Chapter 15, Problem 15.30P

(a)

To determine

To derive: The expression for the frequency of oscillation

Answer to Problem 15.30P

The expression for the frequency of oscillation is ${\omega }_o=\frac{\sqrt{10}}{RC}$

Explanation of Solution

Given:

Consider the circuit shown below.

  

Calculation:

The non-inverting terminal of op-amp is grounded. Hence, the voltage $v_4$ is equal to zero.

  $v_4=0\text{V}$

From virtual ground, the voltage at non-inverting terminal is equal to voltage at inverting terminal

  $\begin{array}{l}{v}_4=v_5\hfill \\ v_5=0\hfill \end{array}$

Use KCL at the node $v_1$

  $\frac{v_1-v_1}{R}+\frac{v_1-v_2}{R}+\frac{v_1}{\frac{1}{sC}}=0$

  $\begin{array}{c}\frac{v_1-v_1}{R}+\frac{v_1-v_2}{R}+\frac{sC{v}_1}{1}=0\\ \frac{v_1-v_1+v_1-v_2+sRC{v}_1}{R}=0\\ \frac{v_1(2+sRC)-v_2-v_1}{R}=0\\ v_1(2+sRC)-v_2-v_1=0\\ v_1=v_1(2+sRC)-v_2\dots \dots (1)\end{array}$

Use KCL at the node $v_2$

  $\begin{array}{c}\frac{v_2-v_1}{R}+\frac{v_2-v_3}{R}+\frac{v_2}{\frac{1}{sC}}=0\\ \frac{v_2-v_1}{R}+\frac{v_2-v_3}{R}+\frac{sC{v}_2}{1}=0\\ v_2-v_1+v_2-v_3+v_{2}sRC=0\\ -v_1+(2+sRC)v_2-v_3=0\\ v_1=(2+sRC)v_2-v_3\dots \dots .\text{ (2) }\end{array}$

Use KCL at the node $v_3$

  $\frac{v_3-v_2}{R}+\frac{v_3-v_4}{R}+\frac{v_3}{\frac{1}{sC}}=0$

Substitute $v_4=0\text{V}$ to obtain the node equation at $v_3$

  $\begin{array}{l}\frac{v_3-v_2}{R}+\frac{v_3-0}{R}+\frac{sC{v}_3}{1}=0\hfill \\ \frac{v_3-v_2+v_3+sRC{v}_3}{R}=0\hfill \\ \frac{v_3(2+sRC)-v_2}{R}=0\hfill \\ \frac{v_3(2+sRC)-v_2=0}{2+sRC}\dots \dots (3)\hfill \\ v_3=\frac{v_2}{2}\hfill \end{array}$

Substitute equation (3) in equation (2),

  $\begin{array}{l}{v}_1=(2+sRC)v_2-\frac{v_2}{2+sRC}\\ =\left(2+sRC-\frac{1}{2+sRC}\right)v_2\\ v_1=\left(\frac{{(2+sRC)}^2-1}{2+sRC}\right)v_2\dots \mathrm{..}(4)\end{array}$

Substitute equation (4) in equation (1),

  $v_i=v_1(2+sRC)-v_2$

  $\begin{array}{l}{v}_t=v_1(2+sRC)-v_2\\ v_l=(2+sRC)\left(\frac{{(2+sRC)}^2-1}{2+sRC}\right)v_2-v_2\\ =\left\{\left({(2+sRC)}^2-1\right)-1\right\}{v}_2\\ =\left(4+s^2{R}^2{C}^2+4sRC-1-1\right)v_2\\ =\left(s^2{R}^2{C}^2+4sRC+2\right)v_2\\ v_2=\frac{v_1}{s^2{R}^2{C}^2+4sRC+2}\end{array}$

Use KCL at the node $v_4$

  $\begin{array}{l}\frac{v_4-v_o}{R_F}+\frac{v_4-v_3}{R}=0\hfill \\ \frac{0-v_o}{R_F}+\frac{0-v_3}{R}=0\hfill \\ \frac{-v_o}{R_F}+\frac{-v_3}{R}=0\hfill \\ v_o=-\frac{R_F}{R}{v}_3\hfill \end{array}$

Substitute $v_3=\frac{v_2}{2+sRC}$ to obtain $v_o$

  $v_o=-\frac{R_p}{R}\left(\frac{1}{2+sRC}\right)v_2$

Substitute $v_2=\frac{v_1}{s^2{R}^2{C}^2+4sRC+2}$ to obtain $v_o$

  $\begin{array}{l}{v}_o=-\left(\frac{R_F}{R}\right)\left(\frac{1}{2+sRC}\right)\left(\frac{1}{s^2{R}^2{C}^2+4sRC+2}\right)v_1\\ \frac{v_o}{v_l}=-\left(\frac{R_F}{R}\right)\left(\frac{1}{2s^2{R}^2{C}^2+8sRC+4+s^3{R}^3{C}^3+4s^2{R}^2{C}^2+2sRC}\right)\\ =-\left(\frac{R_F}{R}\right)\left(\frac{1}{s^3{R}^3{C}^3+6s^2{R}^2{C}^2+10sRC+4}\right)\end{array}$

The transfer function is given by.

  $\begin{array}{l}T(s)=\frac{v_o(s)}{v_t(s)}\hfill \\ T(s)=-\left(\frac{R_F}{R}\right)\left(\frac{1}{s^3{R}^3{C}^3+6s^2{R}^2{C}^2+10sRC+4}\right)\hfill \end{array}$

To find the frequency oscillation, set $s=j\omega$

  $\begin{array}{l}\end{array}$

  $T(j\omega )=-\left(\frac{R_F}{R}\right)\left(\frac{1}{{(j\omega )}^3{R}^3{C}^3+6{(j\omega )}^2{R}^2{C}^2+10(j\omega )RC+4}\right)$

  $=-\left(\frac{R_F}{R}\right)\left(\frac{1}{-j{\omega }^3{R}^3{C}^3-6{\omega }^2{R}^2{C}^2+10(j\omega )RC+4}\right)$

  $=-\left(\frac{R_F}{R}\right)\left(\frac{1}{4-6{\omega }^2{R}^2{C}^2+j(10\omega RC-{\omega }^3{R}^3{C}^3)}\right)$

  $=-\left(\frac{R_F}{R}\right)\left(\frac{4-6{\omega }^2{R}^2{C}^2-j(10\omega RC-{\omega }^3{R}^3{C}^3)}{{\left(4-6{\omega }^2{R}^2{C}^2\right)}^2+{\left((10\omega RC-{\omega }^3{R}^3{C}^3)\right)}^2}\right)$

  $T(j\omega )=-\left(\frac{R_F}{R}\right)\left(\frac{4-6{\omega }^2{R}^2{C}^2}{{\left(4-6{\omega }^2{R}^2{C}^2\right)}^2+{\left((10\omega RC-{\omega }^3{R}^3{C}^3)\right)}^2}\right)+$

  $j\left(\frac{R_F}{R}\right)\left(\frac{(10\omega RC-{\omega }^3{R}^3{C}^3)}{{\left(4-6{\omega }^2{R}^2{C}^2\right)}^2+{\left((10\omega RC-{\omega }^3{R}^3{C}^3)\right)}^2}\right)\mathrm{......................}(6)$

From the Barkhausen criterion, the condition for oscillation is that $T\left(j{\omega }_e\right)=-1$

To satisfy the condition $T\left(j{\omega }_o\right)=-1$ , equate the imaginary component of the equation (6) equal to zero.

  $\begin{array}{l}\left(\frac{R_F}{R}\right)\left(\frac{\left(10{\omega }_{o}RC-{\omega }_o^3{R}^3{C}^3\right)}{{\left(4-6{\omega }_o^2{R}^2{C}^2\right)}^2+{\left(10{\omega }_{o}RC-{\omega }_o^3{R}^3{C}^3\right)}^2}\right)=0\hfill \\ \left(10{\omega }_{o}RC-{\omega }_o^3{R}^3{C}^3\right)=0\hfill \end{array}$

  ${\omega }_{o}RC\left(10-{\omega }_o^2{R}^2{C}^2\right)=0$

  $\left(10-{\omega }_o^2{R}^2{C}^2\right)=0$

  ${\omega }_o^2{R}^2{C}^2=10$

  ${\omega }_o^2=\frac{10}{R^2{C}^2}$

  ${\omega }_o=\frac{\sqrt{10}}{RC}$

Conclusion:

Therefore, the expression for the frequency of oscillation is ${\omega }_o=\frac{\sqrt{10}}{RC}$

(b)

To determine

The condition of oscillation.

Answer to Problem 15.30P

The condition of oscillation is $\frac{R_F}{R}=56$

Explanation of Solution

Given:

Consider the circuit shown below.

  $R=20\text{k}\Omega$ , $f_o=22\text{kHz}$

Calculation:

Substitute $2\pi f_c$ for ${\omega }_o$ to obtain frequency of oscillations.

  $\begin{array}{l}2\pi f_o=\frac{\sqrt{10}}{RC}\hfill \\ f_o=\frac{\sqrt{10}}{2\pi RC}\hfill \end{array}$

The condition for oscillation is that $\left|T\left(j{\omega }_o\right)\right|=1,$ to satisfy this condition equate real part of the equation (6) equal to 1

  $-\left(\frac{R_F}{R}\right)\left(\frac{4-6{\omega }_o^2{R}^2{C}^2}{{\left(4-6{\omega }_o^2{R}^2{C}^2\right)}^2+{\left(10{\omega }_{o}RC-{\omega }_o^3{R}^3{C}^3\right)}^2}\right)=1$

Substitute ${\omega }_o=\frac{\sqrt{10}}{RC}$ to obtain the condition of oscillations

  $\begin{array}{l}-\left(\frac{R_F}{R}\right)\left(\frac{4-6{\left(\frac{\sqrt{10}}{RC}\right)}^2}{{\left(4-6{\left(\frac{\sqrt{10}}{RC}\right)}^2{R}^2{C}^2\right)}^2+{\left(10\frac{\sqrt{10}}{RC}RC-{\left(\frac{\sqrt{10}}{RC}\right)}^3{R}^3{C}^3\right)}^2}\right)=1\hfill \\ \left(\frac{R_F}{R}\right)\left(\frac{56}{{(-56)}^2}\right)=1\hfill \\ \frac{R_F}{R}=56\hfill \end{array}$

Therefore, the required condition for oscillation is $\frac{R_F}{R}=56$

Conclusion:

Therefore, the required expression for the frequency of oscillation is $\frac{R_F}{R}=56$

(c)

To determine

To find: The values of capacitor and $R_F$

Answer to Problem 15.30P

The required values are $C=3.1623\times {10}^{-9}\text{F}$ , $R_F=11.12\text{k}\Omega$

Explanation of Solution

Given:

Consider the circuit shown below.

  $R=20\text{k}\Omega$ , $f_o=22\text{kHz}$

Calculation:

Substitute $20\text{k}\Omega$ for $\text{R}$ to obtain $R_F$

  $\begin{array}{l}{R}_F=56R\\ =56\left(20\times {10}^3\right)\\ =1120\times {10}^3\end{array}$

Thus, the required feedback resistance is $R_F=11.12\text{k}\Omega$

Substitute $20\text{k}\Omega$ for $R,1.12\text{k}\Omega$ for $R_F,$ and $22\text{kHz}$ for $f_0$ to obtain the value of $\text{C}$ .

  $\begin{array}{l}{f}_o=\frac{\sqrt{10}}{2\pi RC}\\ C=\frac{\sqrt{10}}{2\pi f_{o}R}\\ C=\frac{\sqrt{10}}{2\pi \left(22\times {10}^3\right)\left(20\times {10}^3\right)}\\ =\frac{\sqrt{10}}{2.765\times {10}^9}\\ =3.1623\times {10}^{-9}\text{F}\end{array}$

Conclusion:

Therefore, the required values are $C=3.1623\times {10}^{-9}\text{F}$ , $R_F=11.12\text{k}\Omega$