Question

The input ac voltage has magnitude Vs at a frequency of You may assume that the filter, Lbia is sufficiently large such that load <« w (i.e. the ripple in the inductor current can be ignored), and that the diodes are ideal. a. Ldig What is the distortion factor kp and the displacement factor kg for this rectifier? Also find an expression for the power factor (seen by the AC voltage source). Find an expression for the THD of this rectifier. Note that since the source is a voltage source, we care about the THD of the current (i.e. if you get an answer of 0, you are wrong). Simplify this as much as possible (you should only have numbers and fundamental constants in the answer). Evaluate your answer making the approximation that: b. T2 10. Lbig D1 D2 loadload

Answer 1

V_0 = 2v_s/pi average output b a_n = 0 a_0 = 0 as odd function b_n = 2/2 pi [integral^pi_0 I_0 sin n wt dt - integral^2 pi_pi I_0 sin n wt dt] = 2I_0/2 pi nw[(- w_0 n wt)^pi_0 - (- w_0 n wt)^2 pi_pi] = 2I_0/2 pi nw[(1 + 1) + 1 + 1] = 2 times 4I_0/2n pi w = 2 times 4 I_0/2 n pi 2 pi/2 pi = (4 I_0/n pi) so i_s(t) = sigma^infinity_n = 1 4I_0/n pi sin n wt I_1S = (4 I_0/pi squareroot 2) I_S = I_0 \r\n Rms value of source current I_S = squareroot R^2_0 pi/h = I_0 V_0 = 2V_s/pi Input voltage = (V_s/squareroot 2) Input power = V_s/squareroot I_S cos emptyset = V_S/squareroot 2 times I_0 cos emptyset Load power = V_0 I_0 = 2V_S/h I_0 Input power = load power 2 V_s/h I_0 = V_s/squareroot 2 I_0 cos emptyset = 2 squareroot 2/pi = 0.90 \r\n a current distortion factor