Twin injector system?

[jdzura at csc.com]

Remember power is also calculated by current squared times the resistance. This
would apply to the impedence of the coil also.






S.Baly at bom.gov.au on 07/02/98 02:23:58 PM

Please respond to diy_efi at efi332.eng.ohio-state.edu

To:   diy_efi at esl.eng.ohio-state.edu
cc:    (bcc: Joe Dzura/SED/CSC)
Subject:  Re: Twin injector system?




Sandy wrote:
>I have though of just raising the voltage too, but would expect a couple of
>problem, meltdown and destruction of the injector as one ;-). The second,
>is that it might open faster, but not close faster, also I have now to dump
>a lot more stored energy back into the electronics so that may require some
>changes in the protection and related circutry.  I don't really know the
>insides of the injector, if the fuel pressure is used to help close the
>pintle (I think thats what its called), then a combination of increased
>fuel pressure and higher peak voltage might do the trick. An interesting
>thought, would be to take a 24 vdc powersource, and run an injector
>(saturated style) with a 2/1 peak and hold current and see what kind of
>minimum cycle time you can get, and then raise the pressure. I would expect
>that this is all a wash, as with the added pressure the pintle will just
>stick open, worth a ramble however.
I don't think meltdown would be a problem, since power dissipation is a
function of the coil current, not coil voltage. The idea is to drive the coil
hard using the high voltage to ramp the current up quickly, then cut it back to
a lower current, just as a peak and hold driver does. The end current is the
same as using 12V, it just gets there quicker.
The extra voltage will put some stress on the coil insulation, but no more than
what happens when the coil closes.
The stored energy will be the same, because the coil current at the end of the
cycle will be the same as with a 12V system. Garfield said that injectors kick
back like a mutha. All this idea does is to kick the muthas hard first. :-)
Stuart.