A center-tapped rectifier is a type of full-wave rectifier that uses two diodes connected to the secondary of a center-tapped transformer, as shown in Figure (a). The input voltage is coupled through the transformer to the center-tapped secondary. Half of the total secondary voltage appears between the center tap and each end of the secondary winding as shown.
For a positive half-cycle of the input voltage, the polarities of the secondary voltages are as shown in Figure (a). This condition forward-biases diode D1 and reverse-biases diode D2. The current path is through D1 and the load resistor RL, as indicated. For a negative half-cycle of the input voltage, the voltage polarities on the secondary are as shown in Figure (b). This condition reverse-biases D1 and forward-biases D2. The current path is through D2 and RL, as indicated. Because the output current during both the positive and negative portions of the input cycle is in the same direction through the load, the output voltage developed across the load resistor is a full-wave rectified dc voltage, as shown.
Effect of the Turns Ratio on the Output Voltage
If the transformer's turn’s ratio is 1, the peak value of the rectified output voltage equals half the peak value of the primary input voltage less the barrier potential, as illustrated in Figure (9). This is because half of the primary voltage appears across each half of the secondary winding Vp(sec) = Vp(pri). We will begin referring to the forward voltage due to the barrier potential as the diode drop.
In order to obtain an output voltage with a peak equal to the input peak (less the diode drop), a step-up transformer with a turns ratio of n = 2 must be used, as shown in Figure (c). In this case, the total secondary voltage (Vsec) is twice the primary voltage (2Vpri), so the voltage across each half of the secondary is equal to Vpri.
In any case, the output voltage of a center-tapped full-wave rectifier is always one-half of the total secondary voltage less the diode drop, no matter what the turn’s ratio.
Vout = (Vsec/2) — 0.7 V
Peak Inverse Voltage
Each diode in the full-wave rectifier is alternately forward-biased and then reverse-biased. The maximum reverse voltage that each diode must withstand is the peak secondary voltage Vp(sec). This is shown in Figure (d) where D2 is assumed to be reverse-biased.
When the total secondary voltage has the polarity shown, the maximum anode voltage of D1 is +Vp(sec)/2 and the maximum anode voltage of D2 is –Vp(sec)/2. Since D1, is assumed to be forward-biased, its cathode is at the same voltage as its anode minus the diode drop: this is also the voltage on the cathode of D2. The peak inverse voltage across D2, is
P1V = Vp(sec) - 0.7 V
Since Vp(out) = (Vp(sec)/2) - 0.7 V, then by multiplying each term by 2 and transposing,
Vp(sec) = 2Vp(out) + 1.4 V
Therefore, by substitution, the peak inverse voltage across either diode in a full-wave center-tapped rectifier is
PIV = 2Vp(sec) + 0.7 V