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(Kindly if its possible to understand this circuit w/o \$j\$.. I haven't made peace with complex analysis yet... so I beg your pardon using nasty trig expressions... )

Below circuit is NOT a phase shifter. As \$R_4\$ varies from \$0\Omega\$ to \$100\Omega\$, the voltage gain varies from \$-1\$ to \$0\$. I pretty much understand how this circuit works.

schematic

simulate this circuit – Schematic created using CircuitLab

If we simply replace \$R_4\$ by a variable capacitor, this circuit suddenly becomes a phase shifter with constant gain = 1. I see that input becomes a \$RC\$ lag circuit with fundamental frequency \$\omega_0 = \dfrac{1}{ R_3C}\$.

As we change the capacitor value, the phase angle \$\phi\$ across the capacitor voltage varies between \$-\pi/2\$ to \$0\$. So the noninverting input at the op amp is \$V_+ = \dfrac{1}{\sqrt{1+(\omega/\omega_0)^2}}\sin(\omega t + \phi) \$,

where \$\phi = -\arctan(\omega/\omega_0)\$.

Question1 : The input phase \$\phi\$ can only change between \$-\pi/2\$ and \$0\$. So intuitively I expect the output phase also to change with in that window; that is not more than \$\pi/2\$. But my textbook claims that the output phase changes between \$\color{red}{-\pi}\$ to \$0\$. Its almost like the opamp is amplifying the phase difference by a factor of \$2\$. How is this possible ?

Question2 : How can the voltage gain remain constant ? The input voltage \$V_+\$ is a function of \$C\$, it clearly decreases as the capacitor reactance decreases. Shouldn't this disturb the output voltage ?

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  • \$\begingroup\$ For Question1, I believe I just need to show that the phase of below expression is \$2\phi\$ $$ \dfrac{1}{\sqrt{1+(\omega/\omega_0)^2}}\sin(\omega t + \phi) - \sin(\omega t) $$ \$\endgroup\$ – AgentS May 28 '18 at 13:52
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    \$\begingroup\$ For DC Xc = 00; So you have a voltage follower with a gain of +1V ( 0 phase shift). But at high-frequency Xc = 0 hence you have an inverting amplifier with the gain equal -R2/R1 = -1V/V and the phase shift is -180 degrees) \$\endgroup\$ – G36 May 28 '18 at 14:02
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    \$\begingroup\$ Try to add two sinewaves the first one 1V (-180 degrees) and the second one is 1.4V and -45 degrees phase shift (case when Xc = R3). \$\endgroup\$ – G36 May 28 '18 at 14:22
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    \$\begingroup\$ For Vin = 1V and when R3 = Xc the voltage at the output is 1.41V and -45 degrees phase shift. For noninverting amplifier only. And for inverting case one we have 1V and -180 degrees phase shift the at the output. So to get the real output you need to add them. \$\endgroup\$ – G36 May 28 '18 at 14:33
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    \$\begingroup\$ Yes exactly. All pass filter transfer function is $$\frac{\sqrt{1 +(\omega R C)^2} \cdot e^{-j arctg(\omega R C)}}{\sqrt{1+(\omega R C)^2}\cdot e^{ j arctg(\omega R C)}} = 1 \cdot e^{-2 j arctg(\omega R C)} $$ \$\endgroup\$ – G36 May 28 '18 at 14:39
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For the DC current and at "low" signal frequency the capacitor reactance is equal to : \$X_C = \infty \$

Hence you circuit becomes a noninverting voltage follower with the gain of \$+ 1 V/V\$ with \$0^{\circ} \$ phase shift.

The output voltage is superimposed on this two outputs (\$+2 + (-1) = +1V\$) Noninverting amplifier with the gain of \$+2\$ and inverting amplifier with a gain of \$-1\$

schematic

simulate this circuit – Schematic created using CircuitLab

At high the capacitor reactance is \$X_C = 0\Omega\$

And this time your circuit becomes a textbook example of an inverting amplifier with voltage gain equal to \$-\frac{R_2}{R_1} = -1\$

So you have gain one but the output voltage is \$-180^{\circ}\$ out of phase shift.

schematic

simulate this circuit

And the transfer function for your circuit (All pass filter) is:

schematic

simulate this circuit

$$ A_V(s) = -\frac{R_2}{R_1} + (1 + \frac{R_2}{R_1}) \cdot\frac{1}{1+ sR_3 C} = \frac{1 - sRC}{1+ sRC}$$

And the magnitude becomes \$ 1\$ (pole is canceled by the zero)

And the phase shift is

\$\phi = -2arc tg (\omega RC) = -2arctan \left( \frac{F}{F_O}\right)\$

Where:

\$F_O = \frac{1}{2 \pi R_3 C}\$

So for the frequency when \$F = F_O\$ the phase shift is

\$ \phi = -2arctan \left( \frac{1}{1}\right) = -2arctan \left(1\right) = -2 \cdot 45^{\circ} = -90^{\circ}\$

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