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Ignition coil sparks have got shorter :( Some questions about it.

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Cypriotica

Fri May 29 2009, 03:43PM

Registered Member #1811 Joined: Sat Nov 15 2008, 11:52AM
Location:
Posts: 14

I have used a land rover ignition coil and a ignition coil driver from uzzors website. http://uzzors2k.000webhost.com/projectfiles/ignitioncoil/ET_Iggie_Driver.gif. I built the circuit on a solderless board with all the same parts. I powered the circuit off a large car battery which was outputting 11 volts. I could draw 3 cm arcs at first. Then i powered it off a weak transformer (15 volts 1-2A) and the mosfet heated up and the arcs have become shorter and shorter. The mosfet was on a heatsink and i have a replacement mosfet aswell. The coil resistance is 2.4 ohms. What could be my problem? Thanks for any help guys. I can upload pics if necessary.

MOTVolt

Mon Jun 01 2009, 03:39AM

Registered Member #1996 Joined: Wed Feb 25 2009, 03:56AM
Location: Blackburn South, Melbourne, Australia
Posts: 10

Hmm, from what I can see the is circuit is very crude. There needs for be some protection for the MOSFET. Revearse voltage from the coil could have killed the FET. use the diode test function on a multimeter and see if the FET will only conduct one way between the D and S. I would replace the FET and add fast rectifier diodes to protect the rest of the circuit!

1243827498 1996 FT70387 Better Iggie Driver

Proud Mary

Mon Jun 01 2009, 05:44AM

Registered Member #543 Joined: Tue Feb 20 2007, 04:26PM
Location: UK
Posts: 4992

The source impedance of even an old-ish lead-acid accumulator ( what is sometimes called the battery resistance, or the internal resistance of the battery) is very low compared with your 'weak transformer' of one or two amps.

This means that the car battery can produce on demand tens or even hundreds of amps without the output voltage falling flat.

One amp from your power supply is really not enough to run this circuit without the voltage falling.

Does your circuit work properly when you switch back to using the car battery? If so, no other explanation need be looked for.

If it will no longer produce good sparks when connected back to the car battery, then something has changed (broken!

) and must be repaired.

Dr. Dark Current

Mon Jun 01 2009, 09:49AM

Registered Member #152 Joined: Sun Feb 12 2006, 03:36PM
Location: Czech Rep.
Posts: 3384

MOTVolt wrote ...

Hmm, from what I can see the is circuit is very crude. There needs for be some protection for the MOSFET. Revearse voltage from the coil could have killed the FET. use the diode test function on a multimeter and see if the FET will only conduct one way between the D and S. I would replace the FET and add fast rectifier diodes to protect the rest of the circuit!

You know that the inductive voltage kick is what produces the HV output from the coil? If its too high, the FET will catch it and just get hot, but it wont die from that.

Arcstarter

Mon Jun 01 2009, 03:25PM

Registered Member #1225 Joined: Sat Jan 12 2008, 01:24AM
Location: Beaumont, Texas, USA
Posts: 2253

Well, that circuit with a diode and resistors is not a snubber. It is made for slow turn-on and fast turn-off. As Dr. Kilovolt said, you need a spike for the HV output. When the magnetic field collapses once the pulse turns off, the spike on the secondary is what gives the high voltage output, and the primary picks some of this up, but not as high of voltage. That circuit might improve things though.

Most mosfets have an internal anti-parallel freewheeling diode, so reverse voltage should not be too big a problem if this is the case. Also as Dr. Kilovolt stated, if the voltage that is seen across the mosfet is too high, it will hopefully just turn that into heat and keep operating. Not to say that it will withstand a MOT on the gate, but every once in awhile a HV spike *should* be ok, I'd think.

uzzors2k

Mon Jun 01 2009, 03:49PM

Registered Member #95 Joined: Thu Feb 09 2006, 04:57PM
Location: Norway
Posts: 1308

The IRFP450 is avalanche rated, which means it will act as a zener diode if the drain voltage exceeds the breakdown voltage of the device (500V in this case). This allows it to dissipate a small amount of excess energy regularly, the result of which is heat. As long as the avalanche current is limited, which it is by the primary inductance, and the junction temperature is kept under control it can be used to clamp the drain voltage.

1 up on Harry's suggestion, this circuit pulls ridiculous amounts of current and a wall wart doesn't have near the "stiffness" required to power it.

Cypriotica

Thu Jun 04 2009, 09:41PM

Registered Member #1811 Joined: Sat Nov 15 2008, 11:52AM
Location:
Posts: 14

Thank you for all the information guys. I have saved this and will do troubleshooting after my exams are over. May the sparks arc big ^^

Andri

Thu Jun 04 2009, 11:14PM

Registered Member #1533 Joined: Wed Jun 11 2008, 02:13PM
Location: ReykjavÃk, Iceland
Posts: 46

The transistor needn't take all this abuse. Simply placing a freewheeling diode (a big one) over the coil would kill any HV spikes. What MOTVolt suggests won't help. Diodes in series are just as likely to break down as the FET (or even likelier since they mightn't be avalanche rated).

When using a weak transformer, adding a big capacitor between the terminals of the transformer might improve its "stiffness" for short pulses. That way you might be able to get greater pulsed current but the average current will remain the same.

Personally I'd use a push/pull circuit to drive an ignition coil.

Update: Of course a freewheeling diode over the coil is just gonna remove all the energy stored in the system and will thus not do any good.

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