Mumbai University > Electronics ana telecommunication engineering > Sem 3 > Analog electronics 1

Marks: 10M

Years: May 16

Consider circuit operation in each interval,

(0 - t_1): Let the initial voltage on C be zero. The diode D is reverse biased and C charges through is given as, $V_c= V (1- e^\frac{-1}{RC})$

Substituting V = 10 V, t =$t_1$= 0.5 ms. And RC = 0.1

$V_c= 10 (1- e^\frac{-0.5 x 10^(-3)}{0.1})$

= 0.0498 V = App 0 V

$(t_1- t_2)$: The diode D is forward biased and the capacitor charges the D as shown in fig(b). $V_c$= -20-4.3 = -24.3 V and the Load voltage is -20+24.3 = 4.3 V

$(t_2- t_3)$: Output voltage = 10 + 24.3 = 34.3 Volts

$(t_3- t_4)$: Output voltage = -20 + 24.3 = 4.3 Volts