help with displacement current, distributed capacitance

Lots of factors are involved. But I suggest that you forget the "displacement current" thing and just analyze it as you normally would.

Maybe this will help...

Until you can unify general relativity with quantum mechanics, you don't have a clue. So you can relax on that score.

I think you're getting too wrapped around the axle with the idea of displacement current messing things up. After all, in a "normal" high-Q resonator there's displacement current -- it's just all between the plates of the cap.

I'm not buying your explanation, but you might try modeling a tank circuit without any distributed capacitance (as a single L with series resistance and a single cap), and then modeling an equivalent circuit with distributed capacitance (as a whole bunch of ohcrapohcrapohcrap* coupled inductors, each with its own series resistance, and with caps to the adjoining coils).

Assuming that SPICE can deal with all the coupled inductances without blowing a gasket (or you, if you do the work with pencil and paper), you should get some results that bolster or undermine your theory.

• It's not that I'm afraid of _one_ pair of coupled inductors -- it's the idea of _lots and lots_ of them that gives me pause.

--

Tim Wescott 
Wescott Design Services 
http://www.wescottdesign.com 

I'm looking for work -- see my website!

I have seen Dave's video before, but I think he should have said, Before the switch is closed the plates have an equal number of electrons. When the switch closes electrons pile up on one plate creating an electrostatic field the repels an equal number of electrons off of the other plate. Now that I have said that and the word equal came out, I have to question whether my simple explanation might be to simple.

Back on my topic now.

Mikek

Ok, back on your topic--you don't have a clue.

I don't think he knows how to (or does not want to) model anything.

For a single-layer solenoid consider any adjacent pair of turns as plates of a capacitor. What's the voltage between them? Per the dielectric between them, what current *can* flow?

Most importantly, is it real current or reactive?

Mark L. Fergerson

No, not people, such as myself, who have done that.

--
 Thanks, 
    - Win

I don't have any reason to suppose that high Q coils have low distributed c apacitance. High Q just means low resistive losses in the coil. Inductive c urrent flows through the copper. Capacitative current flow through the insu lation, which may or may not be lossy - it usually isn't.

--
Bill Sloman, Sydney

That's the key; displacement current (E-field in a capacitor gap) is sometimes dissipative, meaning it has an in-phase component and losses. That's because of the polarization of the dielectric, where real charge moves around in the E-field.

The high-Q inductors favored by competitive crystal-radio afficianados are air-dielectric. Some very high power tuning capacitors are available in vacuum dielectric.

This paper might be of interest in pointing you in the right direction - depending on what you actually want to do.

I love the use of a mercury gas discharge tube as a probe in his experimental tests of the theoretical models.

--
Regards, 
Martin Brown

Yes and the usually have between 6pf and 9pf of distributed capacitance for an inductor 170uh to 240uh.

My interest is in air core coils.

Mikek

Hi Win, Could you expand on that? Is it your experience that getting the lowest distributed capacitance is important to getting highest Q? What is the mechanism that causes the additional losses?

Thanks, Mikek

As long as there is no loss in the distributed capacitance then it won't change the Q... only the resonance frequency. I would guess (but it's a total guess) that an "air" capacitor, would have low loss... unless there is something going on in the insulation on the magnet wire.

George H.

My theory is that the distributed capacitance causes extra current to flow, increasing I^2R losses. I'm trying to find data or info to back that up. The practical coil will have a styrene form and use 660/46 litz with a frequency range of 540kHz to 1700kHz.

Thanks, Mikek

The above paragraph is not right, it needs modification, but I don't know yet how to explain it.

Oh I see what you are saying... So how about a model of your coil that looks like this... (bad asci art.)

+-RRR-LLL-+-+-RRR-LLL-+-+--etc... | | | | | +--CCC----+ +----CCC--+ +--etc.

Where each of those sections is meant to model one turn of your inductor.

Then as long as there is not much resistance in the capacitive branch of the model, I can take all the caps in series and model them as one cap across the whole coil.

With this model it looks like less current flows through the coil resistance, since there is some flowing through all those caps in series.

(or maybe you have a different model?)

George H.

Thanks George, I don't have a model in mind, I'm running from experience that low distributed capacitance coils have higher Q. But I don't have mathematical proof, nor would I understand it if I did. Looking at your model, I would change it to this. Just my guess.

+-+-RRR-LLL-+-+-RRR-LLL-+-+-RRR-LLL-+-+-RRR-LLL-+-+ | | | | ^ C C C C Note** | | | | +-+-RRR-LLL-+-+-RRR-LLL-+-+-RRR-LLL-+-+-RRR-LLL-+-+

Note** because it's a coil the end goes around and connects to the bottom wire in the model.

I don't know how you come up with a voltage differential between turns of a say, a 20 turn coil.

Mikek

Yes, along with litz wire and other things.

It's the internal high resonant currents that are flowing back and forth between capacitance and inductance sections. You know how the Q of an inductor tanks as frequency is raised past SRF / 10? Exactly the same story.

I made high-inductance coils used at 1MHz with measured Q values in the region of 500. Minimizing capacitance with bank winding was a key.

--
 Thanks, 
    - Win

Thanks for the follow up Win.

Mikek