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Old 09-07-2010, 03:15 PM   #1 (permalink)
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Injection Pulse Sensing on 1973 Citroen DS

I have read the instructions for where to pick off the signal for the injection pulse duration, namely an injector feed wire that is low when the ignition is on but the engine not running.

The electronic injection system on my car is a very early one. Looking at the wiring diagram for my car, the injectors all have one feed wire directly connected to earth with the other being fed from the ECU. I believe the wire from the ECU will go high to open the injector and generate a pulse of fuel. If I follow the standard instructions for sensing the injector pulse, it seems to me this will be tied to earth and not show anything.

Do I need to do something different? Would sensing the positive feed wire work? Would I need to invert the sensed signal, e.g. through an op-amp, as my pulse will be injecting fuel when the sensed signal is high.

I apologize if this is a naive question.

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Old 09-07-2010, 03:47 PM   #2 (permalink)
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do you know what the signal to the injectors looks like? i.e. is it a constant 12 volts applied to open the injectors, or is it pwm and/or peak and hold? If it is constant 12 volts then you can just change the injtrig in the menu to tell it to look for a positive signal.
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Old 09-07-2010, 04:37 PM   #3 (permalink)
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Thank you for the very quick reply. I don't have the waveform, but it is described as 12V applied to open the injectors for a time dependent upon load, temp etc. The diagram I have shows the injection of fuel being timed to start at the beginning of the intake stroke and lasting for a relatively short time compared to the inlet stroke. The ECU uses purely analog electronics; it is an early Bosch system.

If the software allows me to select triggering off a positive edge, then that would seem to cope with my injectors being wired up the other way around and I won't need to put in an inverter. Do you agree I will need to tap off from the wire connected to the ECU and supplying the 12V pulses?

I really appreciate your help.
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Old 09-07-2010, 04:51 PM   #4 (permalink)
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There is a setting you can get to with the buttons to change if the high or low part of the injector signal means that the injectors are open. Can you measure the ohmage on one of those injectors?
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Old 09-07-2010, 05:03 PM   #5 (permalink)
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The nominal value of the injector resistance is 2.4 Ohms. Does this tell you anything?
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Old 09-07-2010, 05:09 PM   #6 (permalink)
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yah, if one injector is 2.4 ohms, you might have peak and hold. we may need to see the signal on a scope to come up with a solution.
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Old 09-07-2010, 05:20 PM   #7 (permalink)
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The injection manual describes the operation as:

"The duration of injection (fuel quantity) is governed
basically by two factors: engine speed and load condition
of the engine. The engine speed is relayed to the control
unit by the distributor contacts I and I I. The load condition
is determined by measuring the absolute pressure in the
inlet manifold. This pressure is converted to an electrical
impulse and relayed to the control unit by the pressure
sensor 0 , which is connected to the common inlet
duct 0 by a hose (A).
The control unit processes this information and gives a
signal for the injection valve to be open for a longer or
shorter period of time (F). The control unit thus permits,
by electrical means, a varying quantity of fuel to be passed
through the injection valves depending on engine load and
speed. This is how the “basic fuel quantity” is governed."

Can you tell from this whether it is peak and hold or saturation?
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Old 09-07-2010, 05:30 PM   #8 (permalink)
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I would not expect that verbiage to capture if it is peak and hold or not. Typically peak and hold injectors are 7 ohms or less (or therabouts), but this is so old I dunno. Maybe if you cross reference the injector part number.
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Old 09-07-2010, 06:23 PM   #9 (permalink)
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I have found this attached image of the waveform of the voltage across the injector in the D-Jetronic system fitted to my Citroen. The description of the waveform and why it has that shape is given below.

The description seems to say the drive transistors are simply switched on and it is the inductance of the injector that controls the waveform.

Do I need to do more that set injtrig?

"The nominal specification value of Rinj is 2.4 ohms. Measurements I've taken on one of my injectors show Linj to be 3.77 mH and Rinj 2.69 ohms. When the driver transistor turns on and current begins to flow through the injector, a magnetic field begins to build in the coil and starts to pull the injector open. Due to the energy required to develop the magnetic field, the inductive reactance (XL - the effective resistance of the inductor to a time-varying signal, measured in ohms) is initially very large. Because XL is initially the largest resistance in the series path of circuit elements (Rload, Rinj, and XLinj), almost all of the voltage is dropped across the injector - that's why right after the injector turns on you see +12.7 V across it.

The rate the magnetic field builds is governed by the time constant, t = Linj / R, where R = Rinj. Therefore, t = 0.00377 / 2.69 = 1.40 msec. As the field builds, XL decreases and less voltage is dropped across the injector - that's why the voltage in the waveform below decays from the peak of +12.7 V. It takes about 2.5 * t for the decay to be complete, about 3.5 msec. Note that this means that for injection pulses shorter than this duration, the magnetic field is still decaying when the injector is turned off. Once the magnetic field is at steady-state, the voltage drop across the inductor is due solely to I * Rinj, where I is the steady-state current. If you use the nominal spec value of R = 2.4 ohms and an assumed battery voltage of +12 V, you would get a sustained voltage drop across the injector of 3 V. When the car is running, the actual system voltage is more like 13.6 V. Accounting for the higher measured injector resistance over the specification value and using the system voltage of 13. 6 V, the actual measured sustained voltage is 4 to 5 V. Note that the calculated drive voltage corresponds exactly with data given for the flow rates of the injectors, which states that the tests were performed at 3 V (data courtesy of Roland Kunz).

At the end of the injection period, the driver transistor turns off, which rapidly changes the current drive to the injector to zero. The rapid collapse of the magnetic field in the coil now has an opposite effect - it induces a high negative voltage across the inductor. The mathematical expression of this voltage is VL = Linj * dI/dt , where dI/dt is the time derivative of current through the inductor. The net effect is that you get a large negative spike of voltage after the injector turns off. I have measured a peak of -27 V, decaying in about 1 msec to zero volts. The inductive voltage spike (and resultant current) is limited by the series combination of a 6.8 mF capacitor and load resistor in parallel with the injector. Note that since D-Jetronic is a grouped injection system, a paired set of injectors experience all of the events described above simultaneously. When the injectors turn off, the inductive voltage spike of both injectors is limited by the 6.8 mF capacitor and load resistor. "
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Old 09-07-2010, 06:50 PM   #10 (permalink)
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interesting. If this is what you are referring to, I *think* if you tap it on the "switch" side of an upper load resistor then you are in business, if that is doable


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