? Inductive Discharge Ignition System ?

[zawminoo]

Hi...

I would like to know about Inductive Discharge Ignition System.

I have known about Capacitive Discharge Ignition System.
Firstly, Battery supply 12V are transformed to about 400V for primary of ignition coil.
For this transform, we use transformer,rectifier,capacitier ect...
Secondary coil induce this primary voltage and produce high voltage(may be 40 kV).

How secondary coil produce high voltage in Inductive Discharge Ignition System?
It use only battery supply voltage(12V).Coil ratio may be only about 1:120.
Can anyone explain this by circuit and equation?
I really want to know it.

Zaw

 

[Buriedcode]

Hi,

I'll give this a go, but don't take my word for it, this is based on what I've read, seen, and a few assumptions.....

Right, in 'normal' ignition, the 12v from the car bettery is 'chopped' (basically turned on/off, to give it an AC component) this is fed into the ignition coil to produce anything from 20-40kV. Now, this is continually providing HV AC from the coil, and the distributor selects which sparkplug gets the HV.

In 'capacitive discharge ignition' exactly when each sparkplug fires, is down to the EMS (Engine management system). Essentially, it steps up the battery voltage to something like 330-500v DC. So now you have a 500v supply with a small current. This supply is used to charge capacitors, one cap per cylinder (sparkplug). When the EMS tells it to (by sending out a pulse) the cap is discharged through another transformer..(again, one transformer per cylinder)..to produce a very quick, sharp HV spike at 40-50kV. This has several advantages:

1. The timing of the sparks can be very accurately controlled, and modified depending on conditions. This can give constant revs when ticking over. Also, can change the performance of the engine while its running.

2. Because we are not simply switching a HV line to the sparkplug, but charging a capacitor over time, and then discharging it VERY quickly (1us?) The power in that spark is alot greater than in standard ignition....if you've done any chemistry you'll know that that more energy used to start a reaction (initiation energy), the quicker the reaction occurs....better fuel economy and more BHP

3. Also, because of the above....we are not producing 40kV constantly, and only using it when the distributor switches it to a sparkplug...instand we use the power form the battery as and when we need it. It uses less current, more efficient (depending on the design, some seem to use MORE power).

I hope this helps, I'm planning on buying an old car and modding it with this idea, just to see how it runs.

BuriedCode.

 

[zawminoo]

Hi,

Thank BuriedCode.
But I would like to know how Inductive type perform.

In Inductive Discharge Ignition System, there are generally about 100 times the number of secondary windings to the primary windings in the coil called the turns ratio. When the points are closed, current flows through the coil, charging the primary windings. The length of time that the points are closed is called dwell time. When the points open, the current flow is interrupted and the magnetic field created by the current in the primary windings collapses across the secondary windings, creating a much higher voltage, but at a reduced amperage. Generally, a stock points-type coil can create as much as 25,000 to 30,000 volts.

I'm not clear that.
Can anyone discuss it step by step.

Zaw

 

[vgg]

Very short description: when contacts is closed, 12 V is applied to primary winding. It results to growth of a current through it. U=L*dI/dt, =» dI=U*dt/L. If U=12V, L=10 mH, dt=2 ms, therefore dI=2.4 A. And if 2.4 A current flows through 10 mH inductance, the energy about 30 mJ collects into that inductance. When net is broken off, that magnetic energy from the inductance must transform to electric fild energy. A capacitor about 0.25 uF is connected in parallel to breaker's gap and the voltage will rise. All the energy from the inductance (28.8 mJ) will pass to capacitor : 28.8 mJ=C*U*U/2 =» U=√(28.8 mJ*2/0.25 uF)=480 V. There are some losses, of course, but some hundreds of volts - absolutely real.