GM ECM injector driver capacity

[Greg Hermann]

Got my foot lined up in the sights here! Ready, aim,

Seems to me that you need to consider the total energy stored in the mag
field of each injector (which is what feeds the flyback pulse). Since the
injector is perfect, lets say that it amounts to .01 joule. If you double
the number of injectors, you now have .02 joules of flyback to dissipate
when they snap closed. If the resistor through which said energy is
dissipated stays the same, either the same current has to run for twice as
long at the same voltage, or the voltage has to rise some (along with the
surrent) in order to get the total (doubled) amount of energy available
dissipated in the same length of time.

Seems to me that if you put a second resistor of equal value and wattage to
the first one in parallel with it, Voltage would remain the same, and same
flyback current as before (for each injector) (more or less) would flow
through each resistor, and time to discharge flyback current would remain
as before, and power dissipation rate for each resistor would remain as it
was for the single one before.

We are still left with the question of  whether a low impedance injector
stores more energy in its mag field than a high impedance one does, and if
so, how and how fast we choose to get rid of it.  Come on, someone elso
must remember more first year stuff than I do--the parallels between fluids
and electricity are failing me here! I don't really even remember whether
it was first year, second year, or what course. (Belated Uurrrrp.)

Regards, Greg

>I don't know about the nubers, but it relates to the heat dissipation of
>the flyback diode. Look at the LM1949 datasheet, it has some rough
>calculations for power dissipation of that. Like I said, might be fine, but
>you now have 9 times the current going back into that diode, and although
>the numbers seem low, I would imagine that it will more 9 times more heat
>as you are dumping much more back emf them one injector. The real question
>is, since they were peak and hold, and the impedance of the injector is
>much lower, thus drawing lots more current the EMF from that type of
>injector will be much higher, how much, I doubt 9 times, but might just
>work fine. The bigger problem _again_ is that if the injector is a peak and
>hold type, and it goes into hold mode will you be driving 10 injectors with
>1/4 of their required current. 10 saturated injectors = 10 amps in parallel.
>
>Anyone that knows electronics jump in at any time ;-)
>
>Sandy
>
>At 12:03 PM 8/25/98 -0400, you wrote:
>>OK, I have one injector, and it is perfectly designed, and does the
>>following
>>
>>14.2 ohms
>>runs on 14.2 v
>>draws 1 A
>>
>>when turned off generates
>>300v
>>.001A
>>.001msec.
>>
>>When I hook up 9 more, wouldn't I expect to see, at turn off
>>300v
>>.01A
>>.001msec
>>
>>Why would more produce more voltage rather than current,
>>or anything more than more current?.
>>TIA
>>Bruce
>>