Sunday, May 20, 2012

Full Wave Recftifier : Part2

1.6 D.C. Power Output ()
       D.C. power output = EDC IDC =IDC2 RL
       Substituting value of Im we get,
Note : Instead of remembering this formula, we can use the expression EDC IDC or IDC2RL to calculate PDC while solving the problems.
1.7 A.C. Power Input (PAC)
       The a.c. power input is given by,
       Substituting value of Im we get,
1.8 Rectifier Efficiency (η)
       But if R2 + R2 < < RL, neglecting it from denominator
       This is the maximum theoretical efficiency of full wave rectifier.
1.9 Ripple Factor (γ)
       As derived earlier in case of half wave rectifier the ripple factor is given by a general expression,
       For full wave IRMS  = Im/√2 and = 2Im/π so substituting in the above equation,
 ...                                        Ripple factor = = 0.48
Note : This indicates that the ripple contents in the output are 48% of the d.c. component which is much less than that for the half wave circuit.
1.10  Load Current (IL)
       The fourier series for the load current is obtained by taking the sum of the series for the individual rectifier current. The two diodes conducts in alternate half cycle, i.e. there is a phase difference of π radians between two diode currents. Hence,
       and                id2 = id1 with replaced by (ωt + π)
       Then the Fourier series for the load current is,
       The first term in the above series represents the average or dc value, while the remaining terms ''ripple''. It is seen that the lowest frequency of the ripple is 2f, i.e. twice the supply frequency of ac supply. The lowest ripple frequency in the load current of the full-wave connection, is double than that in the half-wave connection.
       As seen from Fig. (2 and 3 see previous post) the individual diode currents are flowing in opposite directions through the two halves of the secondary winding. Hence the net secondary current will be difference of individual diode currents.
Thus,          isec  = id1  - id2
       The fourier series of isec  is obtained by the difference between the series of individual diode currents. Using above relations we can write,
                 isec =Im sin ωt
       Hence under ideal conditions, the secondary current is purely sinusoidal. No d.c. component flows through the secondary hence there is no danger of saturation. This reduces the transformer losses and overall size and cost of the circuit. Thus the transformer gets utilized effectively.

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