To: | "[email protected]" <[email protected]> |
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Subject: | LF: 9kHz |
From: | Rik Strobbe <[email protected]> |
Date: | Thu, 25 Feb 2010 10:48:47 +0100 |
References: | <001c01cab575$6bbb0460$0401a8c0@xphd97xgq27nyf> |
Reply-to: | [email protected] |
Sender: | [email protected] |
Stefan, did you have a look at the graph Jim (M0BMU) showed yesterday ? This may indicate that in an open area you might be better off than 1/f, maybe in the 50..100 Ohm range. Last night I was mulling about Alexanders remark that a parallel C would increase the losses. At first though I would say that the parallel C just performs some kind of impedance transformation without inducing additional losses. And this is true if the loading coil would be perfect (no losses). But for a real world situation (ie. assuming a constant Q = 250 for the loading coil) Alexander is right: the transformation caused by a large parallel C causes extreme high currents in the loading coil, and thus higher losses. Simulation (with SIMetrix) confirms this. Taking your antenna (580pF / 50 Ohm) 1. Without parallel C you would need a 550mH loading coil. At Q = 250 the coil loss would be 130 Ohm. Putting 100W into this system would result in an antenna current of about 750mA and antenna voltage of 23kV. 750mA trough the loading coil means a dissipation of 73W. 2. With a 3.2nF parallel C you would need a 85mH loading coil. At Q = 250 the coil loss would be 20 Ohm. But putting 100W into this system would result in only 330mA antenna current, while the current through the loading coil would be 2.1A (= 88W dissipation). The reduction in antenna current (750mA to 330mA) would lower the ERP by 7dB. In addition the parallel C will decrease the antenna bandwidth significantly, making it difficult to tune. 73, Rik ON7YD - OR7T At 01:14 25/02/2010, you wrote: Here i agree. And here i understand "dreamers" in a positive and optimistic way, just like children are dreamers! |
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