Coil charging - Page 1
Forum section: electronics / beginners

author: injb
author_rank: Regular Contributor
timestamp: May 12, 2019, 06:27:44 pm
content: Hi, I have some noobish questions about inductance and current etc.I captured a trace of my car's ignition coil using a scope and current clamp, and what I found was that it charges up to approx 9A in around 3.7ms or so. The alternator voltage is 13.85 and I know the coil's inductance is 5.85mH. But I don't know exactly the resistance of the whole circuit, and the ECU has some circuitry that limits current to 9A.Is there a way that I can calculate the coil charge time mathematically? I found this calculator (https://daycounter.com/Calculators/Inductor-Current-Power-Calculator.phtml) which uses the formula:Ton_max= I_sat * L/Vand 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?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?Thanks!

author: Benta
author_rank: Super Contributor
timestamp: May 12, 2019, 07:26:19 pm
content: Your ECU may limit the current to the ignition coil (to protect itself), but mostly the coils have internal current limiting resistor/resistance.The 'scope plot you've supplied indicates this as well (semi-exponential shape), where magnetic core non-linearity most likely also plays a role.A constant voltage applied to an ideal inductor will result in a linear increase of current in it.

author: injb
author_rank: Regular Contributor
timestamp: May 12, 2019, 07:41:21 pm
content: Quote from: Benta on May 12, 2019, 07:26:19 pmThanks. This coil is an old school coil with nothing fancy in it. It has a resistance of around 0.4 ohms.

author: Buriedcode
author_rank: Super Contributor
timestamp: May 12, 2019, 08:26:08 pm
content: You can play around with LTspice, providing pulses of fixed periods to an inductance, and change its DCR, inductance etc.. It's free and quite handy for playing around to see how things work:https://www.analog.com/en/design-center/design-tools-and-calculators/ltspice-simulator.html#As for inductor "charging" a link I often pull out is this one:https://www.dos4ever.com/flyback/flyback.htmlIt explains how the inductor works, in both a boost converter and a flyback (ignition coils can behave like flybacks, because current doesn't always flow in both the primary and secondary at the same time). It isn't my website, but he explains it remarkably well. The equation I believe you're after is: I=(V/L)*t Which provides the current flowing through the inductor, for a given voltage, after a period of time. It does not deal with the inherent resistance of real-world coils, but like most things in electronics, one deals with "ideal" components first, then dealing with the real-world limitations after.

author: Jwillis
author_rank: Super Contributor
timestamp: May 13, 2019, 05:59:49 am
content: How "old school are you talking" .The ECM will reduce the voltage in the coil from the expected 12 -13 v.This is done to prevent damage to the coil after the alternator comes "online" .The old way was with a ballast resistor .The coils don't normally run at 12 v . So you need to measure the voltage at the coil and not the voltage at the alternator or battery . The ECM also controls the Dwell Or duration that the coil is turned on or the length of time that current flows through the primary winding of the coil.This has a direct effect on the current level and hence the energy stored.

author: mikerj
author_rank: Super Contributor
timestamp: May 13, 2019, 06:56:39 am
content: A modern ECM measures battery voltage and adjusts dwell time so that the peak coil current is pretty consistent (and also compensates fuel injector duration). There will be current limiting in the coil driver for protection, but this is typically only for fault conditions and not used to limit coil current in normal operation.

author: magic
author_rank: Super Contributor
timestamp: May 13, 2019, 06:58:46 am
content: Not really, OP's trace clearly show current raising to a constant value and staying there for a while.I would say it is ballast resistance limited except that the curve doesn't look exponential to me.Perhaps there is active current regulation with the pass transistor in linear region (but does any car actually do that?) or maybe OP's current probe runs out of bandwidth.

author: soldar
author_rank: Super Contributor
timestamp: May 13, 2019, 07:17:13 am
content: Quote from: Benta on May 12, 2019, 07:26:19 pmAre you sure? Or would you like to amend your answer?

author: Benta
author_rank: Super Contributor
timestamp: May 13, 2019, 09:13:00 am
content: Quote from: soldar on May 13, 2019, 07:17:13 amWhy should I? Something about the phrasing?

author: soldar
author_rank: Super Contributor
timestamp: May 13, 2019, 09:21:03 am
content: Quote from: Benta on May 12, 2019, 07:26:19 pmAs 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.

author: xavier60
author_rank: Super Contributor
timestamp: May 13, 2019, 09:28:58 am
content: 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.

author: Benta
author_rank: Super Contributor
timestamp: May 13, 2019, 10:10:26 am
content: Quote from: soldar on May 13, 2019, 09:21:03 amNo. Go back to your books. You are talking about an RL combination, not a pure inductor.

author: soldar
author_rank: Super Contributor
timestamp: May 13, 2019, 10:23:11 am
content: Yeah, no, I thought you were talking about the OP's case.

author: injb
author_rank: Regular Contributor
timestamp: May 13, 2019, 01:11:44 pm
content: Quote from: Jwillis on May 13, 2019, 05:59:49 amBosch Motronic 3.1 from an 89 Porsche 944 TurboQuoteYes 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. I don't have the skills to analyze what's going on with that driver circuit though. 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.

author: magic
author_rank: Super Contributor
timestamp: May 13, 2019, 09:26:36 pm
content: Okay, this thread is getting a bit weird so maybe I'll just answer some of the original questionsQuote from: injb on May 12, 2019, 06:27:44 pmIt 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 pmNope, 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 likeone of thosebut 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.

author: injb
author_rank: Regular Contributor
timestamp: May 13, 2019, 09:38:01 pm
content: Quote from: magic on May 13, 2019, 09:26:36 pmThanks, that answers my questions perfectly. I was thinking of a constant voltage, but I didn't think about how the voltage drop would increase as current increases.

author: langwadt
author_rank: Super Contributor
timestamp: May 13, 2019, 09:39:45 pm
content: Quote from: injb on May 13, 2019, 01:11:44 pmdwell time should be constant with rpm, dwell angle changes