Electronics > Beginners
Coil charging
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xavier60:
Even though the coil current limits for some time before the spark, the ECM is still likely to be calculating an optimum turn on time before the spark is needed to minimize dissipation.

BTW: a constant voltage applied to an inductor will cause a mostly linear current ramp up(or down) so long as any resistive voltage drop is a small fraction of the applied voltage.
Benta:

--- Quote from: soldar on May 13, 2019, 09:21:03 am ---
--- Quote from: Benta on May 12, 2019, 07:26:19 pm --- A constant voltage applied to an ideal inductor will result in a linear increase of current in it.
--- End quote ---


As I am quite sure you know, this is erroneous. As has been already mentioned, a constant voltage applied to an ideal inductor will result in an exponential (not linear) curve.

--- End quote ---

No. Go back to your books. You are talking about an RL combination, not a pure inductor.
soldar:
Yeah, no, I thought you were talking about the OP's case.
injb:
 
--- Quote from: Jwillis on May 13, 2019, 05:59:49 am ---How "old school are you talking" .
...

--- End quote ---

Bosch Motronic 3.1 from an 89 Porsche 944 Turbo


--- Quote ---Even though the coil current limits for some time before the spark, the ECM is still likely to be calculating an optimum turn on time before the spark is needed to minimize dissipation.

BTW: a constant voltage applied to an inductor will cause a mostly linear current ramp up(or down) so long as any resistive voltage drop is a small fraction of the applied voltage.

--- End quote ---

Yes it uses a map for dwell time which is rpm vs battery voltage.

Regarding the current limit, I don't really know how it works, but I've been told that it's there. What the hell, I might as well attach the schematic since I have it. The ground side of the coil is connected to Pin 1, which seems to be controlled by the Darlington transistor. The actual logic timing of the signal comes in from Pin 32[1]. I don't have the skills to analyze what's going on with that driver circuit though.

[1] In case anyone's curious, it leaves this ECU on Pin 21 and goes to another computer that can retard the ignition timing if it detects certain problems, then comes back into this one for the final driver stage.
magic:
Okay, this thread is getting a bit weird so maybe I'll just answer some of the original questions ;)


--- Quote from: injb on May 12, 2019, 06:27:44 pm ---I found this calculator (https://daycounter.com/Calculators/Inductor-Current-Power-Calculator.phtml) which uses the formula:

Ton_max= I_sat * L/V

and it gave me approx 3.8ms for the values above, which is close to the charge time I'm seeing on the scope. But I don't know if I really understand what this formula is saying. Is it "this is how long it takes to reach I_sat, given L and V" ? Is the resistance value implied, because I've specified I and V?
--- End quote ---
It assumes zero (or small/irrelevant) resistance. In such case current rises at the rate of V/L and so you get that particular formula relating time to current.
If your measurement is accurate, it looks like this is indeed happening. The current rises almost linearly and then somehow becomes limited at 9A.


--- Quote from: injb on May 12, 2019, 06:27:44 pm ---I thought the above formula might just be a rearrangement of:

T = L / R

But this formula is supposed to tell me the time it takes to reach 63.2% of maximum current, right? I know that the ECU of the car limits the current to 9A. Does this mean that the current I'm seeing after approx 3.8ms is 63.2% of the maximum that would flow if it wasn't limited by the ECU?
--- End quote ---
Nope, this one is for an inductor with significant series resistance. As current increases, increasing amount of voltage is wasted on overcoming resistance and the voltage that actually "accelerates" current through the inductor decreases and therefore, by V/L, the rate of current rise slows down.
You get an exponential decay curve like one of those but upside down. After each L/R time, current gets 63% closer to its final value, which is V/R.
This doesn't look like what you observe.
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