The "Jones" Super Gainer

[pwdrive]

Thanks Al, yes the regen det in my Regenerodyne is a conventional Hartley (cathode tap) I did find a free google book on the web about oscillators by Gottlieb where he mentions that two non coupled coils can be used in the Hartley but the diagram ref. he refers to does not seem to be in the online version of the book.
I seem to remember that the tap was about 10% up from the earthy end on mine.

Lawrence.

[Skywave]

The oscillator ccts. that I have seen that use two non-coupled coils and no other forms of feedback have all been TPTG oscillators (which rely on anode - grid capacitance). Perhaps that is what was shown; doesn't sound like a Hartley osc. to me. I'll have a wander around the 'Net to see what I can find.

Al.

[pwdrive]

Hi Al, as said, I can't really find a decent description either, I believe the Jones supergainer circuit was changed later to grid leak as opposed to cathode bias, the coil arrangements were the same though. The tuned anode tuned grid is the only one I have seen as well where there is no mutual inductance involved.
Regarding the supergainer circuit, if the bottom end of the capacitor across the sec. of the IF transformer was connected to deck then the circuit would make some sought of sense.

Lawrence.

[Skywave]

I did find a free Google book on the web about oscillators by Gottlieb where he mentions that two non-coupled coils can be used in the Hartley.
Lawrence.

The mode of operation of the Hartley oscillator: I looked it up in my copy of 'Electronic and Radio Engineering' by F. E. Terman. He provides an insightful analysis in which he shows that for zero mutual coupling between the two 'halves' of the coil, the circuit degenerates into a TPTG oscillator. Moreover, for the cct. to behave as a 'genuine' Hartley, there must be a mutual inductance between the two coils.
The 'Radio Engineer's Handbook', 4th. ed., (Langford-Smith), provides an equation which provides the minimum value of µ of the valve for oscillations, in terms of the circuit constants. That equation includes a term for the mutual inductance, M, between the two 'halves' of the coil. Substituting typical values for those constants into that equation, but setting the mutual inductance, M, to zero, it produces a very high value of µ. Moreover, as M is increased from zero, the required value of µ falls - and falls to a range of figures for µ that seem typical.

So, on the basis of those two entries (from gentlemen who know far more about this subject than I do), I submit that for a Hartley oscillator to be a 'true' Hartley osc., the coil must be tapped and a corresponding mutual inductance must exist between the two parts of that tapped coil.

Al.

[pwdrive]

Thanks very much for your time on that Al, what you have said from your two books makes sense and that seems to satisfiy the question.
I don't posses such books these days (used to have all sorts years ago)

Lawrence.