> If the voltage on the output of the coil is higher than that on the
input then
> (for constant power) the current must be less.
> Nick
> G4WHO
Ah. This is one of the great mysteries of loading coils that
has always baffled
me. Does anyone have a good explanation?
Mike, G3XDV (IO91VT)
Are we in a Monty Python farce here ? Are you being serious ?
Nick are you trying to wind us up (and are you coming to the Forum on
Sunday week just up the road) ? Hopefully you're just being convoluted
:-)
Several amps of RF current through a (perfect lossless) coil of a
specified reactance gives a voltage across the coil that is 90 degrees
Out Of Phase with the current through it. 90 degrees phase shift equals
no power. For a properly tuned antenna system the reactance of the
coil equals that of the antenna capacitance in series with it, and hence
reactances cancel, giving purely the resistive losses in the rest of the
system. The network is defined from input terminal, through loading
coil, through antenna capacitance to gound.
No power has been dissipated yet.
eg. 270pF antenna capacitance at 137kHz = 5000 ohms reactance. This
reactance is tuned out with 5mH inductance. For an antenna current of
2.5A passing though this combination a voltage of 2.5 A * 5000 ohms =
7500 V appears at the top end.
For a real system take losses and resistance into account:
Loss resistance of coil - say 15 ohms
Ground resistance 85 ohms (includes proximity effects, but we'll keep it
simple and just call it ground)
Antenna radiation resistance 0.004 ohms
Total resistance is the sum of these ie 100.004 ohms (just call it 100
!) which appear in series with the L/C combination
Total power dissipated = 2.5^2 * 100 = 625 Watts
And purely for interest, power radiated is that in just the radiation
resistance = 2.5^2 * 0.004 = 0.025W
Also, antenna efficiency = Radiation resistance / total resistance =
0.004 / 100 = 0.004% = -44dB
And also, to complete the antenna analysis :
Q = Reactance / Resistance = 5000 / 100 = 50
So, approximate bandwidth of the antenna is F / Q = 137000 / 50 =
2.74kHz
All simple, basic, RAE level calculations and by measuing just a few
values the essential parameters of any antenna system can be easily
determined.
The above figures correspond quite closely to my antenna in dry weather.
The ERP IS about 20mW, the bandwidth IS about 2.5kHz, the other values
measure as above.
---------------
I also suggest that the different current measured at each end of a
loading coil, as mentioned in an earlier posting, could quite easily be
due to the slight mistuning when the extra hardware associated with the
ammeter is connected in circuit on the top side. To illustrate this,
just try touching the antenna connection with (very, very, very, very
well insulated) screwdriver and see what happens to Antenna current - in
a switching PA this will directly be related to PA current which is much
easier to measure. An extra stray capacitance of an estimated 2pF on
my antenna will detune the system by 500 Hz which is enough to shift
current by a few percent.
To properly measure the two currents, you should use two (identical)
meters connected in circuit simultaneously.
Those fortunate to have larger antennas, with say 500pF or more will
have lower Qs, larger bandwidths, lower voltages, less proximity effect,
smaller coils and less fun getting the antenna to work.
Andy G4JNT
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