The topic “Full Wave Rectifier” is covered in Chapter 14 Semiconductor Electronics: Materials, Devices and Simple Circuits of NCERT Class 12 Physics. Read More: Bridge Rectifier Things to Remember The output voltage as well as the output power obtained in full wave rectifiers are higher as compared to that obtained by using half wave rectifiers. The value of the ripple factor in full wave rectifiers is 0.482 whereas in half wave rectifiers it is 1.21. The rectification efficiency of the half wave rectifiers is 40.6% while that of full wave rectifiers is 81.2%.Ī simple filter is required in full wave rectifiers as the ripple factor is low. The rectification efficiency of the full wave rectifiers is double as compared to the half wave rectifiers. Read More: Diodes Advantages of the Full Wave Rectifiers The efficiency of the full wave rectifier is 81.2%. The rectification efficiency can be obtained by the following formula: Peak factor of the full wave rectifier is calculated using the formula: The form factor of the full wave rectifier can be calculated by using the following formula: The following formula can be used for calculating the RMS value of current: V dc = V dc R L = 2/π I max R L RMS Value of Current The average value of the DC output voltage is represented by the following formula: Read More : Junction Transistor DC Output Voltage The peak inverse voltage of a full wave rectifier is double that of half wave rectifier. In the reverse biased direction, peak inverse voltage refers to the maximum voltage a diode can withstand before breakdown. Full Wave Rectifier Formula Peak Inverse Voltage Therefore, in a full wave rectifier, DC voltage is obtained for both positive as well as negative half cycle. This is because the top half of the secondary circuit becomes negative, while the bottom half of it becomes positive. In case of a negative half cycle, the diode D1 is reversed biased and the diode D2 is forward biased. D1 will let the current flow and D2 will block the flow through it. When there is the positive half cycle of the AC voltage then terminal 1 will be positive, terminal two will be negative and centre tap will be at the zero potential.Īt the time of the positive half cycle the diode D1 will be forward biased and diode D2 will be reverse biased. AC voltage is supplied to the input transformer.
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