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fuel system resistors

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Old 12-09-09, 10:12 AM
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fuel system resistors

i have four 3.1ohms resistors
2,1600cc injectors with 4.5ohms
2, 850cc injectors with 2.3ohms

are these 3.1ohms resistors the right ones for my injectors?
Old 12-09-09, 11:32 AM
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anyone?
Old 12-09-09, 01:32 PM
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The short answer to your question is no, you will need higher resistance on the primaries or a peak-and-hold injector driver.

According to the FC Datalogit manual, the PFC can handle a maximum of 3 watts for all the injector drivers, although you probably want to run less than that. With your setup we will need a separate calculation for each injector,

V=I*R (Ohm's law. in this case voltage equals the injector's current multiplied by resistance of the injector itself)

P=I^2*R (Joule's law. In this case, the injector's power/wattage equals the injector's current multiplied by itself, then multiplied by the resistance of the injector driver transistor)

Power consumption - Front Primary Injector

Step 1: V=I*R to find current across the injector.

let V = 14V, max battery voltage you will likely see
let R = 2.3ohms+3.1ohms = 5.4ohms, this is the resistance of the 850cc low impedence injector + the 3.1 ohm resistor

Now we solve for current across the injector: I=V/R , or I = 14 V / 5.4 ohm = 2.59 amps across the injector.

Step 2: P=I^2*R to find how much power this injector is drawing at 100% duty.

I=2.59amps , from step 1
R=.25ohms . This is actually the resistance across the injector driver itself. Where did this come from? It came from the FC Datalogit manual, the section about running 6 injectors.

so P= (2.59amps)^2 * .25 ohms = 1.67 watts for one primary injector

1.67 watts for the front primary
1.67 watts for the rear primary

you're already over the 3 watts the injector drivers can handle and we haven't even started calculating power usage for the secondary injectors.

If it were me I would either

1) get a peak-hold injector driver
2) run the 3ohms on the secondary and run 10 ohms on the primary. The 10 ohm resistors would put them near the stock levels but would slow injector opening. I think injector opening is more important on the secondarys than the primarys. I did a similar calculation in that other thread. You would end up with

Front primary (total 12.3 ohm w/ resistor): .34 watts
Rear primary (total 12.3 ohm): .34 watts
Front secondary total: 1 watt
Rear secondary: 1 watt

that's at 100% duty, but you wouldn't be running that high
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