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As far as I understand, if I'm not wrong there is capacitor discharge ignition and inductive discharge ignition.

Is the difference between two in one of them the current spike is drawn from battery and the other from the capacitor? And which system is used in modern automobiles?

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Your reasoning is that you know a capacitor is like a tiny battery, and it stores energy... But that an inductor is just wires and can't possibly store energy.

Actually, it can. It is storing it as a magnetic field instead of an electric field.

Just as capacitors resist changes in voltage and will flow as much current as needed to correct that...

... Coils resist changes in current and will flow as much voltage as needed to correct that.

This is why relays have a lower rating for inductive/ballast loads than they do for resistive loads.

So if you're looking for a very high voltage, this is a great setup. Interrupt current flow through the primary coil. The collapsing magnetic field will now drive up voltage on both primary coil and secondary coil, infinitely until current flows, i.e. Until insulation fails somewhere, allowing the current to shunt itself. If you set things up right, that happens at the spark plug.

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  • \$\begingroup\$ Will the peak current be driven from the inductor/coil or battery? \$\endgroup\$ – pnatk Dec 2 '19 at 13:58
  • \$\begingroup\$ By the size of the inductor coil. An inductor's whole thing is it wants to keep current constant, or to be more precise, magnetic field. \$\endgroup\$ – Harper - Reinstate Monica Dec 2 '19 at 14:14
  • \$\begingroup\$ Capacitors don't store energy in capacitance any more than inductors store energy in inductance. The energy is in the electric field, just like inductors put it in the magnetic field. \$\endgroup\$ – Hearth Dec 2 '19 at 14:28
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There is a piece of the puzzle you seem to be missing. All ignition systems use an ignition coil. The coil serves as a transformer, stepping up low lovage to high voltage. Transformers do not work with DC so there is some trickery involved. To a transformer, a rapid change in current looks just like AC. (I'm simplifying here) The trick is in creating these rapid current changes. There are two types of ignitions Accelerative and Decelerative.

Accelerative use the proverbial capacitor. They use a boost converter to generate somewhere in the 100V to 300V range. Then dump that voltage into the primary winding of the coil. This generates a sudden rush of current and generates a high voltage in the secondary windings of the coil.

Decelerative ignitions set up the current flow in the primary of the coil ahead of time. Then when they are ready to fire they suddenly stop the current. This abrupt stop in current works just like a sudden rush. The inductance of the primary winding wants to keep the current flowing and generates a high voltage spike. This intern generates a high voltage in the secondary windings.

Modern vehicles tend to use the Decelerative type ignition because it is cheaper. Also, note that the ignition coil configuration is irrelevant to either system.

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  • \$\begingroup\$ Thanks I see. In both systems the inrush current during the spark/ignition is driven from the capacitor(Accelerative ) or the coil(Decelerative) but not the battery itself, is that correct? Im wondering during the current discharge whether the battery involved or not. \$\endgroup\$ – pnatk Dec 2 '19 at 15:52
  • \$\begingroup\$ @panicattack The battery is not directly involved with either one. With decelerative, the current is built in the coil gradually and often the current is restricted by a resistor in older cars or the ignition system itself in the newer cars. With accelerative the charge in the capacitor is built gradually. Then the capacitor is connected across the coil to fire. \$\endgroup\$ – vini_i Dec 3 '19 at 16:43
  • \$\begingroup\$ But in simulation huge amount of current is drawn from the battery \$\endgroup\$ – pnatk Dec 4 '19 at 9:15
  • \$\begingroup\$ @panicattack How are you simulating it? What is the circuit? \$\endgroup\$ – vini_i Dec 4 '19 at 12:22
  • \$\begingroup\$ Please see my edit. As you see both the primary conductor and the battery discharges the same sudden current. You wrote "The battery is not directly involved". Don't you think it is directly involved unlike in capacitor discharge case? \$\endgroup\$ – pnatk Dec 6 '19 at 13:14

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