Ignition coil driver - flyback arrangement - Page 1
Forum section: electronics / beginners

author: Junkers
author_rank: Contributor
timestamp: April 14, 2015, 06:52:48 am
content: I'm looking to build a single coil driver for the sake of a small engine project. I've decided to adopt the circuit schematic illustrated below however I'm a little unclear on the choices of flyback drivers in this circuit.Can anyone shed some light on why they've used a RC circuit instead of a diode to suppress kickback across the coil? Why would you want a high pass filter there? (They use a 0.1uF cap and 10K resistor = 160Hz cutoff)Secondly, why is the flyback diode across the diode useful? I've seen this flyback arrangement in many examples but was under the impression that the coil will produce a positive inductive kickback as opposed to a negative one. Does this have something to do with Lenz's law?

author: HighVoltage
author_rank: Super Contributor
timestamp: April 14, 2015, 07:19:25 am
content: Look up a Fairchild ignition IGBT, they have all protection built in and can be driven by a micro controller directly.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: April 14, 2015, 09:37:50 am
content: I like this better,Assuming free-running is OK. Or you could rewire the 555 for one-shot from a trigger source, etc.Tim

author: max_torque
author_rank: Super Contributor
timestamp: April 14, 2015, 10:41:46 am
content: The answer to your questions is rather in the title! "Flyback" is just that. You suddenly stop the current path through the coils primary windings, and the magentic energy stored in the coil "bounces off" or "flysback" that dead end, building up a high voltage. The ratio of turns on the coil means that the voltage at the output of the secondary is much higher, and when that exceeds the ionisation voltage of the spark gap, current flows out of the coil that way.Hence, if you were to just put a diode around the coil primary, you would never get any "Flyback" because the current would just cycle round the coil at a low voltage (depending on the diodes Vfwd at that current).In the circuit you posted, they use that high pass circuit to take the worst of the "sharp edge" off the extreme di/dt, but it does not provide a long term (DC) path around the coil.The diode across the power switching device helps that device survive the almost certain voltage oscillation that occurs at switch off, and will otherwise pull the coil primary output below ground potential. Generally, you'd also have some form of zener here too, to limit the max primary voltage to below that of the power switch.However, having said that all, there is no real reason to make all this yourself! Either use an off-the-shelf IGBT designed for driving coils, with all the necessary protection/thresholds built in, or use a coil amplifier that takes a logic level input.These are easily and cheaply available, and have the advantage of keeping the high current out of your controller, and can be got with built in dwell control/limitation which makes life a lot easier. eg

author: Junkers
author_rank: Contributor
timestamp: April 16, 2015, 12:32:41 am
content: @HighVoltage Thanks for that, that looks like extremely good value.@T3sl4co1l If you factor out push/pull bjt pair driving the MJE18008 that circuit is almost identical. I'm using a BDX53C darlington pair which hopefully should cater for gain.@max_torque Cheers for answering my question. I did a little further digging around based upon what you have said and believe I may have the answers to my questions.Realizing that the primary coil will create a counter potential when current is initially driven through it helps explain the need for the lower diode. From what I also understand the flyback voltage itself is created when the overlapping fields of the primary coil begin to collapse, yielding a force of their own.I realize that I can just buy a complete system but where would be the fun in that!

author: Jolyon39
author_rank: Newbie
timestamp: September 15, 2017, 01:47:12 pm
content: Bouncing this old topic.I am a beginner but smart and willing to learn.I am trying to understand and then calculate a signal so I can replicate it with a different circuitCan you teach me???BackgroundI have an old ignition system, single coil with distributor using points, that I want to replace with a modern coil pack system. There will be no distributor, just an EDIS controler that controls two High Current coil packs. The programmed controller for the EDIS will be megajolt.My problemThis car has a very very early electronic EFI system and the ECU only starts fully operating when it gets an ignition pulse from the coil negative, via a resistor....... this is where I get fuzzy. I think this pulse is "flyback" because of all the reading and googling I am doing. The ECU is from 1980 and called a 4CU and is used in an early EFI Triumph.How do I calculate this flyback pulse from a cars ignition coil? I have a 12V tack feed from the Megajolt.I have had two theories put to me..... 1. A 5V square signal will do it (Sounds like a resistor on the 12 V tach feed might do this?) or 2. Four diodes off each of the 4 new coil pack feedsCan you explain this flyback to me?Can you help me calculate what is coming off the coil negative? (The resistor is listed as a 6K8 in one publication and in another it is meant to range from 6.5K to 9.2K)I dont know what I don't know so please illuminate me.