Solution:

Given Data-

$ \begin{aligned} &f_0=2 \mathrm{kH} \\ &v_{p p}=7 \mathrm{~V} \\ &v_{\text {sat }}=\pm \mathrm{14KV} \end{aligned} $

Step 1;-

$ V_{\text {o }}[\text { peak }]=\frac{R_2 \cdot V_2}{R_3}\\ $

Now because the Zener diodes are not being used, replace $V_z$ by $V_{\text {sat }}$ moreover as $V_0[p-p]=$ $2 v_0$ [peak], the above equation gets modified to,

$ \begin{aligned}\\ &\therefore V_0[p-p]=2\left[\frac{R_2}{R_3}\right] \times V_{\text {sat }} \\\\ &\therefore\left[\frac{R_2}{R_3}\right]=\frac{V_{0 p-p}}{V_{\text {sat }} \times 2}=\frac{7}{14 \times 2}=\frac{1}{4} \\\\ &\therefore R_2=\frac{R_3}{4} \\\\ &\text { Let }\left[R_2=10 \mathrm{k} \Omega\right] \\\\ &\therefore R_3=40 \mathrm{k} \Omega \text { [use } 50 \mathrm{k} \Omega \text { potentiometer] }\\ \end{aligned}\\ $

Step 2:-

$f=\frac{R_3}{4 R_1 C_1 R_2}$

$ \begin{aligned} 2 \mathrm{KH}_2 &=\frac{40 \mathrm{k} \Omega}{4 \times\left[R_1 C_1\right] \times R_2} \\ R_1 C_1 &=\frac{40 \mathrm{k} \Omega}{4 \times 2 \mathrm{kll}} \times 10 \mathrm{k} \Omega \\ \therefore R_1 C_1 &=500 \times 10^{-6} \\ &=500 \mathrm{rs} \\ &=0.5 \mathrm{~ms} \end{aligned} $

$ \text { Let } \begin{aligned} C_1 &=0.05 \mu \mathrm{F} \\ R_1 &=10,000 \\ R_1 &=10 \mathrm{k} \Omega \end{aligned} $