Problem in generating uniform RF electric field between two metallic plates.

Thread Starter

mandman

Joined Oct 5, 2018
2
Screenshot from 2018-11-27 18-23-09.png
My objective is to maintain uniform electric field oscillating at 13.56 MHz frequency between two circular parallel plates. For this I have a device with block diagram as shown in attachment.

(1) is the RF generator generating 13.56 Mhz current at 250 VRMS with 50 Ohm internal impedance, (2) is the auto-match network (50 A output current) for matching the load impedance with the internal impedance of the generator. (4) and (5) are the aluminium plates placed opposite to each other forming a parallel plate capacitor, with no lossy material between them, inside a metal cage (3), which is not connected to power or ground anywhere. The bottom plate (4) is connected to output of the auto-match(2) , using a short wide copper strip, and the top one(5) is grounded. The diameter of plates is 11 cm and the gap between them can be varied from 1 cm to 10 cm and the corresponding capacitance values range from 0.84-8.4 pF and the reactive impedance range from 13948-1394.8 Ohm. There is no resistive load in line after auto-match. Generator (1) is connected to auto-match(2) via 50 Ohm Coaxial cable.

When I turn on the generator the reflected power is 100% as read by meter, on the front panel of power generator, which means power is not dissipating anywhere in the circuit. Does this mean that there is no displacement current between the capacitor and hence no electric field between the plates?

Till now I have got two suggestions, which I have not yet implemented, from people over the internet and people active in RF area

1. Use an inductor and resistor in series with the plates keeping in mind the oscillation frequency of inductor should match with that of power generator i.e. 13.56MHz. With these elements, power will dissipate in resistor and inductor will boost the voltage to very high value which will cause an appreciable amount of current to pass from bottom plate (4) to upper plate (5).

2. Add an inductor in parallel with the capacitor (formed by plates 4 and 5).

Apart from these, I was thinking why not add a resistor in parallel with the capacitor (formed by plates 4 and 5) to allow the current to flow and power to dissipate. Since the bottom plate will be at matching network output potential and top grounded then there should be electric field between them, but somebody told me that in this way the capacitor will act as open circuit means no E-Field between them. Is he right?

Out of above suggestion which one to follow and before starting what equipment like RF current meter, voltage meter or power meter, should I procure.

Any other workable suggestion is also welcome.
 

MrChips

Joined Oct 2, 2009
22,075
What do you mean by uniform electric field?
There will be edge effects causing non-uniformity at the edges of the plates.

 

nsaspook

Joined Aug 27, 2009
7,728
Yes, there is a electric field between the plates in a matched or unmatched condition. The fictional displacement current term is a non factor because here we have a near-field reactive environment where EM energy is stored and dissipated locally instead of being radiated.
https://forum.allaboutcircuits.com/threads/capacitor-in-coupling-decoupling.147257/#post-1254611

My suggestion is to look at typical RF Generator Driven Resonator systems used in heavy ion accelerators for answers. Usually there is a small acceleration plate element that acts like part of the tuning capacitor of the LC matching/tuning network for the RF generator.

13.56 MHz resonator that delivers ~80KV at 3KW input power to a capacitive beam coupler.
https://patents.google.com/patent/US6262638B1/e
In the circuit of FIG. 3, the complex impedance Z of capacitor CS is proportional to 1/f, with I leading V by 90°; the complex impedance Z of inductor L is proportional to f, with I lagging V by 90°; and the resistive losses RL are generally independent of frequency, with I and V in-phase with each other. At resonance, maximum voltage is achieved at the accelerating electrode 32 for a given input RF signal, the currents in CS and L cancel because they are 180° out of phase, and all power in the circuit is dissipated through resistor RL. To attain a resonant state, ω=2πf=(LC)−½. For example, in the Eaton GSD series, ω=13.56 megahertz (MHz).
Background Theory.
http://www.iuac.res.in/event/rfworkshop/GJoshi.pdf
 
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