Substituting value of I

_{m }we get,**Note**: Instead of remembering this formula, we can use the expression E

_{DC}I

_{DC}or I

_{DC}

^{2}R

_{L}to calculate P

_{DC}while solving the problems.

__1.7 A.C. Power Input (__P

_{AC}

__)__

__1.8 Rectifier Efficiency (η)__

But if R

This is the maximum theoretical efficiency of full wave rectifier.

As derived earlier in case of half wave rectifier the ripple factor is given by a general expression,

For full wave I

Then the Fourier series for the load current is,

_{2 }+ R_{2 < < }R_{L}, neglecting it from denominatorThis is the maximum theoretical efficiency of full wave rectifier.

__1.9 Ripple Factor (γ)__

For full wave I

_{RMS }= I_{m}/√2 and = 2I_{m}/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 (__I

_{L}

__)__

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 i_{d2}= i_{d1}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, i_{sec}= i_{d1}- i_{d2}
The fourier series of i

i_{sec}is obtained by the difference between the series of individual diode currents. Using above relations we can write,_{sec}=I_{m }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.

**Sponsored links :**
## 0 comments:

## Post a Comment