Injector Driver Module
mrvette
mrvette at bellsouth.net
Mon Aug 10 14:19:45 GMT 1998
IF I read your analysis correctly....you forgot that is a 4 cycle engine...so
the on time is only ~40% on a real time basis....NO? GENE
Jose Rodriguez 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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