I have the feeling that the original circuit could also
benefit from differing R and C values in the phase shift
network, but had no luck with just guessing changes that
would increase the loop gain. I have to get past the mental
block about current versus voltage output for the network.
From left to right, label the first section as C2-R2, the second as C1-R1
and the final cap as C. Define a non-existent resistor as R (to be used in
the math). Select a value for R and C, and set C2 = 9*C, R2 = R/9; set C1
= 3*C, R1 = R/3. Then the condition for oscillation is Rf > 3.852*R. This
is with scaling of 9:3:1.
Similarly, with scaling of 100:10:1 the condition for oscillation is Rf >
2.442*R.
How do you see the ratios of feedback to grounding
resistance as gain. It is just an arbitrary resistor ratio.
If you have a voltage controlled voltage source (just a voltage
amplifier), with input Vi and output Vo, the ratio Vo/Vi volts per volt is
called the voltage gain, or just "the gain". Given a current controlled
current source, the ratio Io/Ii amps per amp is called the current gain, or
just "the gain" when the parties to the discussion understand that it's a
current amplifier.
What if you have a current controlled voltage source with a transfer
ratio of Io/Vi amps per volt? Is not the transfer ratio a kind of "gain"?
Anyway, that's what I mean by "gain".
And, in fact, the simple inverting amplifier that you would have if the
phase shifting network in the referenced circuit were replaced by a single
resistor, Ri, between Vi and the minus input of the opamp would have a
voltage gain given by -Rf/Ri, an arbitrary resistor ratio.
The standard voltage in, voltage out, equal R and C, 3-section phase
shifting network has a "gain" (voltage transfer ratio) at the frequency of
oscillation of 1/29. This "gain" doesn't depend on the impedance level of
the network; the output voltage of the (unloaded) network is always 1/29 of
the input voltage at the frequency of oscillation.
The network under discussion:
http://www.digital.ee.eng.chula.ac.th/2102384/download/OSC.pdf
has the property that the current delivered to the minus input of the
opamp, at the frequency of oscillation, varies with impedance level. If
the current delivered to that node is called Is, then the ratio Is/Vi is
1/(12*R), and the "attenuation", if you will, of the network increases with
impedance level. This "attenuation" must be made up by the opamp, and we
see that as R gets larger, then Rf also becomes larger.
If you consider the "canonical" network, with R and C equal to unity,
then that number 12 in the 1/(12*R) expression becomes the current to
voltage transfer ratio of the (opamp + feedback resistor), which is acting
as a current controlled voltage source.
The dependence on impedance vanishes if you take the ratio Rf/R, so this
is a fundamental property of the circuit. To repeat, it's the "gain"
(current to voltage transfer ratio) of the (opamp + feedback resistor)
circuit with unity R's and C's.
I probably should have explained this point of view in more detail in the
first post, but I had been playing with the circuit for a while and it had
become familiar to me.