[TowerTalk] Bazooka 1/4 wave balun

[Jim Lux]

At 09:25 AM 7/7/2005, Michael Tope wrote:
>This discussion brings up a good question. What is the best what
>to make accurate inductor Q measurements? I understand that there are 
>Boonton Electronics Q meters out there. How do those
>compare with other techniques like using a vector network analyzer?  I see 
>people talking about Q's of 400 to 800 for HF loading coils. How easier or 
>hard is it to measure Q's in that
>range?
>Thanks,
>
>Mike, W4EF......................................
Tom will certainly pitch in with his description of his test fixtures, 
which I think are a big part of the challenge.

Here's the problem.. A Q of 500 implies that the resistive component is 
1/500th that of the inductive component, so you're basically faced with 
measuring 0.1% kinds of changes.
Say you put a constant current through the device, and you measure the 
voltage across it (which is how the lower frequency LCR meters work these 
days).  You need to measure the phase of that voltage relative to the phase 
of the current to tiny fractions of a degree, which is VERY 
challenging.  One commercial device I'm familiar with uses a 16 bit A/D on 
the reference and measurement ports and then digitizes the waveform at a 
fairly high rate.    They essentially digitally filter it to get a very 
narrow band measurement (like using a lock-in amplifier). 16 bits gives you 
1 part in 65000 voltage measurement, which is in the ballpark.

If you can put it into a resonant circuit, then you can measure Q by 
measuring 3dB points, turning the measurement problem into one of 
accurately matching levels (at the two points), and measuring frequency 
accurately (which is fairly easy).

A resonant circuit with a Q of 500 at 14MHz would have a 3dB bandwidth of 
28 kHz. The challenge would be accurately knowing when you're really at 
3dB. And, how would you couple in the power in a consistent way? A single 
turn link coupling probably wouldn't cut it.

Probably some sort of reflectometer scheme would be best, with a resistive 
hybrid.  You calibrate with a short, and open, and a load.  At least for 
high Q, the bandwidth is narrow, so instrumental variations will be small.

There's a paper that talks about this, with a discussion of the problems in 
getting good standards (for these guys IF means 10 kHz to 100 MHz) because 
neither lumped nor transmission line models work particularly well:

Maurice G. Cox, et al., "An Interpolation Scheme for Precision Intermediate 
Frequency Reflection Coefficient Measurement", IEEE Trans on Inst and 
Meas., v52, n.1, Feb 2003, pp27-37



>----- Original Message ----- From: "Tom Rauch" <w8ji at contesting.com>
>To: "Jerry K3BZ" <k3bz at arrl.net>; "Jim Lux" <jimlux at earthlink.net>
>Cc: <towertalk at contesting.com>
>Sent: Wednesday, July 06, 2005 11:56 PM
>Subject: Re: [TowerTalk] Bazooka 1/4 wave balun
>
>
>>>One way is if you have an air core inductor that you can
>>measure the loss
>>>of (say you've got it resonated with a capacitor at 14
>>MHz). Shove the
>>>suspect hunk of plastic in the middle and see if the Q of
>>the coil
>>>drops.  If not, you're good to go.
>>Space winding an inductor reduces turn-to-turn capacitance
>>and increases inductor Q. Fill the area between turns in
>>with a dielectric and Q decreases.

Inserting a dielectric in the middle of the coil won't change the current 
distribution in the wires very much, especially if the coil is closewound, 
and I think that the proximity effect/skin effect is what determines the AC 
resistance of the coil (and the Q of a tuned circuit including it.



>>   That's because
>>capacitance and circulating currents increase.

You're saying, then, that without the dielectric, you have some L C and R, 
forming a resonant circuit, with Q = Xl/R, and Xc=-Xr at resonance

Put in the dielectric, so C increases. To keep resonance, the resonant 
frequency would drop.  So Xl will be smaller. R will be about the same 
(assuming a lossless dielectric for now), making the circuit Q 
(Xl/R).  Hmm.. if it were only skin effect, the resistance would change 
roughly as the square root of frequency, so the R would actually be a bit 
lower)

Of course, typically, you'd resonate the coil with an external capacitor, 
so the change in Fres due to the dielectric would be proportionally smaller 
(of course, so would the Q killing loss in the dielectric)

Anyway, it's proposed as a qualitative test to separate OK from hideous..



>>  Knowing that,
>>how do we quantify how much Q loss is caused by an increase
>>in turn-to-turn capacitance as a dielectric is brought near
>>or against the turns and how much is lost through power
>>dissipation in the dielectric? How many people can measure
>>Q?