Fw: [Antennas] Tuner Inductor Q

[Wes (N7WS) and Linda]

At 08:17 PM 1/4/2003 -0600, Sandy and Kees Talen wrote:
>You do need to short the unused end of the inductor to the tap
>for exactly the reason you indicated. I don't know what the 
>effect is on Q for your particular roller inductor, but rather 
>than guessing here are some quick results using a 40m ARC/5 
>coil wound on a ceramic coil form. Q of tuner air variables is 
>generally pretty high so that's not a consideration. 

The Q of capacitors is often ignored in calculations because it is much
higher than that of the associated inductors.  With very high Q inductors,
this isn't the case and the capacitor Q might well be a factor.

>
>Whole coil 14.7uH .....partial coil, tap at approx 5.7uH
>Q of whole coil at 7Mhz is "480" (good as it gets)
>Q of partial coil at 9Mhz, shorted to end "350"
>Q of partial coil at 9Mhz, not shorted to end "370"
>    ....however, because of the floating end, it becomes
>    extremely sensitive to any capacitance (and the point 
>    you made about high voltage). In other words, I'd always
>    short the unused portion for a HF tuner.

A saying leaps to mind: If it sounds too good to be true then it probably is.

How are you determining these Q values?  IMHO they are much too high.

I can't find my exact data, but I did measure (using a calibrated HP Q
meter) an E. F. Johnson roller inductor that was similar in size to the
ARC-5 roller and had the benefit of variable winding pitch. I don't recall
exactly the upper Q value but know that it was around 100~150.  What sticks
in my mind however, is that at the lower inductance settings, the Q was
down to 50 or so.  These numbers are similar to what Karl reported earlier
on Palstar.

>
>The main thing on roller inductors is to make sure the high
>band end has more turn spacing or the Q on 10/15m will drop
>like a rock. That's why using an ARC/5 inductor without an
>external coil for 10/15m is not a great idea.

Yes, that is why some rollers use variable winding pitch.

>
>Q of a coil on a good clean ceramic form is very close to
>air wound (hard to tell the difference, even at 30Mhz). By
>the way, CPVC and PVC is pretty good also. Anything 
>that can absorb moisture is terrible. The biggest factor I've
>found which affects Q is turn to turn capacitance (spacing
>between wires), especially at the higher frequencies where
>skin effect also comes into play). 

The author of the Eagleware modeling program (I know that's a nasty word
with some here) has developed a new model for inductors that refutes the
idea of turn-to-turn capacitance.  For an intutive example consider a coil
that has a lot of close-wound turns.  Must have a lot of turn-to-turn
capacitance, right?

Wrong.  The phase shift from one turn to the next will be small (there are
a lot of turns and the phase shift from end-to-end is fixed).  Therefore,
the voltage and phase difference between turns is also small, which in
effect shorts out the "capacitance" between turns.  (See: "Filters and an
Oscillator Using a New Solenoid Model", Randy Rhea, Applied Microwave &
Wireless, Nov. 2000, pp 30-42)

>
>The other point that I would make is that the tuner will be 
>much more efficient at the higher Q but the tuning will be
>SHARP ...maybe you don't want it to be too sharp.  

Hold on a minute.  A "broad" network can be obtained with high Q components.

Example 1:  A 50 ohm source, a 50 ohm load and a pi-network composed of two
shunt capacitors, Xc = -j50 and a series inductor, Xl = +j50 at frequency Fo.

Note: A pi-net is not necessarily (probably not) the best network to use in
this case but it does serve as a viable example.

Let Fo = 10 MHz, C = 318.3 pF and L = 0.796 uH. 

a)  With Qc = Ql = infinity, the 2:1 SWR bandwidth is 0 to ~12.5 MHz.

b)  With more reasonable values, Qc = 500 and Ql = 100, the BW is still 0
to ~12.5 MHz.

The component Q has almost no influence on the SWR bandwidth. Furthermore,
the efficiency is hardly affected by the change in unloaded Q either.  The
loss of course is zero for the first case and increases to 0.1 dB in the
second case.

Example 2: Change the load to 10 ohm (SWR 5:1). A reasonable match can be
found at 10 MHz with Xc = -j22.4 (C = 710 pF) and Xl = +j22.4 (L = 0.356
uH).  

a)  With ideal components, the 2:1 BW is ~ 8.0 to 11.5 MHz.  

b)  With reasonable components the BW is still ~ 8.0 to 11.5 MHz and the
loss is 0.14 dB.

So with exactly the same network topology, in Example 1a (ideal components)
the 2:1 BW is 12.5 Mhz. In Example 2b (lossy components) the BW is 3.5 MHz.

Lower Q parts do not always make a lower Q network.  In Example 1, the
overall network Q is 2.  In Example 2 it is about 2.7.  Not a lot of
difference, so how come so much BW shrinkage?

Dividing the network in two at the "virtual impedance"; in Example 1, the
virtual impedance is 25 ohm, the network is symmetric and both the input
and output halves have a loaded Q = 1.  

In Example 2, the virtual impedance is about 8.4 ohm, the input Q is ~ 2.2,
the output Q ~ 0.5.  Different parts of the network are operating at
different Q's. This BTW is about an optimum solution.  By improper
adjustment the situation only gets worse.  

The ratio of impedances to be matched and the network topology can have a
much greater influence on network performance than does the component Q.
That said, higher component Q is always desirable.

Wes  N7WS