How are higher voltages achieved in car ignition systems?

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Car ignition systems achieve the required high voltages of 5-17kV through a transformer with a winding ratio of 1:100, which steps up a 12V battery to 1200V. However, the transformer operates differently than typical AC systems; it uses interrupted DC to generate high voltage. The output voltage is determined by the rate of change of current (di/dt) in the primary coil, meaning that faster interruptions in the primary current lead to higher voltage pulses. This principle allows for the necessary voltage to create a spark at the spark plug. Understanding these dynamics is crucial for grasping how ignition systems function effectively.
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Homework Statement


Hey there! I have simple problem:
To start a spark in a car ignition system, voltages of 5-17kV are required. A transformer with winding ratio 1:100 is used with a 12V battery, so we get a 1200V voltage. How are higher voltages achieved in such systems?
Thanks
 
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Noirro said:

Homework Statement


Hey there! I have simple problem:
To start a spark in a car ignition system, voltages of 5-17kV are required. A transformer with winding ratio 1:100 is used with a 12V battery, so we get a 1200V voltage. How are higher voltages achieved in such systems?
Thanks
The transformer turns ratio gives the voltage ratio only for sinewaves, i.e., AC of appropriate frequency. The coil in a car is designed to operate differently. It is supplied with interrupted DC and the high voltage output is defined by the more general induction characteristic

voltage = k. di/dt

The faster the points interrupt the primary current, the greater is di/dt, hence a greater voltage pulse goes to the spark plug.

http://imageshack.us/scaled/landing/109/holly1756.gif
 
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