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LF: Re: Re: LF RX loop

To: <[email protected]>
Subject: LF: Re: Re: LF RX loop
From: "Clemens Paul" <[email protected]>
Date: Fri, 17 Jun 2011 19:49:02 +0200
References: <[email protected]> <A73DB25E1DD9454A910A4747400BE5D7@JimPC>
Reply-to: [email protected]
Sender: [email protected]

Jim,

DK7FC's Q of 8 with 470n may be due to this - it implies Rloss of about 0.3 ohms, which seems much too high for a few metres of copper tubing, so something is definitely wrong here.

I would agree.
For 3,14m copper tubing with a diameter of 10mm
you can expect around 10 Milliohm RF resistance on 137kHz.
Given the C with 470nF the loop inductance is 2,872µH whis is 2,47Ohm.
So the loop itself *without C and connections* would have an unloaded
Q of 252.


73
Clemens
DL4RAJ

----- Original Message ----- From: "James Moritz" <[email protected]>
To: <[email protected]>
Sent: Wednesday, June 15, 2011 11:39 PM
Subject: LF: Re: LF RX loop


Dear Stefan, LF Group,

I did some experiments on single-turn loop Q when designing the "bandpass loops" (BPloops2.pdf at https://sites.google.com/site/uk500khz/members-files/files ). With a 1m x 1m square loop made of 15mm copper tubing, I got unloaded Q well over 100 with C about 400nF at 137k. This was one unusual example of where the capacitor dominates the loss in a tuned circuit - I got a much lower Q using a single, lower voltage metallised polypropylene capacitor than when using 4 x 100n, 1kV capacitors in parallel. I assume this is because higher voltage = greater electrode area connected in parallel = less resistance, and more capacitors = less resistance from the connecting leads and interface to the metallisation. I have noticed similar effects with high-current capacitors in QRO PAs. DK7FC's Q of 8 with 470n may be due to this - it implies Rloss of about 0.3 ohms, which seems much too high for a few metres of copper tubing, so something is definitely wrong here. Obviously, it is also very important to have excellent connections between tubing and capacitors in order to realise milliohms of resistance. Another possibility is that the loop is inductively coupled to something that has high losses - you need to keep it a metre or so away from other conducting objects when doing the measurements.

I assume there are different paths the signal comes from, this is why i cannot eliminate it completely, right?

Apart from what others have already said, the null can be degraded by RF currents induced in nearby conductors acting as parasitic antenna elements - this includes connecting leads, building structures, cables etc etc. 25dB is fairly typical I think.

I used a high mu toroid (ferroxube, blue material) to transform the primary side (=the loop) to 50 Ohm. I have done this by varying the secondary turn number until i achieved a maximum voltage at a 50 Ohm load at a given input signal.

Whatever the value of unloaded Q, the "maximum power transfer" theorem applies, ie. the source resistance of the antenna at resonance should be transformed to equal the RX input resistance to achieve the maximum signal power delivered to the receiver - this is what you did empirically. In this condition, the loaded Q should be half the unloaded Q , so in Stefan's case tuning should be very flat with a loaded Q of 4.

Is it important to terminate the loops transformer output with a 50 Ohm load in that case? On an oscilloscope i found that the level of DCF39 is higher when having no R connected to the output but the waveform looks much better / cleaner!

In my loop designs, I have made a trade-off by increasing the loading (reduced loaded Q ), which increases the bandwidth, but reduces the signal power delivered to the receiver. If you wanted to obtain maximum loop selectivity, you could reduce loading of the loop, which would also reduce the power delivered to the RX, so it is your choice... Whatever you do, the band noise at the output of a loop like this is only a fraction of a uV, so you either need a RX with a low noise figure at 137k (very rare!) or a preamp with a low noise figure. Actually, with the preamp in the BPloops article, simply connecting the 50ohm input directly across the parallel tuned single turn loop is quite a good combination (loaded Q about 15 for my loop, a little higher for Stefan's slightly smaller loop, but only assuming he can improve the unloaded Q to say 50 or more). The preamp has low enough noise to easily hear the band noise with these loops, but beware - some receivers have such bad sensitivity at 137k that further gain will still be needed.

I am quite pleased with the way the single-turn loops have worked out - they work just as well as multi-turn loops, but are almost impossible to de-tune, are less susceptible to unwanted capacitive coupling, easy to make weatherproof, are mechanically simple and robust.

Cheers, Jim Moritz
73 de M0BMU



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