Trying to model loop antenna (why does capcitance go down when adding parallel cap?)

"John Woodgate" wrote in message news: snipped-for-privacy@jmwa.demon.co.uk...

(in

loop antenna (why does capcitance go down when

of the unit. I also measure the DC resistance. I

really strange results. I attached the 10,000 PF

How in the world does the capacitance go down when

all of it, and you would measure just 4.2 ohms

total actual capacitance is 10421 pF and the net

opposite, reactance as 306.1 uH at frequency f =

While I agree with the basic thrust of your answer, I am having difficulty seeing how the OP could have measured both L and Cp at a single frequency. That puzzlement was enough to keep me from answering with a calculation along the lines you suggest. To make sense of his numbers, we need to know how they were obtained. I expect that the LC meter has been used in different ways to get those

2 (or 3) values, or it uses a few frequencies itself. Either way, that calculation is not as simple as you suggest.

Best regards,

--
--Larry Brasfield
email: donotspam_larry_brasfield@hotmail.com
Above views may belong only to me.

The AADE meter is just an oscillator; the "unknown" shifts its frequency. On the "C" range the circuit config is different from "L" measurement, and the test frequency will be different, so the results can be strange.

It's a great little instrument for clean L or C measurement, but it's easily fooled by complex stuff.

John

My suspicion has been that the meter may not be giving accurate values due to the fact that I'm measure both L and C together (I'm not sure it was designed for that).

However, the L and C measured without the external cap results in the correct resonance frequency.

Can someone suggest a better way to build a model for the antenna and arrive at its values?

"John Woodgate" wrote in message news: snipped-for-privacy@jmwa.demon.co.uk...

wrote (in

does capcitance go down when adding parallel cap?)',

seeing how the OP could have measured both L and Cp at

a calculation along the lines you suggest.

expect that the LC meter has been used in different ways

way, that calculation is not as simple as you suggest.

nF, we don't **need** to query his data for the loop

I agree completely, and your explanation of inductance canceling capacitance is quite appropriate in this case.

complications into answers to newbies or less

concern about this has reached the trigger point for

Well, I appreciate your restraint. However, at the risk of taxing it some more: It was you who introduced the idea that the OP's measurement frequency could be deduced and his seemingly anomalous result predicted by doing a calculation based on values he measured. When he gets confused by your invitation to "do the arithmetic", I believe you should share the credit with me for that. The thrust of my post was to suggest that such calculations may be trickier than is apparent from your post.

Taking your point more generally, I can only say that judging what complications are necessary (or maybe helpful) rather than just likely to confuse is difficult. We necessarily tread in the region between what OP's aleady know and what they need to learn, usually in the face of poorly described problems.

Best regards,

--
--Larry Brasfield
email: donotspam_larry_brasfield@hotmail.com
Above views may belong only to me.

My goal is to be able to model the antenna accurately.

I would like to be able to reasonably estimate the amount of capacitance needed to get the thing to resonate at 60 khz. It very easy for me to just start adding capacitors by trial and error until I get there, but that doesn't really teach me much.

The measured values (no capacitor) work pretty well. From what I've studied, resonance should occur when the two impedances equal one another, so F =1 / sqrt(LC) = 438 khz. I measure a resonant frequency of 442 khz. Not bad! This leads me to believe that the L/C meter can indeed give the right answer for L and C in cases where they both exist. Is this a bad conclusion?

So, I'd really like to learn here. In adding the eternal 10,000 PF capacitor.. why did the total measured capacitance decrease? Is it an error in the instrument? or something else?

Should my model consist of:

A) (A capacitor in series with an inductor in series) in parallel with a cap

or

B) (A capacitor in series with an inductor in series with a cap) in parallel with a cap

?

I realize that the six loops of coaxial will add some capacitance due to thier geometry (being next to each other). Should that be considered a series or a parallel capacitance? Is a good model possible, or am I searching for something that will not be found?

Thanks to anyone willing to help here.

Stephen

Are you interested only in modeling it at frequencies near 60 KHz, within maybe +/- 10%? An LCR model is probably reasonable in that case.

If you measure the first resonant frequency and Q, then measure inductance at a much lower frequency, you (or many folks here) should be able to calculate the capacitance needed to tune to 60 KHz. That will be a simple reactance or admittance summing and differencing calculation. (I assume here that you have a simple loop antenna, not the thing you described in your initial post.)

If you want to learn antenna theory, this is not the best place to go, except for some pointers to good sources for that knowledge.

Strictly speaking, the impedances are complex conjugates at resonance. This will matter when you do the tuning calculations.

It is a dangerous generalization. It may not apply to devices operating at other frequencies. I think Mr. Larkin's note on that subject is pertinent.

As others have stated, there is some reactance cancellation going on. Without knowing more about how your LC meter works, it is difficult (and impossible for me) to explain in detail why that result is obtained.

I doubt your antenna looks like a trap. That circuit would.

No, for the same reason. (And the two caps within the parentheses are redundant for a model.)

Given that you intend to operate the loop quite a ways off of its resonance, I think the loop model that does you the most good will most resemble an inductor with series and parallel resistance to reflect radiation and heating losses. Those will probably not be the same values that model the antenna at it natural resonance.

I have to skip that question. Your arrangment of coax is confusing to me. Your shield hookup is unclear, and its purpose a mystery. What are you trying to do that simple wire could not do?

You're welcome, FWIW.

--
--Larry Brasfield
email: donotspam_larry_brasfield@hotmail.com
Above views may belong only to me.

I read in sci.electronics.design that Stephen wrote (in ) about 'Trying to model loop antenna (why does capcitance go down when adding parallel cap?)', on Mon, 25 Apr 2005:

The inductance cancels out some of the capacitance. At resonance, it cancels all of it, and you would measure just 4.2 ohms resistive.

We could even tell from your data what frequency you used to measure. The total actual capacitance is 10421 pF and the net measured value is

7055 pF. The difference is 3366 pF. This has the same, but opposite, reactance as 306.1 uH at frequency f = 1/{2pi*sqrt(LC)}. You can do the arithmetic.

--
Regards, John Woodgate, OOO - Own Opinions Only.
There are two sides to every question, except
'What is a Moebius strip?'
http://www.jmwa.demon.co.uk Also see http://www.isce.org.uk

I read in sci.electronics.design that Larry Brasfield wrote (in ) about 'Trying to model loop antenna (why does capcitance go down when adding parallel cap?)', on Mon, 25 Apr

2005:

To get a newbie to understand why adding 10 nF doesn't give him a measured 10 nF, we don't **need** to query his data for the loop alone too closely. We just need to explain in principle the result he obtained.

I don't wish to suggest that you are the only one introducing unnecessary complications into answers to newbies or less experienced enquirers, still less that you are the worst offender, but my concern about this has reached the trigger point for ranting about it.

--
Regards, John Woodgate, OOO - Own Opinions Only.
There are two sides to every question, except
\'What is a Moebius strip?\'
http://www.jmwa.demon.co.uk Also see http://www.isce.org.uk

I read in sci.electronics.design that Larry Brasfield wrote (in ) about 'Trying to model loop antenna (why does capacitance go down when adding parallel cap?)', on Mon, 25 Apr 2005:

Have a Turkish lira. No, go on, I've got plenty.

The time to introduce that is when/if he comes back to say that the calculated frequency doesn't gel with the actual frequency he used. At that point, we can be reasonably confident that he is receptive to more explanation.

Unless, of course, his response is abusive.

--
Regards, John Woodgate, OOO - Own Opinions Only.
There are two sides to every question, except
'What is a Moebius strip?'
http://www.jmwa.demon.co.uk Also see http://www.isce.org.uk

I read in sci.electronics.design that Stephen wrote (in ) about 'Trying to model loop antenna (why does capacitance go down when adding parallel cap?)', on Mon, 25 Apr 2005:

You appear to be 'thinking complicated'. A diagram is very helpful. Use Courier font:

o----+----+----, | | | Cext Cloop Lloop | | | o----+----+----'

In a previous response I explained why you measured less capacitance than you expected.

--
Regards, John Woodgate, OOO - Own Opinions Only.
There are two sides to every question, except
'What is a Moebius strip?'
http://www.jmwa.demon.co.uk Also see http://www.isce.org.uk

I read in sci.electronics.design that Larry Brasfield wrote (in ) about 'Trying to model loop antenna (why does capacitance go down when adding parallel cap?)', on Mon, 25 Apr 2005:

The self-resonant frequency is very much higher than 60 kHz. It is irrelevant. I'm beginning to think that you, too, can only 'think complicated'.

With such a small inductance tuned with something like 10 nF, the Q will be high enough to disregard, at least for the first try-out.

--
Regards, John Woodgate, OOO - Own Opinions Only.
There are two sides to every question, except
'What is a Moebius strip?'
http://www.jmwa.demon.co.uk Also see http://www.isce.org.uk

resistance. If it is a decent antenna, this will >be the dominant term, and will degrade the

resistance. So no, I don't think Q

This antenna being a small loop (>1/10 wavelength) will have a very small radiation resistance, less than the loss resistance. Stephen may want run this short program named RJELOOP3 located at

It says you need about 50,000 pf to tune to 60,000khz. Calculations are somewhat complicated by the sheild. Mike

Yes. Actually there a few sites that talk about make just such an antenna and they also say that 50,000 pf is about right.

Some have asked about the way the antenna is made.... I use coaxial instead of regular wire so as to have a shield. I only want the magnetic portion of the signal (coming all the way from Colorado). All electrical field pickup will be local noise so I want to get rid of that. To accomplish this I use the outer conductor of the coaxial as a shield. It is broken in the exact middle of the length of cable so that it well not be a loop. I then take both sides of the out conductor and attach them to my ground. If anyone is interested, a better explanation can be found at

So, lets look at some numbers for fun. I add a 9750 pF capacitor in parallel with the antenna and get a measured resonant frequency of

130khz. The calculations say that is should be 1 / sqrt( 306.1uH * (421 pF + 9750 pF) ) = 90,200 kHz. So, I'm either measuring L and C incorrectly or our simple model is wrong. My guess is that its the former. So, I need to think through how to made these measurements without using the L/C meter. That will probably be a good exercise.

I think everybody is missing an important point here. This isn't just a resonant circuit - it is an antenna. That means that there is a major resistive term missing - the radiation resistance. If it is a decent antenna, this will be the dominant term, and will degrade the effective Q of the circuit massively. Certainly it will overpower the loss resistance. So no, I don't think Q will be high enough to disregard.

d

Pearce Consulting

"John Woodgate" wrote in message news:uZlP+ snipped-for-privacy@jmwa.demon.co.uk...

(in ) about 'Trying

cap?)', on Tue, 26 Apr 2005:

resonant circuit - it is an antenna. That means that there

electronics set) has appreciable RR.

John, to mitigate the appearance of nitpicking your posts, please appreciate that I read them for education, not just to find problems. At any rate, this is not a nit, IMHO.

The OP wants to model his antenna. I certainly agree that, from the electrical port, the radiation resistance will likely be buried in the internal losses, maybe to the point of indetectability. But the RR loss, small as it might be, represents the useful power output of the antenna, (or its ability to collect radiation). To me, that seems to be something one would want in an antenna model. If its loss is vanishingly small relative to the internal losses, that is worth knowing as it implicates the noise performance of the receiver/antenna combination. A meaningful value for the RR might motivate selection of a less lossy coax.

--
--Larry Brasfield
email: donotspam_larry_brasfield@hotmail.com
Above views may belong only to me.

I read in sci.electronics.design that Don Pearce wrote (in ) about 'Trying to model loop antenna (why does capacitance go down when adding parallel cap?)', on Tue, 26 Apr 2005:

Gordon Bennett! This is 60 kHz. NO practical antenna (except for the US Govt's electronics set) has appreciable RR.

--
Regards, John Woodgate, OOO - Own Opinions Only.
There are two sides to every question, except
'What is a Moebius strip?'
http://www.jmwa.demon.co.uk Also see http://www.isce.org.uk

I read in sci.electronics.design that Stephen wrote (in ) about 'Trying to model loop antenna (why does capacitance go down when adding parallel cap?)', on Tue, 26 Apr 2005:

What you have described is quite right. It does make a loop with rather high self-capacitance, but since you need a much larger tuning capacitor, it doesn't matter.

Now you've partially justified Larry B's caution. When you measured the inductance, (unless you did it at audio frequency e.g. 1 kHz or 10 kHz), the capacitance you also measured has influenced the result. I suggest you don't worry about this, but bear it in mind for the future. Just add more capacitance and measure again. 15 nF (15 000 pF) seems to be indicated: fmeasured/fwanted = sqrt(Cwanted/Cmeasured). KISS!

--
Regards, John Woodgate, OOO - Own Opinions Only.
There are two sides to every question, except
\'What is a Moebius strip?\'
http://www.jmwa.demon.co.uk Also see http://www.isce.org.uk

You are about the dumbest p.o.s. plaguing NG to date- the OP just told you he used his L/C meter- he put it in L mode to measure inductance and C mode to measure capacitance- what else could he have done, stupid? All your other posts in this thread are pure rot as usual too. Think about going away and staying away.

Okay- so antennas are yet another subject about which you understand absolutely nothing. Keep working your mouth, idiot- that is all you have, sorry-assed p.o.s.