Sciencemadness Discussion Board

Ignition coil drivers

tom haggen - 18-4-2004 at 14:55

Hello everyone, I have read much of the labs on the powerlabs website. I found the ignition coil driver to be really cool. I so far have managed to salvage an ignition coil, a 12v power source, and some capacitators out of a 480v HPS light. I have done some searching and haven't really been able to find and details on assembling the device. I wish I could see a copy of that magazine the guy used as a guide to build his. If anyone has any experience or information on how to construct one of these ignition coil drivers I would greatly appreciate some advice. Heres a link just incase you don't know what i'm talking about.http://www.powerlabs.org/igncoildrivers.htm

Here

axehandle - 19-4-2004 at 10:18

http://www.geocities.com/CapeCanaveral/Lab/5322/coildrv.htm

It has a minimum of components, and is easily built. If you use a high wattage powersource, use a bigger transistor and don't forget to put it in a heat sink.

I've built this one, and can vouch for that it works. The trimpots are good, since they allow you to adjust the frequency to avoid radio interference.

tom haggen - 20-4-2004 at 19:01

I noticed that the 2N3055 NPN power transistor seemed to have 3 pins on the schematic. However, went I went to the store the 2N3055 NPN power transistor only had 2 pins, am I reading something wrong?

Axt - 20-4-2004 at 19:49

The third "pin" is the metallic base of the transistor, the base is likely to be the "collector". This is the one that will get hot!


[Edited on 21-4-2004 by Axt]

tom haggen - 20-4-2004 at 20:49

So what i'm supposed to solder a wire to the base of the transistor? By the way, whats the point of having 2 potentiometers?

[Edited on 21-4-2004 by tom haggen]

Axt - 21-4-2004 at 00:45

Connect it under the bolt you will use to fix it onto the heatsink.

Potentiometers look to be used to let you tune the frequency of the discharge (as axehandle said), but im sure there are more qualified people here to give you a more "technical" answer.

tom haggen - 21-4-2004 at 06:51

I know what a potentiometer does, I just don't know why theres 2 of them in the circuit. Also, i'm using a wireless bread board, so I think I will just solder a wire to the base of the transistor. I will accomplish the same task.



[Edited on 21-4-2004 by tom haggen]

[Edited on 21-4-2004 by tom haggen]

Axt - 21-4-2004 at 07:07

Just know that it will get VERY hot, a heatsink is VERY recommended! Even if solder was to stick to the base (dont think it will), the heat may even melt the solder off. Ive set things on fire with those transistors in simular circuits.

If you twisted a wire through the hole in the base of transistor and soldered the wire together you may get away with short runs.

EDIT: The powerlabs site you linked to shows the transistor in place on a heatsink. That would be the same transistor, correct?

[Edited on 21-4-2004 by Axt]

tom haggen - 21-4-2004 at 08:11

Ya ok that is the same transistor i'm using. Well I don't have a heat sink. I think i'm just going to get my circuit running as I have yet to do that. Once I play around with it, doing some short runs like what you were saying. I guess I will have to get some type of heat sink before I try to enclose the project. I hope my solderless bread board doesn't melt.

[Edited on 21-4-2004 by tom haggen]

improvised heat sink

axehandle - 21-4-2004 at 08:56

You can improvise a heat sink. Take a piece of Al plate, make 4 cuts from the corners in the direction of the center, then bend the outsides in a 45 degree angle upwards. Drill holes for the emitter and base leads, and larger holes for the fastening bolts. Use normal thermal paste between the transistor and Al plate.

It will get awfully hot very quickly without a heatsink.
Quote:

Connect it under the bolt you will use to fix it onto the heatsink.

I'd recommend this as well. We're talking very high currents here.


[Edited on 2004-4-21 by axehandle]

tom haggen - 21-4-2004 at 09:50

Could you recommend some dimensions for the aluminum sheet metal I should use. Also, I thought the 2 leads comming out of the transistor were the emitter and collector. I believe the exterior of the transistor would be the base in this type of transistor.

axehandle - 21-4-2004 at 11:23

Well, my commercial heat sinks seem to be about 2mm thick. But any thickness is better than no heatsink. As for which is B, E, and C, I don't remember.

tom haggen - 21-4-2004 at 14:28

ya i Guess I'll research my particular transistor and try to find out which is which.

hodges - 21-4-2004 at 15:19

Quote:
Originally posted by tom haggen
I know what a potentiometer does, I just don't know why theres 2 of them in the circuit.


In that circuit, two potentiometers are used in order to allow both the on time and the off time of the pulse to be varied. Think of a square wave. If it has a frequency of 1kHz, and has a 50% duty cycle, it will be on for 0.0005 seconds and then off for 0.0005 seconds then on for 0.0005 seconds, etc. Now if you adjust the frequency to 5KHz, the wave will be on for 0.0001 seconds and off for 0.0001 seconds. But the coil may work better with some value other than 50% for the duty cycle. Maybe it wants to current flowing twice as long as it is off. In that case, for a 1KHz signal, it would be on for 0.00033 and off for 0.00066 seconds. There are two potentiometers because one adjusts the on time and the other adjusts the off or total time.

tom haggen - 21-4-2004 at 21:30

Thanks for clearing that up hodges. I'm a little rusty with my electronic skills. One thing that I have another question about is this. On the schematic there is a positive and a negative on the ignition coil. What leads are supposed to create the arc of electricity?

[Edited on 22-4-2004 by tom haggen]

hodges - 22-4-2004 at 14:47

The ignition coil will have 3 connections, +,-, and spark. The high voltage is between the spark (top terminal, which has the recessed contact) and -.

tom haggen - 22-4-2004 at 15:13

Once again, a million thank yous to everyone. I will try to post a picture if I ever get this project going.

Heatsink.

axehandle - 22-4-2004 at 18:27

NP. Just get a huge heatsink. Look at mine:

tom haggen - 22-4-2004 at 19:32

Right on, I'm having trouble trying to get the circuit working properly. I'm hoping to get some advice from one of my old instructors soon.

[Edited on 23-4-2004 by tom haggen]

[Edited on 23-4-2004 by tom haggen]

tom haggen - 7-2-2005 at 11:52

Some how I had problems last time I tried wiring this circuit. My 555 timer was destroyed last time I hooked every thing up. Anyone have any idea what might have caused this?

Basically, I took this same circuit, removed the transistor part. I used 2, 1000 ohm resistors, and also I hooked up an L.E.D. to terminal 3 of the timer. I also switched to a 220uF capacitor. Basically I had a circuit that enabled you to see a blinking L.E.D. Then I removed the L.E.D. and hooked up that transistor and the ignition coil in their correct places. When I hit the power it fried my 555 timer. Any ideas?

[Edited on 7-2-2005 by tom haggen]

[Edited on 7-2-2005 by tom haggen]

tom haggen - 7-2-2005 at 12:14

Here's a picture of the circuit which just has a blinking L.E.D.

IMGA0179.JPG - 136kB

Eclectic - 7-2-2005 at 13:56

Probably it's an inductive kickback voltage spike. When your circuit tries to turn off the current flowing through the coil's primary circuit, the magnetic field collapses and generates high voltage. It acts as if the electricity has inertia and "wants" to keep flowing in the same direction. The way to fix that is with a fast diode across the coil primary that will block the charging voltage and short out the inductive kickback. Also make sure your high voltage terminal is going to ground and not the transistor switch when you draw a spark.

hodges - 7-2-2005 at 14:42

You also should put a large (at least 200uF) capacitor in the circuit across the power supply (battery) leads. I had a similar circuit that would sometimes blow out the chip when power was first connected or disconnected. Adding the capacitor solved the problem.

tom haggen - 7-2-2005 at 15:20

So I need to use to capacitors for the ignition coil circuit?

I should also mention that I was never actually able to get an electrical arc.

[Edited on 7-2-2005 by tom haggen]

Pommie - 14-2-2005 at 16:06

Quote:
Originally posted by tom haggen
So I need to use to capacitors for the ignition coil circuit?

I should also mention that I was never actually able to get an electrical arc.

[Edited on 7-2-2005 by tom haggen]


You need to add a diode to get rid of the back EMF from the coil. I would suggest a 1N4004. It should be connected like this.



You also mention earlier that your transistor has the case as the base. If this is correct then I would guess it's not a high current device and suggest you go get a 2N3055.

HTH

Mike.

Eclectic - 14-2-2005 at 16:26

On the circuit above, the ground for the spark should be whichever coil terminal is connected to the secondary winding.

Twospoons - 14-2-2005 at 19:03

If you kill the back EMF with a diode you wont get anywhere near as much output volts. Adding a 100V Zener in series with the diode means your primary voltage is now effectively 100V, not 12V, so your're output will be correspondingly higher.
The zener will get hot, so use a stud mount type, and give it a heatsink.

The 2N3055 has the collector connected to the body, base and emitter on the two pins. You will not be able to solder a wire to the body as it is usually made of aluminium or stainless steel. Use a bolt, ring terminal, and star washer.

Pommie - 14-2-2005 at 19:16

Quote:
Originally posted by Twospoons
If you kill the back EMF with a diode you wont get anywhere near as much output volts. Adding a 100V Zener in series with the diode means your primary voltage is now effectively 100V, not 12V, so your're output will be correspondingly higher.
The zener will get hot, so use a stud mount type, and give it a heatsink.


I think you'll find that the output voltage is proportional to the rate of change of the primary current and therefore, killing the back EMF is the best course of action.

Also the 2N3055 has a max Vce of 60v, so it would simply melt with a 100V zener and, the 100V produced would also fry the 555.

A better (more energetic spark) could be produced by putting a capacitor between the collecter and emitter of the transistor, making a resonant circuit. But this would also fry the other bits.

Mike.

[Edited on 15-2-2005 by Pommie]

Twospoons - 14-2-2005 at 20:59

Quote:
Originally posted by Pommie
I think you'll find that the output voltage is proportional to the rate of change of the primary current and therefore, killing the back EMF is the best course of action.


[Edited on 15-2-2005 by Pommie]


I think I know how a flyback converter works.

Vce of 60V? OK, so lets use something with a bit more muscle, like a MJ15003, which will do 140V, 20A.

Adding a resonating cap can work really well, but is much harder to design, and really requires a switchmode controller chip.

There's also nothing wrong with the old fashioned trembler/points arrangement. The points are set up with a bit of iron in the magnetic path of the induction coil, so that the points open when the coil energises. I built one using an old magneto coil many years ago, used to light petrol in a paint can with it. I could get 2cm spark in air from it.
There may not be enough stray field from a modern ignition coil though.