Full Wave Center-Tapped Rectifier Design

FULL WAVE CENTER-TAPPED RECTIFIER DESIGN:

This circuit's operation is easily understood one half-cycle a time. Consider the first half cycle, when the source voltage polarity is positive on top and negative on bottom. At this time, only the top diode is conducting, the bottom diode is blocking current, and the load "sees" the first half of the sine wave, positive on top and negative on bottom. Only the top half of the transformer's secondary winding carriers current during this half-cycle as shown in figue 3.70.

Top half of secondary winding conduct during positive half-cycle of input, delivering positive half-cycle to load.

During the next half cycle, the AC polarity reverse. Now, the other diode and the other half of the transformer secondary winding carry current while the portions of the circuit formerly carrying current during the last half cycle sit idle. The load still 'sees' half of a sine wave, of the same polarity as before positive on top and negative on bottom.

During negative input half-cycle, bottom half of secondary winding conducts, delivering a positive half cycle to the load.

Disadvantages:

1. Necessity of a transformer with a center-tapped secondary winding is the major disadvantage of full wave rectifier.

2. The center tap rectifier is only seen in low power applications.

Center tap transformer:

• It is more efficient and supplies current in both half cycles.

• In the first half cycle, D1 is forward biased and conducts. D2 is reverse biased.

• In the second half cycle, D2 is forward biased, and conducts.

• When D, conducts, then full secondary voltage appears across D2 therefore PIV rating of the diode should be 2 Vm.

Average value of DC current IDC:

Average value = Area under half of the curve / Half cycle time period

IDC = 2 Im / π

VDC = IDC RL

= 2 Im/π RL

IDC = 2 Im / π

RMS value of the current IRMS

= √Im2/2

IRMS = Im2/√2

VRMS = IRMS . RL

= Im2/√2.RL

= Vm . R1/√2(Rℓ+Rƒ)

VRMS = Vm/√2

DC output power PDC :

PDC = RL . 4Im2 /π2

AC input power PAC :

PAC = Pa + Pr

Pa = I2RMS Rƒ = Rƒ Im2/2

Pr = I2RMS.RL = RL . Im2/2

PAC = Rƒ . Im2/2 = RL. Im2/2

PAC = Im2/2 (Rƒ + RL)

Efficiency (n):

It is the ratio of the DC output to AC input power. It is denoted as n. it is always expressed in terms of percentage.

η = Pdc/ Pac × 100

= 81.2%

Efficiency, η = 81.2%

The maximum possible efficiency of a half wave rectifier is 81.2%, thus the efficiency of full wave rectifier is twice that of half wave rectifier.

Ripple factor (7):

Ripple factor (γ)  = rms value of AC component / average value of DC component

γ = Iac / Idc

The rms value of total load current is given by

IL = √I DC 2 + I12 + I22 + I42 +.....

IL = √ Idc2 + Iac2

IAC = √ IL - Idc2=

γ = Iac / Idc

We know,  IL / IDC =F, From factor

= ILRMS = Im/√2 ,

 IDC = 2. Im/ π

γ = √(1.11)2-1 = 0.48

Peak Inverse Voltage (PIV):

It is the maximum voltage that occurs the rectifying diode in the reverse bias condition.

PIV = 2Vm

Transformer Utilisation Factor (TUF):

It is the ratio of DC power delivered to the load to the ac power rating of the transformer secondary.

TUF = DC power deliver to the load / AC power rating of the transformer secondary

= Pdc/Pac(rating)

Substitute Im = Vm /RL+RF = 8/π2

TUF = 0.812

Regulation:

Regulation is defined as variation of output voltage with change in load current. It is denoted as R. It is expressed in terms of percentage (%).

Percentage of regulation (R%) = VnL-VL / VL × 100

VnL = no load voltage

VL = load voltage

VnL = 2.Vm/n

VL = IDC.RL = RL . 2Im/π

= RL . 2Vm/π (RL+Rƒ)

% R = Rƒ/RL ×100

Merits:

• High efficiency.

• Low ripple factor.

• High TUF.

Demerits:

• More circuits

• High cost