# How does this opamp phase shifter produce 2x phase shift at the output without changing the magnitude?

My textbook just gives formulas without any proofs and even though it is frustrating I'm really enjoying deriving the proofs with your invaluable help XD Here is one I'm stuck with:

Please see the given circuit. I managed to get the voltage at $$\V_{+}\$$ input of the op amp:

Voltage across $$\C\$$ is given by $$\V_{+}=\dfrac{1/(j\omega C)}{R+\dfrac{1}{j\omega C}} v_{\text{in}}= \dfrac{v_{\text{in}}}{1+j\omega RC}=\dfrac{v_{\text{in}}}{\sqrt{1+(\omega RC)^2}}\angle -\arctan(\omega RC)\$$

Since the cutoff happens when $$\X_C = R\$$, we have $$\RC = \dfrac{1}{\omega_c}\$$:
$$\V_{+}=\dfrac{v_{\text{in}}}{\sqrt{1+(\omega RC)^2}}\angle -\arctan\left(\dfrac{\omega}{\omega_c}\right)\$$

The phase shift is throwing me off. Any help? In particular I'm stuck on two questions:
1) The phase shift at capacitor doesn't have the factor $$\2\$$. How are they getting it at the opamp output?
2) How is the gain of this circuit $$\\pm 1\$$ for all frequencies?

• You've only gone 50% of the way with your derivation. If you went the full way to derive the output expression of the op-amp you'd probably see why. Commented Mar 20, 2020 at 12:17
• Oh so I'm correct so far. Awesome! I would have finished it off if the voltage at $V_{+}$ doesn't have any phase shift. I never dealt with phase shift before :( Commented Mar 20, 2020 at 12:19
• Try to analyze the circuit in two different cases. Case 1 at DC where Xc-->oo, therefore, the voltage gain is +1V/V and in the second case when Xc = 0 the gain becomes equal to -1V/V (180 degrees phase shift). As you can see the magnitude of a voltage gain is constant but the phase which is changing with the frequency.
– G36
Commented Mar 20, 2020 at 16:58

How is the gain of this circuit ±1 for all frequencies?

Using superposition we can immediately say that the gain is: -

$$H(s) = (1+\dfrac{R'}{R'})\cdot (\dfrac{1}{1+sCR}) - \dfrac{R'}{R'}$$ $$= \dfrac{2}{1+sCR} - 1$$ $$= \dfrac{1- sCR}{1+sCR}$$

And, because the magnitude of $$\1 - sCR\$$ is the same as the magnitude of $$\1+ sCR\$$, the gain magnitude is 1.

The phase change in the numerator and the phase change in the denominator go in opposite directions so, given that they are disposed as "numerator" and denominator" any phase angle from the circuit is double what it would be for a single RC network.

• Is $s=j\omega$ ? If so everything makes complete sense. You basically just added nonverting gain and inverting gain(-$\dfrac{R'}{R'}$). I don't know why I converted from rectangular form to polar form in my work. I should have continued working in rectangular form haha. Thank you so much:) Commented Mar 20, 2020 at 13:40
• Yes, s = jw.... Commented Mar 20, 2020 at 13:41
• @beccabecca s is the Laplace domain variable. If you're not familiar with it, I highly recommend trying to learn a bit about the Laplace transform; it's a very useful tool for analyzing these kinds of circuits. Commented Mar 20, 2020 at 14:19

Well, we are trying to analyze the following circuit (assuming an ideal opamp):

simulate this circuit – Schematic created using CircuitLab

When we use and apply KCL, we can write the following set of equations:

\begin{cases} \begin{alignat*}{1} \text{I}_0&=\text{I}_1+\text{I}_3\\ \\ \text{I}_1&=\text{I}_2\\ \\ 0&=\text{I}_3+\text{I}_4 \end{alignat*} \end{cases}\tag1

When we use and apply Ohm's law, we can write the following set of equations:

\begin{cases} \begin{alignat*}{1} \text{I}_1&=\frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}\\ \\ \text{I}_2&=\frac{\text{V}_1}{\text{R}_2}\\ \\ \text{I}_3&=\frac{\text{V}_\text{i}-\text{V}_2}{\text{R}_3}\\ \\ \text{I}_4&=\frac{\text{V}_3-\text{V}_2}{\text{R}_4} \end{alignat*} \end{cases}\tag2

Substitute $$\(2)\$$ into $$\(1)\$$, in order to get:

\begin{cases} \begin{alignat*}{1} \text{I}_0&=\frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}+\frac{\text{V}_\text{i}-\text{V}_2}{\text{R}_3}\\ \\ \frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}&=\frac{\text{V}_1}{\text{R}_2}\\ \\ 0&=\frac{\text{V}_\text{i}-\text{V}_2}{\text{R}_3}+\frac{\text{V}_3-\text{V}_2}{\text{R}_4} \end{alignat*} \end{cases}\tag3

Now, when we have an ideal opamp we know that $$\\text{V}_\alpha:=\text{V}_+=\text{V}_-=\text{V}_1=\text{V}_2\$$. So we can rewrite equation $$\(3)\$$ as follows:

\begin{cases} \begin{alignat*}{1} \text{I}_0&=\frac{\text{V}_\text{i}-\text{V}_\alpha}{\text{R}_1}+\frac{\text{V}_\text{i}-\text{V}_\alpha}{\text{R}_3}\\ \\ \frac{\text{V}_\text{i}-\text{V}_\alpha}{\text{R}_1}&=\frac{\text{V}_\alpha}{\text{R}_2}\\ \\ 0&=\frac{\text{V}_\text{i}-\text{V}_\alpha}{\text{R}_3}+\frac{\text{V}_3-\text{V}_\alpha}{\text{R}_4} \end{alignat*} \end{cases}\tag4

Now, for the output voltage we get:

$$\text{V}_3=\frac{\text{V}_\text{i}\left(\text{R}_2\text{R}_3-\text{R}_1\text{R}_4\right)}{\text{R}_3\left(\text{R}_1+\text{R}_2\right)}\tag{5}$$

Now, applying this to your circuit we need to use (from now on I use the lower case letters for the function in the 'complex' s-domain where I used Laplace transform):

$$\text{R}_1=\frac{1}{\text{sC}}\tag6$$

So, the output voltage will be:

$$\text{v}_3\left(\text{s}\right)=\frac{\text{v}_\text{i}\left(\text{s}\right)\left(\text{R}_2\text{R}_3-\frac{\text{R}_4}{\text{sC}}\right)}{\text{R}_3\left(\frac{1}{\text{sC}}+\text{R}_2\right)}\tag7$$

So, when we use the transformation $$\\text{s}=\text{j}\omega\$$ (where $$\\text{j}^2=-1\$$), we get:

$$\underline{\text{v}}_3\left(\text{j}\omega\right)=\frac{\underline{\text{v}}_\text{i}\left(\text{j}\omega\right)\left(\text{R}_2\text{R}_3-\frac{\text{R}_4}{\text{j}\omega\text{C}}\right)}{\text{R}_3\left(\frac{1}{\text{j}\omega\text{C}}+\text{R}_2\right)}\tag8$$

So, we can write a transfer function:

$$\underline{\mathcal{H}}\left(\text{j}\omega\right):=\frac{\underline{\text{v}}_3\left(\text{j}\omega\right)}{\underline{\text{v}}_\text{i}\left(\text{j}\omega\right)}=\frac{\text{R}_2\text{R}_3-\frac{\text{R}_4}{\text{j}\omega\text{C}}}{\text{R}_3\left(\frac{1}{\text{j}\omega\text{C}}+\text{R}_2\right)}=\frac{\text{R}_2\text{R}_3+\frac{\text{R}_4}{\omega\text{C}}\cdot\text{j}}{\text{R}_3\left(\text{R}_2-\frac{1}{\omega\text{C}}\cdot\text{j}\right)}\tag9$$

So, for the magnitude we will get:

$$\left|\underline{\mathcal{H}}\left(\text{j}\omega\right)\right|=\frac{\sqrt{\left(\text{R}_2\text{R}_3\right)^2+\left(\frac{\text{R}_4}{\omega\text{C}}\right)^2}}{\text{R}_3\sqrt{\text{R}_2^2+\left(\frac{1}{\omega\text{C}}\right)^2}}\tag{10}$$

And, for the phase we will get:

$$\arg\left(\underline{\mathcal{H}}\left(\text{j}\omega\right)\right)=\arctan\left(\frac{\text{R}_4}{\omega\text{C}}\cdot\frac{1}{\text{R}_2\text{R}_3}\right)-\left(\frac{3\pi}{2}+\arctan\left(\text{R}_2\omega\text{C}\right)\right)\tag{11}$$

Now, when $$\\text{R}:=\text{R}_3=\text{R}_4\$$ all this simplifies to:

$$\left|\underline{\mathcal{H}}\left(\text{j}\omega\right)\right|=\frac{\sqrt{\left(\text{R}_2\text{R}\right)^2+\left(\frac{\text{R}}{\omega\text{C}}\right)^2}}{\text{R}\sqrt{\text{R}_2^2+\left(\frac{1}{\omega\text{C}}\right)^2}}=1\tag{12}$$

And:

$$$$\begin{split} \arg\left(\underline{\mathcal{H}}\left(\text{j}\omega\right)\right)&=\arctan\left(\frac{\text{R}}{\omega\text{C}}\cdot\frac{1}{\text{R}_2\text{R}}\right)-\left(\frac{3\pi}{2}+\arctan\left(\text{R}_2\omega\text{C}\right)\right)\\ \\ &=\arctan\left(\frac{1}{\text{R}_2\omega\text{C}}\right)-\left(\frac{3\pi}{2}+\arctan\left(\text{R}_2\omega\text{C}\right)\right)\\ \\ &=-\left(\pi+2\arctan\left(\text{R}_2\omega\text{C}\right)\right) \end{split}\tag{13}$$$$

• So, if R3 = R4 then.... Commented Mar 20, 2020 at 13:06
• @Andyaka I was editing my answer. Commented Mar 20, 2020 at 13:08
• Then $V_{-} = \dfrac{v_{in}+v_{out}}{2}$ @Andyaka Jan I'm still going through the work. Take your time. Really appreciate it:) I'm not good with complex numbers so I'm also revising it now. Thank you so much! Commented Mar 20, 2020 at 13:10
• @beccabecca I'm done now! You're welcome, I'm glad that I could be of any help. I personally also love to use math to analyze these circuits. Commented Mar 20, 2020 at 13:11