From: Chuck Hill ([email protected])
Date: Tue Feb 20 2001 - 19:20:01 PST
You are on the right track. The nature of the losses change depending on
position in the resonator. There are two things you should consider:
First, I guess you are looking into the end of a piece of semi-rigid which
is shorted at the other end. So if the resonator is a 1/4 wavelength, you
are tapped at the highest impedance point on the resonator. One of the
issues you must consider for your oscillator is the impedance match
between the active device and the resonator. That is, what impedance
should the resonator present to the active device. I suggest you tap the
resonator down in impedance by drilling a hole in the side of the
semi-rigid. You could also add a high Q capacitor to the open end of the
resonator to increase the Q. Capacitors come in varying loss, you can get
low loss ceramics or the old time standby, porcelain.
Second, measuring high values of Q is difficult. This can be done with a
network analyzer (directional bridge), an impedance analyzer (VI
bridge)(better), or for high Q resonating with a known high Q capacitor
The method of propagation loss used in transmission lines does not work
since only 1/4 wavelength is used and typically the resonator is
capacitively loaded to achieve higher Q.
At 12:40 PM 2/20/01 +0000, Steve Rogers wrote:
>I have been looking into the use of shorted transmission line 'stubs' as
>synthetic inductors. In particular I hoped that this approach could yield a
>high Q factor inductor at VHF.
>Shorted stubs made using semi-rigid coaxial line seem to be popular for low
>phase noise oscillators at UHF frequencies which suggests a high Q structure
>be possible by this method. I had an attempt at producting a synthetic
>inductor for VHF using a length of semi-rigid coaxial line shorted at one
>end the result was poor to say the least. Details below:-
>Line length 0.7 metre
>Resonant frequency approx 80 MHz
>Q: approx 50
>I wonder why the Q is so poor?
>My analytical attempts.
>My first stab at attempting to justifiy the loss assumes that the cable loss
>given by the manufacturer may be used together with the line length and the
>driving impedance to calculate the loss due to the inevitable multiple
>reflections between the source and the load (Basically use of complex power
>flow with masons rules). At first this seemed to be the way to go but then I
>thought about the loss value for the line. The loss given by the
>manufacturer only applies to a matched line! So what is the loss for a
>shorted line???? The V:I ratio will clearly vary along the line length being
>min V max I at the short. The loss of the line will thus be a function of
>the position of the section of line in question with respect to the short
>circuit (at and near the short where currents are max the i squared R losses
>must be dominant, at and near the V maximum points the dielectric losses
>must be maximum).
>Is my thinking taking me off on the wrong path here (is there an easy answer
>to this problem?)
>IF ANYONE CAN SUGGEST HOW TO SOLVE THIS ONE IT WOULD BE VERY MUCH
>Thanks in advance
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