# Op amp phase shifter with unity constant gain - How does this circuit shift the phase between -180 and 0?

(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.

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 ?

• For Question1, I believe I just need to show that the phase of below expression is $2\phi$ $$\dfrac{1}{\sqrt{1+(\omega/\omega_0)^2}}\sin(\omega t + \phi) - \sin(\omega t)$$ – AgentS May 28 '18 at 13:52
• 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) – G36 May 28 '18 at 14:02
• 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). – G36 May 28 '18 at 14:22
• 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. – G36 May 28 '18 at 14:33
• Yes exactly. All pass filter transfer function is $$\frac{\sqrt{1 +(\omega R C)^2} \cdot e^{-j arctg(\omega R C)}}{\sqrt{1+(\omega R C)^2}\cdot e^{ j arctg(\omega R C)}} = 1 \cdot e^{-2 j arctg(\omega R C)}$$ – G36 May 28 '18 at 14:39

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 $-1$

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.

simulate this circuit

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

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

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