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Calculating Current in Sparkgap

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Coronafix

Sun Jun 29 2008, 06:36AM

Registered Member #160 Joined: Mon Feb 13 2006, 02:07AM
Location: Melbourne, Australia
Posts: 938

I am wondering if there is a formula that will tell me how many Amps are being dumped across
my sparkgap from the capacitor bank.
Maybe HVWiki needs to be updated, because I couldn't find it there.
Apologies if it's a silly question. Thanks.

Sulaiman

Sun Jun 29 2008, 09:13AM

Registered Member #162 Joined: Mon Feb 13 2006, 10:25AM
Location: United Kingdom
Posts: 3140

Once a spark gap 'fires' it can be considered as a virtual short-circuit
(usually a few hundred volts drop which can be ignored relative to the thousands of volts across the resonant circuit)

So the question is, what is the current circulating in the primary resonant circuit?

That's a peak of Vpri/Z, ringing down to zero over many cycles (10 to 50)
where Z = 2 x PI x Freq x Inductance = 1/(2 x PI x Freq x Capacitance)

Coronafix

Sun Jun 29 2008, 12:19PM

Registered Member #160 Joined: Mon Feb 13 2006, 02:07AM
Location: Melbourne, Australia
Posts: 938

It's strange, perhaps I am wrong. I had worked on that principle but ended up with a value that caused me to ask the question because I didn't believe it. Perhaps I am doing something wrong.
If the primary circuit is charged by a NST of 15kV with a capacitance of 17.6pF and a line frequency of 60Hz, then I get 100mA which doesn't seem right to me. Or should I be using the frequency of the tank circuit? That would make more sense and would be in the order of 200A. Is this correct?

Shaun

Sun Jun 29 2008, 02:23PM

Registered Member #690 Joined: Tue May 08 2007, 03:47AM
Location: New Jersey, USA
Posts: 616

Yes, you, use the frequency of the tank circuit. That's the frequency it will be ringing down at.

vasil

Sun Jun 29 2008, 08:59PM

Registered Member #229 Joined: Tue Feb 21 2006, 07:33PM
Location: Romania
Posts: 506

Whole the energy from the primary cap transform in magnetic energy in the primary. For the first 1/4 cycle we have:

(C*Up^2)/2 = (L * Ip^2) + R* I^2

C = capacintance of the primary cap
L = inductance of the primary
R= ohmic resistence
Up = breakdown voltage for the primary cap
Ip = maximum current value in the first 1/4 cycle
R = primary ohmic resistence

Considering R -> 0, we can approximate

Ip = Up * sqrt (C/L)

Coronafix

Sun Jun 29 2008, 11:02PM

Registered Member #160 Joined: Mon Feb 13 2006, 02:07AM
Location: Melbourne, Australia
Posts: 938

Thanks all, I appreciate all the help.
Maybe someone could update HVWiki with this info.
The reason I have been wondering is because if I have a maximum current to work with for a thyratron
then I can work out what size my system can be.

GeordieBoy

Mon Jun 30 2008, 10:41AM

Registered Member #1232 Joined: Wed Jan 16 2008, 10:53PM
Location: Doon tha Toon!
Posts: 881

Primary surge impedance is given by:

Zp = sqrt (Lp / Cp)

Peak primary current is equal to the peak capacitor voltage divided by the surge impedance Zp. A reasonable guess at the peak capacitor voltage is probably about 2.5 times the RMS output voltage rating of the HV transformer if you are using resonant charging.

Expect a peak primary current in the hundreds of amps or low kAmps for a large system.

Of course most of the component heating in the primary is more closely related to RMS primary current, (which is harder to calculate). And it's even harder to model the spark gap losses because of its negative incremental resistance.

-Richie,

Steve Conner

Mon Jun 30 2008, 03:03PM

Registered Member #30 Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706

The RMS current is roughly one-twentieth of the peak value that Richie showed how to calculate. This is because the current is only flowing for about one-hundredth of the time (100 microseconds discharge time, repeating every 10 milliseconds) but you have to take the square root of that duty cycle to see how it affects a RMS quantity. So that's one-tenth.

Then you have the waveform itself, which is the sum of two sine waves, so its RMS value is one half of its peak, as opposed to 0.707 for a single sine wave. Taking the two factors together, one-tenth times a half is one-twentieth.

There are many factors involved in getting a more accurate calculation. For a start, I ignored the log decrement of the primary current envelope. I also assumed the breakrate of your coil was 100Hz and that the gap quenched after 100us. But I think the above is a reasonable estimate. Tesla coils aren't rocket science, they just need to be in the ballpark.

Finally, RMS current is OK for calculating the I2R heating in conductors and capacitors, but is it the right figure for spark-gap heating? A spark has a negative resistance, so is I2R valid?

vasil

Mon Jun 30 2008, 06:21PM

Registered Member #229 Joined: Tue Feb 21 2006, 07:33PM
Location: Romania
Posts: 506

I think not, maibe more correct in the formula above would be I^2 * Rt, where Rt is the total resistence of the primary tank

but Rt = R - r, where
R = ohmic resistence
r = negative resistence of the arc

The intensity of current decay after a sinusoidal law, like as:

I = Io * e^(-Î´t) * sin (2*pi*f*t) where Î´ is the dumping coeficient Î´ = Rt/(2*L)

f = frequency, L = primary inductance so

Î´ = (R - r) / (2* L)

Now we integrated the negative resistence in the current formula, but it does not the things simplier for calculating the current. A better way to approximate the losses in the circuit would be to integrate the area under current envelope and from there to get a RMS value of current. So it would be more experimental than math.

GeordieBoy

Mon Jun 30 2008, 06:42PM

Registered Member #1232 Joined: Wed Jan 16 2008, 10:53PM
Location: Doon tha Toon!
Posts: 881

Steve Conner wrote:

The RMS current is roughly one-twentieth of the peak value that Richie showed how to calculate. This is because the current is only flowing for about one-hundredth of the time (100 microseconds discharge time, repeating every 10 milliseconds)

Cool! That's neat Steve. I like "back of fag packet" calculations and I agree with how you worked it all out. I just never realised you could do it that way until now. I guess using Microsim for too long made my brain get lazy!

It's much more useful to know the RMS current when your primary side switch is made of MOSFETs and less so when it is a piece of conducting air. You mentioned the primary ringdown envelope being an exponential decay (except for the notches.) In an ideal world it would be, but in practice with a spark gap there the bare primary ringdown is almost a linearly decaying envelope. It can be explained two ways:

1. Arc resistance is lowest when most current is flowing and therefore losses are lowest at the start of the bang. Conversely as the current decreases the arc's resistance rises and losses increase. This "straightens out" the textbook exponential decay.

2. The spark gap acts like a fixed voltage-drop in the tank circuit. So on each pass through the spark gap it looses 800V or so. Again leading to a linear decay.

For simulation purposes I used to model the spark gap with a pair of back-to-back zener diodes in series with a small resistance. Strangely the modern trend towards IGBTs as primary switches makes the voltage-drop model appropriate again as their dissipation is closer to being proportional to I than to I-squared.

-Richie,

PS. I used to have a good scope picture showing the linear primary voltage ringdown for a spark-gap TC with no secondary present. It also displayed interesting bursts of VHF oscillation around the points where the primary current oscillates through zero and the spark in the gap presumably goes out momentarily! These intense RF bursts are the cause of much TVI and VHF/UHF hash. Perhaps Terry Fritz is the only guy here that will remember these investigations from the pupman list! I'm showing my age now!

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