Chapter 15, Problem 15.40P

To determine

The loop gain, the frequency of the oscillation and the required ratio of the resistance to obtain the oscillation for the given function.

Answer to Problem 15.40P

  $\begin{array}{l}\text{Transfer funciton:}\\ T\left(s\right)=\left(1+\frac{R_2}{R_1}\right)\left(\frac{sRC}{s^2{R}^2{C}^2+3sRC+1}\right)\\ T\left(j\omega \right)=\left(1+\frac{R_2}{R_1}\right)\left(\frac{j\omega RC}{1-{\omega }^2{R}^2{C}^2+j\left(3\omega RC\right)}\right)\\ \text{The frequency of the oscillation, }{f}_0=\frac{1}{2\pi RC}\\ \text{For oscillation:}\\ \frac{R_2}{R_1}=2\end{array}$

Explanation of Solution

Given:

The circuit is given as:

  

For the ideal operation amplifier, the inverting and non-inverting terminal currents should be 0. The inverting and non inverting node voltages should be equal by the virtual ground concept.

Redrawing the above circuit:

  

Applying Kirchhoff s current law at node $V_x$ :

  $\begin{array}{l}\frac{V_x-V_1}{R}+\frac{V_x}{\frac{1}{sC}}+\frac{V_x-V_0}{R}=0\\ sC\left(V_x-V_1\right)+sC\left(V_x\right)+\frac{V_x-V_0}{R}=0\\ \frac{sRC\left(V_x-V_1\right)+sRC\left(V_x\right)+\left(V_x-V_0\right)}{R}=0\\ sRC\left(V_x-V_1\right)+sRC\left(V_x\right)+\left(V_x-V_0\right)=0\\ sRC{V}_x-sRC{V}_1+sRC{V}_x+V_x-V_0=0\\ \left(1+2sRC\right)V_x-sRC{V}_1-V_0=0\\ \left(1+2sRC\right)V_x=sRC{V}_1+V_0\\ V_x=\frac{sRC{V}_1+V_0}{1+2sRC}\end{array}$

Hence,

  $V_x=\frac{sRC{V}_1+V_0}{1+2sRC}$

Applying Kirchhoff s current law at non-inverting node:

  $\begin{array}{l}\frac{V_1-V_x}{\frac{1}{sC}}+\frac{V_1}{R}=0\\ sC\left(V_1-V_x\right)+\frac{V_1}{R}=0\\ \frac{sRC\left(V_1-V_x\right)+V_1}{R}=0\\ sRC\left(V_1-V_x\right)+V_1=0\\ sRC\left(V_1-V_x\right)+V_1=0\\ sRC{V}_1+sRC{V}_x+V_1=0\\ \left(1+sRC\right)V_1-sRC{V}_x=0\end{array}$

Substituting $V_x$ in the above equation:

  $\begin{array}{l}{V}_1\left(1+sRC\right)-sRC\left(\frac{sRC{V}_1+V_0}{1+2sRC}\right)=0\\ \frac{V_1\left[\left(1+sRC\right)\left(1+2sRC\right)\right]-sRC\left(sRC{V}_1+V_0\right)}{1+2sRC}=0\\ V_1\left[\left(1+sRC\right)+\left(1+2sRC\right)\right]-sRC{\left(sRC{V}_1+V\right)}_0=0\\ V_1\left[1+2sRC+1sRC+2s^2{R}^2{C}^2\right]-s^2{R}^2{C}^2{V}_1-sRC{V}_0=0\\ V_1\left[1+2sRC+1sRC+2s^2{R}^2{C}^2-s^2{R}^2{C}^2\right]=sRC{V}_0\\ V_1\left[1+3sRC+s^2{R}^2{C}^2\right]=sRC{V}_0\\ V_1=\frac{sRC{V}_0}{1+3sRC+s^2{R}^2{C}^2}\\ \frac{V_1}{V_0}=\frac{sRC}{s^2{R}^2{C}^2+3sRC+1}\end{array}$

Therefore, the feed-back gain is given as:

  $\beta \left(s\right)=\frac{V_1}{V_0}=\frac{sRC}{s^2{R}^2{C}^2+3sRC+1}$

Applying Kirchhoff s current law at inverting node:

  $\begin{array}{l}\frac{V_1}{R_1}+\frac{V_1-V_0}{R_2}=0\\ \frac{R_2{V}_1+R_1\left(V_1-V_0\right)}{R_1{R}_2}=0\\ R_2{V}_1+R_1\left(V_1-V_0\right)=0\\ R_2{V}_1+R_1{V}_1+R_1{V}_0=0\\ \left(R_1+R_2\right)V_1-R_1{V}_0=0\\ R_1{V}_0=\left(R_1+R_2\right)V_1\\ V_0=\left(\frac{R_1+R_2}{R_1}\right)V_1\end{array}$

Therefore,

  $A\left(s\right)=\frac{V_0}{V_1}=\left(1+\frac{R_2}{R_1}\right)$

The loop gain is known as the multiplication of the amplifier gain and the feedback transfer function gain as shown below:

  $\begin{array}{l}T\left(s\right)=A\left(s\right)\beta \left(s\right)\\ T\left(s\right)=\left(1+\frac{R_2}{R_1}\right)\left(\frac{sRC}{{\left(sRC\right)}^2+3sRC+1}\right)\\ T\left(s\right)=\left(1+\frac{R_2}{R_1}\right)\left(\frac{sRC}{s^2{R}^2{C}^2+3sRC+1}\right)\end{array}$

Therefore,

The transfer function is $T\left(s\right)=\left(1+\frac{R_2}{R_1}\right)\left(\frac{sRC}{s^2{R}^2{C}^2+3sRC+1}\right)$ .

Put $s=j\omega$ in the above transfer function:

  $\begin{array}{l}T\left(j\omega \right)=\left(1+\frac{R_2}{R_1}\right)\left(\frac{j\omega RC}{{\left(j\omega \right)}^2{R}^2{C}^2+3j\omega RC+1}\right)\\ =\left(1+\frac{R_2}{R_1}\right)\left(\frac{j\omega RC}{-{\omega }^2{R}^2{C}^2+j3\omega RC+1}\right)\\ T\left(j\omega \right)=\left(1+\frac{R_2}{R_1}\right)\left(\frac{j\omega RC}{1-{\omega }^2{R}^2{C}^2+j\left(3\omega RC\right)}\right)\end{array}$

Therefore,

  $T\left(j\omega \right)=\left(1+\frac{R_2}{R_1}\right)\left(\frac{j\omega RC}{1-{\omega }^2{R}^2{C}^2+j\left(3\omega RC\right)}\right)$

For frequency oscillation the real part should be 0:

  $\begin{array}{l}{\omega }_0{}^2=\frac{1}{R^2{C}^2}\\ {\omega }_0=\sqrt{\frac{1}{R^2{C}^2}}\\ {\omega }_0=\frac{1}{RC}\\ 2\pi f_0=\frac{1}{RC}\\ f_0=\frac{1}{2\pi }\frac{1}{RC}\\ f_0=\frac{1}{2\pi RC}\end{array}$

Therefore,

The frequency oscillation is $f_0=\frac{1}{2\pi RC}$

For the oscillation, the imaginary part is equal to 1:

  $\begin{array}{l}1=\left(1+\frac{R_2}{R_1}\right)\frac{j\omega RC}{j3\omega RC}\\ 1=\left(1+\frac{R_2}{R_1}\right)\frac{1}{3}\\ 3=1+\frac{R_2}{R_1}\\ \frac{R_2}{R_1}=2\end{array}$

Therefore, $\frac{R_2}{R_1}=2$