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HIGH Q CAPS FOR VLF LOOP ANTENNA?
Thanks Richard,
I have a bunch of reading to do. Appreciate your time and the explanation that you gave. My receiver has a 2 to 3 ohm input impedance, my hope was to use a series tuned loop that was entirely floating (with no ground anywhere) and feed it to the house with a balanced line. Since it's a short run to the house (in terms of wavelength, I had hoped the impedance mismatch between the 90 ohm impedance twisted pair transmission line would not produce a big loss. The receiver input is also untuned, so the only selectivity available to it will be the antenna selectivity, which I think I can get away with since the antenna is relatively high Q. Again, thanks for the info and for the web resource. I have some more studying to do! Regards, T On Tue, 25 Oct 2005 23:01:01 -0400, TRABEM wrote: No, I don't understand this. I thought a shielded loop meant the loop antenna wire was shielded by the copper (non-ferrous) surrounding the wire. It is not an effective antenna shield if it is wholly continuous - and it is not, it has a gap opposite the mounting point which is generally at ground/reference potential. Part of the point of being "shielded" is to enforce a symmetry and that ground/reference is electrically neutral as long as you guarantee it is equidistant both sides around the loop to that gap. The shield tends to protect the wire from electrical field inputs and allows it to only respond to magnetic field variations. There is no such thing as "only" magnetic fields variations. I thought the capacitance between the wire and the surrounding shield material represented a loss in Q, Q is a simple relation between loss and storage. Lower Q for the same storage (be it in a capacitor or an inductor) can only result from resistive loss of Ohmic conduction or radiation. Any loss attributable to a capacitor is conductive loss - hence the discussion of ESR. You would have to go back to the stone age of electronics with paper and wax dielectrics to find loss BETWEEN the plates. Equivalent Series Resistance for garden variety capacitors, when compared to radiation resistance, is not trivial. That is, unless, you swamp that loss by putting your loop in the closet with your mothballed summer wardrobe or burying it in the garden mud. Design for failure is easily achieved if you need a rationale to ignore simple considerations. Consult: http://www.w8ji.com/magnetic_receiving_loops.htm There are countless horror stories about those attempting to use surplus hardline as shielded loops on LF and VLF, all with disappointing results. Such disasters that arise are one of two possible scenarios: 1. They don't have a gap (short circuit city); 2. They don't guarantee symmetry (poor balance, poor tuning, poor response). The predominate attitude was that the capacitive coupling between the wire and the shielding material was the cause. I don't say the predominate attitude is correct.but, if it is a false assumption, then I am not the only one who needs revision:)) We get that traffic - yes. They suffer the same learning slope. If the copper pipe IS the antenna, then why have the wire inside it at all?? Because you have to have a conductor pair back to the receiver. The grounded "shield" serves as one half of the pair, the other spans the gap connecting to the other half's "shield" (it looks like you are shorting the inner conductor to ground) to thus pick up the opposite potential. The voltage across the gap is thus sensed and it only takes one wire. Look closely at any such standard "shielded" loop. The sense of what is being shielded is THAT conductor which you contrive to keep in a controlled environment (a coaxial shield) and away from the imbalance of nearby capacitive couplings. The "shielded" inner conductor spans a very small distance whose opposite poles' capacity is balanced to all neighboring paths to ground. That is, unless you push one side up against the wall. Stretch out the gap of the shield loop and you have a conventional dipole. A conventional dipole exhibits high Z and high V at its tips. The middle of such a dipole has a low Z and a high I. With respect to both ends, the middle is neutral and strapping it to a conductor does nothing to change that topology (and is a common tower mounting benefit). Being curved into a loop does not change this and allows you to connect your transmission line to both sides without greatly exposing a significant length of the transmission line (and thus forcing an unbalance and upsetting the applecart). This dipole is obviously very small with respect to its wavelength and thus some form of end loading is required. Thus the capacitor arrives on the scene. The circulating currents and potentials become astronomic for progressively smaller antennas. Those currents flow through and to the plates of the capacitor. If you don't choose the right components for that capacitor (and manufacturers of HF loops like to crow how they achieve this) then your design efficiency goes TU. Hence to speak of capacitor Q is not appropriate as the correct term is D (dissipation factor). It is certainly related (an inverse relation) and despite comments to the contrary, D is resolvable with standard bridges (although those bridges are of considerable design sophistication in maintaining balance and their own shielding - not a trivial matter). There are simpler ways of achieving the same thing by building a completely exposed loop with capacitor (still paying attention to the ESR and keeping the whole shebang out of the mud), and simply building a shielded coupling loop. Reg has adequately described this before many times. 73's Richard Clark, KB7QHC |
HIGH Q CAPS FOR VLF LOOP ANTENNA?
I could perhaps scan the relevant pages of the references I
mentioned... How are you planning to couple your 2-ohm load to your loop without doing really bad things to its Q? (And just what sort of detector do you have that represents a 2 ohm load?) Cheers, Tom |
HIGH Q CAPS FOR VLF LOOP ANTENNA?
"K7ITM" wrote in message
ups.com... I could perhaps scan the relevant pages of the references I mentioned... How are you planning to couple your 2-ohm load to your loop without doing really bad things to its Q? (And just what sort of detector do you have that represents a 2 ohm load?) Cheers, Tom He probably measured the DC resistance at the antenna input connector. If there's an inductance path to ground, then that's probably what he measured. The DC resistance is NOT the RF impedance of the input. Cheers!!!! -- Dave M MasonDG44 at comcast dot net (Just substitute the appropriate characters in the address) Never take a laxative and a sleeping pill at the same time!! |
HIGH Q CAPS FOR VLF LOOP ANTENNA?
On Wed, 26 Oct 2005 20:58:26 -0400, "DaveM"
wrote: "K7ITM" wrote in message oups.com... I could perhaps scan the relevant pages of the references I mentioned... How are you planning to couple your 2-ohm load to your loop without doing really bad things to its Q? (And just what sort of detector do you have that represents a 2 ohm load?) Cheers, Tom He probably measured the DC resistance at the antenna input connector. If there's an inductance path to ground, then that's probably what he measured. The DC resistance is NOT the RF impedance of the input. NO. Cheers!!!! |
HIGH Q CAPS FOR VLF LOOP ANTENNA?
On Thu, 27 Oct 2005 00:36:07 -0400, TRABEM wrote:
NO. The group deserves a better specification for the input Z of your amp than that. 73's Richard Clark, KB7QHC |
HIGH Q CAPS FOR VLF LOOP ANTENNA?
On 26 Oct 2005 09:27:44 -0700, "K7ITM" wrote:
I could perhaps scan the relevant pages of the references I mentioned... No, I can get them att he schools library I think. thanks for the offer. How are you planning to couple your 2-ohm load to your loop without doing really bad things to its Q? Is it better to convert the loop to a higher impedance just to feed it into the house? It appears that anything I do is going to knock the heck out of the antennas Q though. I have not decided whether to mount the receiver at the antenna yet, or whether to run the twisted line directly into the house from the antenna (since it's a short run). Most likely it will have a short run of cat 5 cable going from the antenna to a 1 to 1 toroid transformer located in the receiver. The only selectivity for the receiver will be the antenna itself. The receiver is very small, and uses very little power, so it's pretty feasible to mount the entire receiver at the antenna and run a balanced line feed of the audio into the house. (And just what sort of detector do you have that represents a 2 ohm load?) Is it better to convert the loop to a higher impedance just to feed it into the house? It's an analog switch input, modified by my neighbor that gave me one of them. The switch vendor says the switch series resistance should be around 3 ohms, but it measures around 2.5 ohms. Probably is a little lower than expected due to the integrating capacitors (.1 uF) which are hung on the output of each switch. The .1's go to ground. I measured it twice, once with a 1:1:1 isolation transformer and once with a 6:1:1 isolation transformer....The tester looses accuracy at low impedances, so we repeated the measurement with the generator feeding the high impedance side of the a transformer also. I got nearly the same reading after correcting for the transformers impedance step down value, since both readings agree pretty well with the switch vendors ratings, it's very likely that the receiver input impedance is around 2 ohms. (And just what sort of detector do you have that represents a 2 ohm load?) Cheers, Tom |
HIGH Q CAPS FOR VLF LOOP ANTENNA?
On Wed, 26 Oct 2005 22:07:48 -0700, Richard Clark
wrote: On Thu, 27 Oct 2005 00:36:07 -0400, TRABEM wrote: NO. The group deserves a better specification for the input Z of your amp than that. Yes, of course they do Richard. However, the reply was more than Dave deserved after the statement that he made. Maybe that's how he measures receiver input impedance, I certainly don't measure it like that. You should have the schematic in your mailbox by the time you get this message. I can't post a schematic here. T 73's Richard Clark, KB7QHC |
HIGH Q CAPS FOR VLF LOOP ANTENNA?
The point is that the loop's inductive reactance is on the order of 10
ohms in the frequency range you're talking about. When resonated with a capacitor, if the Q is, say, 300, then the impedance at resonance will be about 3000 ohms, resistive, as seen across the capacitor. Reg's program gives you an estimate of what it will be. If you put a low-resistance load across that, the Q will drop drastically. And if you put your 2 ohms (which it won't be at the received frequency, if I understand what you have) in series with the loop and capacitor, it will also drastically lower the Q. So my question remains: how will you couple to the loop and maintain the Q? When you measure the input impedance of your detector, you should do it while the detector is operating, and do it versus frequency. I expect you'll see a large increase in impedance at the operating frequency. Cheers, Tom |
HIGH Q CAPS FOR VLF LOOP ANTENNA?
On Mon, 24 Oct 2005 18:03:49 -0400, TRABEM wrote:
Assume the wire diameter is a conservative thick 2mm. Assume nothing Reg. Well T, reading through the thread you seem to be real short on relevant information (ie you don't adequately anticipate the information people need to answer your questions), then very ready to deal abruptly with people who make the wrong assumptions about the context. Owen -- |
HIGH Q CAPS FOR VLF LOOP ANTENNA?
On Thu, 27 Oct 2005 02:02:53 -0400, TRABEM wrote:
You should have the schematic in your mailbox by the time you get this message. If I am to translate the annotation correctly (and it is obviously in error - R10 and R12 are not the pair being changed), then Input Z somewhere in your frequencies of interest (you've named several) runs around 10 Ohms with the switch itself attenuating your signal by 6 to 10dB. This, of course, says nothing of the abysmal match to the antenna whose Q will be buried in potter's field. I gather that the switch operates as a baseband quadrature/synchronous detector. It looks suspiciously like it will either short the input, or leave one half dangling, but I trust you got the schematic from a good source. I won't bother to try to verify the logic. The roll-off frequency of the amplifier(s) is at the bottom of the AM band, it would do better to track the oscillator frequency about one seventh below so that AM signals are depressed below WWVB instead of competing with it. Increase the caps from 470pF to 3300pF. The reason why you want low front end Input Z is to satisfy the amplifier topology (the gain will be roughly half what it is probably specified at). This could have been done better in half a dozen different ways with the same active parts. The problem here is some bozo marked the input "antenna" and removed the necessary follower amplifier that would have been fed by the antenna which would in turn feed this circuit through the transformer. This sucker, as drawn, is going to be deaf, deaf, DEAF. If you hear WWVB it will be by virtue of their strong power, not by any quality of design here. To unplug this design's ears and make up for the massive goof, add a FET follower. Load the FET drain with a 2.2 Ohm resistor and make sure you couple the signal through a large cap feeding the transformer. Also, bias the FET on with a hi Z divider so you don't wipe out the Q of the antenna. I will bet that even the proposed "I" and "Q" paths are mislabled or missapplied. 73's Richard Clark, KB7QHC |
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