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Old 05-01-2010, 08:52 AM
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Harvey Harvey is offline
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Join Date: Jul 2009
Posts: 1,137
Hi Chris,

D1 is supposed to be a voltage controlled capacitor (varicap) as far as the part specification goes, but I just slapped that in there to represent your adjustable cap. It would need another series capacitor and voltage source at the cathode to function correctly.

Some of the parts are not Spice ready parts (like the relay) and I would have to prep it to do a simulation. Right now I am spending my simulation time on my Big Coil when I get the urge to work on it which is not all that often.

If I understood you correctly, your transformer is an ignition coil. These coils are made to withstand upwards of 30,000V on the secondary. Unless you have the capacitive discharge type, the principle behind their operation is that they support a very fast field collapse in the primary when the point contacts open. This very steep change in magnetic flux produces an extremely high negative BEMF pulse in the primary with nowhere to go (right at the point contacts) and this is stepped up by 100 times by your winding ratio. So a 300 volt negative BEMF on the primary will manifest a 30,000 volt negative on the secondary and that's what jumps the .030 inch gap on the spark plug in the engine.

Your use of a resistor in series with the primary puts a limit to the current that can flow in the primary. This is because there is a maximum imposed due to the voltage drop across it. Even if your primary was a superconductor at zero ohms, there would be a 12V drop across the 40 ohm pot limiting the maximum current to 300mA. This directly effects the inductive charge of the primary - it may or may not be getting a full charge relative to its inductance, or how much the core can hold before it saturates and this plays a big part in how much BEMF is produced when the contacts open. This resistor also plays a part in the Impulse Response of the primary. If you look at the equation in that link you will find that Tau (1 time constant) = L / R. We usually assign 5 Time Constants (or 5 x Tau) to represent a full charge (even though, theoretically it never reaches full charge as each step is a percentage of the previous and just keeps getting smaller and smaller to an infinitesimal value) but 5 gets us to 99.99% full charge. What we see here, is that our time constants get shorter the greater the resistance. For example, let's say L = 200 and R = 40, then Tau = 5. Now lets increase R to 100 and we find that Tau drops to 2. So to put it simply, the primary of the coil responds faster with the resistor in series with it, but the overall energy is reduced (because it is spent on the resistor). That's the trade off. So you may be seeing that if you had a larger voltage source on the primary, then you could boost the energy level and get the benefit of the faster response. A 30V supply, and a 100 ohm setting on your pot would be about 9W dissipated, the current would be the same, but the charge time would be faster. Conversely, a 15V supply and a 25 ohm setting would still give about 9W dissipated in the resistor, but the energy in the spark would be greater as the current is doubled - while the impulse response suffers.

You may also note that the way the diode is in the schematic on the secondary, it allows the positive impulse that occurs during the relay contact closure to pass, but it attempts to block the large negative BEMF induced pulse. It is probably being pushed into its avalanche region and producing the noted 'punch through'. By turning it around, we would maximize the function of this coil, keeping the positive pulse in the core to be used in BEMF collapse, and then allowing that negative spike to conduct fully through the diode. Chances are, in that case, we wouldn't reach avalanche - I haven't looked at the specs on that diode to see what it's limits are.

Now to bring this on topic. To be a MOSFET Heater device, it would need a MOSFET in place of the relay, and it would need an inductive resistor somewhere - perhaps in place of the 10W Pot?

Now - what would happen if we ran the switch common to the inside of a isolated metal sphere instead of the battery negative? And what if we connected some CCFL tubes to the outside of that metal sphere? And then (the part I really like here) we connect the other side of the CCFL tubes to the 12V battery positive. Would we still get heat, light the lights and recharge the battery all at the same time? (Trick Question )


Last edited by Harvey; 05-01-2010 at 09:00 AM.
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