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Figure show a RC phase shift oscillator using Op-Amp with gain 5(Vo/Vi).

My assumption : Circuit is generating a sine wave with a peak to peak voltage of 10v.

Consider positive maximum point (+5V) at the output. A fraction of this voltage is feed to inverting terminal of Op-amp with a 180 degree phase shift. ie -1V ( 180 degree phase shifted value 1/5 of 5 ie barkhausen criteria) reaches negative terminal of Op-amp. This -1V is again amplified by the Op-amp and we will get +5v at the output. That means this is a stable condition.

Then why output voltage is oscillating ?

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There is stable oscillation if the Barkhausen criterion for oscillation is satisfied.

I.e:

  • loop gain is 1
  • (total) loop phase shift is 0°, (or 360° or an integer multiple of 360°):
    in this circuit there is 180° phase shift by the inverting input of the OpAmp and another 180° by the phase shift network which results in 360° for exact one frequency.
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The output is oscillating because it takes time for the changes in the output voltage to propagate through the filter.

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  • \$\begingroup\$ 0 phase shift means there is no delay. Am I right ? \$\endgroup\$ – tollin jose Sep 22 '14 at 15:37
  • \$\begingroup\$ not necessarily, since a 360 degree delay also exhibits zero phase shift. \$\endgroup\$ – EM Fields Sep 22 '14 at 17:24
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Tollin - you cannot pick up one single moment (example: Why do you think that +5V at the output are producing -1V at the inv. input?).

Rather, you have to consider steady-state condition - and now Barkhausen´s oscillation condition comes into play: A circuit with feedback is able to oscillate if the loop gain around the complete loop is unity (magnitude unity, phase=0) for one single frequency only. The feedback network in your circuit produces a phase shift of -180deg at one single frequency - and the damping at this frequency is 1/29. Therefore, when you close the loop with an inverting amplifier (gain of -29) you fulfill the mentioned condition and the circuit oscillates.

Comment: The above mentioned factor of 1/29 applies if R=R1=R2=R3 and C1=C2=C3 and Rfb=29*R

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  • \$\begingroup\$ By Barkhausen criteria loop gain should be 1. I already assumed gain of amplifier as 5, therefore gain of feed back network must be 1/5) since output voltage is 5V, 1/5 of 5 ie 1V will reach input end. Since there is 180 degree phase shift feed back value will be -1V. \$\endgroup\$ – tollin jose Sep 22 '14 at 15:43
  • \$\begingroup\$ ....but it isn`t 1/5 at the frequenncy which produces -180deg phase shift. You cannot simply "assume" something which contradicts electronic rules. In fact, it is 1/29. \$\endgroup\$ – LvW Sep 22 '14 at 16:02

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