Chapter 15, Problem 15.29P

a.

To determine

The expression for the frequency of the oscillation.

Answer to Problem 15.29P

The expression for the frequency of oscillation:

  $f_0=\frac{1}{2\pi RC}\sqrt{\frac{3R+2R_V}{R_V}}$

Explanation of Solution

Given:

The circuit is given as:

  

The circuit is redrawn by labeling the voltages as shown below:

  

Assuming:

  $\begin{array}{l}{R}_{F1}=R_{F2}=R_{F3}=R_F\\ R_2=R_3=R_{A1}=R_{A2}=R_3=R\\ C_1=C_2=C_3=C\end{array}$

Considering $R_1$ as the variable resistance:

  $R_1=R_V$

Nodal analysis at the node $v_1$ :

  $\begin{array}{l}\frac{v_1-v_0}{R_V}+\frac{v_1}{1}=0\\ v_1\left(\frac{1}{R_V}+sC\right)=\frac{v_0}{R_V}\\ v_1=\frac{v_0}{1+s{R}_{V}C}\end{array}$

The amplifier gain of $A_1$ :

  $\begin{array}{l}\frac{v_{01}}{v_1}=1+\frac{R_F}{R}\\ v_{01}=\left(1+\frac{R_F}{R}\right)v_1\\ v_{01}=\left(1+\frac{R_F}{R}\right)\left(\frac{1}{1+s{R}_{V}C}\right)v_{01}\mathrm{............}\left(1\right)\end{array}$

Now, the output of $A_2$ operational amplifier:

  $\begin{array}{l}{v}_{02}=\left(1+\frac{R_F}{R}\right)v_{01}\\ v_{02}=\left(1+\frac{R_F}{R}\right)\left(\frac{1}{1+sRC}\right)v_{01}\mathrm{............}\left(2\right)\end{array}$

The nodal analysis at the node $v_3$ :

  $\begin{array}{l}\frac{v_3-v_{02}}{R}+\frac{v_3}{\frac{1}{sC}}+\frac{v_3}{R}=0\\ v_3\left(\frac{1}{R}+sC+\frac{1}{R}\right)=\frac{v_{02}}{R}\\ v_3\left(2+sCR\right)=v_{02}\\ v_3=\left(\frac{1}{2+sCR}\right)v_{02}\mathrm{.........}\left(3\right)\end{array}$

Applying the nodal analysis at the inverting terminal of the $A_3$ amplifier:

  $\begin{array}{l}\frac{0-v_3}{R}-\frac{0-v_0}{R_F}=0\\ v_0=-\frac{R_F}{R}{v}_3\end{array}$

Solving equation (3):

  $v_0=\left(\frac{R_F}{R}\right)\left(\frac{1}{2+sCR}\right)\left(1+\frac{R_F}{R}\right)\left(\frac{1}{1+sCR}\right)v_{01}$

Now, solving the equation (1):

  $\begin{array}{c}{v}_0=-\left(\frac{R_F}{R}\right)\left(\frac{1}{2+sCR}\right)\left(1+\frac{R_F}{R}\right)\left(\frac{1}{1+sRC}\right)\left(1+\frac{R_F}{R}\right)\left(\frac{1}{1+s{R}_{V}C}\right)v_0\\ 1=-\left(\frac{R_F}{R}\right){\left(1+\frac{R_F}{R}\right)}^2\left(\frac{1}{2+2sRC+sCR+s^2{R}^2{C}^2}\right)\left(\frac{1}{1+s{R}_{V}C}\right)\\ =-\left(\frac{R_F}{R}\right){\left(1+\frac{R_F}{R}\right)}^2\left(\frac{1}{2+3sRC+s^2{R}^2{C}^2}\right)\left(\frac{1}{1+s{R}_{V}C}\right)\\ =-\left(\frac{R_F}{R}\right){\left(1+\frac{R_F}{R}\right)}^2\left(\frac{1}{\left[\begin{array}{l}2+3sRC+s^2{R}^2{C}^2+2s{R}_{V}C\\ +3s^{2}R{R}_V{C}^2+s^3{R}^2{R}_V{C}^3\end{array}\right]}\right)\\ =-\left(\frac{R_F}{R}\right){\left(1+\frac{R_F}{R}\right)}^2\left(\frac{1}{2+sC\left(3R+2R_V\right)+s^{2}R{C}^2\left(R+3R_V\right)+s^2{R}^2{R}_V{C}^3}\right)\end{array}$

Evaluating the frequency of the oscillation, use $s=j\omega$ in the above equation:

  $1=-\left(\frac{R_F}{R}\right){\left(1+\frac{R_F}{R}\right)}^2\left(\frac{1}{2+j\omega C\left(3R+2R_V\right)+{\omega }^{2}R{C}^2\left(R+3R_V\right)+j{\omega }^2{R}^2{R}_V{C}^3}\right)$

Now, equating the imaginary values to 0:

  $\begin{array}{l}\omega C\left(3R+2R_V\right)-{\omega }^2{R}^2{R}_V{C}^3=0\\ \omega C\left(3R+2R_V-{\omega }^2{R}^2{R}_V{C}^3\right)=0\\ 3R+2R_V-{\omega }^2{R}^2{R}_V{C}^3=0\\ {\omega }^2{R}^2{R}_V{C}^3=3R+2R_V\\ {\omega }^2=\frac{3R+2R_V}{R^2{R}_V{C}^3}\end{array}$

Squaring the both sides:

  $\begin{array}{l}\omega =\sqrt{\frac{3R+2R_V}{\sqrt{R_V}RC}}\\ 2\pi f_0=\frac{1}{RC}\sqrt{\frac{3R+2R_V}{R_V}}\\ f_0=\frac{1}{2\pi RC}\sqrt{\frac{3R+2R_V}{R_V}}\end{array}$

Hence, the needed expression for the frequency of oscillation:

  $f_0=\frac{1}{2\pi RC}\sqrt{\frac{3R+2R_V}{R_V}}$

b.

To determine

The condition for the oscillation.

Answer to Problem 15.29P

The condition for oscillation:

  $\frac{R_F}{R}=2$

Explanation of Solution

Given:

The circuit is given as:

  

When, $R_V=R$ .

The condition for oscillation:

  $\begin{array}{l}1=-\left(\frac{R_F}{R}\right){\left(1+\frac{R_F}{R}\right)}^2\left(\frac{1}{2-{\omega }^{2}R{C}^2\left(R+3R_V\right)}\right)\\ =-\left(\frac{R_F}{R}\right){\left(1+\frac{R_F}{R}\right)}^2\left(\frac{1}{2-{\omega }^{2}R{C}^2\left(R+3R\right)}\right)\\ -\left(\frac{R_F}{R}\right){\left(1+\frac{R_F}{R}\right)}^2\left(\frac{1}{2-4\left(\frac{3R+2R}{R^{2}R{C}^2}\right)R^2{C}^2}\right)\\ -\left(\frac{R_F}{R}\right){\left(1+\frac{R_F}{R}\right)}^2\left(\frac{1}{2-\frac{4\left(5R\right)}{R}}\right)\\ =-\left(\frac{R_F}{R}\right){\left(1+\frac{R_F}{R}\right)}^2\left(\frac{1}{-18}\right)\\ 18=-\left(\frac{R_F}{R}\right){\left(1+\frac{R_F}{R}\right)}^2\end{array}$

Taking $\frac{R_F}{R}=2$ .

Hence, the needed condition for oscillation:

  $\frac{R_F}{R}=2$

c.

To determine

The condition for the oscillation.

Answer to Problem 15.29P

The condition for oscillation:

  $\frac{R_F}{R}=2$

Explanation of Solution

Given:

The circuit is given as:

  

  $\text{R=25kΩ,C=0}\text{.001μF}\text{and}{\text{15<R}}_{\text{V}}\text{<30kΩ}$

Now substituting the values in equation 4:

  $\begin{array}{c}{f}_{\max }=\frac{1}{2\pi RC}\sqrt{\frac{3R+2R_V}{R_V}}\\ =\frac{1}{\left(2\right)\left(3.14\right)\left(25\times {10}^3\right)\left(0.001\times {10}^{-6}\right)}\sqrt{\frac{3\left(25\times {10}^3\right)+2\left(15\times {10}^3\right)}{15\times {10}^3}}\\ =\frac{1}{1.571\times {10}^{-4}}\left(\sqrt{7}\right)\\ =6.366\times {10}^3\left(2.646\right)\\ =16.843\text{kHz}\end{array}$

Hence, the condition for the oscillation:

  $13.505\le f\le 16.843\text{kHz}$