"Virgil Smith" wrote in message news:3iNHb.504277

Anyone know of any neat ideas for measuring the Q of a
crystal?

I would welcome any and all suggestions,

-vs-

Let's see if I can describe a method based on W7ZOI's efforts, with
some updates by K8IQY. By the way, the W7ZOI techniques from QST were
added as appendixes into one of Doug Demaw's ARRL books. Demaw
integrated Hayward's various test jigs into one crystal tester. You
can get a board from Far Circuits.

You put the crystal in a specific (typically 50 ohm) environment by
putting 50 ohm attenuators between generator and crystal and between
crystal and detector. The detector might be a diode detector plus
voltmeter or an oscilloscope. You'll also have another 50 ohm 3 dB
pad in series that can be switched in and out.

1. With the pad switched IN, adjust frequency until the peak output
is read (resonance). Note the center frequency and meter indication.

2. Switch the pad OUT and find the frequencies above and below the
center frequency that give the same indication. These are the + and –
3 dB frequencies and the difference between them is the bandwidth.

Now you can calculate Q by dividing the center frequency by the
bandwidth. Pretty low isn't it? This is a loaded Q, not the
crystal's Q. The total resistance in the circuit is 50 ohms source
plus 50 ohms load plus the crystal's yet unknown Rs.

3. Tune back to the center frequency where the output is at its peak.
Unplug the crystal and substitute a pot of maybe 0 to 25 ohms range.
(I actually just substitute fixed resistors since I don't have a
decent pot like this.) When you find a value that gives the same
meter indication as the crystal did, that is the value of Rs. (This
is because we are at series resonance, so Xc and XL have cancelled.)

Alright. Another formula for loaded Q is XL / R. R in this case is
100 ohms plus the Rs you just measured. And XL is 2*PI*Fc*L, where Fc
is the center frequency you measured earlier. You already have the
loaded Q from step 2, so you are down to one unknown. Just solve for
XL and/or L, which is the equivalent inductance of the crystal. In
like manner, you can find the series C, since Xc = XL.

And finally, the Q you wanted (the crystal's Q) is XL / Rs.

Now you know about everything about the crystal and can get one of
your Hayward or Demaw books and start designing some filters. Hayward
has nice software for this in both Introduction to RF Design and
Experimental Methods in RF Design, both from ARRL. If you want to
include holder (parallel) C in your modeling, 5 pF is a good estimate.

Jim Kortge, K8IQY puts his crystal in a 12.5 ohm environment by using
4:1 (impedance) bifilar transformers into and out of the crystals (50
ohm attenuators are still used though, to assure known generator and
load resistances). This has the advantage of making the resistance
measurement easier, but it makes the loaded bandwidth smaller, so you
need a good stable generator that you can read to the Hz. Those
little DDS units are nice. Be sure to use filtering if necessary for
signal purity. Jim designed a nice little VXO (variable crystal
oscillator) that uses one of the crystals from the set you are
measuring in the oscillator circuit. See it on NJQRP's page:

http://www.njqrp.org/pvxo/index.html

I followed Jim's lead but used resistive 50 to 12.5 ohm matching pads
instead of transformers. That way I got rid of any reactive effects
(which are minimal) of the transformers, but required a lot more
driving power due to the loss in the matching pads.

OK, that was probably more than you wanted to know. Hope I remembered
this stuff right.

73—Nick, WA5BDU
in Arkansas