**1 Answer**

written 8.4 years ago by |

## Procedure for design of Schmitt trigger circuit :

1) Determine hysteresis voltage from UTP and LTP $V_H$ = $V_{UTP}$ - $V_{LTP}$, if UTP = LTP then follow following steps.

2) Assume $V_{SAT}$ =13V and -$V_{SAT}$ = -13V

3) Assume $R_2$ = 10K

4) Find the value of $R_1$ from hysteresis voltage formula

$$V_H = \frac{2R_2}{R_1 + R_2} × V_{SAT} $$

5) If UTP≠LTP then most of the things remains same but there will be two different values for $R_1$ and use two diodes for each $R_1$ in opposite direction. It is found using $V_{UTP}$ and $V_{LTP}$ formula. $$V_{UTP} = \frac{R_2}{R_1 + R_2} × V_{SAT}$$ $$V_{LTP} = \frac{R_2}{R_1 + R_2} × -V_{SAT}$$

6) Draw designed circuit diagram.

Design :

## Since UTP = LTP

## Step 1 : Determine hysteresis voltage

$$V_H = V_{UTP} - V_{LTP}$$ Given, UTP = 0.5V and LTP = -0.5V

$$V_H = 0.5 – ( - 0.5) = 1$$

## Step 2 : Determine $R_1$

$$V_H = \frac{2R_2}{R_1 + R_2} × V_{SAT}$$ Assume $V_{SAT}$=13V and $R_2$ = 10KΩ $$1 = \frac{2 × 10K}{10K + R_1} × 13V$$ $$\boxed{R_1 = 250KΩ}$$