I recall seeing references in the 1930s ARRL Handbooks but I thought they applied to using crystals in high-power triode oscillators.
Xtal fracture could easily happen in the Tritet oscillator.
Lawrence.
(26-08-2014, 08:05 PM)Skywave Wrote: [ -> ]A bit more research into this topic has revealed that the current flowing through the crystal - and thus through the indicator bulb in series with it - is an r.f. current, not d.c. (Obvious, in retrospect 
) Therefore, the d.c. grid current of the oscillator valve is quite a different current altogether. And that explains why it appears that the grid current of the osc. valve appears to be so very and contradictory high when first meeting this technique.
Al.
Still a bit puzzled, why should it make so much difference? The r.f. current is still a current with presumably an RMS value comparable to a dc one, sufficient to light the bulb and perhaps burn out the crystal, the anode will be supplying an r.f current as well which could be of similar magnitude.
John
Yes, point taken, John. My mind was wandering down that route when I wrote all that. However, this provides for some interesting reading . . . .
[
attachment=11087]
Al.
I admit that I'm not well up on these things but I'd assumed that the Bulb was there because of its large Positive Temperature Coefiicient and hence would limit the Crystal Current.
Alan
What's the bias arrangement for the oscillator in that schematic?
Lawrence.
Err, what schematic, Lawrence?
Al.
Consider an X-cut crystal operating in series-resonant mode.
Typically, it will have the following electrical characteristics:
Series capacitance, Cs: 0.04 pF.
Series inductance: Ls: 3.5 H.
Q-factor: 25,000.
Now for any L/C cct., at resonance, we have:
Q = (1/R).√(L/C).
Upon a re-arrangement, that becomes:
R = (1/Q)√(L/C).
Plugging in the quantities from the above, and doing the arithmetic, we then get:
R = 125 Ω.
The peak-peak r.f. voltage across the crystal will be in the order of 15 v., so the R.M.S. value is approx. 5 v. And 5 v. R.M.S. and 125 Ω → 40 mA.
That 40 mA is the r.f. current in the crystal. The 15v. p-p signal will be sufficient to bias the grid of a triode oscillator (whose cathode is at 0 v.) to Class-C operation. If one-third of the waveform voltage contributes to grid current, and a 50 kΩ grid leak is installed, then 5 v. peak and 50 kΩ →100 µA.
Note: Yes, there a few rough-'n'-ready approximations in all that, but it does give an idea of the magnitudes of the voltages and currents in that cct.
Al.
(27-08-2014, 08:49 PM)Skywave Wrote: [ -> ]Err, what schematic, Lawrence?
Al.
The schematic that John posted in his original post.
Lawrence.
Thanks for that Al, I'm glad you did the maths not me, I hope I didn't put anyone to too much trouble on this as it is not a circuit I chose to use in the end.
I've just finished building an 80M AM transmitter with a 6AG7 driving an 807 final, after looking at a lot of ideas on the web I ended up with the 6AG7 in a 'electron coupled oscillator' circuit, this works very well with miniature HC49 crystals providing more than enough drive for the 807.
The modulator is a 'Reference Shift Modulator' taken from June 1956 issue of 'CQ'.
Thanks
John