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Peak primary current calculation

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rulmismo

Sun Jan 08 2012, 10:16PM

Registered Member #4187 Joined: Fri Nov 04 2011, 08:08PM
Location: Spain
Posts: 43

JavaTC gives me, with V1=8kVrms C1=40nF L1=5uH, about 1000A of peak current.

,anybody knows or have a reference of the formulae or derivation of that calculation?

I want to assure that my primary gets lower than 500-600A to avoid stress in my metalized film caps (that will support about 1000A nominal)

(see

Regards!

Uspring

Mon Jan 09 2012, 04:26PM

Registered Member #3988 Joined: Thu Jul 07 2011, 03:25PM
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Posts: 711

Reactance of your tank is sqrt(L1/C1) i.e. about 11 ohms. Peak voltage is V1*sqrt(2) if V1 is given as rms, i.e. about 11kV. 11kV/11ohms gets you 1000A.

Mads Barnkob

Mon Jan 09 2012, 05:24PM

Registered Member #1403 Joined: Tue Mar 18 2008, 06:05PM
Location: Denmark, Odense C
Posts: 1968

Maybe you can find some use of this mmc calculator:

rulmismo

Mon Jan 09 2012, 06:47PM

Registered Member #4187 Joined: Fri Nov 04 2011, 08:08PM
Location: Spain
Posts: 43

Probably I forgot my circuits lessons, but donÂ´t get really what does it mean the "tank impedance" sqrt(L/C) here, and why it can be used as (Vc / Z) to calculate the Ipeak.

I see that it is the same value than the "characteristic impedance" of a line; is there is any relation?

I googled for LC circuit equations to get some info on the subject, but donÂ´t got any clue about it (just the usual LC formulae about resonances etc.)

Doing a spice sim with primary data as series RLC, R=100m L=7u and C=40n, charged to 12000V i get a current of about 120kA!

this makes also wonder what is the order of the spark-gap resistance, once fired

as you can see, I am quite lost here, please help!

thanks

zzz_julian_zzz

Tue Jan 10 2012, 03:39AM

Registered Member #3964 Joined: Thu Jun 23 2011, 03:23AM
Location: Valenzuela City
Posts: 332

how did you get 120kA in your primary?

Uspring

Tue Jan 10 2012, 04:47PM

Registered Member #3988 Joined: Thu Jul 07 2011, 03:25PM
Location:
Posts: 711

Probably I forgot my circuits lessons, but donÂ´t get really what does it mean the "tank impedance" sqrt(L/C) here, and why it can be used as (Vc / Z) to calculate the Ipeak.

The tank oscillates at its resonant frequency given by f=1/(2*pi*sqrt(L*C)). In a way the capacitor can be seen as an AC voltage source connected to the inductance L. L is much like a resistor for AC, except, that there is a phase shift between the voltage and the current. The AC "resistance" of L is given by 2*pi*f*L and you can apply Ohms law to derive the current from the voltage.

In your simulation you probably looked at a series LCR tank driven by an input voltage. At resonance the L and C "resistances" cancel and your are left with a 100m resistance, which gets you the enormous currents. A SGTC works differently. Here the capacitor is charged up until the gap fires creating the tank. There is no input voltage apart from the initial charge on the cap.

rulmismo

Tue Jan 10 2012, 06:48PM

Registered Member #4187 Joined: Fri Nov 04 2011, 08:08PM
Location: Spain
Posts: 43

Thanks for the reply upspring,

In the serial RLC sim I just set the initial condition as Vcap=10kV and then run it. There is no voltage source apart from the charged cap.

Your argument could be ok, but in first post is said that "tank reactance" is sqrt(L/C), and this is the commonly used formulae seen, so I assume that is correct.

Now you just use L reactance for calculate the current?, this has not to be similar to sqrt(L/C) in general case, isnÂ´t it?

Maybe someone can give some light to this topic

Uspring

Tue Jan 10 2012, 07:25PM

Registered Member #3988 Joined: Thu Jul 07 2011, 03:25PM
Location:
Posts: 711

Note that the equation for the L reactance 2*pi*f*L also contains the frequency f. If you use the resonant frequency f=1/(2*pi*sqrt(L*C)), you'll get sqrt(L/C).

My response to your simulation was just a wild guess. I have no real idea, what went wrong with it. 120kA don't make any sense to me.

Steve Conner

Tue Jan 10 2012, 08:47PM

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

In a resonant circuit, energy oscillates between the capacitor and the inductor. If you charge the capacitor to voltage V, one quarter cycle later V is 0 and all the energy is a current I in the inductor.

The ratio V/I is the characteristic impedance.

rulmismo

Tue Jan 10 2012, 08:55PM

Registered Member #4187 Joined: Fri Nov 04 2011, 08:08PM
Location: Spain
Posts: 43

Um, :-o, that f substitution trick seems to make sense

Steve, LC working is clear but from that is not straighforward that Ip=Vinitial / sqrt(L/C)

i adjoint the sim, I will try to sim with secondary to check if that could be the reason of the difference
http://www.megaupload.com/?d=AV6MWIBT

Update: sim with
primary c=40n L=7u R=100m
secondary K=0.13, L2=12mH c2=18pF (data from my coil in progress..)

same about 100kA peak primary result, maybe spark gap impedance could be the real difference, what do you think?
http://www.megaupload.com/?d=KE0CDXIZ

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