Hello Alexander,
Let us see what will happen if you cancel out that 100nF Cp:
The overall C in the circuit is the C of the antenna, 1nF, so you need about
100* the L before. If we assume now, the resistance of L is 100 Ohm, the
overall resistance in the circuit is abt 1100 Ohm. Applying the same power than
before (Pin=1,1W (ground+coil loss before)) will give a current of 32mA
=sqrt(10) * I antenn before and thus the "radiated" signal is 10 dB stronger
than before, although the coil losses are 100* more, right?
73, Stefan/DK7FC
________________________________
Von: [email protected] im Auftrag von Alexander S. Yurkov
Gesendet: Mi 24.02.2010 22:35
An: [email protected]
Betreff: Re: AW: LF: VLF_8.79 kHz_grounding systems
>
> So, the coil losses decrease with it's L (assuming a given wire diameter,
> coil diameter and spacing and so on) and thus it is good to decrease the
> needed L.
You can decrease L only by condenser in parallel with a antenna. But then
only part of coil current will flow thrue the ant (and ground stake also).
Let's describe some numerical example. Let antenna capacity (denoted
as Ca) is 1 nF (large antenna!). Resistance of grounding we assume 1000
Ohm (huge resistance!). Parralel C (Cp) we assume to be 100 nF. Then L=3.2
mH. Let coil resistance is 1 Ohm (large coil!) and coil current I is 1 A.
Then coil loss is 1 W. Antenna current for Ca<<Cp is (Ca/Cp)I = 0.01 A.
Grounding loss is (0.01)^2 * 1000 =0.1 W. With such a huge ground
resistance (unreally huge!) ground loss is only 10% of coil loss.
Regards,
Alexander
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