From the Author:
This material is only an attempt to explain the ongoing processes and does not claim to be “the ultimate truth”.
The photo below shows the oscillogram of the operation of the nozzle of the GDI system:
In a first approximation and analysis of circuitry, I figured out a circuit that could implement this
(i.e. get this result on the oscilloscope screen).
And this is a bridge amplifier circuit.
It consists of 2 pairs of transistors (they can be field, but most likely IGBT transistors, bipolar are drawn for simplicity. A 12 volt source, connected via a diode, and a 100 volt converter as a separate unit charging the storage capacitor C1 are conventionally indicated in the circuit. Nozzle winding included in the amplifier bridge as a load.
How can it work
At the initial moment, the capacitor is charged to a voltage of 100 volts from the converter 12 volts to 100 volts.
All transistors are closed, the control voltage at the bases is zero.
The initial peak of 100 volts is when the voltage C 1 is applied to the nozzle winding L 1 .
This happens when VT 1 – VT 2 is opened at the same time. The current flows through the red line, measured at point A. Section a-b is the discharge of the capacitor through the winding. Given that the inductance is very small, and the resistance of the winding is less than 2 Ohms, the current reaches a significant value and this leads to an instant rise of the needle. Why instantaneous – because, judging by the waveform, the entire duration from a to c (from burst to burst) is less than 0.7 ms, the engine clearly runs on xx. when the opening time is about 0.6 ms. If you take this into account, then everything is correct and the real time of opening the nozzle is about 0.6 ms. Taking into account that at such a pressure there is a sufficiently powerful return spring in this nozzle, the closing pulse can be made with a smaller amplitude (summing with the spring force), therefore, when VT 3 – VT 4 is opened, the current in the winding is reversed (blue line) and the nozzle quickly closes . Locking is obviously necessary, since high pressure in the rail is constant and it is impossible to lock the nozzle by depressurization (as in diesel engines with mechanical nozzles), at the same time, pressure from the liquid flowing out of the nozzle presses on the needle from below, therefore, the locking mode is entered. In such a switching circuit, the locking mode is elementary feasible as reversing the direction of the current through the winding.
Now about the waveform.
The first surge of 100 volts is the initial moment of opening VT 1 – VT 2 and the subsequent decline – discharge of the capacitor .
The residual voltage of 12 volts is formed as holding by the control circuit or converter (it is not known exactly but not important) most likely the control .
Since the machine still works with a faulty 100 volt converter, albeit in emergency mode. Therefore, I indicated an external source of 12 volts, which is isolated by a VD 1 diode from a 100 volt converter. Section b – B is the retention section. the first surge B is the switching of the switch that controls the converter (obviously, the converter must be disconnected from the load to be able to recharge the capacitor. In this case, during switching VT 1 – VT 2, as well as power control circuits, the holding voltage is removed from the winding, VT 1 – VT 2 are closed , which leads to an inductive emission B – C. It is seen that the voltage at this moment tends to zero. At moment C, VT 3-VT 4 opens, while the converter has already charged C 1, a sharp change in current in the winding nozzles and the needle under the action of the return spring and electromagnetic force close sharply, cutting off the fuel supply.
The moment of stress relief B – C, the fuel continues to flow out of their nozzles, possibly due to the high pressure on the nozzle seat of the atomizer, the residual magnetization of the core and inertia. A pulse of a smaller amplitude during closure may be associated with a short time (insufficient) for a full charge of C 1 to 100 volts, and may be dictated by the need for a lower current, since the nozzle return spring helps in closure.
The section after C is the aperiodic discharge of the capacitor through the inductance of the winding. Since the transistors are open, the quality factor of the circuit is low and self-oscillating processes (emissions) do not occur. The capacitor is discharged to zero, so switching VT 3-VT 4 is not visible at this level. This once again confirms that the surge B is due to interruption of the current through the winding due to the closure of VT 1-VT 2 (as in the ignition system). Perhaps in the amplifier there are damping diodes to limit inductive emissions, parallel to the transitions of K-e transistors, which leads to one half-period of limitation.
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