Question

Problem 3 (30 points): For the circuit in Fig. 3, v, has an RMS value of 9.0 V and a frequency of 60 Hz. Assume that R-20 Ω and that V,-0.7 V for each diode. Vs R o Figure 3: Circuit for Problem 3 (10 points) 1. Find the maximum value of Vo 2. Find the value of C to guarantee a ripple voltage of 0.25 V or less. (10 points) 3. For what fraction of the time is the diode on? 4. Find the PIV rating for each diode. (5 points) (5 points)

Answer 1

Given dat: V_s = 9 V (RMS) V_s (max) = V_s times squareroot 2 = 9 times squareroot 2 V_s (max) = 12 . 728 V frequency f = 60 Hz R = 20 ohm V_r = 0 . 7 v (cut in voltage) (1) Minimum value of V_0: rightarrow During positive half cycle, if V_s is just greater than V_r (0 . 7 V) . D_1 starts conduction, till 180 degree rightarrow During Negative half cycle, if V_s is just greater than V_r (0 . 7 V) , D_2 starts conduction till 360 degree . rightarrow Diodes D_1, D_2 in state drop V_r = 0 . 7 V rightarrow Therefore, output voltage at any instant is always less than input voltage by 0 . 7 v