Mumbai University > FE > Sem 1 > Basic Electrical and Electronics Engineering

Marks: 4 M

Year: DEC 2012

Center Tapped Full Wave Rectifier

As shown in the figure, an AC input is applied to the primary coils of the transformer. This input makes the secondary ends P1 and P2 become positive and negative alternately. For the positive half of the ac signal, the secondary point D1 is positive, GND point will have zero volt and P2 will be negative. At this instant diode D1 will be forward biased and diode D2 will be reverse biased. As explained in the Theory behind P-N Junction and Characteristics of P-N Junction Diode, the diode D1 will conduct and D2 will not conduct during the positive half cycle. Thus the current flow will be in the direction P1-D1-C-A-B-GND. Thus, the positive half cycle appears across the load resistance RLOAD.

During the negative half cycle, the secondary ends P1 becomes negative and P2 becomes positive. At this instant, the diode D1 will be negative and D2 will be positive with the zero reference point being the ground, GND. Thus, the diode D2 will be forward biased and D1 will be reverse biased. The diode D2 will conduct and D1 will not conduct during the negative half cycle. The current flow will be in the direction P2-D2-C-A-B-GND.

Centre-tap Full-wave Rectifier-Waveform

When comparing the current flow in the positive and negative half cycles, we can conclude that the direction of the current flow is the same (through load resistance RLOAD). When compared to the Half-Wave Rectifier, both the half cycles are used to produce the corresponding output. The frequency of the rectified output voltage is twice the input frequency. The output that is rectified, consists of a dc component and a lot of ac components of minute amplitudes.

Peak Inverse Voltage (PIV) of Centre-Tap Full Wave Rectifier

PIV is the maximum possible voltage across a diode during its reverse biased period. Let us analyze the PIV of the centre-tapped rectifier from the circuit diagram. During the first half or the positive half of th input ac supply, the diode D1 is positive and thus conducts and provided no resistance at all. Thus, the whole of voltage Vs developed in the upper-half of the ac supply is provided to the load resistance RLOAD. Similar is the case of diode D2 for the lower half of the transformer secondary.

∴ PIV of D2 = Vm + Vm = 2Vm

PIV of D1 = 2Vm

Centre-Tap Rectifier Circuit Analysis

1. Peak Current

   The instantaneous value of the voltage applied to the rectifier can be written as

   Vs = Vsm Sinwt

   Assuming that the diode has a forward resistance of RFWD ohms and a reverse resistance equal to infinity, the current flowing through the load resistance RLOAD is given as

   Im = Vsm/(RF + RLoad)

2. Output Current

   Since the current is the same through the load resistance RL in the two halves of the ac cycle, magnitude of dc current Idc, which is equal to the average value of ac current, can be obtained by integrating the current i1 between 0 and pi or current i2 between pi and 2pi.

   $\boxed{SoI_{dc}=\dfrac{1}{\pi \int^\pi_0i1d(wt)}=\dfrac{1}{\pi \int^\pi_0 I_{\max} \sin wtd(wt)}=\dfrac{2I_m}{\pi}}$

   Output current of center Tap rectifier

3. DC Output Voltage

   Average or dc value of voltage across the load is given as

   $\boxed{SoI_{dc}=\dfrac{1}{\pi \int^\pi_0i1d(wt)}=\dfrac{1}{\pi \int^\pi_0 I_{\max} \sin wtd(wt)}=\dfrac{2I_m}{\pi}}$

   DC Output Voltage of center Tap Rectifier

4. Root Mean Square (RMS) Value of Current

   RMS or effective value of current flowing through the load resistance RL is given as

   $\boxed{I^2rms=\dfrac{1}{\pi \int^\pi_0 i 12d(wt)}=\dfrac{I^2m}{2}OR I_{rms}=\dfrac{I_M}{\sqrt2}}$

   RMS Value of Current of centre Tap Rectifier

5. Root Mean Square (RMS) Value of Output Voltage

   RMS value of voltage across the load is given as

   $\boxed{V_{LOAD}rsm=I_{rms}R_{LOAD}=\bigg[\dfrac{I_M}{\sqrt2}\bigg]R_{LOAD}}$