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I have to simulate the electric field within a gas filled discharge gap generated by a radio frequency voltage generator. The circuit, provided to me by the experimenters somewhat far away, is given in the second picture below.

Now the program I use expects me to feed it the time depended voltage curve and three values do describe the circuit: R, L, C.

How to match the more complicated real circuit to the model? In any case, I'll have to plug in some values for the three parameters R, L, C eventually.

(Numerical values: $C1 = 22\ nF=2.2\cdot 10^{-9}, R1 = 2\ Om, L1+L2 = 8\ \mu H=8.0\cdot 10^{-6}, C2 = 3.4n\ F=3.4\cdot 10^{-9}$ [transformer parasitic capacitance]$, R2 = 100\ MOm=10^8\ Ohm, R3 = 1\ Om, L3 = 100\ nH=10^{-7}\ H.$ Moreover, it's running at 1MHz, sinusoidal, about $10kV$.)


Edit.:

So computing the resistence of the "real" model, setting $R_2=\infty$, I get

$$Z_\text{tot}=Z_{C_1,L_1,R_1,L_2}+\frac{1}{\frac{1}{Z_{C_2}}+\frac{1}{Z_{R_3,L_3,R_\text{Gap}}}}$$

$$=\left(R_1+i\ \omega\ (L_1+L_2)-i\frac{1}{\omega\ C_1}\right)+\frac{1}{\frac{1}{-i\frac{1}{\omega\ C_2}}+\frac{1}{R_3+R_{Gap}+i\ \omega\ L_3}}$$

$$=\left(R_1+i\ \omega\ (L_1+L_2)-i\frac{1}{\omega\ C_1}\right)+\left(R_3+R_{Gap}+i\ \omega\ L_3\right)\frac{1}{1+(i \omega\ C_2)\left(R_3+R_{Gap}+i\ \omega\ L_3\right)}$$

with

$\omega\ C_2\approx 2\cdot 10^{-2}F\ Hz,$

$\omega\ L_3\approx 6.8\cdot 10^{-1}H\ Hz,$

$R_3=1\Omega$

enter image description here

I tried to ask this on the electrical engineering page, but with no responses.


I have no idea if it's of relevance, but I can post the text of the manual of the generator, which I also have, but which isn't really written in my language. I have no idea how the picture even corresponds to the question above. The letter symbols seem a little off. So if the following doesn't make any sense, then ignore it and just take the question above this at face value.

enter image description here

"This generator simplified block diagram is presented on Fig. 5*' (the third picture). Clock signal generator, protect and control circuits are realized by using programmable logic device(PLD) Xilinx XCR32256XL. This chip configuration(and generation parameters) could be changed in-system by special JTAG (Joint Test Action Group)-programmer. Control signal are boosted by gate drivers and gone to the power switch gates. The converter is designed as half-bridge circuit using MOSFET(Metal-Oxide-Semiconductor Field Effect Transistor) switches and fed by voltage up to 400 VDC. Power oscillatory circuit, included elements C1 and high voltage step-up transformer having loss inductance L and parasitic capacitance C2, forms output voltage. In fire mode this circuit part can be simulated as two coupled oscillators; one of them consists of L and C2 and has resonance frequency about 1 MHz. As a result, output voltage will increase fast, have sinusoidal-like shape and reach breakdown level of load. After discharge gap breakdown, in limit mode, capacitance C2 is shunted by small plasma impedance and we can ignore its influence on power oscillation. Resonance of power oscillatory circuit in this mode will be defined by elements L and C1 and resonance frequency value will be less significantly than clock signal frequency(~1MHz). Output voltage will decrease a lot, its shape is changed to near triangle and power supply passes to limit mode. Powerful 1.3 kW capacitor charging module is used to feed RF converter circuit. It has galvanic isolation from mains and adjustable output voltage. Control circuits and gate drivers are fed by auxiliary AC/DC and DC/DC supplies. External synch signal is transferred by digital isolator, it has standard 50 Ohms impedance, active level is “HIGH”(+5 Vampl). Control signal is inactive if input connector is not jointed."

I also don't actually know where the voltage they measure (which is supposed to be 10kV) is taken from in the circuit. Maybe the pictures plainly say it, but I can't read it.


Lastly, I want to post this, because I find it pretty amusing:

"High output voltage (up to 25 kV) and high level of electromagnetic interferences require highly experienced personnel to operate with generator. Careless use may be cause of electrical shock, health hazard, malfunction or even damage of nearby equipment."

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Cross-posted from electronics.stackexchange.com/q/70364 –  Qmechanic May 25 '13 at 17:50
    
>homework, haha I wish. –  NiftyKitty95 May 25 '13 at 23:24
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1 Answer

up vote 1 down vote accepted

I'm surprised the "real" schematic doesn't include a transformer turns ratio (actually, a model for the high voltage transformer, which has a primary winding with $N_p$ turns connected to the power oscillatory circuit and a secondary winding with $N_s$ turns (with $N_s > N_p$) connected to the discharge gap). Maybe everything is referenced to its primary?

$R_2$ is a negligibly large parallel bleed resistor (and so is neglected from here on).

Pre-spark:

Prior to sparking, the circuit is applying a square wave of voltage (caused by power switches 1 and 2 alternately connecting C1 to +400V and ground) to a resonant circuit to give a boosted output voltage across the gap. By "boosted" is meant: one expects the transformer secondary voltage to be $N_s/N_p$ times its primary. However, because the driving frequency is near the resonance of the L-R-C circuit, the actual voltage at the transformer secondary is higher than expected.

  • $R_3$ and $L_3$ don't come into play until the gap starts to conduct, and so can be ignored. That is, their branch of the circuit conducts no current and is treated as open. The remaining elements form a series R-L-C circuit. The current in such a circuit (and so the voltages across its elements) doesn't depend on the order of the elements.
  • $C_1$ (a DC blocking cap which keeps DC current from saturating the transformer), is much larger than the capacitor $C_2$ and can be ignored, or, more accurately, you can replace them by their series combination $C_s$, where $1/C_s = 1/C_1 + 1/C_2$.

You're left with $L_1+L_2, R_4,$ and $C_s$ forming a series LRC circuit, with the capacitor voltage being the gap voltage. This last statement is a bit inaccurate, to the extent that $C_1$ is not much larger than $C_2$, because the voltage across the effective element $C_s$ is shared by $C_1$ and $C_2$, inversely proportionally to their capacitances, with the smaller capacitance having most of the voltage.

I don't see how your sim model works pre-discharge, because its gap is in series, not in parallel with its capacitor.

Post-spark:

  • The plasma impedance (+ $R_3 + L_3$) "shorts out" C2, which can then be neglected.

You're left with the series LRC comprising $C_1$, $L_1+L_2+L_3$, and $R_3+R_4$, +plasma impedance. This circuit is still being driven by the 1MHz switcher, but its resonant frequency is now substantially lower.

The last paragraph you cite is especially intended for potential operators who find it amusing...

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Thanks for the answer. Some questions: In a series configuration, is does the order of the elements not matter? And what is a transformer turns ratio, a square wave of voltage, a boosted output voltage and a blocking cap? Lastly, why can $C_1$ be neglected if it's the bigger capacity and also in series (while $C_2$ is parallel)? –  NiftyKitty95 Jun 3 '13 at 13:05
    
You're welcome. I've attempted to answer your questions with my edit. –  Art Brown Jun 4 '13 at 4:20
    
Okay thanks again. If I neglect the parallel $R_2$, and add up $R_4+R_3$ (the $R_4$ is a $R_1$ really, but whatever), then I only get a value of 3ohm. That is several powers under the default value of the simulation - does it seem strange to you, this low resistance value in that situation? –  NiftyKitty95 Jun 4 '13 at 8:17
    
oops, sorry about the resistor designator error... This circuit is designed for power (1.3 kW is mentioned in the manual) so low resistance is necessary to keep losses reasonable. 3 ohms doesn't seem out of line. –  Art Brown Jun 5 '13 at 3:31
    
There will be additional frequency-dependent losses in the high voltage transformer (especially at this high frequency of 1MHz), so the Q will be lower than you'd calculate from the DC resistances alone. The characteristic impedance $\sqrt{L/C}=52 ohms$, implying a Q of 50/3=17, but I'd be surprised at an actual Q above 5 or so. A lot depends on that transformer... –  Art Brown Jun 5 '13 at 3:38
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