Question

Consider the circuit shown with ideal op-amp:



If the saturation voltage of the output is $\pm 15V$ and the hysteresis width of the given circuit is 5 V, then the value of $R_2$ is

• A. $2.8k\Omega$
• B. $5.2k\Omega$
• C. $1.6k\Omega$
• D. $3.6k\Omega$

Answer 1

The correct option is C $1.6k\Omega$

Width of hysteresis loop,

$V_H=\frac{R_2}{R_1+R_2}(V_{sat}^{+}-V_{sat}^{-})$

$\frac{R_2}{R_1+R_2}=\frac{5}{15-(-15)}=\frac{5}{30}=\frac{1}{6}$

$6R_2=R_1+R_2$

$5R_2=R_1$

$R_2=\frac{8}{5}k\Omega =1.6k\Omega$