To calculate the resonance frequency of the filter inductance and C-prim is very simple: the resonance frequency = 1 / (2 PI x SQRT(LC)) The problem is: what is L and C in this case, as there is a HV transformer in between. You can approach this from either side, the HV end or the LV end. Let's take the HV end. On the HV end, C is simply the actual capacitance of C-prim, but what inductance do you see from this end? A low current on this side will translate to a higher current on the other side of the transformer, meaning that more current will flow through the inductance and thus the inductance will show a higher impedance. How much higher? Assuming that your transformers have a 120V primary and 14.4KV secondary, the current on the primary side will be 120x higher. Therefore L in the above equation should be 120 x the inductance of your filter inductors. PS. I ran into this problem myself and it cost me my ceramic primary capacitors. (which still saddens me today, as they were magnificent) PSS: Don't worry too much about insulating the secondary coil. Just wind it tight (windings close together) and tune the circuits at low power.
Man I’m 22 from NY. I’ll pay you and pay my motel and flight just for u to teach me how to make one better yet how to understand it. name ur price I want a teacher
Lineman? This guy is an engineer!
Hats off to you sir. That a work of art
That's such a beautiful system, I'm thoroughly impressed! Thank you for taking the time the explain everything and really show the different parts!
Great video. You're clearly a natural and knowledgeable instructor!
That coil is a beast! Thanks for the run through - wise move on the ballistics lexan - true you don't want bits inside becoming mini missiles :)
Excellent presentation. That’s a Big Beautiful Monster. Work safe⚡️
Why stop at linesman man,electrical engineering.Frickin awesome man!Would have loved to see the r&d testing! Just awesome!
To calculate the resonance frequency of the filter inductance and C-prim is very simple: the resonance frequency = 1 / (2 PI x SQRT(LC))
The problem is: what is L and C in this case, as there is a HV transformer in between.
You can approach this from either side, the HV end or the LV end. Let's take the HV end.
On the HV end, C is simply the actual capacitance of C-prim, but what inductance do you see from this end? A low current on this side will translate to a higher current on the other side of the transformer, meaning that more current will flow through the inductance and thus the inductance will show a higher impedance. How much higher? Assuming that your transformers have a 120V primary and 14.4KV secondary, the current on the primary side will be 120x higher. Therefore L in the above equation should be 120 x the inductance of your filter inductors.
PS. I ran into this problem myself and it cost me my ceramic primary capacitors. (which still saddens me today, as they were magnificent)
PSS: Don't worry too much about insulating the secondary coil. Just wind it tight (windings close together) and tune the circuits at low power.
Gigacoil new Collab
That's a beautiful coil mate😎 and that topload is fantastic...
Good work cant wait to see it in person at some point 😁
Steve
Nice build man thanks for sharing
Hi, have you considered using a three phase motor controlled by a frequency drive?
The FCC loves this man
thats fenomenal
Wow beuty! gnarly arcs
Very nice👍⚡⚡
Hi sar,,, I have one qestion..
How much kva is rated of your 14 kilovolt transformer??
Genious waw great video congratulation amazing keep up the great work
Very well explained !
keep up the work
Nice coil but at this point it would be easier and cheaper to just built a DRSSTC that size. And it could also play music.
Spark gap coils are og.
Man I’m 22 from NY. I’ll pay you and pay my motel and flight just for u to teach me how to make one better yet how to understand it. name ur price I want a teacher
I have to have help to shut off a nanotech bmi mkultra implanted at Omaha nebraskaunmc please help!!!