A repost

[Bruce Bowling]

> 
> Looking at it from a purely electrical point of view:
> 
> Since the current supply to the injector when directly hooked to a 
> 12V supply is about 5A (assuming 2.4Ohm injectors), a fifty percent 
> duty cycle gives 2.5A which is enough to open them very hard. Perhaps 
> that will bounce the pintle off its stop, giving the appropriate 
> vibrations.

I was looking at the above in a time-domain manner. If it takes 1 ms
to move full range, and I apply only .1 ms pulse, the pintle will move
only 10% open (assuming the opening process is linear). If opening 
time is close to the closing time, then if the injector is left alone
(i.e. no current) for .1 ms, then the pintle comes back to rest, the same
position as t=0. Then repeat the above sequence. The solenoid applies 
a force to the pintle, the electrical effects are just a means
of applying this force. There may be some screwey solenoid-related charging/
discharging effects - I am not accounting for them in this simplistic
model.

If opening time is shorter than closing time, then the pintle will
not fall all the way back to the resting point, and this will be the
new starting point for the next pulse. This will cause the pintle
to "walk" towards an open position as time marches on.

If opening time is longer than closing time, then the pintle is assured to
return to the resting point before the next pulse hits.

For those who want to be safe, then use the 30% duty cycle, and this
should ensure the pintle moves back home each pulse (I assume).

- Bruce

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                 Bruce A. Bowling
                 Staff Scientist
   Thomas Jefferson National Accelerator Facility
    12000 Jefferson Ave - Newport News, VA 23606
                 (804) 249-7240
bowling at cebaf.gov  http://devserve.cebaf.gov/~bowling
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