[ARC5] QST articles.

[Ian Wilson]

Transform the series Ca, Rr to an equivalent parallel
combination. Add Cp to the transformed C. Transform
the result back to a series RC. Now you have a simple
LCR circuit. The L and C disappear at resonance. Rr
is transformed into a smaller equivalent value that provides
the correct magnitude load that the ARC-5 TX wants.

I have measured the output into a 50 ohm load using this
(and other) matching technique(s). I also have (somewhere)
a SPICE model including the variable mutual inductance
of the link coupling; this gives reasonable agreement with
my measurements (all at around 7050kc).

73, ian K3IMW

On 12/15/08, J Forster <jfor at quik.com> wrote:
> I would think so. Look at the circuit this way:
>
> Tx ------CCCCCCC-----o---------
>                      |        |
>                      |        |
>                 Cp =====    ===== Ca
>                      |        |
>                      |        Z Rr
>                      |        Z
>                      |        |
> GND -----------------o---------
>
> Cp is parallel capacitor
> Ca is antenna capacitance
> Rr is radiation resistance
>
> The capacitors form a Current Divider. The Roll-a-ductor resonanates the
> parallel combination of Cp+Ca; . Cp >> Ca
> Little gets out, IMO.
>
> -John
>