Design square wave generator using opamp to have output voltage $=\pm 5$ volts, frequency $1 \mathrm{khz}$, with $70 \%$ duty cycle? Assume $\mathrm{VCC}=\pm 12$ volts.

Solution:

$ \begin{aligned} &F=1 \mathrm{KHz}_2 \text { duty cycle }=70 \%, \quad \mathrm{VO} \pm 5 \mathrm{~V} \\ &+\mathrm{Vsat}=12 \mathrm{~V},-\mathrm{Vsat}=-12 \mathrm{~V}, \ldots \text { (given) } \end{aligned}\\ $

Assume $R_1=R_2=10 \mathrm{~K}-2$

When $v_0=+5 v$ then ' C ' charges through $R_3$ via $D_1$ when $v o=-5 v$ then ' C' discharges through $R_4 v i a P_2$

For $70 \%$ duty cycle,

$ \text { TON }=\frac{7}{10} \times T=\frac{7}{10}\\ $

$ \begin{aligned} &\text { TON }=700 u s \mathrm{sec} \\ &V_{V T}=\frac{R_1\left(+V_{\text {Sat }}\right)}{R_1+R_2}=6 \mathrm{~V} \\ &V_{L T}=\frac{R_1\left[-V_{\text {Sat }}\right]}{R_1+R_2}=-6 \mathrm{~V} \end{aligned}\\ $

Let $c=0.1 \mathrm{MF}$

chargirg $e q^n$,

$ \begin{aligned} &V_{U T}=+V_S \text { at }+\left[V_{L T}-V_{5 a t}\right] e^{-T \text { ToN } / R_3 \times C} \\ &6=12+[-6-12] \times e^{-\frac{700 \times 10^{-6}}{R_3 \times 0.1 \times 10^{-6}}} \end{aligned} $

solving for $R_3$

$ R_3=6.3 K \Omega $

$ \begin{gathered}\\ \text { Assume } R_4=70 \% R_3 \\\\ \therefore R_4=4.4 \mathrm{k} \Omega\\ \end{gathered}\\ $