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Calculating Primary Current / Impedance

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LAN_403

swordfish

Tue Jul 24 2018, 02:42AM

Registered Member #3218 Joined: Mon Sept 20 2010, 10:42AM
Location:
Posts: 29

Hi All,

I have a quick question I hope someone can help with about primary current, or maybe more correctly, primary impedance calculations.

Calculations I see for primary current use the formula I = V/sqrt(L1/C1). However, literature for series LC circuits seems to state that at the resonant frequency Xl = Xc, and therefore the impedance drops to that provided by pure resistance. Is anyone able to explain why we use the above formula rather than assuming purely resistive losses? I know I'm missing something here!

Many thanks,

Greg

Sulaiman

Tue Jul 24 2018, 03:47AM

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

the primary capacitor is in parallel with the primary inductance

swordfish

Tue Jul 24 2018, 04:01AM

Registered Member #3218 Joined: Mon Sept 20 2010, 10:42AM
Location:
Posts: 29

Ah yes, I see now. Makes more sense looking at it as a parallel resonant circuit! Thanks!

Dr. Slack

Tue Jul 24 2018, 06:25AM

Registered Member #72 Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659

Resistive losses are 'small', therefore the current is controlled by energy considerations, at least initially.

The formula I = V/sqrt(L/C) is just another way of saying I^2L/2 (energy stored in primary due to current) = V^2C/2 (energy stored in capacitor due to voltage).

Uspring

Tue Jul 24 2018, 09:37AM

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

Is that a SGTC or a DRSSTC you are asking about? In a SGTC primary inductance and capacitance are in parallel, while in a DRSSTC they are in series.

Fumeaux

Tue Jul 24 2018, 03:54PM

Registered Member #61890 Joined: Sun Nov 05 2017, 05:57PM
Location:
Posts: 8

I think I have an useful tool for you: http://www.classictesla.com/java/javatc3d/javatc3d.html

There you can look at an array of varius (some are useful) calculations.

klugesmith

Tue Jul 24 2018, 08:36PM

Registered Member #2099 Joined: Wed Apr 29 2009, 12:22AM
Location: Los Altos, California
Posts: 1714

The series/parallel question is a good opportunity to discuss duality.

A simple LC tank circuit (figure A) is series, because both components share a current. It's also parallel, because the components share a voltage. As Dr Slack said, the voltage/current magnitude ratio is the tank impedance Z = sqrt(L/C).

If you add an impedance-vs-frequency meter in series (figure B), the reading matches "series circuit" formula. High Z at DC.
If you add an impedance-vs-frequency meter in parallel (figure C), the reading matches "parallel circuit" formula. Low Z at DC.

[edit] Cases B and C are equally disruptive to the original circuit. One has a new voltage element sharing the common current; the other has a new current element sharing the common voltage.

Now back to question in OP. The log |Z| vs log f chart at the bottom of figure B has a wide V shape. Its asympotes are |Z| = 1/wC below resonance and |Z| = wL above resonance. (here using w in place of radian frequency Ï‰ = 2 pi f.) The asymptotes cross at the resonant frequency, where 1/w = sqrt(LC), and |Z| for the capacitor and inductor are both sqrt(L/C).

What about that narrow impedance dip at resonance, you asked? It goes down to the circuit's resistance value, far far below sqrt(L/C). Well that means that at resonance, the driver (our Z meter) needs to force only a tiny voltage to see a large current. Or conversely, if the Z meter is forcing a fixed current, it sees a very small voltage between the terminals. The voltages across the capacitor and inductor are still sqrt(L/C) times the current, either way.

In other words, at resonance, the third voltage element in figure B has a very small value with respect to the current and the tank voltage. The third current element in figure C has a very small value with respect to the voltage and the tank current.
Does that help? [\edit]

The tank in a DRSSTC could theoretically be driven either way, though component limitations seem to favor an AC source in series. In SGTC we charge the cap through inductor in series (switch open). Then close the switch (spark gap), so the LC resonates as in figure A. Same as figure B, with the forcing voltage source stepped to 0 volts.

swordfish

Tue Jul 24 2018, 10:21PM

Registered Member #3218 Joined: Mon Sept 20 2010, 10:42AM
Location:
Posts: 29

Thanks very much everyone for the advice, I think I understand it sufficiently now. Appreciate the effort some of you have gone to to help!

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