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It was my pleasure... you are welcome always... wish you good luck.
Added:
Just to be sure, I like to point out the following points concerning the circuit of interest here:
(1)
In the steady state, the output power device and the ignition transformer are on by a rather high current. In case the input switch fails to continuously trigger the circuit, the power (heat) dissipation would be considerable:
P_total = Vcc * I_coil
(2)
The voltage range of the supply (the car battery) is rather wide. But there is no current regulation at the output if a constant optimum spark is desired.
(3)
The pulse duration varies with the supply voltage. It is better to check if this variation is acceptable or not for the whole system.
(4)
Practically, we consider the range 10V to 15V for the car battery. If its voltage goes above 15V, it means that the car charger is both overcharging it and the charging current is relatively very high at the end of charge. When the battery voltage is below 10V, one of its 6 cells (or more) is damaged or it is almost empty, keeping it running will likely damage one or more cells (hence cannot be recharged properly). This circuit should monitor the low battery voltage in the least, in order to stop working if it is lower than 10V by assuming the voltage drop on the wirings is negligible or the sensor voltage is taken directly at the battery two terminals (actually one terminal since the other one is common via the car chassis). But the circuit should be made to work properly from 8V for example (not 10V), because the same wire usually drives both the control unit and its heavy load. Therefore a voltage drop on the supply wires could be relatively high (up to 2V for example).
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