# Ignition voltage in a waste spark system.

by CHICAGO
Tags: ignition, spark, voltage, waste
 P: 50 Hi all I have a question concerning the voltage required to initiate the ignition spark in a combustion chamber. I learnt this voltage depends on the dielectric in the plug gap. The more pressure the higher voltage, the richer gas mixture the lower voltage,....and so on....there are many variables which condition the required voltage to "trigger" the spark. Now, my doubt. I own an old 2-cylinder car (citroen 2CV) with a points ignition system without distributor and a dual output ignition coil, so that is a waste spark system. While one cylinder is compressed for ignition (useful spark), the opposite is in the exhaust stroke (waste spark). If we need, say, only 8000 volts for sparking in the exhaust cylinder, because that chamber is unpressurized. How can the coil still go up to, say 12000 volts for initiating the spark in the opposite highly pressurized chamber? It is supposed that just when a spark is stablished, the voltage falls down dramatically to that value just to maintain that igniting initiated spark, which may be between 3000 and 4000 volts. So, just when the waste spark initiates, the secondary coil voltage drops and there is no option to get the 12000 volts for generating the useful spark. What am I thinking in wrong? Any input is greatly appreciated. Thanks.
 Sci Advisor P: 4,044 I own an old 2-cylinder car (citroen 2CV) with a points ignition system without distributor and a dual output ignition coil, so that is a waste spark system. While one cylinder is compressed for ignition (useful spark), the opposite is in the exhaust stroke (waste spark). Does that mean there are really two secondaries producing sparks? If so, wouldn't they be independent of each other? You could check by measuring the resistance between the two HT leads. If it is not almost zero, the outputs could be from two magnetically isolated coils.
 P: 50 Hi, vk6kro Thanks a lot for your quick interest. What I call a dual output ignition coil is that from we got two outputs at the same time, each one from each coil end. One has a reversed polarity respect to the other. The xformer ratio is in the range of 1:100 This is the ignition coil and the resistance readings from both primary and secondary coils. By chicago49 at 2010-03-20 By heating the terminals we can unsolder the coil leads and this is what is inside. Just to let you know that all coils are submerged in a refrigiration oil. By chicago49, shot with COOLPIX L3 at 2010-03-20 Then, I have splitted the 2 HV sections which are internally connected by those copper contacts. Here you can see what I mean. By chicago49, shot with COOLPIX L3 at 2010-03-20 This is ignition diagram. Using a graphite pencil I could see that while one plug sparks in the right polarity the other one does in “reverse polarity”. To me, this indicates that both HV coils are internally connected in series. By chicago49 at 2010-03-20 There is not any electrical connection between the primary and the secondary (HV) coils. If you want more details I would try to search for them. Thanks a lot in advance. .
 Sci Advisor P: 4,044 Ignition voltage in a waste spark system. It looks like both spark plugs are then in series and they have to both fire together or neither can fire. So, this appears to answer your question. Since they are in series, it doesn't matter that they fire at different voltages. There is apparently enough voltage to jump the gaps in both spark plugs.
 P: 50 . I see what you mean, vk6, and that was my initial idea about this issue. But after reading and checking something else I learnt that in this type of circuitry, which we call in-series, there is no need of sparking both for having a spark in one plug. In fact, if I remove the cable going from one coil output to its plug I can start the engine. Obviosly the car runs really badly with only one cylinder, but it does. I think there is no "real current" through the plug lead when the spark jumps between its electrodes, but that "current" is only present in the plug gap ionized dielectric. That is to say we cannot apply Ohm´s Law in the HV circuit. I am really confused about this issue. Any more ideas ? Thanks again.
 Sci Advisor P: 4,044 Maybe the coil is sparking internally or right at the low voltage connections. There is certainly current flowing in the gap of a spark plug and there has to be a complete circuit for it to flow through one spark plug, if the coil is as you describe. Curious about your test with a graphite pencil. How were you able to tell the polarity of a spark with the pencil?
 P: 50 The polarity test using a graphite pencil consists of locating the pencil tip in a previous open gap of about 2 cms while the engine is running. Then, two types of sparks can be seen. One is a neat blue brilliant arrow and the other (others) several yellowish arrows jumping from the pencil tip. By chicago49 at 2010-03-21 I guess that spark (negative charges) flows from that open end where we see the blue and unique arrow to the pencil tip. Then the graphite spreads those feathery sparks to the other open end. By chicago49 at 2010-03-21 I once took some pictures with my cell phone, but their quality are very poor. If I find some other with higher resolution I will upload them. It is a really curious experiment. But going back to my question, I do not think of an internal coil spark when using only one plug. This will probably cause serious damage in the coil and I know of friends who have driven hundreds of Kms with this problem without damaging it. I agree both sparks must jump at the same time, but because of different dielectric conditions, the voltage needed for each one will differ. What really causes the spark to jump is the highly ionized gas between the plug gaps due to a some Kvolts electric field. I do not think we have electrical current in the sparks cables when they jump. In fact, we have EMI suppresion cables and plugs with thousands of ohms which in the case of a real current, this would be almost zeroed. I am still thinking there is something I am missing in my assumptions. Regards.
 Sci Advisor PF Gold P: 2,731 Nice drawings CHICAGO. What program did you use?
 Sci Advisor P: 4,044 What really causes the spark to jump is the highly ionized gas between the plug gaps due to a some Kvolts electric field. I do not think we have electrical current in the sparks cables when they jump. In fact, we have EMI suppresion cables and plugs with thousands of ohms which in the case of a real current, this would be almost zeroed. Whatever happens in the spark gap, there is certainly current flowing in the wires. The RF suppression resistors are chosen to allow plenty of current to flow in the spark gap, but control the rise-time of the spark voltage which is what contains harmonics and causes RF interference. I have wondered about that "right" or "wrong" spark voltage. If there is a spark, the gasoline will ignite regardless of polarity, surely. I saw a diagram on this Forum (by Bob S, I think) that showed the spark voltage as a positive then a slightly smaller negative then a slightly smaller positive waveform. This damped waveform went on for many cycles before it vanished. I found it: http://www.physicsforums.com/attachm...6&d=1250393723
 P: 50 . Thanks vk6kro: I totally agree with Bob_S waveform. But that one is what we have when we disconnect the secondary coil out of the circuit and leave the LC resonant circuit formed by a primary coil and the capacitor. (first ghraph) But once we connect the HV coil with plugs and so on... the real wave form is that I have drawn (I have omitted some LC ripple).(second graph) By chicago49 at 2010-03-22 What I think it happens. Lets imagine the plug gap needs 15Kvolts to set the spark (150v at the primary coil) and, once the spark is igniting, it needs between 3 and 4kvolts to be maintained. Once the coil EM field is not able to keep more than 3kvolts between the plug electrodes, the spark goes off. So, although, our LC resonant circuit is able to support up to 200v (with a 1:100 xformer ratio 20,000 volts in the plug electrode) our spark initial discharge will only take that voltage necessary to break the dielectric, always depending on gas pressure, temperature, lean or rich mixture, etc....and a lot of more variables. This "voltage saving" is then converted into "more stored energy" and the spark duration will be longer than that spark which needs 20,000 volts to start jumpimg. This is an overall idea about what I think this ignition works. Am I wrong on this general overview ?. ************************************************************* Ops !! hello, digoff. Nothing special, for these drawings I have used the normal Paint software. Thanks a lot, you guys for your interest in helping me on this matter. .
 Sci Advisor P: 4,044 This is an overall idea about what I think this ignition works. Am I wrong on this general overview ?. Yes, you are wrong. Sorry. What happens is that the high voltage initially forces a few electons to ionize the gas in the gap. Once this ionization happens, the resistance of the gap drops and a large current can flow. The voltage across the gap falls, but not to zero, so that when the coil voltage drops below this gap voltage, the current stops flowing and the arc vanishes. The arc you see across the gap of a spark plug, or anywhere else, is an actual flow of current, not voltage. Nothing special, for these drawings I have used the normal Paint software. How did you draw the coils?
 Sci Advisor Thanks PF Gold P: 12,251 For a spark to ignite fuel there must be a certain amount of Energy. That means Current X volts X time. The voltage across the spark gap under compression is higher than the other one so there is more energy in the cylinder that needs it each time. I must say, I bought an electronic ignition system for my 2CV, several years ago. The only problem I ever had, after that, was when the lead fell out of the 'push fit' connection. A fairly common fault, I learned. What a Manic piece of Engineering the 2CV was! Charming, though - and I never came across such a trusting and trust-worthy set of car owners. Actual Cheques were offered and accepted for goods and services and no one seemed to need to wait for them to be cleared. That's how life should be! (United in adversity, possibly).
P: 50
 Quote by vk6kro [COLOR="Blue"]....Once this ionization happens, the resistance of the gap drops and a large current can flow. The voltage across the gap falls, but not to zero, so that when the coil voltage drops below this gap voltage, the current stops flowing and the arc vanishes. ...
I agree that a spark is created by a current and that we need energy for that. A jump from a catode to an anode of electrons, photoelectrons, or whatever they are called. But, is all that current flowing from the HV coil windings trhu the plug cables?

Now, I cannot recall where, I read something about the nature of a spark, and it sounds to me that most electrons jumping in a spark are coming from the same gas molecules
which are contained in the plug gap and the coil cables are only used to "carry" that high voltage to the plug electrodes in order to create that high electric field in the gap. I will try to search for that article.

 Quote by vk6kro How did you draw the coils?
Just using Paint you can draw two ellipses with the same major axis but different eccentricity, then match the right half of one of them with the left half of the other. And by "copy" and "paste" multiply the number of turns.

By chicago49 at 2010-03-25

But a more realistic coil can be obtained if we first save, open with the Office Picture manager and tilt those ellipses, say 5 or 7 degrees. Here you can see the process:

By chicago49 at 2010-03-25[

 Quote by sophiecentaur For a spark to ignite fuel there must be a certain amount of Energy. That means Current X volts X time. The voltage across the spark gap under compression is higher than the other one so there is more energy in the cylinder that needs it each time. I must say, I bought an electronic ignition system for my 2CV, several years ago....

[/QUOTE]

I also agree with what you say, sophiecentaur.

As you did, I mounted the electronic ignition in my 2CV, too, the 123ignition and no problem at all. But I am preparing an article about the anatomy of ignition in a 2CV article for a 2CV enthusiast forum and, as most of us I have some questions not answered yet.

 Quote by sophiecentaur (United in adversity, possibly).
Yes, you are right, not possibly but for sure.

Thank you both for replying my probably wrong arguments, but I am not convinced that all the current flowing between electrodes causing the spark comes from the coil windings.

Best regards.
P: 4,663
 Quote by vk6kro [COLOR="Blue"] I saw a diagram on this Forum (by Bob S, I think) that showed the spark voltage as a positive then a slightly smaller negative then a slightly smaller positive waveform. This damped waveform went on for many cycles before it vanished. I found it: http://www.physicsforums.com/attachm...6&d=1250393723
The discharge shown in my simulation was the primary side of the ignition coil; the secondary (high voltage) can have either polarity, depending on how the coil is hooked up. The RF waveform is a damped LC oscillator. As soon as a spark occurs, the stored energy in the LC waveform is quenched. In fact, some (many?) modern transistorized ignition circuits may have a series rectifier (or SCR) in the primary, which limits the waveform to one half-cycle. I looked at the schematic for the Mark Ten CD ignition circuit I have. The energy stored in the capacitor is dumped into the coil primary by a SCR triggered by the points (or by a Hall Effect sensor). As soon as the voltage on the SCR reverses, the SCR stops conducting and the spark stops.

Bob S
P: 50
 Quote by Bob S .....As soon as the voltage on the SCR reverses, the SCR stops conducting and the spark stops.....
That would mean that the spark duration is about half the wavelenght of the LC resonant frequency.

As a normal spark durantion is between 1 and 2 milliseconds that means a resonant frequency of only 250 to 500 hz.

What do we do with the energy contained in the rest of oscillating cycles?

There something I do not understand, Bob S, Is it not the normal LC oscillating frequency in the range of 10 to 40 kilohertz ?.

Thanks a lot for joining us in this matter.

Regards.
P: 4,663
 Quote by CHICAGO That would mean that the spark duration is about half the wavelength of the LC resonant frequency. As a normal spark durantion is between 1 and 2 milliseconds that means a resonant frequency of only 250 to 500 hz. What do we do with the energy contained in the rest of oscillating cycles? There something I do not understand, Bob S, Is it not the normal LC oscillating frequency in the range of 10 to 40 kilohertz ?.
You are right. I recall once measuring the natural resonant frequency of an ignition coil and "condenser", and determined it to be ~40 kHz. The "condenser" was 0.02 uF, so the coil must have been ~1 mH. At 40 kHz resonant frequency, a half cycle is ~12 microseconds. If a spark occurs on the first half cycle of the HV pulse, it will quench (discharge) any further HV, like you showed in your previous post #10.

Bob S
P: 50
 Quote by Bob S You are right. I recall once measuring the natural resonant frequency of an ignition coil and "condenser", and determined it to be ~40 kHz. The "condenser" was 0.02 uF, so the coil must have been ~1 mH. At 40 kHz resonant frequency, a half cycle is ~12 microseconds. If a spark occurs on the first half cycle of the HV pulse, it will quench (discharge) any further HV, like you showed in your previous post #10. Bob S
Thanks Bob S. So it is the very first half cycle which causes the initial spark. Even more, depending on the dielectric characteristics this half cycle does not need to reach its maximum voltage to create the spark, because probably this only requires a percentage of that voltage to get established.

What I mean, even when that half cycle peak may reach -250 volts to produce a 25000 voltage at spark plug, the spark may jump with only 15000 volts, and in that case that primary half cycle voltage drops as soon as it reaches -150 volts.
Do you agree with this?