Mumbai University > Information Technology > Sem3 > Analog and Digital Circuits

Marks: 6M

Year: Dec13

Duty Cycle (D) = 40%=0.4

Frequency, $f = 10^3$ Hz

$T_{ON} = 0.693 R_AC$ …(1)

$T_{OFF} = 0.693 R_BC$ …(2)

D = $T_{ON}/(T_{ON}+T_{OFF})$

From eq (1) & (2)

D = 0.693 $R_AC/(0.693 R_AC+0.693 R_BC)$

I.e. $D = R_A/(R_A+R_B)$

Since D=0.4 we get,

$0.4R_A + 0.4R_B = R_A$

Thus $R_B = 1.5R_A$ …(3)

Since D < 0.5

$f = 1/(0.693C (R_A+R_B))$

I.e. $103 =1/(0.693C (R_A+R_B))$

Assume $C = 0.1µ_F$

Thus $R_A+R_B$ = 14400 …(4)

From eq (3) & (4) we get,

$R_A = 5.76KΩ$

$R_B = 8.64KΩ$

Since Duty cycle is less than 50% the circuit is modified as shown below, the diode is assumed to be ideal.