I'm having problems understanding this exercise:

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The question asks that I find the "oscillation conditions" (which I assume is the value of \$K\$) as well as the frequency of the oscillation for C = 100 nF and R = 1 kΩ.

The approach I used for the problem was based off of other course examples:

The circuit consists of 3 op-amps in the inverting configuration.

Op-amps 1 & 2 have a transfer function $$ H(s) = -\frac{1}{sRC} $$ each. Op-amp 3's TF is $$ H(s) = -K $$. Together their TF should be $$ H(s) = -K\frac{1}{(sRC)^2} $$ which is always a positive real number.

My issue is that when I try to apply the Barkhausen criterion where \$ \arg(H(s)) = 0 \$ -- it always applies regardless of frequency. If that's true (which I suspect it isn't) then what frequency does it oscillate at?

I think I've either missed something (and I'm going to be feeling very silly for asking) or I just don't understand the oscillators properly.

The question doesn't specify, so I must assume all op-amps are ideal.

  • \$\begingroup\$ Is your book modeling the op-amps as ideal, or as having a frequency-dependent voltage gain (the simplest "accurate" model of which is \$A(s) = \frac{G}{s}\$, where \$G\$ is the gain-bandwidth product in radians/second). Please edit your question to include this information. \$\endgroup\$
    – TimWescott
    Feb 5, 2022 at 17:20
  • \$\begingroup\$ Maybe the question asked needs copying and pasting verbatim because, as it stands, it seems a badly composed exercise question. \$\endgroup\$
    – Andy aka
    Feb 5, 2022 at 17:32
  • \$\begingroup\$ Unfortunately the question is in greek. I am confident however that I have given all relevant information \$\endgroup\$ Feb 5, 2022 at 17:50
  • \$\begingroup\$ Well, you can find an oscillator condition where the overall loop gain is unity. Maybe that is what is expected of you. \$\endgroup\$
    – Andy aka
    Feb 5, 2022 at 18:38
  • \$\begingroup\$ oscillation conditions might just mean they want you to solve for when the system would produce stable, repeating oscillations of consistent frequency and amplitude. so assume V3 is of the form Asin(wt) where A and w are constant. Run it through the circuit and you should have two equations for V3 to constrain your system. \$\endgroup\$
    – Abel
    Feb 5, 2022 at 18:40

1 Answer 1


The shown circuit is one of the "Double Integration Oscillator" alternatives.

Quote: My issue is that when I try to apply the Barkhausen criterion where "arg(H(s))=0" it always applies regardless of frequency. If that's true (which I suspect it isn't) then what frequency does it oscillate at?

Please note that the Barkhausen criterion consists of TWO requirements:

  • The magnitude of the loop gain must be unity at the frequency f=fo

  • The phase of the loop gain function must be zero (or 360deg) at f=fo.

  • The shown circuit can meet the phase condition for all frequencies only under the unrealistic assumption of ideal amplifiers. In reality (real amplifiers), the loop gain phase function crosses the 0-deg line at one single frequency fo_p only. That means: The circuit can oscillate for real amplifiers only and not for idealized models (all other known oscillator circuits do not show such a restriction).

  • The loop gain magnitude also crosses the 0dB line at one single frequency fo_m only.

  • It can be shown that the (parasitic) phase shift of the opamps helps to enable oscillations because we always have fo_p<fo_m. Therefore, when the loop gain phase is zero, the loop gain magnitude will always be above unity - and the circuit is able to start oscillations with rising amplitudes.

  • The final oscillation frequency fo will be between fo_p and fo_m.

  • Recently, it was shown that clipping effects (due to the loop gain magnitude above unity) will cause a POSITIVE phase shift which allows that BOTH criteria can be fulfilled at f=fo. This positive phase shift cancels the additional negative phase shift introduced by the real opamps.

  • For this reason, the quality of the produced sinusoidal is rather good and nearly no clipping effects can be observed - even without any additional means for amplitude control.

  • Remark: This works best when the integration time constants for the two integrators are NOT chosen to be equal. In this case, the output signal of the stage with the larger time constant should be used (because the other one reaches the limitation first).

  • Remark/Quote: „Curiously, this is an exceptional oscillator which does not appear to fail to start in spite of this apparent difficulty“ (R. Senani et al: „Sinusoidal Oscillators and Waveform Generatorsusing Modern Electronic Circuit Building Blocks“, Springer Int. Publishing, 2016 ].


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