1. Heating Coil Calculation

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I have a 12V battery connected to a circuit driving an ignition coil. Carry weight console command fallout 4. The circuit requires 2A to run. The ignition coil produces 40,000V and the spark is measured at 50mA. By simple calculations the input power is 24W the output power is 2000W, how?

Analyze your electromagnet to determine its dimensions and the amount of current you will be running through it. For example, imagine you have a magnet with 1,000 turns and a cross-sectional area of 0.5 neters that you will operate with 10 amperes of current, 1.5 meters from a piece of metal. How to Calculate the Henrys in a Coil. Either 12V across a 3-ohm primary coil or 6-9V across 1.5 ohms in a ballast coil. Either way, somewhere in the 4-6 amp range while the car is running. If you have a ballast coil (e.g. Runs on 6 or 9V) that bypasses the ballast resistor when starting the engine, you could get higher amps (12V over 1.5ohms = 8Amps).

If I could harness the output power and my battery is 200Wh does this mean I can run a 2000W device for 8.3 hours, as the input is only 24W from the battery?

This does not seem right, what am I missing?

Or am I actually drawing 2000W power from the battery (over time) to charge the capacitors in the circuit etc, meaning that it will only last 6 minutes?

Thanks for the help.

Heating Coil Calculation

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2 Answers

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You need to read up on spark ignition systems and could perhaps start with something like this.

You are correct in thinking that the output power is in kW, but unfortunately you left out an important element ..time. Spark gap time is in the 1 - 2 ms range for most systems depending on inductor values and total series resistance for the circuit.

Most Kettering systems have input currents in the 10 - 20 A range (I've not seen current in the 2 A range) at 12 V and produce about 40 kV in the secondary.

The currents that flow in the secondary (in the spark gap) can also very high (in the hundreds of mA). Initial breakover currents are typically in the 3 - 5 mA range, and increase to the 100 - 200 mA range as the gap voltage rapidly decreases.

In older systems with long spark plug leads the lead resistance may be several 10's kOhms and along with the higher resistance of the secondary coils this limits the current that will flow. With these systems spark line current may be design limited (by the series resistance) to 200 - 500 mA.

The power dissipated in the spark gap is limited both in time and voltage/current. At breakover (zero current) the voltage at the gap will rapidly rise to 40 kV, but once the plasma is created the instantaneous voltage will drop to only a few hundred volts. In older systems with a distributor gap, you had to create two plasma connections, one for the distributer and one in the cylinder spark plug. Losses abounded.

Modern systems typically use a coil-on-plug design which get's rid of one of the spark gaps (double the delivered spark energy), reduces series resistance and increases the plasma current (much 'hotter' sparks). Spark line current in these is still limited by design, typically to under 200 mA.

Read the document I linked to, it'll give you a good idea on modern systems.

HearthJack CreaseyJack Creasey$endgroup$$begingroup$

Power (WATTS) is often miss-understood. Power is defined as a moment of time thing. e.g. when on, a 100W light-bulb is using 100W all the time. A more real measure is power * time.. and you end up with the standard you use kilowatt-hours you see on your electric bill. So a lightbulb uses 100w x 1Hour = 0.1KW.Hrs of electricity in an hour.

The image above shows a simplified version of what your coil does.

When your engine is running your coil is drawing 2A from the battery MOST of the time through the shown switch which is normally closed when the ignition switch is on. Your coil will draw 2A, so it is running at 24W most of the time.

When the spark is required, the a 'switch' is opened. Since the coil is just a large inductor, the current still wants to flow through it so the voltage at the switch and attacked spark plug suddenly jumps to thousands of volts. A voltage high enough to produce a hot enough spark across the gap at the business end of spark-plug.

This happens in a fraction of a second. Most of the energy that is effectively stored in the coil is released in that very small time. Because the energy is released all at once, the Watts is very high for that very short period.

The switch closes after and the spark is generated and the coil is recharged with energy from the battery.

Because the spark time is small, on average the coil uses 2A, or 24W continuously.

Or perhaps more importantly 0.024kW.hrs.

You can think of it as the tipping bucket principle..

With the switched closed energy is poured into the bucket at a slow rate. When the switch opens the bucket tips over and releases all it's energy at once.

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