# Fullwave Center Tap Transformer Rectifier

A single-phase, half-wave rectifier is not very practical due to its low average output voltage, poor efficiency, and high ripple factor. These limitations can be overcome by full-wave rectification. Full-wave rectifier are more commonly used than half-wave rectifier, due to their higher average voltages and currents, higher efficiency, and reduced ripple factor.

## With Resistive Load

Figure 1(a) shows the schematic diagram of the full-wave rectifier using a transformer with a center-tapped secondary. The source voltage and load resistor are the same as in the half-wave case. During the positive half-cycle (Figure 1(b)), diode D_{1} conducts and D_{2} is reverse-biased. Current flows through the load causing a positive drop.

During the negative half-cycle (Figure 1(c)), diode D_{2} conducts and D_{1} turns off. Current flows through R, maintaining the same polarity for the voltage across the load (see Figure 1(d)). Therefore, the load voltage waveform consists of successive half-cycle of a sine wave, resulting in a higher average value and higher ripple frequency.

Average and RMS values are similar to those for the half-wave case:

*V _{o(avg)} = (2 V_{m}) / *

*π*=

*0.636 V*

_{m}Note that the full-wave average is twice the half-wave average this is obvious by inspecting the two graphs of voltage versus time. Similarly, the average load current is given by the same factor.

*I _{o(avg)}* =

*(2 I*

_{m}) /*π =*

*0.636 V*

_{m}/ R.

The RMS output current is given by

The graph of the voltage across the diode in Figure 1(d) shows that each diode must withstand a reverse voltage equal to 2V_{m}. the PIV rating for the diodes used in this circuit is therefore given by:

PIV rating for diodes ≥ 2 V_{m}

The average diode current is

*I _{D1(avg)} = I_{D2(avg)} = I_{m} / *

*π*

The RMS diode current is

*I _{DRMS} = I_{m} / 2*

The average or DC power delivered to the load is given by

*P _{o(avg)}* =

*V*X

_{o(avg)}*I*

_{o(avg)}*= (2Vm / **π) X (**2I _{m} / *

*π)*

*= (4V _{m} X V_{m}) / (*

*π X R)*

*= (4V ^{2}_{m})/* (

*π*

^{2}X R)The AC power input is given by

## With an Inductive Load (RL)

Adding an inductance in series with the load resistance changes the voltage and current waveform. Note that the load current continues to flow for a period after the diode is reverse-biased, and this results in a decrease in the magnitude of the average output voltage.

Figure shows a center tap full-wave rectifier with an inductive load and its associated voltage and current waveform.

The load current is at its maximum when the source voltage (V_{S}) is zero. When V_{S }increases in magnitude during the interval from 0 to π/2, the inductor opposes the flow of current and stores energy in its magnetic field. At π/2 when V_{S} has reached its maximum, the load current is at its minimum. In the interval between π/2 and π, where the source voltage decreases in magnitude, the induced voltage across the inductor opposes any decrease in the load current by aiding the source voltage. Therefore, the load current increases to a maximum value when V_{S} = 0. The process continues for every half-cycle of the rectified sine wave. The load current never reduces to zero since, the energy stored in the magnetic field maintains the current flow.

The equations are similar to those for the center-tap rectifier with a resistive load. The average value of the voltage is:

*V _{o(avg)} = 2V_{m} / *

*π*

*= 0.636*

*V*

_{m}The average value of the load current is

*I _{o(avg)} = 2V_{in} /*

*π*

*R = 0.636 V*

_{m}/ RIf the load inductance is sufficiently large, the load current is nearly constant, as shown in Figure 3

The RMS value of the load current is:

*I _{o(RMS)} = I_{o(avg)} = V_{o(avg)} / R*

*I _{D(RMS)} = I_{o(avg)} / 2*

.

**Example**

The full-wave rectifier shown in Figure 2(a) is supplied from a 115V source. If the load resistance is 100Ω, find

- The output DC voltage
- The average load currents
- The power delivered to the load
- The AC input power
- The rectifier efficiency
- The ripple factors
- The form factors

**Solution**:

The peak voltage is

*P _{L} = V_{o(avg)} X I_{o(avg)}*

* = 103.4 X 1.03*

* = 107 W*

- The output DC voltage is
*V*_{o(avg)}= 0.636 V_{m}= 0.636 X 162.6 = 103.4 V *The average load current is**I*_{o(avg)}= 103.4/100 = 1.03 A*Power delivered to the load**P*_{L}= V_{o(avg)}X I_{o(avg)}*= 103.4 X 1.03**= 107 W**AC input power**P*_{AC}= V_{RMS}X I_{RMS}*= V*^{2}_{m}/ 2R*= (162.6)*^{2}/ 2 (100)*= 132.2 W**Rectifier efficiency**η = P*_{L}/ P_{AC}= 107 / 132.2 = 0.81 or 81%*Ripple factor**= 0.48**Form factor**FF = V*_{RMS}/ V_{o(avg)}= 115 / 103.4 = 1.11