http://www.youtube.com/user/Venturecaplaw#p/u/52/GXcxswDcUbIhttp://www.youtube.com/user/Venturecaplaw#p/u/51/-vznuNkEBtohttp://www.youtube.com/user/Venturecaplaw#p/u/54/HGgfOqlxWqUhttp://www.youtube.com/user/Venturecaplaw#p/u/55/V6xa76IshbE In these videos you will learn these two time constant (TC) formulas
Inductance TC = L / R
Capacitance TC = R x C
And it takes a total time of 5 TC's to fully charge or discharge a inductors current and a capacitors voltage.
Now since we are dealing with a leaky capacitor the time it take my electrode pair to discharge is:
5(3778.83 pf x 78.54 ohms) = 2.967893^-7 sec x 5 = 1.48395^-6 sec to fully discharge my capacitor. Not a lot of time, huh?
Now by adding in resistance you gain more time before the capacitor fully discharges.
Example if I add in 40k ohms, which would be all coils added up together, I'd get this: 5(3778.83 pf x 40078.54 ohms) = 7.5724995^-4 sec
And this takes place at the same time the inductor is dumping current into the capacitor as it losses current giving even more time before the capacitor fully discharges.
So now you can see that adding in resistance has two purposes, to prevent amp leakage, and to add more time to the capacitors fast discharge time so it isn't too fast.
So the bifilar chokes will go to slow this leaky capacitors discharge time down and the resistance will go to slow this leaky capacitors discharge time down. Wow, I learn something new everyday!
We need to calculate the total resistance of the circuit, and due to the way the circuit works we only get to add one resonant value of a choke with this resistance.
So that gives us all resistances that are in-line with the capacitor plus the XL of the choke plus the Xc of the capacitor at the resonant frequency. This value is then added to the Time Constant so see if the choke size will give a time that is greater than the frequency time count per second. Remember the maximum pulse rate is always twice the resonant frequency, that means you divide the one pulse time by two.
Unfortunately I can't add all the resistance of the wires due to I don't know the resistance per foot data. I only know the 0.125 mm wire is around 24 ohms per foot. I will only have all the data on the chokes after it is built and I put a resistance meter to it to take a reading. So here we go .
Planned resonant frequency for me is 1.83k Hz which gives a single pulse time of 5.4645^-4 seconds divide that by two and I get 2.7322^-4 seconds. This is my charge time and it must be less than the capacitor drain time.
So starting with the secondary I will be using 263.6 ft of wire at 24 ohms per foot. Which gives me 6326.44 ohms of resistance.
The chokes @ 2 H gives me, 2 x pie x 1830 Hz x 2 H = 22996.46 ohms
The capacitor gives 1 / 2 x pie x 1830 Hz x 3778.83^-12 f = 23015.04 ohms plus 78.54 gives 23093.58 ohms
Now to use the TC = R x C formula
TC = (6326.44 + 22996.46 + 23093.58) x 3778.83^-12 = 1.9807^-4 seconds is for one time constant. So it is multiplied by 5 giving me 9.9036^-4 seconds for full discharge time of the capacitor.
This shows me that the 2 H choke will be enough for me to charge the capacitor since 2.7322 is less than 9.9036. With this size choke I will have a voltage drop just over 36.8% between charges so I will go to a larger choke size to improve those numbers.
Now at least all of you know how to calculate the value of your chokes so you get a working VIC Matrix Circuit the first time around,
no more guessing! Enjoy!
h2opower