Deciphering this ignition coil driver - Page 1
Forum section: electronics / beginners

author: nsled
author_rank: Contributor
timestamp: May 11, 2024, 10:03:06 am
content: Firstly, yes I am aware this can be done with the same (relay, capacitor) components in another configuration, but I'm really trying to decipher this one.I spent yesterday trying to figure this one out but the description and schematics (as attached) are headache inducing. The video shows it working but doesn't give a good look at the wiring, and the schematics don't translate well into the real world. It is from this videohttps://youtu.be/1EabqJJnWhc?si=JXMnAcphuc4Rh27L&t=293If it was made as in the schematic it will just short straight to ground through the primary coil, as ignition coils have a single negative terminal.Can anyone work out the actual wiring of this?

author: aliarifat794
author_rank: Regular Contributor
timestamp: May 11, 2024, 10:44:00 am
content: This one has a clearer diagram.https://www.instructables.com/super-simple-ignition-coil-drivers/

author: nsled
author_rank: Contributor
timestamp: May 11, 2024, 01:01:18 pm
content: Thankyou, that one makes a lot more sense, and I just tested it to be true. Although I don't think its oscillating for the reason they say its oscillating, it's a much simpler play between the resistances of the relay and transformer coils.

author: Manul
author_rank: Super Contributor
timestamp: May 11, 2024, 09:59:50 pm
content: Quote from: nsled on May 11, 2024, 10:03:06 amThat is certainly not the best coil driver in the world, but what exactly puzzles you? What do you mean "shorts to ground"? Ignition coil is driven by short pulses, so yes, for a short moment primary negative gets shorted to ground by relay contacts. It's called low side switching. It must be a normally open (NO) relay of course.

author: Benta
author_rank: Super Contributor
timestamp: May 11, 2024, 10:51:28 pm
content: Quote from: Manul on May 11, 2024, 09:59:50 pmUmm, no.Ignition coils are normally driven by NC contacts, where opening the contact produces the spark (think flyback).

author: Benta
author_rank: Super Contributor
timestamp: May 11, 2024, 10:59:26 pm
content: Quote from: nsled on May 11, 2024, 10:03:06 amThe wiring is completely normal, it's not unusual for an ignition coil primary to draw 10 A. But many have a limiting resistor in addition to the inherent primary resistance.

author: langwadt
author_rank: Super Contributor
timestamp: May 11, 2024, 11:43:03 pm
content: Quote from: Benta on May 11, 2024, 10:51:28 pmthe coil gets charged while the contact is closed, spark strikes when contacts open

author: Manul
author_rank: Super Contributor
timestamp: May 12, 2024, 12:51:25 am
content: Quote from: Benta on May 11, 2024, 10:51:28 pmI was talking specifically about the OP schematic, where NO contact pair is needed.Electrically speaking, I understand driving as actively supplying current. Although you could say that it's the flyback phase which drives the secondary to high voltage and thus makes a spark, I would argue, that opening of contacts is not driving. Driving is supplying current before they open. That's why I wrote "driven by short pulses". In a more broad definition driving means control algorythm, which then would include "drive current - quickly interrupt current - wait for flyback phase to end - repeat" or something like that. Anyway, that is all very debatable and potentially a messy territory of exact definitions, but we know that we are talking about the same thing here.

author: RJSV
author_rank: Super Contributor
timestamp: May 12, 2024, 01:56:16 am
content: (Interesting post trend, there are a couple of threads now, in 'General Technical' section...including my questions.)Had been thinking along same lines, where the release of low side of a coil will cause field collapse and self or auto-transformer generation of potential.My questions about how to present the numbers, formally; Like what would a peak coil output be, when there's no load, and what form of waveshape happens, on discharge.

author: nsled
author_rank: Contributor
timestamp: May 12, 2024, 08:27:58 am
content: Quote from: Manul on May 11, 2024, 09:59:50 pmI meant the circuit as presented there is no ground wire off the "secondary" in reality, it's an autotransformer with a common ground (primary-secondary ground wires are linked) so if its wired as shown it will short to ground as soon as it's switched on and practically no current will flow through the relay. Essentially it just needs that wire from secondary to ground removed or at least specified as a spark gap.

author: nsled
author_rank: Contributor
timestamp: May 12, 2024, 09:13:00 am
content: Changed the wiring and relay orientation to make more sense, the relay implied a NC contact. This is what was confusing in OP.

author: MrAl
author_rank: Super Contributor
timestamp: May 12, 2024, 10:17:44 pm
content: Hi,A couple points to consider.First, with the circuits that use a transistor (collector) to drive the coil, it looks like one thing that is overlooked is the primary coil acts as a boost inductor in a boost converter when the transistor CE opens. That means the transistor can see very high CE voltages and blow out. This can blow out the other parts of the circuit also.For either the transistor or relay circuits, again the primary coil acts as an inductor in a boost circuit and therefore can generate a high voltage all by itself. This means a high voltage on the primary side also. This can be used to advantage though because the coil is effectively a transformer with more turns on the secondary which means it inherently steps up the voltage by a relatively large factor. With the increased kick back of the primary however, that secondary voltage can go even higher, so it makes sense to drive the primary in the "open the switch to generate high voltage" mode rather than the "close the switch to generate high voltage" mode (note for now I avoid the use of 'flyback' and 'forward' mode nomenclature due to some misunderstandings that arise from those descriptions).In a forward mode converter (close the switch to generate high voltage) the highest output you can get is due only to the turns ratio of the transformer (the ignition coil here). In a flyback mode converter (open the switch to generate high voltage) the highest output can go much higher because the voltage goes higher on the primary as well as on the secondary. With the same turns ratio and the primary voltage going higher than the power source voltage (like 24 volts instead of the power source 12 volts) the output voltage can go much higher. If the kickback in the primary is two times the power supply voltage, that means in theory the output will shoot up to two times what it would have been in the forward mode. It can actually go up even higher though because the primary winding acts as a boost winding all by itself, and that would actually make up a regular boost converter. So, in effect we get a boost converter (where the voltage can go up high) followed by a step-up transformer (which boosts the voltage even more) and that provides the highest voltage.The high voltage is the benefit, but the length of the pulse depends on the inductance of the coil because energy is stored in the coil before the spark is generated. That means the pulse length depends highly on the inductance.There could of course be anomalies where the switch transistor does not provide a fast enough fall time and thus the 'boost' from the primary is not as good as it could be. The transistor has to turn off fast enough because the boost depends on the wave shape. A ramp would not produce a voltage as high as a theoretically perfect square edge. That would have to be tested. I would think in most cases this would not be too much of a problem though.With the transistor though there should be a snubber of some kind to prevent the transistor from blowing out. If a single capacitor does it, that's great, but it is rare to see a capacitor get shorted out by anything including a relay or transistor. It is more typical to see a resistance in series, even a small resistance, to limit the peak current either the relay contacts or the transistor collector. When the transistor opens, the cap + resistor may resonate well enough (or should I say not as good as possible) to keep the CE spike low enough, but that's something that is always tested on the bench not in simulation by powering the circuit with an increasing supply voltage while watching that spike amplitude with a decent oscilloscope. The spike peak follows the DC supply voltage input level, so you can watch the peak rise up little by little as you increase the input voltage. If it gets too high, stop and modify the circuit snubber (or whatever), then repeat the test.Best of luck