Distributor for pulse?

[Erik1n4007]

Hello and I appologize if this post is in the wrong area as I am new to this medium. Here goes my question. I have been working with the Bendini Circuit and have had no success up to the moment. I am working with some large coils and am trying to create a system of high frequency resonance but need to pulse energy in my coils that are each bifilar, about 5" in diameter toroidal iron core with 13 guage magnetic coil wire, and configured into a triangular setting. Since I have had no luck with the bendini circuit as my circuit wires end up vaporizing due to the incredibly high voltage that I am producing, I am wondering if I could use the same system that a car uses. that is Battery to ignition coil, ignition coil to a condensor (Capacitor), distributor to the coil enteries instead of the spark plugs, then a return to the negative terminal of the 12V battery (small 12V 4 amp.). If heat or high resistance is an issue but it would work, can there be a system of cooling that could be implemented? Any help would be greatly appreciated. Thank you.
Erik1n4007

[Harvey]

Hi Erik,

A typical ignition is wired this way:


Battery(+) -------> IGN(+)[Primary Winding] IGN(-) -------> [Contact Points and Condensor] ------> Battery(-)

Battery(-) -------->[Secondary Winding]--------->[Distributor Center Post]----->[Distributor Rotor]---------[High Tension Cylinder Wires]------>[Spark Plug]------->Battery(-)

From this diagram two things should be noted:
1. The condensor sits across the points to minimize burning the contacts by arcing

2. The energy for the secondary path is provided solely by the field collapse of the primary winding. This should be obvious because the secondary path starts with the Battery(-) and ends with the Battery(-). Therefore all the energy is supplied magnetically in the ignition coil.

In your case you are melting wires - presumably the high voltage wires from your explanation. Wires do not melt due to voltage, but instead due to internal heating caused by forcing current through them beyond what they can tolerate. The engineering rule of thumb for copper wire is 700 circular mills of wire mass per ampere. If this is exceeded, the wire will heat up.

This is where the voltage applies: The current is the result of Voltage being applied across a resistance - in this case it would appear to be that resistance of your wires. Using #13AWG (note that British wire gauge is different here) you get exactly 2.00 Ohms per thousand feet of wire. The circular mils of that wire (5180) only supports up to 7.4 Amps without heating the wire. This means that you can only safely have 14.8V applied for every 1000 feet of wire. The formula is E = I x R where E is electromotive force (voltage) and I is intensity (current) and R is resistance.

Using a distributor is a great way to switch High Voltage, but it will not prevent the burning of wires. If you intend to use short lengths of wire that have High Voltage on them then you should have a load between them and ground to limit the current. A spark plug is a good load because the spark itself has a reasonably high resistance. But you will still want to measure the current through your secondary circuit to keep it below the rated values.

I hope that helps

[Harvey]

l)

r = 0.25
N = 32
l = 4pi = 12.56

This gives us approximately 64/(2.25 + 125.60)

Which comes to 0.50 microhenries for each coil.

We already established that your coil wire can withstand 7.4A and you are using a 12V battery. So we are looking for an impedance of about 1.6 ohms max on each coil.

There is a specific frequency that those coils will exhibit a 1.6 ohm impedance. The formula for that is f = XL

[troylee]

s post is in the wrong area as I am new to this medium. Here goes my question. I have been working with the Bendini Circuit and have had no success up to the moment. I am working with some large coils and am trying to create a system of high frequency resonance