(27-08-2014, 08:49 PM)Skywave Wrote: [ -> ]Err, what schematic, Lawrence?
Al.
(28-08-2014, 07:40 AM)pwdrive Wrote: [ -> ]The schematic that John posted in his original post.
Lawrence.
Ah! O.K. - thanks.
Al.
(28-08-2014, 07:49 AM)John M0GLN Wrote: [ -> ]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.
John
No trouble at all, John. The maths required only a bit of basic arithmetic and as for the questions arising from this topic, its gave me the opportunity to start thinking about a common and elementary building block in vintage radio that, until know, I've largely taken for granted.
Al.
I'm still a bit puzzled, presumably if there is 40 ma flowing through the crystal then that same current must be flowing via the grid of the valve as the valve is the only thing that completes the circuit that the current can flow through ignoring the reactance of the shunt capacitance and the value of the grid resistor which will both be be high in relative terms.
As John originally said it seems a lot.
Lawrence.
The currant will be flowing through the cathode/grid capacitance.
It is RF after all.
My understanding is as follows.
To maintain oscillations, there must be a 360° phase-difference between anode and grid. (180° phase-difference due to internal capacitance anode-to-grid + 180° phase difference anode-to-grid on account of a predominately inductive load in the L/C cct. in the anode = 360° net phase difference = +ve feedback). In the case of a triode, that internal capacitance is furnished by the grid-anode capacitance of the valve. In a tetrode or pentode, although that capacitance will be less, a capacitance from anode to grid will still be required: an external 'real' capacitor may be necessary. In either case, some of the r.f. current flowing in the crystal will flow through that capacitor. The r.f. load on the oscillator is usually taken from the anode, so if that load increases, the r.f. current through the crystal increases also. If that load becomes excessive, apart from freq. stability and waveform distortion problems, the excess current through the crystal could cause it to fracture. For those reasons, it is usual to insert a buffer 'amplifier' between the crystal osc. and the r.f. load.
Al.
(28-08-2014, 12:48 PM)Refugee Wrote: [ -> ]The currant will be flowing through the cathode/grid capacitance.
It is RF after all.
Yes, I understand that it is RF. For 40 ma to flow through the cathode grid capacitance then the value of that capacitance must be quite high.
Eg: At 3.5 mhz with 5v rms across ckg and 40 ma current flow via ckg then ckg would have to be around 360 pf...
Lawrence.
The Miller effect, when using a triode, will supply much of that capacitance.
Al.
I had considered that Al but as I understand it if the oscillator valve is operating in class C then the gain won't be all that high, the miller capacitance is proportional to gain, plus the valve in Johns schematic is a pentode which would make the miller effect less than what it would be in triode, having said all that I'm not an expert in this field, just trying to get a handle on it all.
Lawrence.
(28-08-2014, 07:08 PM)pwdrive Wrote: [ -> ]I'm not an expert in this field, just trying to get a handle on it all.
Lawrence.
Lawrence - then that makes two of us!
Just one further thought. Despite the reference to using a bulb to monitor crystal current, which repeatedly appears in successive editions of the RSGB H/B and the ARRL H/B (up to 1974: I don't have a later edition), in "Radio Amateur Techniques" (4th. ed., 1974, Pat Hawker, G3VA), refers to that practice as dating from the 1930s when the old and chunky 10X crystals were available. Now looking at the cct. that John has presented, (which is clearly an American drawing), the cct. refers to 'pin2' and 'pin 4' of the crystal. Modern crystals don't have 4 pins in my experience. Therefore . . .
However, be that as it may - as soon as I can find a bit of spare time, I shall build a simple crystal oscillator as close as I can to the one that John has presented in his OP. Then I can take some measurements, which, hopefully, will throw some light on this matter.
Al.
Yes Al, that would be an interesting experiment.
So far as I can make out a relatively large grid current results in a dramatic drop in the grids input impedance WRT the cathode due to the diode action of the grid and cathode.
I'm thinking the large xtal current is due to a combination of the above and the internal valve capacitances including any effective capacitive increase due to any Miller effect.
Here's a thought, if the Miller effect was indeed large how would a xtal osc circuit designer specify the load capacitance as regards to ordering a xtal specified for parallel mode operation, so far as I can make out load capacitance is typically only around 20 to 30 pf according to various xtal manufactures web sites.
Lawrence.