A singlephase halfwave diode rectifier powered by 230V AC ma

A single-phase half-wave diode rectifier powered by 230V AC mains produces 80 V (average voltage) across an inductive load. What is the load current zero duration per fundamental cycle? A three-phase rectifier shown in Fig. 3 has purely resistive load of R. Justify the following equalities: I_d, avg = 1/3 I_o, avg I_d, rms = 1/squareroot 3 I_o, rms I_s, rms = squareroot 2/3 I_o, rms

Solution

V_m = maximum value of transformer secondary voltage

V_s =rms value of secondary voltage

V_LM =maximum value of load voltage = V_sm – diode drop – secondary resistance drop

V_L = rms value of load voltage

I_L = rms value of load current

V_L(dc) =average value of load voltag

I_L(dc) = average value of load current

I_LM = maximum value of load current

R_L = load resistance

R_S = transformer secondary resistance

r_d = diode forward resistance

now,

R₀ = R_S + r_d

I_LM = V_sm – V_B /(R_S + r_d) + R_L

        = V_sm –V_B

_{VLM = ILM . RL}

V_L(dc) = V_LM/ = 0.318V_LM

I_L(dc) = I_LM/ = 0.318I_LM

I_L = I_LMn/2 = 0.5I_LM = 0.5V_LM/R_L