[ARC5] ARC-5 TX output circuit modeling [long]

[Ian Wilson]

[Note: cross-posted to Glowbugs -- ian].

The following applies to a BC-459 (7-9.1MHz)ARC-5 TX. This is the one
without the RFC in the plate lead.

The relevant elements are as follows:

  - pair of 1625's in parallel, Class C operation
  - tuned tank circuit: inductance = 1.8uH (resonates with 287pF at 7MHz)
  - link circuit: 1-turn, inductance = 36nH
  - coupling: position of link is variable relative to tank coil
  - roller inductor: in series with link coil. About 18uH max inductance

The total 1625 plate current is something like 135mA at 600V, giving a 'beam
resistance' of 4.4k. Applying the ARRL rule gives a Class C plate resistance
of Rbeam/2 or 2.2k.

The tank + link transform the plate resistance down to a much lower
impedance. The link has reactance. The (variable) link position varies the
coupling coefficient between tank and link coils. If the coupling
coefficient were 1, and the link winding were tuned, the impedance
transformation would be given by the square of the turns ratio. I would
guess that the actual coupling coefficient ranges from 0 (link coil
orthogonal to the tank coil) to perhaps 0.75 [handwaving].

The antenna for which the ARC-5 was designed is electrically short. This
means that the radiation resistance is low, and the antenna looks like a
capacitive reactance. The roller inductor is provided to resonate out this
capacitance. The additional inductance of the link winding can be absorbed
by reducing the inductance by 36nH. (In practice, of course, you would just
tune for peak antenna current).

In order to drive a 50 ohm resistive load, we need to do two things:
  - match 50 ohms to the TX output resistance
  - resonate out the inductance of the link + roller inductor

The resistive part of the output impedance is around 4 ohms as far as I can
guess. We can transform 50 ohms to 4 ohms by adding a parallel capacitance
of about 1500pF. This has a series equivalent of 4 ohms in series with
1629pF.

In order to get rid of the unwanted inductances in the output circuit - the
link coil and the roller inductor - we need to add series capacitance. A
fixed capacitor will do, since we can adjust the roller as required. The
output circuit then looks like all these in series:
  - link coil
  - roller inductor
  - fixed capacitance
  - 1629pF
  - output load (50 ohms)

A fixed capacitor of 50-100pF will resonate within the range of the roller
inductor.

It is interesting to simulate this (so far lossless) network. A couple of
things become evident:
  - if the coupling coefficient is greater than about 0.5, the two resonant
    circuits become overcoupled, and you get a dip at the centre frequency
    (NOT what we want!)
  - the output network is a bandpass filter. The output at the second
    harmonic is about 60dB down from the fundamental

There are two primary sources of loss:
  - finite Q of the tank coil and roller inductor
  - series resistance of the [added] fixed capacitance, relay contacts, etc

A Q of 150 for the tank coil corresponds to a parallel resistance of about
12k. 25w output with infinite Q drops to 21w with Q = 150.

A Q of 150 for the roller inductor (resonating with 50pF, i.e. about 10uH)
corresponds to a series resistance of about 3 ohms. 25w output with infinite
Q drops to 13w with Q = 150.

A couple of conclusions:
  - the output resistance is so low that losses due the roller inductor and
    added fixed capacitor are important. [Perhaps this explains why the
    originally supplied fixed capacitor (in the antenna relay box, 50pF) was
    a vacuum type]. Use a low-loss capacitor of a relatively high value so
    that the roller inductance (hence series R) can be kept low
  - the roller inductor/fixed capacitor/parallel capacitor combination forms
    a pretty good bandpass filter for harmonic suppression

[I still have some tuning to do before publishing actual performance
numbers, but thought I would post this while it's still fresh in my mind].

73, ian K3IMW