9472Views33Replies

Author Options:

Ignition Coil Answered

Yes, I just posted something on a flyback. I was interested in that, until I saw this. It got me interested in ignition coils, so I got one new from the autoparts store for $20 (USD). I looked on the site, for the schematic, and realized that I didn't know what the rectangle was. Is it a resistor? I'm new to electronics, what size resistor do I need (if it even is a resistor)?
What about the capacitor? I have a 100uf 200v electrolytic one, will that work?
Also, part of the core is not in the coil, it's square shaped, could I rig it to work like this flyback circuit?Or this one? (simple driver circuit)?

Sorry about the never ending beginner electronics questions, I just really don't have anyone else to ask.

33 Replies

user
john mantova (author)2014-04-18

I think auto systems RING to reduce induced current an radiation currents

Select as Best AnswerUndo Best Answer

user
tesla coil (author)2008-05-09

Yes that rectangle shaped thing is a resistor and you will have to find the size that stops any residual charge from flowing out it is there to protect you so it is not exactly nessecary to have it in the circuit but just to be safe I would keep it in the circuit.

Select as Best AnswerUndo Best Answer

user
Goodhart (author)2007-09-16

Caution, ranges from 15000 to 20000 volts to 120000 volts or more can be gotten and the last thing you need to feel is the "hammer" of 120,000 volts entering your body at some point or another (if you have felt this already, you know what I mean by "the hammer").

Select as Best AnswerUndo Best Answer

user
Patrick Pending (author)Goodhart2007-09-16

It did occur to me that you really should have a good grounding(pun) in electronic basics before building HV circuits. I've had more than my fair share of being an external component in a circuit. Although electric shocks do aid concentration, I don't often recommend them for their therapeutic properties. Pat. Pending

Select as Best AnswerUndo Best Answer

user
> ... external component in a circuit ... aid concentration ... therapeutic properties ... . ROFLMAO! I _love_ the way you write!

Select as Best AnswerUndo Best Answer

user
Goodhart (author)NachoMahma2007-09-16

I have become (all TOO often) the "most conductive route to ground" myself a number of times, with both DC and AC, but thankfully never "hand to hand" (through the heart) nor above the bare minimum of amperage. Ever grab a faulty coil wire while someone was cranking over the engine? THAT is where I got "hammered". About 5-6 poundings (sparks, shocks) from hand to elbow (resting on the frame) and you think someone actually broke your arm. Landing after being thrown backwards is no picnic either, but I don't remember that so well. BTW: anyone that is not careful with an automobile battery can get "burnt" (because of the amperage). A friend of mine was taking the positive terminal off the battery of his car, when the wrench slipped and he came down hard, gripping the wrench in his left hand, his wedding ring completed the short between wrench and negative terminal. He had never had occasion to remove his wedding ring THAT FAST before. It nearly melted through it and it got (a-hem) really hot.

Select as Best AnswerUndo Best Answer

user
killerjackalope (author)Goodhart2008-01-03

Strangely I have found medicinist and octors to be liars, I took a hit of mains electricity across the arms, and yes for about half an hour my heart was pounding like a drum but I was fine, well apart from twitching a bit. i was changing an outside lightbulb and my brother turned the switch on, the bulb was broken so I had to touch contacts to get the bulb out, unfortunately I was wearing thick boots and holding the outer metal surround of the light housing (very good grounding point because it's stuck in the ground. I took one hit of coil across the my hand once, hammer is right but I just lost the ability to use my hand at all for a few days no flying away. What I fail to understand is how the coil manages to have a good bit of current despite high voltages or is it simply because it has been stored then fired.

Select as Best AnswerUndo Best Answer

user
Goodhart (author)killerjackalope2008-01-03

What can occur is a momentary "short" amplifies current to the nth degree. Everything has it's conditions, and position is another factor. The reason I was "thrown" backwards? While the DC line made my hand grasp the coil wire tighter, the kneeling position I was in, created a tension point in my legs, and they involuntarily flexed (straightened) and that reaction catapulted me backwards.

Select as Best AnswerUndo Best Answer

user
killerjackalope (author)Goodhart2008-01-03

Ah right I was leaning down with my back bent (I used to be a good bit shorter) needless to say I got out of there pretty sharpish... just not with flight...

Select as Best AnswerUndo Best Answer

user
Goodhart (author)killerjackalope2008-01-03

Yeah, I became a human "jack-in-the-box" as it were. :-) One other time, when I leaned against a pair of 220 AC lines, I was thrown a bit, but the arm received the worst of the incident.

Select as Best AnswerUndo Best Answer

user
killerjackalope (author)Goodhart2008-01-03

No I always seem to end up stuck, but I've now developed a reaction to use my still working muscles to escape, usually involves standing up as violently as possible, sprained my wrist doing that twice but having the direct out put of a generator through your arm does that to a man...

Select as Best AnswerUndo Best Answer

user

Well, it did make me pay attention VERY quickly...

Select as Best AnswerUndo Best Answer

user
John Smith (author)Goodhart2007-09-16

Once, with a small battery, nothing near normal voltages. It IS like a hammer, while camera caps are more like a BZZZZ- (bleep).

Select as Best AnswerUndo Best Answer

user
jossewiet (author)2008-01-03

The 2n3055 is of course a npn transistor, my bad.

Select as Best AnswerUndo Best Answer

user
jossewiet (author)2007-12-31

Actually there are a lot of ways to drive your ignition coil, but i think this is one of the best and easyest ways to do so : http://www.geocities.com/CapeCanaveral/Lab/5322/coildrv.htm
It's a simple 555 timer driver circuit with a 2n3005 pnp power transistor, I use a 12V / 20A pc psu as power source,(it can also be driven with 17v or 24v) (search the net for explanation) and it works great! I use this coil assembly to drive my 150000V tesla coil ( 15cm sparks!)
( I use a 20kohm potmeter (higher is also possible) instead of one of the 10kohm potmeters to lower the frequency, to get longer sparks, but that might be different for different ignition coils so just experiment)
good luck ! (and be sure to buy enough transistors and timers because they dislike high voltage spikes and can easely be zapped in the process of experimenting)

Select as Best AnswerUndo Best Answer

user
NachoMahma (author)2007-09-16

. BTW, most coils for point-type ignitions are actually made to run at about 9V (that's what the ballast resistor is for). The only time they see full battery voltage is during starting (when the battery voltage drops to about 9V due to the load from the starter). Running one at 12V for extended periods will burn it up.

Select as Best AnswerUndo Best Answer

user
John Smith (author)NachoMahma2007-09-16

Ok, that's good, I've been using a half-dead 6v lantern battery. You don't think I'd burn it up if I wired two new 6v batterys in parallel, would I?

Select as Best AnswerUndo Best Answer

user
NachoMahma (author)John Smith2007-09-16

. I'd find a junk car and rob the ballast resistor, then wire the 6Vs in series. Running the coil "under-voltage" won't hurt anything, but the output will be less. . A 9V "transistor radio" battery should work, but I doubt if it will last long. . If you are just running it for a few seconds and then allowing it to cool off for a few minutes, 12V will work. Come to think of it, as long as you don't leave the switch on (it only takes milliseconds to build the field), it shouldn't be that big of a deal. If you were switching it constantly, as in a car, it would be a bigger problem. . . As others have mentioned, you will be working with voltages that can be lethal. IIRC, even the primary side will have over 100V when the field collapses. Even if it doesn't kill ya, it will make you hurt yourself getting away from it. ;)

Select as Best AnswerUndo Best Answer

user
NachoMahma (author)2007-09-16

. Another BTW, the condenser serves the same purpose as a clamping diode across a relay coil. Why they don't use diodes on cars (or caps on relays), I dunno, but suspect it has something to do with the fact that an auto ignition circuit "rings." I don't really understand the hows and whys of ringing, but it looks cool on a 'scope.

Select as Best AnswerUndo Best Answer

user
John Smith (author)NachoMahma2007-09-16

So... what's a clamping diode?

Select as Best AnswerUndo Best Answer

user
NachoMahma (author)John Smith2007-09-16

. Whenever a magnetic field collapses around a coil, a voltage/current/EMF is produced - happens every time you turn a coil off and is reverse polarity to what generated the field. To protect transistors/&c;, a diode is put in parallel with the relay coil (reversed polarity, so it is blocking when the coil is energized). When the reverse pulse is generated, the diode shorts it out.

Select as Best AnswerUndo Best Answer

user
NachoMahma (author)2007-09-16

> Sorry about the never ending beginner electronics questions . You seem to have done quite a bit of research before asking, so I don't mind at all. I feel pretty safe in saying that most of the others here feel the same way.

Select as Best AnswerUndo Best Answer

user
Patrick Pending (author)2007-09-16

My understanding of the circuit is this:

Applying the 12V supply to the circuit the capacitor will charge to 12V via the primary winding and the resistor (this is purely incidental)

closing the switch causes two things to happen:
1)current will flow in the primary winding
2)the capacitor discharges via the switch and the resistor (this does not contribute to current in the primary winding and is also purely incidental)

When the switch is opened the magnetic field in the primary winding collapses inducing a high voltage spike (back-emf) in the primary and a much higher voltage in the secondary (enough to create a spark across the gap).

The capacitor will have an effect on the output voltage but its primary role is to prevent HV arcing across the switch contacts as the switch is opened.

This circuit is in fact very similar to an automotive distributor type ignition. The switch is the contact breaker, the gap is the spark plug, and the capacitor is the condenser used to prevent burning of the contact breaker points.

Just noticed NachoMahma posted similar explanation

Cheers,

Pat. Pending

Select as Best AnswerUndo Best Answer

user
NachoMahma (author)2007-09-16

. The dotted rectangle is the coil (a transformer) itself. The switch is called points and the capacitor is called a condenser. The "lightning bolt" is the spark plug. Not sure what the solid rectangle is (don't remember one being shown on other auto schematics), as the ballast resistor is usually in series with the switch and coil (on old Mustangs, the resistor is in the supply wire; on some cars, it's built into the coil; some use a large resistor mounted on the firewall). . The condenser's main purpose is to absorb the counter-EMF produced by the collapsing field of the coil, which can arc across the points, eroding them.

Select as Best AnswerUndo Best Answer

user
whatsisface (author)2007-09-16

Yes the rectangle is a resistor, and judging by the lack of a value, it doesn't mater what size resistor you use either. As for the capacitor, you need anything with a rating of more than 12V, otherwise they may tend to explode, so yes, your capacitor should work.

Select as Best AnswerUndo Best Answer

user
John Smith (author)whatsisface2007-09-16

OK, thanks. I guess you don't know about being able to rig it to work like the flybak circuit? If you don't know, thats ok.

Select as Best AnswerUndo Best Answer

user
whatsisface (author)John Smith2007-09-16

Nah I'm not sure, probably better to get someone with a better level of understanding to help you out.

Select as Best AnswerUndo Best Answer

user
John Smith (author)whatsisface2007-09-16

That would be... (I really don't have anyone to ask, except maybe a high voltage electronics forum, and then I'll probably get an overcomplicated answer)

Select as Best AnswerUndo Best Answer

user
whatsisface (author)John Smith2007-09-16

It's ok, just wait for some nice instructables user to come along and help ;-)

Select as Best AnswerUndo Best Answer

user
John Smith (author)whatsisface2007-09-16

Well, I hooked it up, and, using a power transistor (same one(s) i got from the dvd player, see my last forum post), I got a little less than 1 cm sparks from a 6v lantern battery. It normally would only give 1mm or less sparks, alone. I have no idea of how i got it to work, but I loosely followed the flyback circuit, and SOMEHOW it worked. I attached the schematic, maybe you can help me understand HOW I got it to work.(I think it's right, I got confused- It was hooked up with alligator cables.)

Select as Best AnswerUndo Best Answer

user
John Smith (author)whatsisface2007-09-16

Also, do I need a diode to prevent damage to the battery or cap? If so, where?

Select as Best AnswerUndo Best Answer

user
LasVegas (author)2007-09-16

The RC (Resistor/Capacitor) circuit in the schematic is intended to supply a longer pulse to the transformer (coil) when the momentary contact switch is released. Your 100uf capacitor is way overkill, since you don't need much of a pulse and you really don't want to wait a half hour or so for the capacitor to recharge... For those two parts, use a 0.1uf metal film capacitor (don't use ceramic in this case) and a 4.7K ohm resistor (quarter watt would be fine) Note that this will produce one spark each time you release the switch. Use a push-button switch if you want to make it easier to use.

Select as Best AnswerUndo Best Answer