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Design an inverting Schmitt trigger to achieve hysteresis of 7 Volts. Assume voltage swing $=\pm 12$ Volts.
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Solution:

\begin{aligned} \text { Given: }-+\mathrm{VSat} &=12 \mathrm{~V} \\\\ -\mathrm{VSat} &=-12 \mathrm{~V} \\\\ V_{H V}=7 \mathrm{~V} .\\ \end{aligned}\\

The hysteresis voltage $V_{1+v}$ is given by,

$V_{H V}=\frac{2 R_2}{R_1+R_2} \cdot V_{\text {sat }}\\$

\begin{aligned} &\therefore 7=\frac{2 R_2}{\left[R_1+R_2\right]} \times 12 \\\\ &\therefore \quad \frac{R_2}{R_1+R_2}=\frac{7}{24} \\\\ &\therefore \quad R_2=\left[R_1+R_2\right] \times 0.29 \\\\ & \end{aligned}\\

\begin{aligned}\\ &\quad 0.71 R_2=0.29 R_1 \\\\ &\therefore R_e=[0.29 / 0.71] R_1 \\\\ &\text { Let } R_1=10 \mathrm{k} \Omega \\\\ &\therefore R_2=\frac{0.29 \times 10}{0.71}=4.08 \mathrm{k} \Omega\\ \end{aligned}\\

$R_2=4.08 \mathrm{k} \Omega$

\begin{aligned}\\ \therefore \text { RoM } &=R_1 11 R_2 \\\\ &=10 \mathrm{k} \Omega 114.08 \mathrm{k} \Omega \\\\ \text { RoM } &=2.92 \mathrm{k} \Omega\\ \end{aligned}\\