Rectification is a process of converting AC waveform (voltage and current) into DC (voltage and current). The other two well-known rectification techniques are half-wave rectifiers (it uses single diode rectifiers and rectifies only one half of the input signal) and bridge rectifiers (it uses 4 diodes and a little bit more complex circuit in terms of several components).
This type of full-wave rectifier uses two diodes and a centre tap transformer. In this type of transformer, the secondary winding is split into two parts. The voltage that appears on the two secondary windings is the same but with opposite polarity. See figure below:
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| Figure 1: Full wave rectifier without the smoothing capacitor |
The voltage at primary winding or applied voltage signal (peak) = 20V
Turns Ratio: NP:NS = 1:5
The peak voltage at the secondary winding = 4
The voltage at the secondary winding is divided into 2 parts because of the centre tapped transformer = VP(secondary) = ( 2 + 2 )V
Each diode receives 2V of peak voltage at the input.
Both diodes act as a half-wave rectifier. So, in this network, there are two half-wave rectifiers. While the positive half cycle appears at the diode D1, a negative half cycle appears at the diode D2. As a result, D1 is forward biased and D2 is reverse biased. The output of the circuit is given in figure 2.
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| Figure 2: Output is a pulsating DC in the absence of a smoothing capacitor |
After rectification, the pulsating DC voltage is obtained. This DC voltage is of no use in practical situations. Also, have a look at the peaks. The peak value of these pulses is equal to1.3 V.
Voltage appears at D1 & D2 = 2V
Voltage after rectification (each peak is equal to 1.3V) = (2 - 0.7)V = 1.3 V
In practical circuits, consider diode (Silicon) drop of 0.7V.
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| Figure 3: Full wave rectifier with a capacitive filter of an appropriate value |
In the next step, add a capacitor for filtration, this will result in a smooth DC voltage. The appropriate value of the capacitor is selected with the help of the formula given below.
\[Vr = \frac{I_{load}}{fC}\]
\[C = \frac{V_P}{V_r*f*R_{load}}\]
Vr = ripple voltage
Iload = current through the load resistor
VP = peak voltage
For this formula, Vr should be less than 20% of the peak voltage.
For R = 1000Ω, and Vr = 0.06V the value of C = 470μF
For R = 100Ω, and Vr = 0.06V the value of C = 4mF
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| Figure 4: Output is a smooth DC voltage after the addition of capacitor |
For practical applications, up to 100mV are acceptable. All practical power supplies have ripples and noise figures as well.
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