Bri Bro

Here is how I see it.
All the injectors are driven by a single Darlington transistor and are connected in parallel. There are two wires going to the injectors to reduce voltage drop. The injector drive circuit turns on hard to set the injector solenoid, then it oscillates at a lower voltage to hold the injector open. This means the average current is less the DC calculations, but you always will see the max current when the injector first turns on.

Resistance:
Stock == 4.5 ohms / 4 =1.225 Ohm
Other == 2.5 ohms / 4 =0.625 Ohms

Current:
(Voltage across injector is 10V due to driver loss)
Stock == 10 / 1.225 = 8.16 Amps
Other == 10 / 0.625 = 16.0 Amps

Now my cross to the driver transistor shows a max current rating of 16 amps. So, you are at the upper limit went you first turn on the injector. There also is a de-rating that should be applied for the heat generated by this transistor. At 16 amps, there is no more room for this de-rating.

So to Danno point, if you run it hard, put a lot of heat into this transistor, it will be running above it max rating. If your transistor came from a good batch of transistors, they may take this. If you get a normal batch, they may not.

RobNL

beab951,

Are all 4 injectors in parallel? I thought that the injectors are in parallel of 2 each. There are two wires that come out the DME or are they inside the DME connected in parallel to a single transistor?

Based the calculations and the fact that two wires need a resistor would mean that two 1 Ohm ballast resistors will do?

Bri Bro

Yes, the two wire are connected together.
Yes...but watch your wattage rating...it's a lot.

Lorenfb

Beab951 is correct. There is only one injector
driver for all four injectors. The driver that
was mentioned is also right-on, 16 amps peak.

No one mentioned that it is a Darrlington type
with a max. voltage of 100BVceo. I use this
type in my rebuilts from SGS Semiconductors.

Nice to know that you guys know your stuff.

Good luck
Loren
Systemsc.com

Lorenfb

Followup to my last post:

The power rating for the ballast/added resistors
was not calculated. The following should be
helpful;

Power=IxIxR X Duty Cycle (approx. RMS value)

For I=5 amps for each two injectors, the power
for a 1 ohm resistor is:

Power=25 x 75% (D.C.)= 18.75 watts
(Duty cycle=75% assumes RPM>4000 and under accel
or heavy load.)
Use a 25 watt for derated application.

I didn't mention that two other approaches other
than changing the injectors would be to;

1. increase the fuel pressure
2. adjust the AFM spring for a richer mixture
(This could also increase throttle response
if done properly.)

Good luck
Loren
Systemsc.com