Apologies in advance for this esoteric question (and source). I've been studying wideband LC oscillators recently and my search for an amplitude stabilized configuration led me to the Vackar Oscillator (wiki).

This document (published in 1949!) "LC Oscillators and their Frequency Stability" describes the circuit and I've been studying his analysis of a generic LC oscillator presented in the first few pages.

However, one crucial part of the analysis has me stumped. If you look on page 4, derivation (5) you see this relation:

$$ V_{0}=V_{2}\sqrt{\frac{R_{d}}{Z_{{2}'}}} $$


V0 is the voltage developed across the tuned circuit

V2 is the voltage developed across the input impedance to the tuned circuit

Rd is the dynamic resistance of the tuned circuit

Z2' is the input impedance of the tuned circuit

Vackar describes this formula as "the well-known impedance transformation". But I can't find this relation described in any of my texts. It looks vaguely similar to formulas related to tapped capacitors (and in particular how they behave as ideal transformers), but I'm not sure.

Can anyone provide a derivation of this formula?

Vackar generic oscillator


I have not opened the links, but the general impedance transformation is described in any RF textbooks. I used Thomas Lee's and Razavi's books, because I am in IC domain.

Note: One generally uses LC oscillators because the Q factor of the resonator is very high. This results in a low phase noise for the same power. See Lesson's formula. On the other hand its bandwidth depends on the varactor tuning range and the varactors relative capacitance to the fixed capacitances. Usually the varactors Q is smaller then the other parasitic caps. In short, if one needs high tuning range, then usually not an LC oscillator is the solution. Ring oscillators or RC oscillators are better suited for this.

  • \$\begingroup\$ Apologies for the delay. Yes, I found a decent derivation in the end. I think the key to understanding the transform is that vackar is referring to a capacitive divider to transform the impedances. He doesn't say this explicitly, although in retrospect it's obvious. Thanks for the tip re the topologies! \$\endgroup\$ – Buck8pe Oct 30 '17 at 21:46

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