Re: Pulse Width for Ignition Coil

["Tesla list" <tesla-at-pupman-dot-com>]

Original poster: "Jim Lux by way of Terry Fritz <twftesla-at-uswest-dot-net>" <jimlux-at-earthlink-dot-net>

You got it... except watch out for the fact that the voltage across the
inductor may not be the supply voltage (IR drops in the rest of the
circuit, for instance).

Tesla list wrote:
> 
> Original poster: "Finn Hammer by way of Terry Fritz
<twftesla-at-uswest-dot-net>" <f-hammer-at-post5.tele.dk>
> 
> Tesla list wrote:
> >
> > Original poster: "Jim Lux by way of Terry Fritz <twftesla-at-uswest-dot-net>"
> <jimlux-at-earthlink-dot-net>
> 
> Snip..
> >
> > Basically, you want to limit the current to around 5A. Knowing the nominal
> > inductance (around 8mH), you can figure out how long it will take to get
> > there (Ldi/dt = V... 12V = 8E-3 * di/dt.. 1500Amp/second, 5A is 1/300
second
> > (3.33 msec))
> 
> snip
> 
> Oh Boy, would that be the formula that I`ve been looking after for so
> long, the one that describes the charging inductor`s behaviour in a DC
> system, during gap conduction? So that it is possible to determine how
> high a current that will flow in the gap, by the time the ringdown is
> completed, and it is time to quench?
> 
> Let`s see if I`ve got it right:
> 
> Voltage across inductor = 20000V
> Inductor size           = 5H
> 
> 20000/5=di/dt ........  =4000 Amp/Second
>  after a ringdown time of 350µS, the current will rize to
> (4000*350)/1000000 = 1.4 amp. So if the gap can quench at 1.4A, then the
> charging inductor has a good size.
> 
> Is this right?
> 
> Cheers, Finn Hammer