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I noticed that a lot of designs of phase-shift oscillators use a sequence of high-pass filters to get the phase-shift. Is there any reason why one couldn't use three low pass filters instead? I realize that the maximum phase-shift for a single high-pass filter is +90 and for a low-pass filter -90. Does this difference influence the design?

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  • \$\begingroup\$ Think again, what is the criteria for oscillation using the Barkhausen Criteria? Then what is the criteria to make it a stable Sine wave wave output? Then what is the phase shift in each RC with 3 stages. \$\endgroup\$
    – D.A.S.
    Commented Nov 11, 2019 at 5:19
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    \$\begingroup\$ the high-pass filters perform DC_blocking, thus a simple single transistor Common_emitter amplifier provides the power gain. \$\endgroup\$
    – analogsystemsrf
    Commented Nov 11, 2019 at 5:20
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    \$\begingroup\$ Here is one benefit of the high-pass version (assuming an opamp) : The resistor of the last C-R element can fulfill two tasks at the same time: It works (a) as an input resistor Rx as part of the inverting amplifier (gain: - Ry/Rx) and (b) as a grounded R (virtual ground) as required for a simple C-R highpass stage. This simplyfies the design of the whole oscillator. In contrast, when we are using a three-stage lowpass, the resistor Rx (as part of the amplifier) works as an - unwanted - load for the last R-C stage. \$\endgroup\$
    – LvW
    Commented Nov 11, 2019 at 8:41
  • \$\begingroup\$ @TonyStewartSunnyskyguyEE75 to answer your question: "what is the phase shift in each RC with 3 stages" Up to 90 degrees for each one, so a max of 270. At a 45-degree shift, both are equivalent in terms of amplitude changes. Hence I was wondering if there was any advantage to one or the other. I've seen both designs but the high-pass one seems to dominate. Thinking more about it I wonder whether it because it's easier to get minimal amplitude change with a high-pass filter at larger frequency shifts. A low pass filter to exceed 180-degrees would result in a fairly reduced amplitude. \$\endgroup\$
    – user84299
    Commented Nov 11, 2019 at 18:32
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    \$\begingroup\$ @Tony Stewart Sunnyskyguy EE75 I liked that simulation you cited in your response. I don't think I was aware of that site. \$\endgroup\$
    – user84299
    Commented Nov 11, 2019 at 20:28

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