The full wave rectifier conducts during both positive and negative half
cycles of input a.c. supply. In order to rectify both the half cycles of
a.c. input, two diodes are used in this circuit. The diodes feed a
common load R

_{L}with the help of a center tap transformer. The a.c. voltage is applied through a suitable power transformer with proper turns ratio.
The full wave rectifier circuit is shown in the Fig. 1.

Fig. 1 Full wave rectifier |

For the proper operation of the circuit, a center-tap on the secondary winding of the transformer is essential.

__1.1 Operation of the Circuit__

Consider the positive half cycle of a.c. input voltage in which
terminal (A) is positive and terminal (B) negative. The diode D

_{1}will be forward biased and hence will conduct; while diode D_{2}will be reverse biased and will act as an open circuit and will not conduct. This is illustrated in the Fig. 2.Fig. 2 Current flow during positive half cycle |

The diode D

_{1}supplies the load current, i.e. I_{L}= I_{d2}. This current is flowing through upper half of secondary winding of the transformer carries no current since diode D_{2}is reverse biased and acts as an open circuit.
In the next half cycle of a.c. voltage, polarity reverses and terminal (A) becomes negative and (B) positive. The diode D

_{2}conducts, being forward biased, while D_{1}does not, being reverse biased. This is shown in the Fig. 3.Fig. 3 Current flow during negative half cycle |

The diode D

_{2}supplies the load current, i.e. I_{L}= I_{d2}. Now the lower half of the secondary winding carries the current but the upper half does not.**Note**: It is noted that the load current flows in both the half cycles of a.c. voltage and in the same direction through the load resistance.

Hence we get rectified output across the load. The load current is sum
of individual diode currents flowing in corresponding half cycles. It is
also noted that the two diodes do not conduct simultaneously but in
alternate half cycles. The individual diode currents and the load
current are shown in the Fig. 4.

Fig. 4 Load current and voltage waveforms for full wave rectifier |

Thus the full wave rectifier circuit essentially consists of two
half-wave rectifier circuits working independently (working in alternate
half cycles of a.c.) of each other but feeding a common load. The
output load current is still pulsating d.c. and not pure d.c.

__1.2 Maximum Load Current__

Let R

_{f}= Forward resistance of diodes
R

_{s}= Winding resistance of each half of secondary
R

_{L}= Load resistance
e

_{s}= Instantaneous a.c. voltage across each half of secondary**.**e

^{.}._{s}= E

_{sm}sin ωt

ω = 2 π f

E

_{sm}= Maximum value of a.c. input voltage across each half of secondary winding
Hence we can write the expression for the maximum value of the load
current, looking at equivalent circuit shown in the Fig. 5.

Fig. 5 |

where I

_{m}= Maximum value of load current I_{L}.__1.3 Average D.C. Load Current (I__

_{DC})
Consider one cycle of load current I

_{L}from 0 to 2π to obtain the average value of load current.Fig. 6 |

I

_{L }= I_{m}sinωt 0≤ ωt≤ π
But for π to 2π, the current is again positive while sint term is
negative during π to 2π. Hence in the region π to 2π the positive I

_{L}can be represented as negative of I_{m}sin (ωt).**.**I

^{.}._{L }= -I

_{m}sin ωt π ≤ ωt ≤ 2π

But cosπ = -1

For half wave it is I

_{m}/π and full wave rectifier is the combination of two half wave circuits acting alternately in two half cycles of input. Hence obviously the d.c. value for full wave circuits 2I_{m}/π.__1.4 Average D.C. Load Voltage (__E

_{DC}

__)__

The d.c. load voltage is,

__1.5 R.M.S. Load Current (__I

_{RMS}

__)__

The R.M.S. value of current can be obtained as follows :

Since two half wave rectifier are similar in operation we can write,

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