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Do we decrease the the feedback fraction and increase the amplification fraction if we increase the value of L2? (similar to Colpitts oscillator where we by changing C2 capacitor) enter image description here

Amazing Random Guy
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  • Precisely, what do you mean by *amplification fraction*? – Andy aka Dec 28 '20 at 16:08
  • [This answer](https://electronics.stackexchange.com/questions/517675/how-does-the-colpitts-oscillator-reach-a-loop-gain-of-1/517814#517814) does a tear-down of the maths for a Colpitts Oscillator and you should be able to do the same for the Hartley if you follow that answer. It delivers the gain equation you might be seeking. – Andy aka Dec 28 '20 at 16:18
  • ... Tank circuit gain magnitude is \$\frac{L2}{L1}\$ at the oscillation frequency for a Hartley. For a Colpitts it's \$\frac{C1}{C2}\$. – Andy aka Dec 28 '20 at 17:41

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if we increase the value of L2? (similar to Colpitts oscillator where we by changing C2 capacitor)

Almost -- but you want to decrease the value of L2. If you swapped the capacitors and inductors, then you would, indeed, decrease the value of the corresponding capacitor -- but inductive impedance goes up with increasing inductance, while capacitive impedance goes down with increasing capacitance.

Do we decrease the the feedback fraction and increase the amplification fraction if

It's more complicated than that. For any semi-reasonable starting point you decrease the feedback. But you don't necessarily increase the amplification -- you just increase the amplification that is needed. Just willy-nilly cutting the value of L2 (or adjusting the ratio of L2/L1 while maintaining the same resonance frequency) will eventually leave you with a dead lump of circuit that rings a bit when lightning strikes nearby but never breaks into continuous oscillation. Before that happens, you'll have an unreliable oscillator, or one with poor operating qualities.

TimWescott
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  • Tim if I recall correctly the dominant part in the equation of amplification of the Collpits oscillator Xc of the capacitor connected directly to GND is dominant. – The Force Awakens Dec 28 '20 at 16:44
  • @TheForceAwakens you might want to clarify what your comment means. It doesn't make sense to me. – Andy aka Dec 28 '20 at 16:46
  • I am saying that the load impedance is equal to Xc^2/Re where Re is the parasitic resistance of the inductor and Xc is the capacitive reactance of the C2 capacitor which is connected to ground for a Collpits oscillator. – The Force Awakens Dec 28 '20 at 16:51