Solucionario Daniel Hart Electronica De Potencia Checked B1 -

Then ( I_{avg,load} = 280.8/20 = 14.04 A ). Diode RMS current: ( I_{D,rms} = I_{load} / \sqrt{3} = 14.04 / 1.732 = 8.1 A ).

Formula: ( V_o = D \cdot V_{in} ) ( D = 25/50 = 0.5 ) (Correct in all versions) solucionario daniel hart electronica de potencia checked b1

Frequency squared: ( f^2 = (20,000)^2 = 4 \times 10^8 ) Denominator: ( 8 \times 125e-6 \times 100e-6 \times 4e8 ) First: ( 125e-6 \times 100e-6 = 1.25e-8 ) Then ( 1.25e-8 \times 4e8 = 5 ) Times 8 = 40 So ( \Delta V_o = 12.5 / 40 = 0.3125 V ) (or 1.25% ripple). Then ( I_{avg,load} = 280