27-08-2014, 10:45 PM
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.
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.
