When I read about the secondary ignition system (high voltage circuit), I read about the 5,000 to 30,000 volt spark coming form the ignition coil, to the distributor, through the rotor and the high tension spark plug wires to the spark plugs, and jumping to ground making a spark. What I do not get is I thought an electrical “circuit” was just that, a circuit, a circle, a complete loop. This does not sound like a complete loop. It starts one place, and ends another. What am I missing?
Tom
I suspect that if you remove your engine ground strap and isolated your transmission mount you’ll see 
Jumping to a ground completes the circuit.
so is the ground strap simply for charging? I guess it may be since pull start mowers don’t have batteries. So the circuit is back to the coil?
in other Brit cars that’s how you melt the throttle cable.
1 Like
Particularly Alfas. 
I don’t get the question. The coil itself is effectively grounded via the primary circuit, connected to +12 (the coil case isn’t in the circuit, although its usually grounded as well). So it’s a complete circuit, including the part where the spark is traveling in ionized air.
But, beyond that, a transient, or changing, voltage can supply current into a capacitor, putting charge on its plates. The earth itself is one, your body is another, an engine block is one too. With enough capacitance, a spark can jump to a capacitor without a return circuit.
So, isolate a spark plug in air–no spark probably; the metal threaded part isn’t big enough to serve as a capacitor. Touch the plug to an engine block of another car or hold it in your hand–you get a spark (and if in your hand you will feel it), probably weaker than normal. Touch the plug to the engine block grounded to the battery whose + terminal supplies the coil–best spark of all. IMHO.
Paul, I do understand it jumps to ground, but I do not see that completing the circuit.
Look at my (very sloppy) diagram. I see the headlight circuit as starting at one point, the battery,
and going around the entire loop and returning to the battery. To me, that is a complete circuit. I do not see the 10,000 volt secondary circuit completing the loop.
Robert, I think you do get the question. Although the coil case may be grounded, the case typically is not connected to any wires inside the coil, so it is not a circuit. But please complete the rest of what I believe you are saying. It may be so obvious to you (which is why you may not understand why the question), but I have found that is not so obvious to everyone, at least not to me.
Tom
Correct–case isn’t grounded. What is grounded is the other end of the secondary coil of wire. That coil is where the secondary voltage comes from–it’s like a battery that has two terminals, + and - . So if I can convince you that the other end of the coil is grounded, I hope I can convince you that it’s a complete circuit. So here goes:
The other end of the coil secondary is connected to the coil primary. That makes the coil an autotransformer. One side of the primary is connected permanently to 12V (assuming ignition is switched on). That’s the + terminal of the battery. The other side of the battery is grounded. So the spark travels from the “hot” terminal on top of the coil, through the dizzy cap to the plug, through the plug gap to the engine block, from the block via its grounding wire to the battery negative, through the battery to the battery positive, into the coil primary, and through the coil primary to the other end of the coil’s secondary winding.
Operationally, the coil acts as a transformer as you probably know. But it’s a pulse transformer, not the same as one used for AC power, like a transformer that powers a doorbell. The points (or electronic ignition) generate a current in the coil primary, which flows from it’s + terminal (12V) to it’s - terminal and then to ground via the ignition amp or via closed points. When the points (or amp’s output transistor) open, the coil primary’s inductance induces a voltage of about 300 V in the primary, which is “transformed” via a 1:100 turns ratio to about 30,000 V in the secondary. The polarity is - at the coil tower and + at the end of the coil connected to the primary (which winds up at chassis ground).
2 Likes
Thank you, that is more what I was looking for, although to my mind, some of your terminology, while probably 100% correct, seems confusing.
I would not say the coil + is grounding, grounding sounds to me as chassis ground. And I try to leave the idea of 12V out of the secondary completely. I try to simply say the 10,000 volt circuit is completed by returning through chassis ground to the negative post of the battery to the positive post of the battery to the ignition switch and any fuses to the coil + to the coil secondary output.
This has not been how most articles I have seen discuss this.
Tom
1 Like
Here is a wiring diagram.
The low tension side of the coil builds a magnetic field in the high tension side of the coil.
When the distributor cuts the current, the magnetic field in the high tension side of the coil collapses and the energy must go somewhere so it induces a very high voltage which causes energy to spark across the plug
Dennis 69 OTS
1 Like
So are you OK with the explanation. You’ve got it right, but not clear you’re happy with it.
I’ll mention the engine/car capacitance again. That is enough to allow the plugs to spark, but over time charge would build up (excess electrons). These slowly migrate back into the coil circuit so that the charge is dissipated. At the high tension wire/spark plug insulator, the current is a very fast pulse. But flowing through the battery back into the coil, it is a slow trickle–having been smoothed or “filtered” by the capacitance of the engine/car.
Even a doorknob is sufficient to “filter” the spark current. Back in the day, I used to trick-or-treat carrying my trusty Model T spark coil and a lantern battery. I would touch the “hot” coil terminal to the knob when I rang the bell. I thought it was great fun–the other side of the knob gave what I thought was a mild and enjoyable shock. I finally realized that some folks don’t enjoy being even mildly shocked, even on Halloween. I quit after a year or two…I was getting less candy than without it.
Here’s how I’d say it…
The low tension side of the coil builds up a magnetic field in the coil’s iron core.
When the distributor cuts the current, the magnetic field in the core collapses
The primary (low tension) inductance resists the rapid change (decrease) in current (to zero) by inducing a “back voltage” of ~300V in the primary. That voltage, via transformer action, induces a very high voltage in the coil secondary. Because it’s wired as an autotransformer, BTW, the 300V primary voltage is added to the secondary voltage to give a slightly hotter spark. Fun fact: You lose this (the primary 300V subtracts, not adds) if you wire a negative ground coil backwards for use in positive ground applications in order to keep the high tension voltage negative.
1 Like
OK, now I see. Explain it, Wiggie. I don’t have time, just about to leave for a cruise.
LLoyd
Wherever space and time interact, there is information, and wherever information can be ordered into knowledge, and knowledge can be applied, there is intelligence.
Pavel Mirsky, mid 21st Century Russian General
Wow, I sure like the way you explained that. The part that always confuses people is the fact that the secondary ground is not ground, but instead 12V
volts above ground. As explained earlier it grounds through the primary winding. So its not ground but it is (just about) Great discussion. Definition of autotransformer or close enough.
1 Like
Nicely Written Robert!
1 Like
Egghead that I was, I built a breadboarded ignition system for my 7th grade science project, showing how electromagnetic induction worked.
Thanks, propellor-headed Dad!!!
Robert, I am always happy! Thank you for your thoughts.
To review- the purpose of my original post is that I believe some do not understand that the secondary circuit does not stop at the engine block ground. I have found that they have not followed the circuit back to the battery, ignition switch and fuses, and back to the primary post on the coil. That is what I was looking for and Robert, thank you for providing it.
However, now that we got that far, I will request a clarification for me on the 12v part of your description. In my mind’s explanation, I do NOT include any relevance to the 12v battery source in the secondary system. I visualize the 12v battery as only being used as a conductor. I visualize the secondary of the coil as being the sole power source in this circuit. Are you saying that the 12v adds to the 10,000 volts, even if it is insignificant? And do you feel I am wrong or confusing the issue, in that I do not want to call any part of this circuit a ground? To me, a ground is a chassis ground. To me, ground is not negative.
Tom
So, Robert, or anyone else of course, if the battery, ignition switch, and associated wires, etc. are indeed part of the secondary ignition circuit, could they contribute to poor ignition? In other words, if I have a weak, yellow spark, should I consider, in addition to all the usual suspects, a bad ignition switch?
Tom
Yes. You start at 12V (one side of primary) then add roughly -300V (other side of primary; “back” voltage in primary when points open) then add roughly -30,000V (other end of secondary, connected to spark plug). So you get -30,288V total.
The 12V in fact subtracts because the spark voltage is negative in our older cars but the battery voltage is positive. The 12V adds on still older positive ground cars. Some newer cars have half the plugs negative and the other half positive. Capacitive discharge (or multispark) ignition is different still–the 12V is neither subtracted nor added.
Not in the sense you’re implying, Tom. You just need a path to ground, and it doesn’t even have to be low resistance. And that’s ignoring the capacitance effect.
But in another sense, yes. With the points closed, you need to supply the proper current to the coil. Any fault in the switch, wires etc will reduce that current. The current magnetizes the coils iron core, and the amount of magnetization will help determine the voltage the coil primary produces when the current is broken (what I’ve been calling ~300V). But with the points open, the coil is on it’s own and doesn’t depend on integrity of the switch, wiring etc.