ok design?

To: "'Tesla List'" <tesla-at-pupman-dot-com>
Subject: ok design?
From: Tesla List <tesla-at-stic-dot-net>
Date: Thu, 5 Mar 1998 22:41:36 -0600
Approved: tesla-at-stic-dot-net

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From:  Malcolm Watts [SMTP:MALCOLM-at-directorate.wnp.ac.nz]
Sent:  Thursday, March 05, 1998 10:04 PM
To:  Tesla List
Subject:  Re: ok design?

All,
     Using knowledge of switchmode power supply operation:

> From:  Bill Lemieux [SMTP:gomez-at-netherworld-dot-com]
> Sent:  Thursday, March 05, 1998 2:06 PM
> To:  Tesla List
> Subject:  Re: ok design?
> 
> > > I'm trying to build my first coil...this is the design I plan to use;
> > > powersupply: auto ignition coil powered by a motorcycle battery, I estimate
> > > the auto coil to be about 45 watts, 30kv 1.5 ma. frequency: adjustable from
> > > 1 Hz- 2khz
> > 
> > I'll believe the 45 watts when I see (measure) it! I've never seen an
> > automabile ignition coil yield over 25 watts!
> 
<following reformatted>
> For an ignition coild to draw 45 watts would only require about
> 3.5 amps at the 13 volts or so it is designed to run on.  Remember,
> an ignition coil has a very low resistance primary and will
> cheerfully draw many times the design current if it were not for the
> ballast resistor.  On the other hand, it will also cheerfully burn
> itself out if you try to run it for long at a high duty cycle
> without the ballast resistor.

First thing: the current does not go to 3.5Amps immediately. It ramps 
up after the voltage is applied according to i = V x t/L assuming 
resistance to be negligible.

Second thing: Duty cycle is less than 100% as power has to 
be disconnected for the coil to dump its load.

Suppose duty cycle = 100%. To put through 45W, mean current = 3.46 A
which means peak current must hit close to 7 Amps. A duty cycle 
reduction of 10% means Ipk must get to 7/0.9 Amps = 7.8A

Last thing: for the current to get to 7.8A, winding resistance must 
be less than 13/7.8 Ohms = 1.67 Ohms roughly. But if it approaches 
this value, an even longer period would be required as an RL time 
constant is also factored in. Then it is a matter of knowing L to 
know how long V must be applied for the current to hit the required 
value.

One can work through it in other ways too. If L is known, you can use
E/cycle = W x t   where E = 0.5L x Ipk^2  and t = time per cycle

Malcolm