I've built a state-variable filter, to this schematic:

state variable

R6 and R7 are a dual gang potentiometer (with the 3rd terminal unconnected, so variable resistor), and R2 is a single gang potentiometer (again as a variable resistor). R6/7 controls the filter cutoff and R2 the Q-factor.

To a certain extent it works as expected (as my analysis expected), however, when I bring the potentiometer to the position such that R6 and R7 are zero, the circuit starts behaving as a square wave oscillator, with R2 controlling the frequency.

I understand that my analysis breaks down when R6/7 = 0 - I end up with the cutoff frequency being infinite because I'm dividing by zero, but can anybody explain why the circuit starts oscillating? Why is it a square wave, and can I mathematically work out the frequency?

Also, simulating the circuit doesn't (at least for me) end up with it acting as an oscillator

  • 1
    \$\begingroup\$ Because the phase shift is not an integral multiple of 360°. \$\endgroup\$ Aug 14, 2015 at 17:23
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    \$\begingroup\$ Check out Barkhausen criteria \$\endgroup\$ Aug 14, 2015 at 17:26
  • \$\begingroup\$ @IgnacioVazquez-Abrams Barkhausen criteria states it will oscillate when it IS a multiple of 360? \$\endgroup\$
    – ACarter
    Aug 14, 2015 at 17:31

1 Answer 1


When you make R6 and R7 equal to zero ohms the whole circuit operates under conditions that are not "normal" and not predefined. As you have seen the circuit behaves like a Schmitt trigger oscillator (relaxation principle), however without determined frequency.

But you can easily transfer the circuit into a sinusoidal oscillator if you remove the negative feedback and its damping properties: Remove R3. In this case, you have one of the classical oscillator topologies fulfilling the Barkhausen condition.

UPDATE: The oscillation frequency is wo=1/SQRT(R6C1*R7C2). This oscillator structure is - for my opinion - one of the most interesting oscillator principles because it does not need any additional amplitude limiting circuitry (in case T1=R6C1 is NOT equal to T2=R7C2). For unequal time constants only one of both integrator stages is limited (supply rail) - and the other integrator block provides a rather good sinusoidal signal.

(By the way: It is not self-explaining why the oscillation frequency is at 1/SQRT(T1T2). That is another point worth to discuss.)

  • \$\begingroup\$ I don't think the oscillation frequency can be that, because the frequency changes when I change R2. That's the cutoff frequency, but the oscillation I get is not at the cutoff frequency \$\endgroup\$
    – ACarter
    Aug 14, 2015 at 20:03
  • \$\begingroup\$ If you remove R3, the resistor R2 has no influence on the loop gain and it cannot have any influence on the oscillation frequency. It can be set to zero. More than that, R1 can (and should) be removed as well. That means: The whole circuit consists of two integrator stages and one inverter (gain A=-R5/R4). In most cases A=-1 with R5=R4. Note that the given formula for wo applies to A=-1 only. For other values we have under the SQRT (T1*T2*|A|). This allows easy control of the oscillation frequency. \$\endgroup\$
    – LvW
    Aug 15, 2015 at 7:58

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