[time-nuts] time-nuts Digest, Vol 127, Issue 5

[Fuqua, Bill L]

 OOPS. what was I thinking,
  I was trying to minimize the number of transformers and
slipped up. The inputs should be in opposite phase and
output in parallel with a bifiler wound toroid transformer to
provide balanced outputs, one to the crystal and one to the
neutralizing (phasing) capacitor. The advantage of the crystal filter
is that it provides much higher Q and to some degree, depending
on it's selectivity reduces sideband noise. The crystals are cheap
and if you like you can add more stages. No tuning required except for
the phasing capacitor. If you happen to have a source of precisely 
cut series resonant 10MHz crystals you could easily go to 100Hz 
bandpass or even less by using a small value  loading resistor. 
  A ladder filter could be used  but there is still coupling far off 
resonance thru a capacitive ladder network consisting of the 
crystal holders' capacitances and discrete capacitors. 
  An complex LC filter could be constructed but it requires 
a number of stages along with careful selection of components.
73
Bill wa4lav

 
Message: 4
Date: Wed, 04 Feb 2015 06:57:48 -0500
From: Charles Steinmetz <csteinmetz at yandex.com>
To: Discussion of precise time and frequency measurement
        <time-nuts at febo.com>
Subject: Re: [time-nuts] 5>10 doubler
Message-ID: <20150204145756.vsAmCHQL at smtp2m.mail.yandex.net>
Content-Type: text/plain; charset="utf-8"; Format="flowed"

Bill wrote:

>Push-Push Jfet amplifier with parallel inputs and a Toroid output
>transformer, no secondary along with a simple filter using a 10 MHz
>series resonate crystal connected to one drain and an adjustable
>capacitor connected to the other would work fine. You connect the
>other ends of the two together and a loading resistor to
>ground.  The capacitor is used to neutralize or null out the shunt
>capacitance of the crystal so that a capacitive path for the other
>frequencies , 5, 15, 20, etc is eliminated.

I concur with what Bruce said regarding crystal filters (or any
narrow bandpass filter) at the output frequency.

More fundamentally, I'm not sure I understand your description of the
circuit.  You say it is a pair of FETs with parallel input and a
transformer (autoformer) output.  To me, that suggests the circuit
pictured below (one feeds the sources in parallel, the other feeds
the gates in parallel -- it doesn't make any difference in how the
circuit operates).

The usual push-push doubler feeds the FETs differentially, and takes
the common-mode output.  The diagrammed circuit reverses this -- it
feeds the FETs in parallel (common-mode) and takes the 10MHz output
differentially.  As drawn, the circuit would have essentially no
output at the input frequency or any of its harmonics (only that due
to the mismatch between the FETs).  The only signals it would amplify
are uncorrelated signals -- i.e., the FETs' input noise voltages.  A
quick simulation confirmed no significant output at the input
frequency or its harmonics for matched FETs.  Simulating mismatched
FETs produced a 5MHz signal rich in harmonics, but at a very low
level and with no suppression of the 5MHz and its odd harmonics.

I assume I misinterpreted your description and that you had a
different circuit in mind, or that if you did have this circuit in
mind I'm missing something about its operation.  Can you please
describe again what you had in mind, and how it generates 10MHz?

Best regards,

Charles

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