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I am trying to find an equation for the feedback factor of an Armstrong oscillator in terms of the transformer inductances and I can't find one in books.

(c) Floyd

Am I correct if I say that the feedback fraction of the oscillator is the ratio of turns of the tickler coil to the primary coil? If so, according to this equation,

enter image description here

and assuming other inductance dimensions to be equal for the primary and secondary, is the feedback fraction equal to the square root of the ratio of the tickler inductance to the primary inductance?

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is the feedback fraction equal to the square root of the ratio of the tickler inductance to the primary inductance?

Yes, that is the correct formula. Transformer inductance-ratio is the turns-ratio squared and this assumes perfect magnetic coupling. I think it's a fair assumption for two coils closely wound.

Background - the primary voltage sets up a magnetic flux that is shared between the two coils and the formula of note is Faraday's law of induction: -

\$V=-N\dfrac{d\phi}{dt}\$

Because the flux is shared by both coils, you can see that the input-output voltage ratio is also the ratio of the turns.

This means that the gain of the transistor circuit has to be bigger than the step-down factor introduced by the transformer's turns ratio. If the two inductors have a primary:secondary ratio of 9:1 then the number of turns is \$\sqrt{9:1}\$ = 3:1 and therefore the transistor circuit must have a gain greater than 3 to begin oscillations.

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  • \$\begingroup\$ Thanks for confirming. I'm sorry I can't upvote just yet with 10 reputation. Anyway does it mean that the open-loop gain for the transistor amplifier is equal to sqrt(Lpri/Lsec) to satisfy closed-loop gain unity? \$\endgroup\$ – DorkOrc Jul 18 '17 at 7:12
  • \$\begingroup\$ The gain of the transistor stage is the inverse of the turns ratio. Turns ratio is also voltage ratio so if the transformer steps down by 3:1 then the transistor must have a gain that is higher than 3:1 (when loaded by the transformer primary). \$\endgroup\$ – Andy aka Jul 18 '17 at 7:18
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Armstrong oscillators were developed using vacuum tubes, with high input impedance because of the grid's behavior.

This circuit, with bipolar device and its heavily-bypassed emitter, will load the collector (through the turns-ratio-squared).

Include that load in your modeling.

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