Coil output voltage risetime

[upinthehills]

The energy in the magnetic field will dissipate somehow when the IGBT is turned off. There are two coils in this field and if either or both of them offer a way to expend this energy they will do so. The proportion of energy transferring to each coil depends on the relative loads, it will follow the easiest path. That's likely to be a combination of flow thru the IGBT and thru the spark plug gap. Raising the IGBT breakdown voltage makes the path thru the IGBT more difficult, so you will dissipate more energy into the spark.

If your worried about the cap taking too much energy out of the coil before the coil fires, just put in a little more current to account for that. The point is that you will then have energy you can feed into the coil when the IGBT is turned off.

[user-23911]

The energy in the magnetic field will dissipate somehow when the IGBT is turned off. There are two coils in this field and if either or both of them offer a way to expend this energy they will do so. The proportion of energy transferring to each coil depends on the relative loads, it will follow the easiest path. That's likely to be a combination of flow thru the IGBT and thru the spark plug gap. Raising the IGBT breakdown voltage makes the path thru the IGBT more difficult, so you will dissipate more energy into the spark.

If your worried about the cap taking too much energy out of the coil before the coil fires, just put in a little more current to account for that. The point is that you will then have energy you can feed into the coil when the IGBT is turned off.

Like I said, you're confused and don't know WTF you're on about.
When the IGBT turns off, the current flow through it is zero and will remain zero. The voltage clamp is there to protect the IGBT from excessive voltage which will kill it. as in the case of an open circuit secondary or just high cylinder pressure/ large spark gap.

The "condenser" used with points ignition, the normal failure mode is insulation breakdown, they test good until you apply a high voltage.

[Circlotron]

Have you looked up the data sheet yet?
What IGBT?
Do you know how to read a data sheet?

I expediently used the term "IGBT" for the benefit of many of those reading this thread that would be familiar with OEM manufacturers using an IGBT for an ignition coil switching device. In actual fact I used a MOSFET - Infineon SPW47N60. That's 47 amp, 650V, 70 milliohms on resistance. TO-247 package. And an external voltage clamp of a string of TVSs totalling 480V. But the fundamentals are identical as far as this discussion is concerned.

[upinthehills]

Do you know how to read a data sheet?

Like I said, you're confused and don't know WTF you're on about.

You need to go back to school.

I don't claim to be an Electrical Engineer and I don't claim to always be right. I do claim the right to learn things though. Your responses here are really crummy and don't encourage either of us to learn anything. Why you think it's worth your time to even type this stuff is beyond me, that you expect other people to spend the time to read it - i don't know what to say about this really.

At high frequencies the relative differences of the primary circuit impedance and the secondary circuit impedance can have more to do with the output voltage than the turns ratio. This is not an area I've really worked in, I'm just curious because what engineer doesn't like things that go bang?

CIrcletron, the reason I was interested in the cap is because I'm curious about ways to drive the spark plug more directly than a big iron core spark coil. Using high frequency to drive a much smaller transformer and generating a more continuous spark that lasts longer. If our goal is put a plasma on the plug for a certain amount of time, modern parts give us more options. If you also want to do things like measure the plasma in the combustion chamber after the spark, that requires a high voltage power supply. The cost tradeoffs are probably different for us compared to the major OEMs.

[user-23911]

If you're going to try to do electrical engineering type stuff, then you need to learn some basic laws of electricity.

Faraday's laws

https://en.wikipedia.org/wiki/Inductance

V = L (di/dt)

Voltage is proportional to the rate of change of current as well as the inductance.
Adding capacitance anywhere in the circuit slows the rate of change of current and so lowers the voltage.

[Circlotron]

If you're going to try to do electrical engineering type stuff, then you need to learn some basic laws of electricity.
Faraday's laws
V = L (di/dt)

Now you've hung yourself with your own rope.
That equation proves what I said in the spark plug wire thread viewtopic.php?f=15&t=45640 , that coil secondary current with an inductive discharge ignition remains the same despite greater or lesser voltage demanded from it. Greater voltage demand just makes the spark current run down faster, comparable to the primary current running up faster with higher battery voltage applied to it.

Look at these three graphs, especially the third one -> https://en.wikipedia.org/wiki/Flyback_c ... _trace.png

And this - real life proof of my statement and the V = L (di/dt) equation.

Okay here it is:
https://youtu.be/DNgMJN5jLi0
Yellow = spark current.
Blue = coil primary volts.
You can clearly hear when the spark gap is getting larger or smaller, and the coil primary trace from 0:27 sec onward rises up as the arc length is increased so it is a real spark gap.
The fact is, the output current intensity stays much the same over a wide range of spark gaps, only the duration shortens because the higher voltage drop of the larger gap runs the energy out quicker.

Thank you for providing that equation joe 90.

[upinthehills]

Voltage is proportional to the rate of change of current as well as the inductance.
Adding capacitance anywhere in the circuit slows the rate of change of current and so lowers the voltage.

Yes, of course. Faraday was big in capacitors too.

If your only interest in this is voltage, by all means leave out the cap.

[user-23911]

Now you've hung yourself with your own rope.
That equation proves what I said in the spark plug wire thread viewtopic.php?f=15&t=45640 , that coil secondary current with an inductive discharge ignition remains the same despite greater or lesser voltage demanded from it.

You're still confused.

V = L(di/dt)

That's the induced voltage that you get when the coil current is interrupted.
There's 2 windings sharing the same magnetic field, as I already tried to explain, the turns ratio determines the voltage in each coil.
Any current flowing is determined by voltage and resistance.
Lets say we open the plug gap up to the point where the secondary side will reach 25kV before there's a flow of current
100 to 1 turns ratio
The primary side reaches 250volts


The secondary current at that instant is determined by secondary resistance, typically there will be 10 to 15K ohms coil resistance , maybe 10K ohms in the leads and 5 K ohms in the plug. It's got to jump the air gap too.
Using a figure of 25 K ohms total resistance and the plug gap shorted out for ease of measurement the peak current can't be any more than 1.0A. If we double the resistance, the peak current will halve.
If we completely open circuit the secondary, the secondary current will be zero, there will still be current flow but all in the primary side which will flow through the protection device (the voltage clamp) or in the case of a points type ignition, the condenser (capacitor), it's excessive voltages in cases like this that kills the condenser.

Primary voltage gives primary current which makes the magnetic field.
Interrupting primary current causes the magnetic field to collapse which induces voltages into both coils, the ratio of the voltages is determined by turns ratio. The induced current in the 2 coils is determined by the circuit resistance in each coil.
Adding primary capacitance affects di/dt because current flows into the capacitor. Adding secondary capacitance does the same thing but the effect is multiplied by the turns ratio.




Back to post 1.
Inductive rise time is 12 microseconds
CDI risetime is 7 microseconds

And guess what? The rate of rise of voltage is pretty much identical in both cases!

no it's not.

[Circlotron]

Back to post 1.
Inductive rise time is 12 microseconds
CDI risetime is 7 microseconds

And guess what? The rate of rise of voltage is pretty much identical in both cases!

no it's not.

I did take back that statement in the first post on page 2.

[Circlotron]

As regards the rest of the above, you never did try the following experiment, did you?

As an aside, suppose you put a reverse biased diode across the coil primary the same way as is often done with a relay coil. Run the primary up to 6 amps and switch off. How many amps is suddenly diverted into the diode? Then remove that diode and put a suitably rated diode across the secondary. At primary switch-off, assuming a 1:100 coil turns ratio, how much current suddenly flows through this diode? I double-dog-dare you to try this experiment.

I cannot remotely affect the outcome of this when you do it on your own test bench. The results will be the results for anyone that performs it. That is real science. And don't try and say you don't need to do it because you think you already know what the results will be. By your reckoning, because either side of the coil discharges into only a forward-biased diode so low voltage drop then heaps of current ought to flow. But it won't, I guarantee you. If interrupt 6 amps on the primary then 6 amps will flow in the primary diode and 60 milliamps in the secondary side diode. (one diode connected at a time).
Go on.
Do it!
We are all watching.

[jalai]

Some example of RiseTime.

Primary (100x probe) / Secondary (Tektronix P6015 probe = 3pF) voltages. About 5mm free air spark gap.
First, IGN-1 coil, 6ms Dwell.

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Like first + 22pF capacitor connected from secondary to GND.

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Like first + 100nF capacitor connected primary - to GND (parallel with IGBT)

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Crane PS-92 coil 3ms Dwell.

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Accel 140001, 7ms Dwell.

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And same with MSD #8230 coil, 0,9 ms Dwell.

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[jalai]

And more, now with CDI.....
Not much difference.

Yellow = Primary voltage
Cyan = Secondary voltage
Pink = Primary current
Blue = secondary current

MSD6A + Crane PS92

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AEM TwinFire (Twin Trouble) + Crane PS92

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AEM TwinFire (Twin Trouble) + MSD8230

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[Circlotron]

Thank you for your posts jalai.

[jalai]

Thanks for thanks ....

One more, now example, that High Voltage probe can show a fast Risetime too.

Mercury 339-832757A coil 230us Dwell (~100A peak current)
Yellow = Primary voltage
Cyan = Secondary voltage
Blue = Secondary current

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