RF output and plate

The bottom device in the left image is an RF generator with ratings 230V, 15A (max), 13.56 MHz and variable controllable output power up to 1250 W. The top device is an auto match.

The metal strip with holes is the output of auto match, which has to be connected to one of the plate (shown in right image) of parallel plate capacitor made of aluminium. The dimensions of plates are 12 cm diameter and 1 cm thick with adjustable gap between both from 1 cm to 10 cm.

My problem is: How to make a connection between the output of auto-match and the plate of capacitor, so that there are minimum power losses? Somebody suggested me to use short wide flat copper strip and fasten it using nuts and bolts on both ends. Is it a good idea or are there any standard connectors available that can be used here?

  • 2
    \$\begingroup\$ This case is unsolvable for ordinary mortals, who do know nothing about frequency, dimensions, materials and the circuit behind the visible output. Sorry. Provide something relevant info or you will collect only downvotes. \$\endgroup\$ – user287001 Nov 12 '18 at 13:52
  • \$\begingroup\$ Does this system not already function? I know a guy who experimented with plasma-etch RF generators and matching (transferring energy) into the plasma; he was guided by a PhD in E&M so not a trivial question. \$\endgroup\$ – analogsystemsrf Nov 12 '18 at 15:09
  • \$\begingroup\$ @user287001 Isn't it still comprehensible. What's the confusion? I don't have electrical background and usually get down votes when I ask questions in this area. Talking about this question am I not clear in terms of conveying or the data is still insufficient? Your suggestion would be helpful. \$\endgroup\$ – user98179 Nov 12 '18 at 16:15
  • \$\begingroup\$ @user98179 I have no idea what this thing is/does but I'm curious - what exactly is an "auto match"? (Normally it is necessary to know the output impedance of a source in order to be able to tune the circuit for maximum power transfer.) Please tell us more about this device. What do you want to do with it? \$\endgroup\$ – Stefan Wyss Nov 12 '18 at 16:28
  • \$\begingroup\$ @StefanWyss Purpose is to generate uniform electric field oscillating at 13.56 MHz. To achieve this we are using two plates in a configuration as parallel plates, one connected to current oscillating at 13.56 MHz and the other grounded. Yeah you are right we need to know the output impedence in my case gap between plates is variable and impedence varies between 7300 to 1200 Ohm. Auto match is a device that matches the variable impedence on the load side of generator to the internal impedence of power generator using feedback loop. \$\endgroup\$ – user98179 Nov 12 '18 at 19:20

Frequency about 13,5 MHz and the dimensions prove that there's a reactive load until something lossy appears between the capacitor plates. Obviously (=assuming your generator+the auto-match are designed to work with capacitive load) no matching circuits are needed, only a conductor, as short as possible and about the same size as your output terminal. You need also the ground side connection as good. Be sure that the connections are 1 meter or less or all kind of matching problems start to appear due the inductance and capacitance of the wires. There's no steep limit for them.

I guess the shown plate is closed into a metal chamber to where the body of your transmitter is solidly joined, so the ground side connection is no problem.

Actually you seem to have already a good suggestion.

Obviously you need some insulation around your conductor in certain places. That material must stay solid. Glass and teflon probably are ok.


If your generator happens to demand well matched 50 Ohm resistive load to stay alive and your load is the capacitor with no lossy filling between the plates, you must dissipate the output of the generator somewhere. One possiblity is to design a voltage boosting resonant circuit to get the needed E-field strength with reasonably low actual dissipated power. The simplest form of it is an inductor, a 50 Ohm resistor and your plates, all in series. The inductor must be adjusted to be in resonance with the capacitor just at the operating frequency. Your "auto match" obviously can fix small misadjustments.

At full power you will get 5A current which will generate 6000V between your plates when their distance is 1cm. That's a hefty electric field as volts/meter when the frequency is this high. If you happen to touch the parts, you will remember the smell of your burned flesh quite long time.

With Tesla coil type system you can have much stronger fields. Actually you only need to insert a tap to the coil.The resistive losses in the coil become substantial and they need to be taken into the account when you calculate the extra resistor for right resistive load

  • \$\begingroup\$ Thanks for your reply but impedence of the parallel plate capacitor varies between 1200 - 7300 Ohm and with 50 Ohm internal impedence of generator matching network is necessory evil. Infact these plates are placed inside the isolated metal cage to avoid interference. There is a small hole at the bottom of cage through which power input to one plate and earthing to the other has to be done. \$\endgroup\$ – user98179 Nov 12 '18 at 19:26
  • \$\begingroup\$ @user98179 Is the case that you have a radio transmitter which is intended to have well matched resistive 50 Ohm load, a tuner (=auto match) which can handle some range of mismatches, but not enough when you connect your capacitor and now you expect something which transforms your capacitor to the range of the tuner? Or are your electronic boxes already designed to be together able to feed your capacitor without any smoke? \$\endgroup\$ – user287001 Nov 12 '18 at 19:59
  • \$\begingroup\$ @user98179 Your written impedance range 1200..7300 Ohms happens to fit with the reactance range of your variable capacitor in a metal cage at 13,5 MHz. Is there any actual resistive load between your plates? \$\endgroup\$ – user287001 Nov 12 '18 at 20:47
  • \$\begingroup\$ yeah it is in fact reactance only, no resistive load. \$\endgroup\$ – user98179 Nov 16 '18 at 17:12
  • \$\begingroup\$ As per the suggestion the output of the matching network was connected to one of the plate of capacitor using short wide copper strip and the second plate is earthed using short wide copper strip. Now the forward power is only 1 percent. Does it mean that there's no electric field generated between plates or this power thing has nothing to do with development of electric field between plates, it's so low just coz there is no actual resistive load in which power loss occurs. \$\endgroup\$ – user98179 Nov 16 '18 at 17:31

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