Chapter 10, Problem 75P

In the op amp circuit of Fig. 10.118, find the closed-loop gain and phase shift of the output voltage with respect to the input voltage if C₁ = C₂ = 1 nF, R₁ = R₂ = 100 kΩ, R₃ = 20 kΩ R₄ = 40 kΩ and ω = 2000 rad/s.

To determine

Calculate the closed-loop gain and phase shift of the output voltage of the op amp circuit in Figure 10.118 using MATLAB.

Answer to Problem 75P

The closed-loop gain and phase shift of the output voltage are $0.12499$ and $180°$ respectively.

Explanation of Solution

Given data:

Refer to Figure 10.118 in the textbook for op amp circuit.

The value of angular frequency $\left(\omega \right)$ is $2000\frac{\text{rad}}{\text{s}}$.

Formula used:

Write the expression to calculate impedance of the capacitor.

$Z_C=\frac{1}{j\omega C}$ (1)

Here,

$\omega$ is the angular frequency, and

$C$ is the value of capacitor.

Calculation:

Let us assume that source voltage $V_s=10\angle 0°\text{V}$

Substitute $2000\frac{\text{rad}}{\text{s}}$ for $\omega$ and $1\text{nF}$ for $C$ in equation (1) to find $Z_C$.

$\begin{array}{c}{Z}_C=\frac{1}{j\left(2000\frac{\text{rad}}{\text{s}}\right)\left(1\text{nF}\right)}\\ =\frac{1}{j\left(2000\frac{\text{rad}}{\text{s}}\right)\left(1\times {10}^{-9}\frac{\text{s}}{\Omega }\right)}\left\{\begin{array}{l}\because 1\text{nF}=1\times {10}^{-9}\text{F}\\ 1\text{F}=1\frac{\text{s}}{\Omega }\end{array}\right\}\\ =-j500\text{k}\Omega \end{array}$

The frequency domain representation of given figure is shown in Figure 1.

Apply Kirchhoff’s current law at node $V_1$.

$\begin{array}{l}\frac{V_1-V_s}{-j500\text{k}}+\frac{V_1-V_o}{100\text{k}}+\frac{V_1-V_2}{-j500\text{k}}=0\\ \frac{V_1-V_s}{-j500}+\frac{V_1-V_o}{100}+\frac{V_1-V_2}{-j500}=0\\ \frac{V_1-V_s}{-j5}+\frac{V_1-V_o}{1}+\frac{V_1-V_2}{-j5}=0\\ \frac{V_1}{-j5}+\frac{V_s}{j5}+V_1-V_o-\frac{V_1}{j5}+\frac{V_2}{j5}=0\end{array}$

Substitute $10\angle 0°\text{V}$ for $V_s$.

$\begin{array}{l}\frac{V_1}{-j5}+\frac{10}{j5}+V_1-V_o-\frac{V_1}{j5}+\frac{V_2}{j5}=0\\ j0.2V_1+V_1-V_o+j0.2V_1-j0.2V_2=j2\\ \left(1+j0.2+j0.2\right)V_1-V_o-j0.2V_2=j2\end{array}$

$\left(1+j0.4\right)V_1-V_o-j0.2V_2=j2$ (2)

Apply Kirchhoff’s current law at node $V_2$.

$\begin{array}{l}\frac{V_2-V_1}{-j500\text{k}}+\frac{V_2-0}{100\text{k}}=0\\ \frac{V_2-V_1}{-j5}+\frac{V_2-0}{1}=0\\ j0.2V_2-j0.2V_1+V_2=0\\ -j0.2V_1+\left(1+j0.2\right)V_2=0\end{array}$

$V_1=\left(1-j5\right)V_2$ (3)

Apply voltage division rule at node $V_b$.

$\begin{array}{c}{V}_b=\frac{20\text{kΩ}}{20\text{kΩ}+40\text{kΩ}}{V}_o\\ =\frac{20\text{kΩ}}{60\text{kΩ}}{V}_o\end{array}$

$V_b=\frac{V_o}{3}$ (4)

According to the properties of ideal op amp, the voltage at the input of the non-inverting terminal of the op amp is equal to the voltage at the input of the inverting terminal. Hence, $\left(V_2=V_b\right)$.

$V_2=\frac{V_o}{3}$ (5)

Substitute equation (5) in (2).

$\begin{array}{l}\left(1+j0.4\right)V_1-V_o-j0.2\left(\frac{V_o}{3}\right)=j2\\ \left(1+j0.4\right)V_1-V_o-j0.0667V_o=j2\end{array}$

$\left(1+j0.4\right)V_1+\left(-1-j0.0667\right)V_o=j2$ (6)

Substitute equation (5) in (3).

$\begin{array}{l}{V}_1=\left(1-j5\right)\frac{V_o}{3}\\ V_1=\left(0.3333-j1.6667\right)V_o\\ V_1-\left(0.3333-j1.6667\right)V_o=0\end{array}$

$V_1+\left(-0.3333+j1.6667\right)V_o=0$ (7)

Represent the equations (6) and (7) in matrix form.

$\left[\begin{array}{cc}\left(1+j0.4\right)& \left(-1-j0.0667\right)\\ 1& \left(-0.3333+j1.6667\right)\end{array}\right]\left[\begin{array}{c}{V}_1\\ V_o\end{array}\right]=\left[\begin{array}{c}j2\\ 0\end{array}\right]$ (8)

The MATLAB code to solve equation (8) is shown below.

A = [1+0.4*i -1-0.0667*i; 1 -0.3333+1.6667*i];

b = [2*i; 0];

x = A\b

The MATLAB result is shown below.

x =

  -0.4166 + 2.0833i   -1.2499 - 0.0000i

The polar representation of obtained results is shown below.

$\begin{array}{l}{V}_1=2.125\angle 101.31°\text{V}\\ V_o=1.2499\angle 180°\text{V}\end{array}$

Write the expression for closed loop gain.

$G=\frac{V_o}{V_s}$ (9)

Substitute $10\angle 0°\text{V}$ for $V_s$ and $1.2499\angle 180°\text{V}$ for $V_o$ in equation (9).

$\begin{array}{c}G=\frac{1.2499\angle 180°\text{V}}{10\angle 0°\text{V}}\\ =0.12499\angle 180°\text{V}\end{array}$

Conclusion:

Thus, the closed-loop gain and phase shift of the output voltage are $0.12499$ and $180°$ respectively.