteries must be wired in series to deliver a higher voltage). Connecting them in parallel will merely create a bigger capacitor, which will still be charged to the same voltage as the component capacitors.




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[*] posted on 10-6-2004 at 07:02


If you want voltages in the 1MV range, you should feed a Tesla coil from an MOT or an NST. As others have said, transformers aren't made for such high voltages. Powerlabs have many nice pointers.



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[*] posted on 10-6-2004 at 09:49


Preferably, you charge the capacitors in parallel, then discharge in series, thus avoiding the need for a high voltage charging supply, and associated corona losses. The switching devices are traditionally spark gaps. This setup is known as a Marx generator.

Be very careful when taking apart CRTs. They can retain deadly voltage for weeks, not only in the obvious capacitors but also in the tube itself. While we're at it, microwave ovens are bad, too. Read the appropriate bits of the sci.electronics.repair faq conglomeration before proceeding.
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[*] posted on 10-6-2004 at 15:30


Microwave oven transformers are dangerous because they are capable of supplying relatively high currents. They can easily kill you. I've made fulminating silver before, which is something that is supposed to be foolish to do, but I'll never mess with a microwave oven transformer.

The safest thing to play with is a "negative ion" generator. These are often found in air cleaning appliances. They have needles attached to them to electrify the air with charge. You can find an old air cleaner, or else check electronics junk dealers and you should find one for under $10. Some will even run off a battery. They produce about 10KV but the current is low enough that it can't seriously injure you.

Lacking that, or if you do need slightly more current, look for circuits using either an automobile ignition coil or a TV flyback coil. I just did a quick search and found this circuit:
http://members.misty.com/don/igcoilhv.html
These circuits are not really that hard to build, even if you have not done much with electronics before, and they are fairly safe. If you just want to play around a bit and don't want to build a circuit you can always just buy an automobile ignition coil from a junk dealer. Connect it through an ignition condenser to a 12 battery or power supply. By shorting/unshorting the capacitor you can generate high voltage sparks of an inch or more in length.
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[*] posted on 10-6-2004 at 16:03


Why don't you use your vandegraaff generator?

What's that, you don't have one?!:o:P

http://www.amasci.com/emotor/vdg.html




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[*] posted on 10-6-2004 at 20:49


i tried to make a van degraff generator b4... didnt have the correct materials. Anyone made an emp before?
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[*] posted on 11-6-2004 at 02:25


Try Tesla coil?



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[*] posted on 11-6-2004 at 05:18


Wiring 2 transformers as you described wouldnt work and not just because of insulation problems. It simply wouldnt output the voltage you calculated, they arnt black boxes that multiply voltage, you have to take into account the impedence of the inputs and outputs. Voltage rises with the turns ratio of the transformer, impedence rises with the square of the turns ratio (required by conservation of energy). The output of the first transformer will be very high impedence, the input to the second one will be low impedence. For a working circuit the impedence of the primary coil of the second transformer must be at least as high, and preferably a lot higher than the secondary from the first. Simply wiring 2 MOTs together, I think youd be lucky to get your 4kv back out the other side, not to mention burning out the first transformer from power issues after a short time.

All of the suggestions are workable ways of getting high voltage, but they all have different applications. You need to ask yourself what you want to do with it. If it just big sparks, the best way would be with a van der graaf. You can buy something slightly bigger than a tabletop version that will get you half a million volts.

Teslas are quite tricky to build, and the math requirement is heavy if you want it to work properly. You also need a fair amount of test gear to tune everything.
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[*] posted on 11-6-2004 at 06:22


Hmm, I realise what you are saying here, Marvin. I had this thought myself on occasion, but never had problems with it in practise.
I used just such a system to generate a voltage range from 500-1500 V, to charge up the capacitor system described above. In fact, I sometimes used 3 (!!) transformers (or two, with one 220V-->15kV transformer), one bringing down the voltage to i.e. 5 Volts (but high A), and then two/one reverted transformers to bring the voltage up.
Presumably the reason why this worked was because the amount of power flowing through secondary coil of the first transformer was restricted.
The final output was high V, and low A - which was fairly un-dangerous and still useful for charging the capacitors...:)




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[*] posted on 11-6-2004 at 08:12


does anyone know a good way to calculate the distance a spark will travel with a given humidity and voltage? I need to find a way to calibrate my spark gaps on a mosfet (?) generator. Will wiring the caps. in series lower their working life?
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[*] posted on 11-6-2004 at 16:42


Depends on how sharp the electrodes are (sharper = longer spark) but I think about 1kV/mm is ballpark.
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[*] posted on 11-6-2004 at 16:48


Sounds about right, hodges. I've gotten a 10mm spark between electrodes with a voltage of 9kV. That would support the math, w.r.t. the resistivity of air.



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[*] posted on 11-6-2004 at 18:05


Hmm, and I got a continuous spark with DC current, at 15 kV, about 3-4 cm long!
And the 15 kV is not something I make up! :P
To be honest, I think humidity, air pressure etc are going to matter.... but it's interesting to see there are some ball park figures to this!




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[*] posted on 11-6-2004 at 19:55


I always thought that sharper electrodes would yield SHORTER sparks!

When you have sharp electrodes the electricity tends to leak off as corona leakage, when you have spherical electrodes the electricity comes off as sparks! Thats why to test voltage by measuring spark length spherical electrodes should be used.

At least thats the case for vdgs, I'm not sure if that is the case everywhere.

Chemoleo, what did your electrodes look like?




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[*] posted on 11-6-2004 at 20:24


lol, htat used to be simply the tips of wires...



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[*] posted on 12-6-2004 at 07:35


If you use a constant-current source like a van de graff, which has very limited power, sharp electrodes will just mean that due to corona leakage, the voltage will never be able to climb as high as it would for smoth-surface electrodes (basically think a capacitor with a resistor bleeding the charge away, unless you have enough power you won't be able to get it to full charge). But for a flyback, nst, mot, etc the power is high enough that the sharper electrodes simply have the effect of lowering the breakdown voltage of the air.

Remember, you should be measuring the point where the sparks/arcs start, not how far you can draw them out. For sparks (off VDGs, capacitors, etc) they should be the same, but you can usually draw out an arc a lot further than the starting length. The ultimate length is dependant on current as well as voltage, while the breakdown distance is purely a function of voltage (withstanding humidity, etc, of course).
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[*] posted on 16-6-2004 at 13:37


For large spherical electrodes dry air at 1atm breaks down at about 30kv/cm.

chemleo, Its certainly possible to wire transformers together, no doubt, its how the national grids of every country I know reduce pylon losses. They design the transformer impedences specifically to do the job, if you arnt trying to do anything massivly spectacular you usually get away with it at home. Stepping down 220 to 5 and then stepping up works very well because the low impedence part of the first transformer is driving the low impedence primary of the secondary transformer. For the specific case of 2 identical MOTs wired secondary to primary though, it certainly cant work and I'm almost certain the second transformer could not generate a higher voltage over its secondary than the first transformer would unloaded. If a very large MOT was driving a much smaller MOT some stepping up would be observed, but not a lot.

MOTs arnt things Id recommend people use unless they had to, they are rather unforgiving compaired to neon sign transformers.
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[*] posted on 17-6-2004 at 12:31


Provided that you are not trying to run so much current that the core saturates; or so much voltage that the insulation breaks down; you can run transformers in cascade (running them in series is different).
The impedance looking into a transformer is equal to the product of the load impedance connected to the secondary divided by the turns ratio squared.
If you connect no load to the output then real part of the input impedance is (for an ideal transformer) infinite. For the setup that you are talking about ( a pair of MOTs ) this would work for low voltages at the input. Unfortunately, the losses of the second transformer load the first so its output is reduced.
(and, by theway, the arithmetic is wrong in the first post 120/4000 = 0.03 so the output would be 133KV)
You might be able to get away with running a volt into the input to test the idea.
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[*] posted on 20-6-2004 at 21:16


I think that explanation is oversimplified to the extent the answer is wrong. I'm bothered by terms such as 'ideal transformer' because inevitably the device follows some unstated laws of physics and ignores others, and which it follows and which it ignores seems to vary according to who is doing the math and what they expect the answer to be. I'll try to build up a realistic ideal transformer as I go.

The thought experiment, expanded somewhat, that lead me to the no greater output voltage conclusion runs as follows..

Assuming an transformer is 2 coils of resistanceless wire wrapped around a common non saturating core,
Assuming its a 33:1 transfomer and we have 2 of them (they are identical) and assuming the power source is mains and has zero impedence.

Wire the first tranformer primary to 120V mains as a step up transformer. The primary coil excites the core and the field induces a voltage in the secodary of the expected 4kv or so. So far so good.

Same experiment, only this time the mains is 4kv and I'm interested in the difference. The voltage over the same primary is higher, so the change in current over time is also higher resulting in a higher change in core field over time and the field ultimatly reaches a much higher values. This must be true in order to induce the higher voltage in the same secondary which depends on rate of change of flux over time.

Ok, final stage. We wire them in cascade step up - step up again with no load.

The classical voltage we expect is a little over 130kv in this example.

The first core should excite to the level we had before assuming its output is going to be the 4kv or so we expect.
In order to output more than this, let alone the 133kv though, ignoring how the secondary of the first is wired to the primary of the second.... The second core has to be excited to a much higher flux in order to induce the higher voltage in the second core - yet the energy for this to happen comes entirely from the first core.

Given the cores are *identical* to reach the same flux levels requires the same energy, so for the second core to be experiencing a higher change in flux/peak flux than the first core violates conservation of energy.

In what circumstances does this not apply...

We can make the second transformer much smaller, so that the same amount of energy from the first transformer results in a much larger change in flux over the smaller volume of core, and a larger number of turns (while keeping the same ratio) and get out the expected 130kv. The objection only applies because we know the 2 units are identical.

Though I had nothing in the way of maths to backup my argument this was enough evidence for me to post that I doubted the output voltage could exceed 4kv under conditions described. I should point out that reason I had no maths was because nothing in books or in education covered this directly. While I mentioned before that a transformer output impedence rises with turns ratio squared, its not relevent to our 'ideal' example, because our mains input impedence is zero, and our output impedence is infinite.

In my second post I was talking about impedence, unionised, youve pointed out that the real part of the impedence for an ideal transformer is zero and I agree. However the windings are massive inductors and this I think is the clue to why the conservation of energy argument and our simplified transformer maths do not marry.

Going back to a single transformer, and imagine it being attached to an impedenceless DC source. Lets assume the primary coil has a massive reactance, say 10H. Since the secondary is wound to the same core, and the number of turns are in proportion in our ideal setup, the reactance must be in proportion, giving our secondary a reactance of 330H.

When we turn on the juice, it reacts just like an electromagnet, the change in current behaves exactly like a 10H inductor would (open circuit secondary, so thats all it is), and the change in flux is correct for inducing the expected voltage in the secondary.

Now lets say we add a choke to the primary circuit, say 10H. When we turn on the power the current will rise at half the rate it did before (becuase the same voltage is now driving 20H total), This means the flux in the core will raise at half the rate and therefore only half the voltage will be induced at any given point in time. Its equivalent to saying half the voltage is dropped over the primary and half over the choke even though there is no real componant to the impedence.

Right, going back finally to the cascade setup. The primary has a reactance of 10H, inducing the 4kv or so in the second winding. The seondary is closed circuit with the primary of the next transformer. If the primary of transformer 2 was being powered from an impedenceless 4kv supply the current would rise at the right rate to produce 130kv over its secondary. The total reactance of the secondary circuit though is both coils in series. 330 + 10 = 340 H. We expect the current in the secondary to rise 10/340 or 1/34 what it would be, and thus the field and thus the voltage in its secondry will be 1/34 what simple multiplication would get us. 120x33*1/34*33, or slightly less than the output of the first transformer on its own with no load.

Its mismatched impedence causing a power factor problem. This as I understand it, is the meat to my generalisations about transformer impedence. Its also as best as I can see it an intrinsic problem, nomatter how well you make the transformer, no matter how high the reactance (inductance) of the windings, the ratio will still be exactly the same as the step up ratio and thus cascade of 2 identical transformers will still fail to increase the voltage.

Rule of thumb, for this system to stand a hope in hell of working, the impedence (including reactance) of the secondary winding of the source tranformer must be less, and preferably much less than the sink primary winding. You may assume that all the AC values in this post are RMS, not that the values matter at all, just for illustation. In a real transformer, the core would be designed so it was just short of saturating for its working voltage, which keeps down core size for a given power, but also makes it useless for overdriving even when you ignore insulation issues.
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[*] posted on 27-6-2004 at 21:43


Ok, put simply, I screwed up.

2 identical transformers in series can develop a higher voltage than the first one. In a test done by a friend 2 1:10 signal transformers were wired secondary to primary, and a small input produced a little less than half of the expected output. On furthur testing insulation failed on the second transformer.

My first argument on energy grounds is invalid because the argument is only true for no current in a transformer secondary. The current in the secondary creates a field that reduces the field - while increaseing the current in the primary. While there is no power drawn by the second transformer it is drawing current from the first and thus the idea breaks down.

I do not yet properly understand why the second argument fails, only that it does.

My apologies.
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