Hmm; where to start...

Did not go into minute details in consideration of audience; integrals over time are inferred. Since you're an EE we can go a little deeper..


This is/was my main contention with your post #18 in this thread and the main reason for my post #19. Let me address this in another subsequent separate post for clarity.



Not suggesting there's any active regulation. Only suggesting (strongly) that energy is conserved between the primary and secondary winding. Get into that more in subsequent post..

1. Electronic ignition: basic inductive coil design with points (remember them?) replaced by solid-state electronic pick-ups for timing; not CDI.
2. Constant current? What about this system is constant current?
3. So the voltage is flexible but the current is constant? Are there any other examples of passive devices that exhibit this behavior?

Do you mean the LMS theory? I think I've already stated what I've disagreed with on its assumptions.

I'm not sure why your engine stalled after it started and ran for a few seconds. This is why I suggested you ck the compression before doing anything else in post #2; that would have provided some diagnostic info that may have illuminated this issue more.

 

Standing by for your next post.

Couple of things in the meantime.

1. It's my opinion the engine had low compression when I tried to restart it. My basis for this is simply the fact that despite it being a cold morning (30F), the engine was cranking faster than I'd heard before.

2. I said the secondary has mainly constant-current characterics. Obviously, it's not a constant current source. Based on the fact that a spark discharge has a very low resistance (often characterized as negative resistance actually), the resistance of the secodary coil and plug wires dominate in calculating the current. By mainly constant current characteristic, I mean a circuit where say, reducing the load impedance by a factor of 2, causes a current increase of <<2. IOW, the load current is nearly constant with changes in load impedance.

3. Just before the switch in the coil's primary opens, the primary current has been established at some value. Afterwards, the current in the secondary attempts to rise to a value consistent with the turns ratio, without regard to the load impedance.

4. Yes, energy is conserved across the primary/secondary interface in the coil. But, that doesn't mean all the energy stored in the primary ends up in the spark itself, beacuse of resistive losses.

In double thinking this (and re-reading Lee's post) I'm doubting that condensing water vapor played a role. More on this later if I feel like it...

 

Sorry Brick for the slow response; tis the season to be busy..

Above is the statement that I can't abide. Others have already documented this better than me so I will cite some sites at the end of this post that relate to this topic. Some EE factoid/assumptions that will be helpful to remember:

• Current through a coil cannot change instantaneously (relatively) but voltage across coil can.
• Opposite of above for capacitors; pretty sure i remember these correctly; hopefully we won't have to dig up the calculus derivations ().

Here's my synopsis of the spark process in a typical automotive induction coil.

1. Ignition Coil assembly has a primary and secondary winding (or coils); latter typically has 100x more turns than primary. They have strong mutual inductive coupling. Both windings are referenced to common ground.
2. When dwell (period that battery voltage is applied to primary) starts current begins to flow into the primary winding. Keep in mind this is a coil, i.e. an inductive winding. This means it has inductance and therefore resists instantaneous changes in current through it. This means when battery voltage is first applied, it looks like and open circuit to the battery (i.e. no load). This of course only lasts for an instant, the applied EMF (voltage) begins to push the electrons through the primary and current ramps up until it reaches its terminal (aka saturation) current the magnitude of which depends on the resistive component of the primary's impedance (remember: impedance has a real (resistive) and reactive (capacitance or inductance) component. This ramp up in primary current takes on the order of a few milliseconds.
3. At the end of the dwell period the current in the primary is abruptly stopped (as quickly as possible) by removing the battery voltage supply. Remember, this being a coil (with inductance), it will resist changes in current flowing through it. Because of this a much larger than 12V back EMF (voltage) is generated in the primary circuit; up to 300-500V (the magnitude of this back EMF is determined primarily by the speed at which the primary current decreases). This voltage is high enough to jump ignition points (old technology) which is one reason why they had a capacitor across the points to quickly absorb this back EMF energy from the primary (analogous circuitry is employed on modern electronic ignition systems). It's critical that the current in primary is squelched to 0 as quickly as possible. It's the speed at which the primary current goes from saturation to zero that creates the relatively large and rapid rise-time back EMF of several hundred volts in the primary, and it's this voltage that is stepped up roughly 100x (primary to secondary winding turns ratio) to 30K volts or more.
4. Some people are helped by thinking of voltage as an analogue of pressure; a voltage this high is desperately looking for a path to ground, to complete the circuit and dissipate its energy into a load. In this circuit the spark plug acts as the load. But it doesn't have a constant resistance; its initial resistance is very high and varies by variety of things but some of the main factors are spark plug air gap and the pressure in the combustion chamber. Both larger air gap and higher pressure act to increase the resistance path for the secondary voltage across the plug gap. The voltage will build in the secondary until it's high enough to jump the air gap across the plug so it can complete the circuit path to ground; it's at this instant that lots of things happen at once:
5. The gasses around the plug electrodes are ionized causing the resistance to drop dramatically; the voltage drops to a point just high enough to maintain the spark, maybe a few thousand volts if at 1 atm (14.7 psi) until the energy is depleted from the secondary to a point where it cannot maintain the voltage necessary to maintain this spark; the flame then goes out.
6. The energy induced into the secondary is finite and correlates to the energy in the primary. Note there is minimum dwell period that will saturate the primary; any additional dwell period will not add any additional energy and is wasted as heat. As the secondary voltage necessary to initiate and maintain the spark changes (e.g. due to changes in combustion chamber pressure and/or plug gap) then spark duration also changes inversely to the secondary voltage. i.e. if the pressure goes down (say due to low compression) then the voltage necessary to initiate a spark will go down (as does the spark maintenance voltage) so the spark duration time will correspondingly go up. This fact is well known among automotive techs who use secondary ignition scope patterns to diagnose everything from a fouled spark plug to low compression.

So to your point about a spark event under lower combustion chamber pressures; the voltage is reduced but since spark duration will go up (other factors being the same); hence energy is conserved and more or less the same as a higher combustion chamber pressure spark, which will have a higher voltage but shorter duration spark.



Let me know if I left out any key details. Additional links that go into more details below.


Excellent explanation of “How Ignition Systems Work”
How Ignition Systems Work


Ignition Systems - Basics to High Performance


Anatomy of a Waveform
Anatomy of a Waveform Part 1 | Scope It Out


Faraday's Law and Auto Ignition
Faraday's Law and Auto Ignition


Ignition Coil


Induction coil - Wikipedia, the free encyclopedia


Electronic Ignition Overview


Capacitor discharge ignition - Wikipedia, the free encyclopedia


Tech Page


Spark blowout or is it my birthday?


Spark Plug Wire Test


Multi Spark Ignition


Spark plug - Wikipedia, the free encyclopedia


Spark Ignitions Systems operation on BMW motorcycles


MoTeC > About Ignition Systems > Overview


Ignition Coils Application Note | Documents | McLaren Electronics

 

By “backwards” you’re asserting that a low voltage spark results in higher spark energy. In looking through your links, I concede there is a mechanism by which this could happen. But you haven’t shown it (higher energy spark) does happen.

As you point out, a mechanism that could act to increase spark energy in a lower-than-normal compression atmosphere is the lower quench voltage, resulting in a longer duration spark. Honestly, I hadn’t thought of that effect.

But, although this mechanism does play a role, it’s not the whole story. I think we’d agree that the spark energy is the integral of the Volt-Amp product over the spark’s duration. And while your quench voltage mechanism acts to increase the spark time, I’m not willing to ignore the fact that the spark voltage is lower.

So, there are multiple mechanisms at work, influencing the spark energy. And it would seem they are pulling in opposite directions, if you will.

My recent experience in this matter showed me the ignition system failed to light off the charge in the cylinder, and the addition of some oil in the cylinder changed this for the better. Until I hear a more plausible explanation, I’m sticking by my low voltage spark=low energy in spark theory.

From one of your links:

“The Kettering ignition is hardly as simple as it appears.”

Hopefully, that's something we can agree on…

 

I retract the "backwards" remark in the context of spark energy; again, my main point is that a lower voltage spark is not necessarily a lower energy spark.

Here's what I had meant by that remark: you seemed to be inferring that because the spark voltage was lower, that the charge was less likely to be lit, when actually the converse is true. Since a lower voltage spark is longer in duration, it is more likely to ignite the air/fuel mixture in the chamber; not less.

 

Right, but neither is a long duration spark necessarily a higher energy spark. It goes back to the VI integral over t.