***1. The capacitor now has a 200 volt, ~40mJ(***2), reverse charge. This also forward biases the rectifying diode(s) but the XL of the inverter secondary constrains the rise of current flow for a rapid discharge so no second spark can be had.

So what happens to this 40mJ charge...

1. The inverter restarts...?

2. The inverter doesn't restart until the reverse charge current flow through the secondary has abated.

Note: The circuit is almost totally reactive, no substantive resistive element to dissipate the 40mJ as heat.

***2. Actual charge level is dependent on the level of energy expended in the previous arc.

west.
why wont you answer the original question

HOW MUCH CURRENT DOES THE CD DRAW?

I realize that this discussion has evolved into banterings about the theory of operation of these units. However, the 'original question' was answered by Bob Kontak on the first page of this thread. He said "OK so a range of 2 amps up to 3.5". Jonny H also answered the question.

Backing that up, observing hundreds of these units I have repaired, they typically draw about 1.2 - 1.6 A at idle. At 8000 rpm, they usually draw just under 3A, although occasionally I see around 3.2A

I do recommend adding a 5A fuse in-line with the power feed to these units. Although the discharge capacitor is the most common failure (due to age/degradation), the next common problem is shorted components and burned-open pc-board traces due to (a) over-voltage events in the vehicle caused by faulty charging system or removal of battery while running and (b) reverse-voltage events caused by accidental reverse polarity connection of the battery on jump-start. This is an 'instant killer' for these units. The fuse is cheap insurance against an otherwise expensive repair of the unit.

Hope this was helpful.

the thing is WEST has been spouting off at the mouth about his knowledge of these things, even insulting people, but has yet to answer the original question, which BTW he disputed when it was answered, which started this off topic argument.

so west

how much current does the CD draw?

In the case of a reverse battery a solid state component will blow long before a 5A fuse can melt open. And in the over-voltage case the incoming power shifts, concentrates, in the 82V high wattage zener which overheats and fails, allowing the flyback voltage to exceed the 2N3055 rating.

I really don't know, don't remember, and my '78 is out on long term loan, wife's nephew in Montana.

And I haven't disputed anyone's input on this matter, just added the FACT that it rises with engine RPM and engine load, the latter statement being the reason for the follow-on posts.

http://www.sqlsoldier.com/wp/wp-cont.../ITDepends.jpg

OK, I've contributed to this with good and helpful information. West has now offered his opinion that my fuse recommendation does not work. Well, big news: It has been tested. It works for me and my customers and that's what is important. I will not contribute any more to this thread as I can see where this is going.

Bob, it's just the entertainment factor I get out of this..... real-world data more often than not gets contradicted by arm-chair engineers with all their outdated pre-school knowledge and wild theories based on stuff they found online. Pretty funny.

It's always refreshing when someone confirms what really is going on. In the end data doesn't lie regardless of all the numbo-jumbo that some come up with.

Cheers,
Ingo

how many times has someone said THAT about a west thread, even me.

I'm by no means saying that having a 5A inline fast/quick blow fuse to the CDI isn't a good idea. Obviously there can be CDI failures that result in blowing the fuse, which would undoubtedly prevent "cascade" failures.

So, you have customers/others that have installed the 5A fuse... Okay..

Now, how many of that "pool" have had that 5A fuse blow and the CDI subsequently found not be damaged?quently

Jonny answered it two posts before. I just parroted it back to lead into my dummy question of why does the current rise. What is the "deficit" that has to be overcome and why doesn't the unit take all current available.

James Brown in post #19 mentioned resistance which I took to be an ok answer, even if he was not answering my question directly (not sure - really does not matter) All I was looking for was a simple explanation for all circuits. wwest answered it as well but rolled in complexities of the CDI box. In fact he answered it again later and specifically used the light bulb example James did.

http://ep-us.mersen.com/fileadmin/catalog/Literature/Application-Guidelines/ADV-P-Application-Information-Semiconductor-Protection.pdf

Since this thread has detoured south of the border anyway..... Any chance someone with a test setup and an oscilloscope could record the current into (and out of) this thing and post it?

I have always been curious. I would suspect there is some current coming out of the box when it gets charged and also when the spark fires.

You said it rises. Plus wrestling and personal degradation down the road.

I asked my simpleton question and you mentioned higher compression as a contributor. Thank you for responding. No one else did.

I think (compression) was more of a shiet storm starter than "rising" voltage/current.

Regrettably so...

"..."rising" current (with RPM & engine load).." supply voltage remains fairly constant.

Apparently compression pressure affects mostly at WOT and low RPM, engine pumping efficiency declines with RPM.

Just so you know, your '88 has a modern day Kettering (inductive discharge) ignition system.

IIRC I did that with the '78's factory CDI some years ago and posted the results, maybe on another forum. Long discussion with Loren at the time.

Fixed it for you. Current in a diode goes from anode to cathode. The bottom one I guess you assume the cap is charged by the magnetization and leakage inductance before the SCR turns off.

I'm thinking that red arrow to the left may go back to the battery.

A digital scope capture of the input current would answer a lot of questions.

typical