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| Chrysler Electronic Ignition - explained https://slantsix.org/forum/viewtopic.php?t=56150 |
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| Author: | WagonsRcool [ Sun Sep 07, 2014 4:23 pm ] |
| Post subject: | Chrysler Electronic Ignition - explained |
Often I read posts from members who have trouble with the ignition system on their slant (or think they do). It can be difficult for those who want to help because they have to try & explain "how it works" every time, & then go back & forth with the OP to determine what's going on, what tests to perform, what the OP's skill level is, etc. This here is my attempt to provide an explanation of the Chrysler electronic ignition system. Anyone- please feel free to correct me if what I'm putting out here is wrong. The ignition system produces a high voltage spark needed to ignite the fuel/air mixture inside the the engine's combustion chambers. At the heart is the ignition coil- it's a transformer that takes low battery voltage (8-12V) and steps it up to 8,000-30,000 V. The coil has a small number of low voltage windings (primary circuit) & a large number of high voltage windings (secondary circuit)- both are wrapped around a steel core. It produces a spark by having electrical current flowing through the primary windings which builds a magnetic field- when the current is suddenly turned off the field collapses. This induces high voltage in the secondary windings. The primary circuit is turned on & off by the ignition module- it has a power transistor inside- just a fancy electronic switch. The module is triggered by a pulse from the pickup coil in the distributor. The reluctor is a 6 point wheel on the dist shaft. As each point passes the pickup, it produces a brief Alternating Current voltage pulse. The ignition module triggers (turns "off" the ignition coil) about 1/2 way throught that pulse -as the signal goes from positve spike to negative. It takes 720* crankshaft rotation to complete a 4-stroke cycle for a cylinder, & each cylinder in the firing order starts its cycle 120* after the previous one. So in 2 engine revolutions you will have 6 firing events- 1 for each cylinder. One of the distributor's functions is to "connect" the coil high voltage lead to the correct cylinder's spark plug wire. The other function is to trigger the ignition- at the correct time. The "ideal" goal for a typical gasoline engine is to achieve peak cylinder combustion pressure at about 12-15* After Top Dead Center on the power stroke. This is the best compromise between minimum combustion chamber size (= more force applied to the piston) & "maximum" leverage from connecting rod to crank (= more rotational force to spin the crank). In the real world, timing is a compromise between power, tailpipe emissions, driveability, fuel economy, & cost. If ignition is too early; the piston, connecting rod & crankshaft main journal are lined up straight- most of the combustion force just "compresses" the rod/bearing- very little can "spin" the crank. If ignition is too late; the piston has traveled down the bore, so "peak" pressure occurs in a larger volume= less overall force applied to the piston. I'll give some general assumptions here: the car's battery is fully charged (about 12.6V) & has passed a load test recently, the battery to engine & body grounds are clean & tight, the user has a decent Digital Volt Ohm Meter (DVOM). I used a 4-channel Digital Storage Oscilloscope to capture these ignition voltage/current waveforms. Ch A (blue) is ignition primary current in amps. Ch B (red) is ignition voltage measured at the coil (+) terminal, Ch C (yellow) is distributor pickup coil voltage "+" tapped into the black pickup wire, (the pattern for the dist pickup doesn't appear as a true AC waveform because my scope has a common ground which "mucks up" the capture). Ch D (green) is ignition voltage measured at the coil (-) terminal. A scope lets you SEE changes in voltage in a circuit over time. In comparison a DVOM takes a number of voltage samples, averages them together, then displays the result as a number. For example, on a running engine my scope showed that voltage at coil (+) is 8.7V while the coil charges & spikes briefly to 14.0V when the coil fires. A DVOM shows this as a steady 9.7V. Coil (-) is 1V with brief spikes when coil fires- a DVOM might show 2-3V. Similarly, the pickup coil produces 4-5V peak to peak while cranking the starter with 0V between- a DVOM will average this to .3 to .5V AC (300-500 mV) or so- it also can vary with different types of DVOM's. [img][img]http://i1375.photobucket.com/albums/ag442/photousernam3/six-ign1_zpsa7d0ae61.jpg[/img][/img] Everything begins when you turn the ignition switch to the "run" position- battery voltage is supplied to the ballast resistor- from there it goes to the ignition coil (+) terminal, flows through the primary winding, from the coil (-) terminal to the ignition module (which is turned "on"- providing a ground). See point "A" on my first scope capture image. The ballast resistor drops battery voltage down to about 8 V at the ignition coil (+), & limits primary current flow to 3 amps. It's difficult to see in the image but voltage at the coil (-) rises slightly from 0V to just under 1V- this is mostly due to voltage drop through the switching transistor in the module. An important note here, don't leave the key in the run position with the engine off for very long, as the module is default turned on- grounding the coil (current is flowing) & the coil can overheat after a short time. Next, as you twist the key towards the "start" position, several things happen- first the "bypass" circuit supplies battery V direct to the coil (+) side of the ballast (at point "B). (Actual voltage for this car was 11.1V- mostly due to voltage drop through the firewall connections & the ignition switch itself). Primary current rises to about 4.5 amps. Twist the key further and the starter engages (at point "C")- Battery voltage drops to 10V or so, causing a drop in primary voltage & current- back down to about 8V & 3 amps. As the engine rotates, so does the distributor, & a tooth on the reluctor passes close to the pickup coil- inducing a brief AC voltage pulse. You can see this at point "D"- the pickup produces about 3V +peak to -peak; the coil (+) goes up briefly, and coil amps drop as the module switches the primary current "off". The coil (-) shows a very high spike of 270V- this is fed into the secondary windings- the actual kilovolt output from the coil depends on the total secondary circuit resistance. In order for a spark to "jump the gap" at a plug, there has to be enough voltage potential to ionize the gases between the electrodes, & form plasma which allows electrons to flow. Once there is flow, less voltage is needed to maintain the arc. Things like higher cylinder pressure, leaner mixture, wider plug gap (or rotor gap or open circuit plug wire), or low density mixture (ie, lots of EGR) will all raise the secondary resistance. The thing is, the ignition coil can only produce a "fixed" amount of energy - the actual amount depends on coil type, primary circuit voltage/current, & charging time- aka dwell time. If it takes more energy to ionize the plug gap (ie, higher kV), there is less energy left to sustain the spark "burn" duration. At some point if demand increases the coil either can't ionize the plug gap, or the spark "runs out of juice" & extinguishes- both will result in a lack of proper combustion- ie, a misfire. In this respect Chrysler EI has similar performance to the older points system. They both use similar if not the same coil & ballast- so primary current & coil output is about the same (EI has some longer dwell times). The EI system is much more reliable- less maintenance. This is why GM's HEI works better- it has different style ignition coil with lower primary resistance & the module can handle the higher current flow- about 5-6 amps at full battery voltage . HEI can put alot more energy into the coil than Chrysler's EI, so you can produce higher kV output (up to 45kV or so) and/or have a much longer spark duration- both will result in fewer misfires. Now back to the images. [img][img]http://i1375.photobucket.com/albums/ag442/photousernam3/six-ign2_zps9fc07b50.jpg[/img][/img] Here we can see that the engine is running, but I haven't released the key yet. Ignition (+) volts & amps are higher; the engine is "overrunning" the starter& the starter now doesn't draw very much current- so battery V is 11V or higher. I release the key at point "E" & the starter disengages- if you look closely you can see that causes a small glitch right then in the pickup signal- that triggers an incorrect coil firing. At point "F" the bypass circuit opens & ignition volts & amps go down again. After a few seconds the charging system starts producing- battery voltage goes up to 14.0V- ignition amps is up some to 3.4amps (while charging the coil), volts at coil (+) is 8.5V (about 9.7V averaged on a DVOM). Enough for now, will update with trouble shooting tips later. Chris |
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| Author: | wjajr [ Mon Sep 08, 2014 4:56 am ] |
| Post subject: | |
This is great info. Very well written and easy to follow even with just one half cup of half caffeine Maxwell House Lite* coffee down the gullet this morning. I recommend artical to be added to Electrical Sticky department. * Wife dislikes too much caffeine, soooo stuck with Lite Joe, lite cream, lite fake butter, etc., and with lite anything one must double-up on the stuff. Bill |
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