CDI ignition - gross vs nett spark energy

[Circlotron]

We all know the concept of before-tax and after tax as far as your pay packet is concerned. Some gets lost.
We all know the concept of flywheel horsepower vs rear wheel horsepower. Some gets lost.
Well guess what? Some CDI ignitions it seems have the same problem regarding advertised vs actual spark energy.
Did a little test tonight on a red 3-letter CDI with a "6" suffix and the results were interesting. Previously I thought the coil introduced very high losses but this may not be the case. It seems that of all the energy pumped into the coil primary, fully 50% gets bounced back out again into the battery feed wires and doesn't go toward making a spark. Look at the first pic. Yellow is the (simulated) plug voltage while arcing (~1600V). Blue is the current through the plug - 370mA for about 140uS. Purple is the amount of energy (mJ) as the arc progresses. it finishes at 40.8mJ - that's not very much! Where did it all go?

1600V 360mA 40.8mJ.png

The second pic shows primary voltage (yellow) peaking at -450V and primary current (blue) hitting 36 amps. Note also the purple energy trace. It peaks at 80mJ but then slides all the way down to 37mJ. Why? Look at the yellow primary voltage trace. See how after it has been pushed down to -450V it then wants to go upward but a diode connected to the battery wires clamps it to about +15 volts(just above the line) and at this point it dumps off the remainder of it's energy back to the battery rather than transferring it to the spark.

Prim V, prim A, prim mJ.png

So in round figures the box has an advertised spark energy of 100mJ (1uF capacitor charged to 450V) , 80mJ gets fed to the coil, and about 38mJ gets fed to the spark gap.

The interesting thing about all this is, a good inductive ignition can put out between 2-4 times this energy.

[RevTheory]

Subscribed. I've read various things regarding CDI vs a good inductive set-up but the pickins are slim for actual proof. I hope this thread progresses well and thank you for sharing.

[clshore]

I'm afraid that you are confusing the maximum energy capability of the CDI with the actual energy consumed by the particular spark plug during an ignition event.
As well, you don't seem to appreciate the details of how a charged capacitor discharges, which explains the shape of the current curve.

My house electrical panel is rated at 200 Amps.
When I use my stove, it consumes only 20 Amps.
Doesn't mean my panel is defective, that's just what the stove uses.

Once the spark ignites and current starts to flow, it uses what it uses, determined by the gap and mixture being burned.
You can't somehow cram more energy into it.

The big advantage of a CDI is the very fast risetime of the pulse, which will fire a spark in the gap under conditions
that a Kettering ignition may not.

Try increasing your spark gap to 0.100, and note the results.

[bmcdaniel]

Subscribing.

[dieselgeek]

Never got this far in testing, all I needed were dyno numbers to show me that Inductive ignitions always score more EMC points than capacitive.

[Tuner]

What coil was used?

[MadBill]

I don't trust anything I can't see, so to me electricity is a black art, but (I think this is what clshore is saying in more accurate tech-speak) won't the delivered energy depend on the demand? At 1600 v., that simulated plug must be about a 0.050" gap in air. A real plug firing in an engine is several thousand at idle and can easily be 20k plus at full load. I think it would take a spark gap of more than 1/2" in air to simulate this..

[Rick360]

The second pic shows primary voltage (yellow) peaking at -450V and primary current (blue) hitting 36 amps. Note also the purple energy trace. It peaks at 80mJ but then slides all the way down to 37mJ. Why? Look at the yellow primary voltage trace. See how after it has been pushed down to -450V it then wants to go upward but a diode connected to the battery wires clamps it to about +15 volts(just above the line) and at this point it dumps off the remainder of it's energy back to the battery rather than transferring it to the spark.

In your comments about the second graph your hi-lited comment makes no sense??? Energy (measured in Joules or kw-hours etc) is not an instantaneous measurement that can be shown on a graph that rises and falls. It can be shown as an accumulated total amount, but it would only increase over time or stay the same. How are you calculating that?

As others have posted the impedance of the load will affect the power delivered as well as the total mJ of energy. A resistor doesn't simulate a plug gap very well.

Rick

[JohnP]

MadBill:

Yes, it takes several kv to fire. The 1600v he is referring to is called the sustaining voltage. This is the voltage required to maintain the arc once struck.

[Circlotron]

MadBill:

Yes, it takes several kv to fire. The 1600v he is referring to is called the sustaining voltage. This is the voltage required to maintain the arc once struck.

That's right. 1600V was chosen because it takes 1600V to *maintain* the arc on one of these -> https://img1.fastenal.com/productimages/0245717.jpg

[Circlotron]

As others have posted the impedance of the load will affect the power delivered as well as the total mJ of energy. A resistor doesn't simulate a plug gap very well.

Rick

I used a long string of Zener diodes, not a resistor. Notice how the voltage drop across the "gap" remains more or less constant even as the current decreases. This would not be the case if it were a resistor, but it does approximate what happens with a spark gap.

[Circlotron]

Once the spark ignites and current starts to flow, it uses what it uses, determined by the gap and mixture being burned.
You can't somehow cram more energy into it.

That would mean that a more powerful ignition setup would give the same results as a less powerful one, would it not? Why do top fuel guys use a magneto instead of a plain old factory HEI ?

[Circlotron]

In your comments about the second graph your hi-lited comment makes no sense??? Energy (measured in Joules or kw-hours etc) is not an instantaneous measurement that can be shown on a graph that rises and falls. It can be shown as an accumulated total amount, but it would only increase over time or stay the same. How are you calculating that?

Rick

Charge a battery and Joules go in. Discharge it and Joules go out. The amount of Joules in the battery doesn't ratchet up higher and higher with each charge-discharge cycle but does in fact go up and down.

"How are you calculating that?"
Integral of (channel 1 x channel 2) as per purple text on screenshot.

[Tuner]

Did you try different coils?

[bmcdaniel]

Once the spark ignites and current starts to flow, it uses what it uses, determined by the gap and mixture being burned.

And cylinder pressure.