Bridge rectifier

·        The circuit diagram of a bridge rectifer is shown above. It uses four diodes and a transformer.

·        During the +ve half-cycle, end A is +ve and end B is –ve thus diodes D₁ and D₃ are forward bias while diodes D₂ and D₄ are reverse biased thus a current flows through diode D_1, load R_L ( C to D) and diode D₃.

·        During the –ve half-cycle, end B is  +ve and end A is –ve thus diodes D₂ and D₄  are forward biased while the diodes D₁ and D₃ are reverse biased. Now the flow of current is through diode D₄ load R_L ( D to C) and diode D₂. Thus, the waveform is same as in the case of center-tapped full wave rectifier.

Advantages

• The need for center-taped transformer is eliminated.
• The output is twice when compared to center-tapped full wave rectifier.

           for the same secondary voltage.

• The peak inverse voltage is one-half(1/2) compared to center-tapped full wave rectifier.
• Can be used where large amount of power is required.

Disadvantages

·         It requires four diodes.

·         The use of two extra diodes cause an additional voltage drop thereby reducing the output voltage.

2.5.6 Efficiency of Full-wave rectifier

Let V = V_msinθ be the voltage across the secondary winding

       I = Imsinθ be the current flowing in secondary circuit

        r_f = diode resistance

        R_L = load resistance

dc power output

                -----------------------------(1)

input ac power

Squaring both sides we get

The efficiency will be maximum if r_f  is negligible as compared to R_L.

Maximum efficiency  =  81.2 %

This is the double the efficiency due to half wave rectifier. Therefore a Full-wave rectifier is twice as effective as a half-wave rectifier.

Comparision of Rectifiers

Particulars	Half wave rectifier	Centre-tapped Full wave rectifier	Bridge rectifier
1. No. of diodes 2. Idc 3. Vdc 4.Irms 5.Efficiency 6.PIV 7.Ripple factor	1       Im / Π Vm / Π Im / 2 40.6 % Vm 1.21	2 2Im /Π 2Vm / Π Im /√ 2 81.2 % 2Vm 0.48	4 2Im /Π 2Vm / Π Im /√ 2 81.2 % Vm 0.48

Note:

• The relation between turns ratio and voltages of primary and  secondary of the transformer is given by
• N₁ / N₂ = V_p / V_s

• RMS value of voltage and Max. value of voltage is related by the equation.
• V_rms = V_m / √2  ( for full-cycle of ac)

• If the type of diode is not specified then assume the diode to be of silicon type.

• For an ideal diode, forward resistance r_f = 0 and cut-in voltage , V_γ = 0.