Question

In the figure below, let the diode threshold voltage be vTH = 0.6 volts. Also let the load
R0 = 50
and the voltage supply vS(t) = 5 cos 100 t − 180 ◦:
v t
S( ) ~ +
− −
+
v t 0R ( ) 0
(a) Compute t0, the first time after t = 0 when the diode starts conducting.
(b) Plot the load current i0(t) over 0 t 0.125 sec.

Answer 1

Given data: - V_th = 0.6 v R_0 = 50 Ohm a) Assume the diode like open circuit with 't' reference term final on pride and '-' terminal on N - side then when V_oc greaterthanorequalto V_th, then diode is ON 5 cos (100t - 180 degree) = 0.6 + V, then diode is ON. \r\n after t = 0, V_s (t) is equal to 0.6 V in the terminal (T_0/4, T_0/2) = (15.707 ms, 31.416ms) now, 5 cos (100 t- pi) = 0.6/5 100 t = pi t cos^-1(0.6/5) t = pi/100 + 1/100 cosa^-1(0.6/5) = T_0/2 + E_1 but actually at t = T_0/2 - t_1, V_s(t) = 0.6V for the first time Therefore t = pi/100 - 1/100 cos^-1 (0.6/5) = 17.53 ms at t = 17.63ms, the diode is for the first time. \r\n b) when diode OFF, diode acts like open circuit . Therefore i_0(t) = 0 A and when diode is ON, is replaced with 0.6V