[ARC5] QST articles.

[J Forster]

The current divider ratio does not transform out as you say.

-John

============

Ian Wilson wrote:

> 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
> >