User:CPES/WIP 02

Transformer

A recent unsourced change has replaced transformer with "inductor with a step-up winding". Why? Andy Dingley (talk) 10:53, 4 March 2016 (UTC)

And an unsourced change has also switched it back to a transformer!

It is incorrect to describe an ignition coil as any form of transformer, including a pulse transformer- it is an inductor with a step up winding, just as in switch mode power supplies. Also most ignition coils are wound like- I hate to say this- an auto transformer, with the low voltage winding common to the high voltage winding, unlike the schematic currently shown in the article at the moment.

A transformer, at a very basic level, has the characteristic Vp*N1= Vs*N2.

Where N1= number of primary turns, N2= number of secondary turns, Vp= primary voltage, and Vs= secondary voltage.

Another characteristic of a transformer is that even when a primary voltage is applied, if no current is flowing in the secondary, no current will flow in the primary (assuming a perfect transformer that is. In practice a small magnetizing current will flow and will be lost as heat).

For this reason, a transformer is designed to have a high primary inductance. If a coil had the same inductance as a similar voltage transformer, little current would flow, and only a relatively low voltage would be produced and a low energy, insufficient to ignite a fuel air mixture. Also switching at engine speeds would be difficult.

A typical turns ratio of a car (auto) coil is 1:50 to 1:100. This means if a 100:1 ratio coil were functioning as a transformer, and taking the battery voltage to be 12V, the high voltage winding would produce 100*12V= 1.2KV. This voltage would not be sufficient ignite a petrol air mixture reliably and, in any case, it is at odds with a typical HT voltage for a car (auto) of between 12KV to 30KV. In theory, the coil in a Kettering ignition system does not need a step up winding. It is not fundamental to the functioning of the circuit and is only there for practical considerations.

From the moment that the points (or switching element) closes and applies 12V DC to the coil, the current builds aiming towards infinity. The rate of current increase is proportional to V/L. and the energy stored in an inductor is (I↑2*L)/2 joules, where I is the final current in the inductor in amps and L is the inductance in Henrys. If the points remain closed for a long time, the coil low voltage current is limited by the ballast resistor (if fitted), and the self resistance of the low voltage coil. This current is typically 4A.

When the points open the current in the coil drops to zero (not absolutely true) and the instant change in current generates a voltage heading toward minus infinity (-L*di/dt). Before this voltage can be reached the plug arks over at a high voltage depending on the plug gap, temperature, pressure, and fuel air mixture. This dissipates much of the energy in the coil which previously had no where to go. The coil then oscillates at a frequency defined by its inductance and parasitic capacitances.

Normally though, a capacitor of between 100nF and 470nF is connected across the switching element to protect it and to lower the resonant frequency of the coil (the capacitor is effectively in parallel with the low voltage winding on the coil). The lower resonant frequency gives a longer spark duration to ensure a good burn.

In view of the above, at present this Wikipedia article gives a completely misleading description of how a Kettering ignition system works. It is especially misleading because of the superficial similarity between an inductor (coil) and a transformer; you see this error frequently in electronics, and when I was a newbee electronics design engineer I needed to learn about ignition systems. It took me a while to untangle all the miss information before I could start designing.

Worse still, the article at present says that an AC signal is applied to the primary of the transformer. That is absolutely not true: a DC voltage is applied to the low voltage winding which causes a current to grow in the low voltage winding.

This Wikipedia article should be changed to give a correct description of the normal spark ignition system. CPES (talk) 11:53, 4 March 2016 (UTC)

(1) Reply to (Andy Dingley)

Agreed, the image is wrong. The coil internals need fixing.

That's good

You have confused me though.

I don't see how; typos notwithstanding, I have explained everything at length and quite clearly.

You cite a simple turns ratio for voltage and then talk about "the points apply 12V DC to the coil"

I can't see any problem with either statement. The closing points do apply 12V DC to the coil low voltage winding.

for what is absolutely a pulse transformer and needs to be treated as such (i.e. considering current, not voltage), more than a simple AC transformer.

You don't seem to appreciate the difference between an inductor and a transformer even though I have gone to some length to explain this very basic principle. Just insisting something is so does not make it so. It is important to differentiate between form and function. There is no fundamental difference between a pulse transformer, current transformer or any other kind of transformer for that matter. Quite simply a transformer does not store energy in it's core like an inductor does. It supports a flux. But if you drive a transformer like an inductor it is then functioning like an inductor, and in terms of circuit function, is an inductor.

Yet you have some electronics background?

Not sure if that is a genuine question. If it is, you haven't read/understood a word of my long and comprehensive post. But just to say, I am an electronics design engineer of 35 years standing and have designed, developed, and trialed a number of electronic ignition systems for road vehicles. All used standard ignition coils and capacitor discharge where, you will be pleased to know, the coil is used as an auto transformer with a 400V or so pulse being fed to its primary.

The Kettering system is a system, and it just doesn't work as isolated components (the condenser isn't simply there as a snubber).

I have no idea what you are getting at. The Kettering Ignition system is a... system and I didn't say otherwise. The important thing is that it functions by storing energy at one VI ratio in an inductor and discharging it at another VI ratio. This is not a transformer action. Further more, I said that the high voltage winding is only there for practical reasons and is not theoretically necessary. This can be proved by drawing a simple schematic of the Kettering system with a simple one coil inductor and analyzing that with a 12V pulse. It will still produce a theoretically infinite voltage when the switching element opens As there is no secondary involved even you should see that no transformer action is involved either.

I didn't say that the capacitor is only a snubber. I specifically said that it protects the switching element and lowers the resonant frequency of the inductor with it's parasitic capacitances, to provide a longer spark. I get the impression that you have not read my post above.

Why is this not a transformer? It's not an auto-transformer as the two winding are connected, not shared.

The question is, why is it a transformer. There has been no statements to support this. Sure, it looks like an auto transformer and also as an adjunct uses inductive coupling as both transformers and inductors do, but that does not make it a transformer when used in the Kettering ignition system.

I do not see how, in one breath, you can claim that a coil is a transformer and in the same breath claim it is not an auto transformer. The low voltage winding and the high voltage winding are connected and they are both on the same core and their voltages sum. That is the definition of an auto transformer- except when it is an inductor with an inductively coupled winding that is.

What's the definition of a transformer? How is this not meeting that? (talk)

I have already explained that in some depth.

(2) Reply to Elektrik Fanne (but this Wikipedian seems to have resigned)

You have got the principles wrong, and you have multiplied the wrong voltage.

I have not got the principle of a transformer wrong, as you can quite easily see from my subsequent calculations about secondary voltage and turns ratio. You are being deliberately obtuse and trying to score brownie points. I do admit to a typo though which, had you been objective, you could have simply pointed out and cleared up the issue. By the way, I have corrected the error in the original post.

The circuit consists of a transformer (often called a coil for historic reasons), a contact breaker (often called points) and (very importantly) a capacitor (often called a condensor for historic reasons). The contact breaker is in series with the primary winding of the transformer and the capacitor is in parallel with the contact breaker - you already knew this. The system will not work well, if at all, without the capacitor.

A transformer has never been called a coil, except maybe prehistoric times.

I can't see the point of you telling me something I already know and have already stated.

You are making statements without any supporting explanations.

When the contact breaker closes, the current through the primary builds up to a value limited by the resistance of the primary and any other resistance in the circuit. The inductance does not affect this final current, only the speed with which it builds up. The magnetic circuit of the transformer is now storing energy (0.5 x L x I² Joules) - your formula was wrong.

Only at low revs or when the engine stops is the current limited by resistances. As the revs increase there is less and less time for the inductor current to grow and hence hence the stored energy drops radically especially as the stored energy is proportional to the square of the current. This is one drawback of the Kettering system, which can more easily be combated by a capacitor discharge (CD) system, where incidentally you can claim that the coil is acting as an auto transformer.

Yes, my formula for energy stored in an inductor was missing the square on the current, but you surely can see that it was only an omission and did not affect the underlying principle anyway. Once again you are being deliberately obtuse.

When the contact breaker opens, the current does not fall to zero. This is impossible with an inductance in the circuit. The inductance makes the current continue to flow (it has to in order to disipate the stored energy). For the inductor to cause the current to continue to flow, it induces a driving EMF as result of Lenz's law that is much larger than the original battery voltage and it attempts to oppose the decrease in current (this means that the induced EMF is of the opposite polarity to the battery voltage as seen at the transformer primary).

You are correct about the current not falling to zero. As you quite rightly point out that would be impossible- my error (I do like your explanation). But I was correct in principle as supported by the formula shown. By the way, you have just described an inductor perfectly- just as I did.

It is this much larger EMF that is stepped up by the transformer action to 400 times this value. The current charges the capacitor until it stops. You have also called the 'transformer' an inductance because it is necessary for your explanation.

I have no idea what this means. What larger EMF? By the way the coil turns ratio is 50 to 100, not 400 as I originally said. Are you saying that the coil resonates with the capacitor. If so I agree- that is exactly what an inductor does. As I have said before, practical aspects aside, the Kettering circuit will generate the same high voltage without a coil with a high voltage winding, just as, in principle, boost switch mode power supplies do.

The energy then flows back and forth between the capacitor and the inductor in a decaying sinusoid oscillation. As noted, this gives a better spark and more reliable ignition. The article does not make this clear and I shall clarify the point.

I already said that the inductor will resonate with the capacitances. Yes, providing there is no back connected diode across the switching element, there will be an exponential decaying sinusoid, after a fashion. But if a diode is fitted the waveform will normally be suppressed on the first negative excursion (in a +12V system).

In a practical ignition coil one terminal of the primary and one terminal of the secondary is usually a common connection, but that does not stop it being a transformer.

It does not make it a transformer either.

(3) Summary

Apart from insisting that a coil is a transformer neither of you have explained how the high voltage of around 18Kv is generated. I have done this. As I have explained, the turns ratio of the coil, treated as a transformer won't do it, especially as the high voltage winding is not theoretically necessary, and a simple two terminal inductor will generate the required high voltage necessary to spark the plug.

The worry is that, at present, this Wikipedia article, perpetuates a fallacy about the basic functioning of the the Kettering ignition system, which is used on practically all spark ignition engines, and will lead the average reader down the wrong path when trying to get a simple understanding of its modus operandi.