Half wave Rectifier

 Half wave Rectifier

Fig. Half wave Rectifier

The line voltage 230 Volt 50 Hz is stepped down by using step down transformer and secondary of transformer voltage is given as output to the half wave rectifier using single diode.

The load resistance RL is connected at the output where, we get the rectified D.C. Voltage.

Working : During positive half cycle at point A of the input voltage, diode D1 becomes forward biased and current flows through the diode and load resistor RL producing a voltage drop across RL.

The output voltage and current waveforms are shown in fig.

During negative half cycle of the input voltage, D1 is reverse biased and no output voltage across the load RL.

The D.C. voltage developed across the load RL is given by,

 





The undesired A.C. components present at D.C. output is known as ripples.

Ripple Frequency of HWR is fr = fi = 50 Hz

The effectiveness of a rectifier depends upon the magnitude of A.C. component in the output, which is determined by ripple factor.

The ratio of r.m.s value of A.C. component to the D.C. component in the rectifier-output is known as ripple factor.

 


Thus ripple factor of HWR is 1.21. That means HWR is somewhat ineffective to convert A.C. to D.C.

Another important parameter of rectifier diode is PIV i.e. Peak Inverse Voltage.

It is the maximum reverse bias voltage applied to the pn junction diode upto which it can withstand without any damage.

In HWR, PIV is the maximum transformer voltage Vm across the secondary coil.

Hence for a HWR, the PIV = Vm

Also, Transformer Utility Factor (T.U.F.) is important, it is defined as “ The ratio of D.C. power delivered to load to the A.C. rating of the transformer.”

T.U.F. = P dc / P ac(specified)

For a HWR, T.U.F. is 0.287



Fig. Input and output waveforms of HWR

 

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