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A half wave rectifier has a secondary emf of 15.3 Vrms with a resistance of 0.4 ohm. The diode has forward...

A half wave rectifier has a secondary emf of 15.3 Vrms with a resistance of 0.4 ohm. The diode has forward resistance of 0.8 ohm and a forward voltage of 0.6V. If load resistance is 12 ohm, determine (i) dc load current (ii) dc load voltage (iii) voltage regulation (iv) efficiency. Solution.

Answer

  1. DC Load Current:

    The DC load current (Idc) can be calculated using the formula:
    Idc = Vdc / (R + r + Rf)
    where Vdc = Vm/π, Vm = Vsm - Vd, R = load resistance, r = secondary resistance, Rf = diode resistance, Vsm = √2 * Vrms, and Vd = forward voltage of the diode.
    Vsm = √2 * 15.3 = 21.63 V
    Vm = 21.63 - 0.6 = 21.03 V
    Vdc = 21.03/π = 6.69 V
    Idc = 6.69 / (12 + 0.4 + 0.8) = 0.52 A

  2. DC Load Voltage:

    Vdc = Idc * R = 0.52 * 12 = 6.24 V

  3. Voltage Regulation:

    Voltage regulation = [(Vnl - Vfl) / Vfl] * 100%
    Vnl = Vm
    Vfl = Vdc
    Voltage regulation = [(21.03 - 6.24) / 6.24] * 100% = 237%

  4. Efficiency:

    Efficiency (η) = (Pdc / Pac) * 100%
    Pdc = Idc^2 * R = (0.52)^2 * 12 = 3.24 W
    Pac = Vrms^2 / (R + r + Rf) = (15.3)^2 / (12 + 0.4 + 0.8) = 18.8 W
    η = (3.24 / 18.8) * 100% = 17.23%

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