Full-wave Rectifier
The circuit of full-wave rectifier consists of a center-tapped transformer,
followed by the rectifier formed by two diodes D1 and D2, and finally the load R with a capacitor
filter C. The circuit is designed such that the current through the load is always in the same
direction during both the half cycles.
Assume that the capacitor is not connected initially in the full-wave rectifier circuit. Due to the center tap rectifier, during the positive half cycle of the input Vs, A is positive with respect to B, hence diode D1 will be forward biased and D2 will be reverse biased. This results in the current flowing from A-D1- R-B-A. In the negative half cycle, C is positive with respect to B. This makes C positive with respect to B causing diode D2 to get forward biased making the current flow from C-D2-R-B-C. Thus in both the half cycles the current through R flows in the same direction resulting
Assume that the capacitor is not connected initially in the full-wave rectifier circuit. Due to the center tap rectifier, during the positive half cycle of the input Vs, A is positive with respect to B, hence diode D1 will be forward biased and D2 will be reverse biased. This results in the current flowing from A-D1- R-B-A. In the negative half cycle, C is positive with respect to B. This makes C positive with respect to B causing diode D2 to get forward biased making the current flow from C-D2-R-B-C. Thus in both the half cycles the current through R flows in the same direction resulting
in pulsating DC across R.
With capacitor filter
During the positive half cycle, capacitor C will charge to
the peak of the input waveform
while the load R is being supplied current through D1. When the input starts to go below its
peak value, the voltage across C will cause D1 to be reverse biased, thus disconnecting the
rectifier from the load. The capacitor will then provide the necessary current for the load. The
rate of the potential drop across C will be based on the values of R & C.
while the load R is being supplied current through D1. When the input starts to go below its
peak value, the voltage across C will cause D1 to be reverse biased, thus disconnecting the
rectifier from the load. The capacitor will then provide the necessary current for the load. The
rate of the potential drop across C will be based on the values of R & C.
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