I've been playing with oscillators a bit here, and I wanted to ask if the way
I'm intuitively thinking about the operation of a tapped capacitor tank is the
way other people do:

Say we have an inductor L in parallel with tapped capacitors, C1 and C2. If
the loaded Q of the tank is reasonably high (say, 10 -- which you'd want for
any decent oscillator anyway), we can pretty much ignore the "input" and
"output" of the tank, as their effect on what happens within the tank isn't
significant. As such, we can just look at the current circulating around the
tank, and from the "top" of the tank to the "bottom" you get 90 degrees phase
shift going from C1 to ground and then another 90 degrees from ground through
C2. Hence, from the top to the bottom you have 180 degrees of phase shift,
with the values (ratios) of C1/C2 just setting the magnitude of the output
voltage vs. the input. Sound good?

I started going through the math involved and determined that you can readily
confuse yourself if you don't make the "high Q" assumption. For instance,
something like a 1 ohm source and 1 ohm load feeding a tank of 159.2mH (j1
ohms) and 2 318.3mF caps (-j/2 ohms) has such a horribly low Q that there
isn't a 180 degree phase shift top to bottom, nor is the resonant frequency
1Hz.

Thanks,
---Joel Kolstad