Question

Consider the ideal comparator circuit shown in the figure shown.

If input signal is a sinusoidal signal with peak to peak value of amplitude equal to 8 V. Then the duty cycle of the output wave is equal to
(Assume $-V_{sat}=0V$)

• A. 50%
• B. 2.33%
• C. 33.33%
• D. 12.33%

Answer 1

The correct option is C 33.33%

$V_{P-P}=8V$

Thus, $v\left(t\right)=4\sin \left(\omega t\right)$

$V^{-}=2V$

$\therefore \text{Duty cycle}=\frac{T_{on}}{T_{total}}=\frac{{\theta }_{on}}{{\theta }_{total}}$

now, ${\theta }_1={\sin }^{-1}\left(\frac{2}{4}\right)$

${\theta }_1=\frac{\pi }{6}$

and ${\theta }_2=\pi -\frac{\pi }{6}=\frac{5\pi }{6}$

thus, ${\theta }_{\infty }=\frac{5\pi }{6}-\frac{\pi }{6}and{\theta }_{total}=2\pi$

$\therefore \text{Duty circle}=\frac{\frac{5\pi }{6}-\frac{\pi }{6}}{2\pi }\times 100=33.33\%$