Injector Driver Module

[peter paul fenske]

Howday

Try running the injectors in series parallel arrangement.

later:peter

At 09:12 AM 8/6/98 -0700, you wrote:
>  I am designing an injector driver module. I would like your input, to see
if my calculations are correct, particularly in my assumption that all eight
injectors could be on at the same time, for up to 85% duty cycle. Thanks in
advance for your answers!
>
>  The current specs I have are:
>
>  Eight injectors.
>  Injector coil resistance: 1.5 ohms (it could also be 2.3 ohms)
>  Injector current: 4A peak, 1A hold.
>  Duty cycle: up to 80..85%
>  Operating ambient temp: -40C..125C
>  Operating voltage: 4.5V to 16V.
>
>  I did some calculations, to estimate how much power would I need to
dissipate, worst case, ASSUMING ALL EIGHT INJECTORS COULD BE ON, FOR 85% OF
THE TIME (is this assumption correct?)
>
>  First, I set Vbatt=14V. In order to have 4A circulating, the driver's
output voltage would be (14-4*1.5)=8V. Injector's coil voltage = (14-8) = 6V.
>  Instantaneous power at the driver: (8V)*(4A) = 32W. Instantaneous power
thru injector's coil: (6V)*(4A) = 24W.
>  The 4A will only circulate for a brief period of time, so average power
over one cycle will be much less.
>
>  One amp, sustaining, automatically means (14*1)=14W have to be dissipated
between the injector driver and the injector coil (100% duty cycle).
>  In order to have the 1A sustaining, the driver's output voltage would be
(14-1*1.5)=12.5V. Injector's voltage =(14-12.5) = 1.5V.
>  Inst. power at driver: (12.5V)*(1A) = 12.5W. Inst power at injector:
(1.5V)*(1A) = 1.5W. (total=14W, as predicted).
>
>  Thus, since the Duty Cycle can be up to 85%, and disregarding the initial
4A power peak, and the switching power losses, then the average power for
each injector driver is (12.5W)*(0.85) = 10.625W. The average power for each
injector coil is (1.5W)*(0.85)=1.275W.
>
>  If all 8 injectors could be on 85% of the time, then the maximum average
power at Vbatt=14V (again, disregarding some power losses) would be
(8)*(10.625W) = 85W at the injector driver side. Obviously, at 16V the power
dissipation would be higher (98.6W!).
>
>  As you can imagine, dissipating 98.6W+ into an under-the-hood ambient of
125C, with no expected air flow, is a formidable task. The heatsink's
surface temperature should not be allowed to go above 150C; that would mean
the thermal resistance from the heatsink to ambient would be (150-125)/98.6
= 0.25C/W. I can not see any feasible heatsink doing that, without the need
for forced cooling.
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