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Old 04-16-2008, 08:39 AM   #4 (permalink)
needs more cowbell
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Thx John. I need to talk through this circuit a bit.

for the injector:
when it is closed, there will be voltage applied to the emitter of about 1 volt, which should turn on the transistor and pull pin 2 HIGH.

when the injector is open, .7 volts is applied to the 100k resistor. The voltage at the emitter should be even less (like 0.00009 V) given the resistors, so the transistor is basically off, and pin 2 is pulled LOW.

So what about that diode? When the injector ground circuit opens (closing the injector), the injector coil creates, or induces, a large spike in voltage in the opposite direction (positive spike on the ground side of the injector). So this spike will be attenuated by the 100k resistor and dropped another .7 volts by the diode, before (hopefully) being absorbed by the 5v voltage regulator. But the transistor should have already been off so it should not trigger additional interrupts on pin 2. Of some concern however is the oscillations that occur after the injector circuit is opened and if they will trigger false interrupts on pin 2:

Note: maybe the software can reject pulses that are "obviously" too short and/or disable interrupts on pin 2 for a brief period after getting a good looking pulse.

Last edited by dcb; 04-16-2008 at 08:46 AM..
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