Thank you Dave. You were the first to point this out way back in post #21. It is the crux of the argument. No charge is saved.

On a V8 engine COP "dwell" time is, can be, 8 X that of an older common coil pak inductive system, so much so that the dwell time, dwell initiation, results from a computation in accordance with RPM.

I believe that a modern day CDI system could be designed using COP and have equal reliability and functionality of any modern day COP system. But to what purpose?

"no charge is saved...."

Okay, for your benefit, let's assume the capacitor does discharge fully, then we're left with "what happens to the energy still in the coil's now collapsing field?"

Any argument/objection to the subsequent half-cycle current flow path I've drawn in RED in the 3 diagrams?

With the 12 volt buss "clamp" diode removed the third diagram most closely replicates the OEM Bosch CDI circuit. How do you not store at least a portion of the coil's remaining energy storage in "the" capacitor??

PS: Your own scope images indicate a significant positive voltage excursion following the negative firing pulse, does this go toward firing the plug a second time or does the power return to the capacitor, or some of both?

http://forums.pelicanparts.com/uploa...1449949724.jpg

Wishing you hadn't ask this question..??

Heh-heh. The catalyst of turmoil. The unknowing axis of evil, I am.

Let's take a step back. I got no skin in this game. I am ignorant but still curious.

The initial post/question related to amperage, not voltage.

James Brown posted early on, in-short, "resistance" is what has to be overcome. Amps do that, I think, yes? I believe the answer to my question is that "current has to increase" when resistance increases. Voltage remains constant as there is a cap on that, yet there are fused amps up to ten (let's say). That is what varies.

Jonny says max is around three. That tells me that as RPM's increase, resistance increases but the CDI box only needs to use a max of three to do it's job.

If it required four, it would take four.

wwest, you are able to tell me if this is conceptually in the ball park.

So it can keep up at high RPM's where Kettering grew tired., yes?

My little bit of reading says the CDI delivers a squillion volts but only for a micro second.

So IDI has a degree of lower voltage at the spark plug tip, but delivers a nice sustained spark all day regardless of rpm.

The traditional problem with the Kettering/inductive system is that as RPMs climb there is less and less time to overcome the inductive impedance (***1) and store enough charge. The coil design therefore must be a compromise between not overcharging it at low RPM but having the inductive impedance still remain low enough that it can be fully charged at maximum RPM.

Even with the most optimal design compromise most inductive system's charge level will decrease with RPM, eventually reaching a point where no spark is struck.

So power draw decreases with RPM.

The distinct advantage of an inductive system is FAST initial HV rise-time

CDI, on the other hand, has a relatively slow HV rise-time, but can be designed to maintain maximum power storage beyond what most ignitions required historically.

***1: Inductive Impedance: a coil of wire's "resistance" to allowing a change in the amount of current flowing through it.

The "400kV is probably a typo, 40kV is more likely.

All About Ignition System: Direct Ignition.

Except for the resistances "increase", yes.

The 12 volt supply buss "sees" a decreasing resistance with increasing RPM and/or engine loading.

'For my benefit'? Seriously?

I explained this in detail in post #62. I will try to do so again:

After the initial negative -360V pulse, the positive voltage rise to around 200V DOES charge the cap and would most likely cause a 'smaller' second spark. So yes, the energy from the coil went into the capacitor. Unfortunately though, the energy reverse charged the cap and then the SCR switched off.

What happens when the inverter tries to positively charge a reverse charged cap? it effectively short circuits it and thus it discharges. What happens to the energy? It is converted to heat.

Therefore the energy is not saved, it is wasted as heat.

Oops, sorry, I was wrong to be using the 2nd half-cycle to recharge the capacitor, it's the second half-cycle that reverse charges the capacitor and thereby keeps the SCR fired until the coil field fully collapses.

But what happens once the SCR stops conducting....??

And does ELI the ICE man come into play here? SmileWavy

Yes, thanks, I have neglected to note that the peak current flow co-insides with highest rate of rise/fall of the voltage, 90 degrees out of phase. I should have been referring to the second half-cycle of the current flow waveform.

Obviously I should spend more time reading to comprehension the material I link as reference.

I just said in the last post. The inverter starts charging the cap but it is reverse charged so it effectively gets shorted out. The negative charge is goes back into the inverter, thus is wasted as heat.

Anyhow, why are you asking me, I 'don't know beans' remember? ;)

"..capacitor voltage progresses again below zero to about -100V."

"...about -100V...."

"About..." The variable, energy returned to BOOTSTRAP the next charge cycle.

http://forums.pelicanparts.com/uploa...1449965699.jpg

"...wasted as heat..." NOT! read above post...

I am a simpleton when it comes to this. Can you take two seconds and ramp it down to simpleton level so I can get a rough handle on it?

I say resistance rises, I think. You say nope and then carry on.

Why can't you drop it down a notch and help folks who want to "sort-of" understand? This has to be simple stuff for you. Take a minute for the team. Why does it decrease?

In this case resistance is a holding back of current flow, like water flowing in a 2" pipe with a 1" tap. Pressure ("E", voltage) constant, but 1" restriction to flow.

100W 120 VAC bulb, ~100 ohms resistance, 1 amp current flow.

200W, 50 ohms resistance, 2 amps flow at 120 VAC

The more plug firings/second the more often the power used to fire the plug must be replenished. The resistance to current flow that is "seen" by the 12 volt buss goes lower and lower as the firing rate increases.

So the pipe is not as pinched off.

Let me do some research on the light bulb example you provide. Best I swim there. Shallower waters.

Thanks.

Just ran across my last interview of and last question set to wwest.

Sums it up. It is what it is, right or wrong.

Check it out. wwest could be an educator. He is not. Why not?

Lack of persuasion. 55 seconds in.

<iframe width="560" height="315" src="https://www.youtube.com/embed/LRitr8RYsh4" frameborder="0" allowfullscreen></iframe>

woooooow - amazing reading through this - my head is spinning and I only skimmed page #1 and #4.

To answer the OP question:

At 6500 RPM I have seen anything from 2.3 Amps to 3.3 Amps of current draw for different boxes on the bench. Most boxes do ~2.& Amps and the current goes down slightly as the box comes up to operating temperatures. A typical reduction is about 0.2 Amps.

The differences are due to "........" (insert your smarts here).

My observation is that differences in DC/DC converter frequencies and differences in losses at the main cap are the two main components that "move the needle". What's interesting is that with raising temperature a poor capacitor becomes less ideal (more losses) yet current consumption goes down. The loss (heat) is mostly observed in T1 - the main transistor of the DC/DC converter.

My data is based on more than 300 boxes I tested. No difference between 3-pin and 6-pin. That is expected as their DC/DC portion is identical.

Ingo