# Simple Ignition Coil Circuit Driver

Discussion in 'The Projects Forum' started by Sirisian, May 14, 2010.

1. ### Sirisian Thread Starter New Member

May 14, 2010
2
1
My goal with this project is to build a very simple ignition coil circuit for generating a spark across a 1 mm gap to ignite butane like from a lighter. (Trying to make a small motor out of boredom).

First I'll list some parameters I want. The power source I'm planning to use is a 9V battery and the ignition coil is a very small pot core.
http://www.surplussales.com/Inductors/FerPotC/FerPotC-1.html
(The H6A3P26/16-E32147M one with AL=250, I'm thinking now I might have wanted the 400).

I have a spool of 34 gauge wire that I was planning to use, but it could depend on the current and such if that's practical.

For the switch I was planning to use a transistor (wrong symbol drawn probably) with a 5V input (if I can find one).

Also imagine that I only have the battery and no external ground.

Assuming my wires for the spark are 1 mm apart then for a 6 million V/m I only need 3000 V to jump the air gap (butane might be different). So I need a voltage higher than that.

None of the electrical engineering people I know have a clue how to analyze analog circuits so I came here to ask if that circuit makes sense ignoring all the values I put for the cap and the number of turns.

The idea is that you turn on the transistor and it creates a magnetic field in the primary coil and when you shut off the transistor the magnetic field breaks down and inducts a current with a high voltage in the secondary coil which jumps the spark gap. C1 protects the transistor and c2 and the resistor are ideally supposed to cause a very fast oscillation allowing for the spark to last more than an instant.

My main question is does this circuit do what I think it does or will it not work? I noticed there's another circuit online that looks like:

It connects the secondary coil with the primary coil which I noticed a lot of ignition coil circuits do. When the switch is turned on the primary coil gains a magnetic field and then when it turns off the primary coil oscillates with the capacitor or is it designed to protect the switch so that a spark doesn't jump across it? I'm confused about that part.

After analyzing that I'm wondering something else about the ignition coil. I've been told the voltage in the primary inductor (in the magnetic field I think?) will be much higher than 9V. So in order to calculate that I found the meaning behind AL here which has it as L = n^2 * AL, where L is the inductance, n is the number of turns and AL is stated as "The relationship between the inductance with a given core and the number of turns on it is called its AL value."

I know the AL value and I'll have a certain number of turns on the primary coil presumably.

I also have this equation from Wikipedia for finding the voltage over time which is less obvious how I'd use it.

Once I find the peak voltage output though I want to find the number of turns on each coil in order to step the voltage up to a large value. The number of turns is based on the the peak voltage which in turn seems to be based on the number of turns which causes a recursive problem. I assume this can be worked out by picking a set number of turns and then calculating the values.

Ideally I want to find all of the math behind the ignition coil and be able to mathematically prove that it works before I build it. (Don't want to hurt the battery or wind 5K+ windings by hand just to have the wire burn). If anyone can help clarify anything about this I'd be grateful.

(I've only taken EM aka Physics 2 so RLC, LC circuit type stuff I am somewhat familiar with, but basically treat me like I know nothing since it makes things easier for me).

A big help would be to explain to me how to read an analog circuit and step through everything that occurs as the switch is switched on and then off for the circuits since my EE friends and the people I talked to gave me some very different interpretations.

Last edited: May 14, 2010
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2. ### jpanhalt AAC Fanatic!

Jan 18, 2008
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You are asking quite a complex question. Have you reviewed the STMicroelectronics Application Note AN819? (see: www.st.com )?

Is this a school project or just a hobby? If the latter, there are many circuits in the Internet for homemade ignitions.

John

3. ### Sirisian Thread Starter New Member

May 14, 2010
2
1
Yeah, I didn't want to deal with trying to generate a HV source or use SCRs. Trying to keep this very minimalistic. It's an interesting diagram they have though.

Just a personal project since I've always wanted to build something powered by butane. Yeah I saw a few circuits online but most of them just used a 12V and a car ignition system. I need something much smaller. (Trying to figure out how small I can actually make this).

Anyway I have another inquiry. Is the c1 even necessary in my circuit. I've been told it probably isn't doing anything meaningful.

4. ### jpanhalt AAC Fanatic!

Jan 18, 2008
5,698
905
Does your circuit work with C1? If so, then test it without the capacitor. If your circuit doesn't work, then it is an unanswerable question. Every CDI circuit I have seen uses two primary coils. One in the collector circuit to power the secondary and one in the base circuit for the oscillator. Some people have experimented with mosfets driven by microcontrollers. The circuits I refer to for models all operate on 4.8 V or so. A typical 9V battery will probably not last long as an ignition source for a motor.

You can go to the start of this thread to see the original design: http://www.rcuniverse.com/forum/m_4344316/mpage_50/key_/tm.htm

Unfortunately, it has now gotten quite long and veered off course. Go back about 2 months to see discussion of some alternative approaches, including TCI. There are way too many of basically the same circuit on the Internet to give you any "best" design.

Here is a schematic comparing the two basic designs that were prevalent a couple of years ago:

John

5. ### SgtWookie Expert

Jul 17, 2007
22,182
1,728
A few tidbits...

The generally accepted voltage breakdown of dry air is 3,000,000V per meter, or 3kv per mm, at sea level (one standard atmosphere).

So, with your spark gap sitting on a table, a 1mm gap (about 0.039") will require 3kv to ionize the air between the gap.

But, if your engine uses compression, the insulation of the air will be multiplied by the compression ratio. For example, if you have a 10:1 compression ratio, you will have 10x as much air between the gap, so you now have effectively a 10mm gap at sea level pressure, and you will need 30kv to ionize the air in the gap.

You are limited as to how small the gap can be in order to reliably ignite the gases. Going below about 0.7mm (about 0.027") will likely result in misfiring, or no ignition at all. This is in part due to the cooling effect the electrodes have on the gases. Ignition at idle speeds (closed throttle) will be the most troublesome, as the stratified fuel/air charge will be much harder to ignite. That is why many newer auto ignition systems use plugs with wider gaps (perhaps 1.5mm, or 0.060")

So, for a 0.7mm gap, you would need 2.1kv to ionize the gap at sea level air pressure, and 21kv to ionize the air at 10:1 compression ratio.

One item you'll have trouble with is when the current through the transistor is turned off, the voltage on the transistor's collector will rise to a very high peak voltage. There are HV transistors available that are specifically designed for automotive ignition systems.