On Sun, Dec 05, 2010 at 05:06:16PM -0500, Dave wrote:
> That's simple!  :-)

Dave, please be aware that the first circuit at:

http://www.rmcybernetics.com/projects/DIY_Devices/homemade_ignition_coil_driver.htm

is seriously oversimplified.

If a 2N3055 is used, as shown, the 100v collector breakdown voltage
(BVcbo) won't get the best out of an ignition coil. It is common for the
primary back-emf to be 400v in a kettering system, so limiting it to
100v is only 25% of what's usual. (Something like a BUW76 (350v),
2SC2333 (500v), or 2SC3153 (900v), would be better.)

Relying on the snubber (RC1) to protect a low voltage transistor like
the 2N3055 is very risky, because the back-emf from an arbitrary
ignition coil will be unknown. Even if using a BUW76, I'd connect a
transient suppression diode across its collector and emitter. (Somthing
like a 1.5KE250, or 300v if you're game.)

If a MOSFET is substituted for the 2N3055, as suggested there, it will
not be turned off, because there is no pull-down on the emitter of T1. A
low resistor value would be needed to provide a high di/dt in the coil,
to generate a high flyback voltage. Better would be to reverse the
drive, with a pull-up resistor, and the transistor pulling the gate low
through a small gate resistor. A decent gate driver circuit needs a few
more components.

If not confident of high voltage design skills, I'd google for an
electronic ignition circuit or two, to see how they protect their power
transistor. (OK, IIRC, many of them use a small inverter to generate 400
Vdc, then use something like an SCR to dump into the ignition coil
primary. Still, there might be some which give good ideas.)

A flywheel diode is enough protection when fitted across an inductive
load such as a relay coil, but that would defeat the generation of high
voltage in this case, because a high di/dt would not be achieved.

To obtain something which works from that site, it appears necessary to
buy. They say "We sell an ignition coil driver module which has built in
protection against most spikes that would damage a driver." Implicit in
that is an admission that the simplistic circuits displayed do not do
that. (At least, AIUI.)

That's just some thoughts, which might help save some wasted time. ;-)

Erik

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