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Stepping 12 to 800 volts: Arc-over on my transformer, is this even a good approach?

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Proud Mary

Thu Sept 03 2009, 01:56AM

Registered Member #543 Joined: Tue Feb 20 2007, 04:26PM
Location: UK
Posts: 4992

There is a German-language site (address, sadly forgotten) giving a practical example showing that quite a low primary to secondary turns ratio can produce lively sparking in disruptive discharge mode.

Think of the disruptive discharge transformer like this:

You'll know that when a direct current is applied to a coil or inductor that it becomes surrounded by a magnetic field. I say "becomes surrounded" because the maximum magnetic field strength does not occur at once, but takes time to build up as a result of (a) the inductance, L, of the coil and (b) the resistance of the coil, R. This combination of L and R results in what is called the time constant, which you must now Google for further explanation.

The time constant, known by a funny looking letter T, is absolutely set in concrete for a given inductance with a given resistance. (It is one of the building blocks of analogue electronics) This means that if it takes 1 second for the magnetic field of a given inductor to reach its maximum strength, driving it with a faster frequency will only result in incomplete development of the field below its possible maximum, which is why you should always take claims of high frequency driving of ignition coils with a pinch of salt.

Now, imagine your magnetic field at its maximum strength, after charging for the time constant period. What will happen if the voltage supporting the magnetic field is very rapidly switched off? The energy stored in the magnetic field cannot simply vanish (as I said before, "Energy Can Neither Be Created Nor Destroyed" - a fundamental law of physics) so, in a sense, it has no choice but to revert to the electric current that gave rise to it in the first place.

Now, the faster the magnetic field can be made to collapse, the faster the voltage induced in the inductor builds and builds. You are stuck with the time constant, T as a fact of life in the charging up of the magnetic field around the inductor.

But if the power is suddenly disconnected, as in the vibrating 'buzzer' contacts of the original Victorian induction coil, the sharp switch-off of the rotary distributor in vintage petrol/gasoline engines, and all the ways of imitating this make-break action using semiconductors, or electron tubes, then all the energy stored in the magnetic field is, put simply, reconverted back into the electrical energy that created it in the first place.

Imagine a situation wihere a grossly overweight inductor has a time constant of 1 second. Again, to put things rather basically, if that one second's worth of energy used in charging up the magnetic field is suddenly released in its entirety in one millionth of a second - 1us - then you will have a pulse of very high peak power, even though the average power may only be ~60 - 100W (typical small-car 12V ignition coils need between ~5A to ~8A.

So if you remember that "Time is of the Essence" that should help you to remember the basic principle of the disruptive discharge transformer, that it is the rate of change of current in the inductor, and the value of the inductor itself, which determine the voltage induced by magnetic field collapse.

Of course, in practice, other factors come into play to inflluence the behaviour of this simple model - a real-world inductor has not only inductance, but capacitance too, primarily the capacitance between adjacent windings. This combination of R, C, and L . together with the magnetic properties of the transformer core (if any) results in it having an impedance, Z, which may limit or constrain the current flow in the circuit in such a way that ultra-fast magnetic field collapse may be impossible.

Test your knowledge. True or False? If the collpase of the magnetic field surrounding an inductor could be made to happen in zero seconds, the voltage induced in an ideal inductor would be infinitely large.

I hope this helps.

Harry

DerpBrap

Thu Sept 03 2009, 02:12AM

Registered Member #2320 Joined: Fri Aug 28 2009, 06:22AM
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Posts: 7

Ohhhh I see. Makes sense!
I'll keep looking for information and stuff. Thanks for the clarification on the concept.

Proud Mary

Thu Sept 03 2009, 02:25AM

Registered Member #543 Joined: Tue Feb 20 2007, 04:26PM
Location: UK
Posts: 4992

DerpBrap, I added a little edit to the end of my explanation, whilst you were replying to mine yourself. H

DerpBrap

Thu Sept 03 2009, 02:35AM

Registered Member #2320 Joined: Fri Aug 28 2009, 06:22AM
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Posts: 7

True because faster collapse mean more voltage put out. Right?

Proud Mary

Thu Sept 03 2009, 02:43AM

Registered Member #543 Joined: Tue Feb 20 2007, 04:26PM
Location: UK
Posts: 4992

DerpBrap wrote ...

True because faster collapse mean more voltage put out. Right?

Let's see what other folk have to say - remember it's an 'ideal' inductor without resistance or other encumbrances.

DerpBrap

Sun Sept 06 2009, 06:36AM

Registered Member #2320 Joined: Fri Aug 28 2009, 06:22AM
Location:
Posts: 7

Well all my problems turned out to be caused by a lack of grounding. I had the output of the transformer rectified and isolated from everything else. My voltmeter didn't like it; while the caps were charging the voltmeter would spazz out but when turned off it'd read a normal voltage. Upon connecting a the ground of the caps to the ground of the rest of the circuit I'm getting completely predictable voltages out of the transformer and can follow it while the circuit is powered on. I'm kicking myself for not noticing the voltmeter wasn't reading properly when it should have been.

Everything regarding stepping 12 volts to the ~800 volts I wanted is working correctly (and I assume had been the whole time).

Now I'm fiddling with an LM741 op amp (that's all I have) as a voltage comparator to shut off the 555 timer and stop the charging. The new challenge is trying to find transistors or mosfets in my parts bin that'll agree with it. Circuit simulations on the computer work great assuming the op amp outputs 0 or 12 volts, but in practice it's outputting 1.8 or 10 and it leaves things partially on instead of full off. Gotta read up on how to deal with the funky signal.

Steve Conner

Sun Sept 06 2009, 09:52AM

Registered Member #30 Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706

That's normal behaviour for a 741, and indeed most op-amps. It's rare to find one with an output voltage that includes both rails. The LM324 will go down to the negative rail, IIRC.

Or a couple of diodes in series with the op-amp's output, then a pull-down resistor to ground, will drop those remaining few volts ghetto style. I show the technique in this circuit

where I use the spare half of an op-amp as a comparator driving a 555's reset pin. The circuit didn't work at all without D1, R7 and R8.

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