Physics · Semiconductor Electronics : Materials, Devices And Simple Circuits · NEET
In each input cycle there is a positive half and a negative half. In a full wave rectifier, one diode handles the positive half and the other diode handles the negative half. Both halves are flipped to the same (positive) side of the load. So one full input cycle gives TWO output humps. Two output pulses in the time of one input cycle means the output repeats twice as fast, so f_out = 2 x f_in. For 50 Hz mains, output = 100 Hz. For 60 Hz, output = 120 Hz.
The standard centre-tapped full wave rectifier uses TWO diodes (D1 and D2). During the positive half cycle, terminal A is positive, so D1 is forward biased and conducts while D2 is reverse biased. During the negative half cycle it reverses: D2 conducts and D1 is off. Only one diode conducts at a time, but the load current keeps flowing in the same direction in both halves. (A bridge rectifier does the same job with FOUR diodes and no centre tap.)
The centre tap gives two equal AC voltages that are out of phase (opposite in sign) with respect to the centre point. This is what lets one diode be forward biased while the other is reverse biased at the same instant. Without two opposite voltages you could not steer the current so that both halves push current the same way through the load.
Yes. The output is pulsating DC (a series of humps), not smooth DC. It still contains an AC part called ripple. A filter CAPACITOR placed across the load removes most of the ripple: it charges up at the peaks and slowly discharges through the load between peaks, smoothing the output. NEET 2023 asked exactly this: the capacitor removes the AC ripple, not the transformer or the diodes.
A half wave rectifier throws away one half of every cycle, so you get only ONE output pulse per input cycle. The output pattern repeats once per input cycle, so f_out = f_in (50 Hz stays 50 Hz). A full wave rectifier keeps both halves, giving TWO pulses per cycle, so f_out = 2 x f_in (50 Hz becomes 100 Hz). This is the single most tested difference.
Yes. Any full wave rectifier, whether centre-tapped (2 diodes) or bridge (4 diodes), uses BOTH halves of the input. So it always produces two output pulses per input cycle and the output frequency is always 2 x input frequency. The number of diodes does not change the output frequency.
A full wave rectifier circuit consists of two p-n junction diodes, a centre-tapped transformer, capacitor and a load resistance. Which of these components removes the ac ripple from the rectified output?
A full wave rectifier circuit with diodes D1 and D2 is shown in the figure. If the input supply voltage is V_in = 220 sin(100 pi t) volt, then at t = 15 ms, which is true?
Try the real previous-year questions from this chapter — each with the answer and a full solution.
100 Hz. A full wave rectifier gives two output pulses per input cycle, so the output frequency is double the input: 2 x 50 = 100 Hz.
A centre-tapped full wave rectifier uses TWO diodes. A bridge type full wave rectifier uses FOUR diodes but does the same job and gives the same doubled output frequency.
It uses both halves of the AC input, so it delivers more DC power, has higher efficiency, a higher output frequency (easier to filter), and lower ripple than a half wave rectifier.
The maximum theoretical rectifier efficiency of a full wave rectifier is about 81.2 percent, which is twice that of a half wave rectifier (about 40.6 percent).
No. The filter capacitor only smooths the humps and reduces ripple. The pulse rate, and therefore the output frequency, stays at twice the input frequency.
For an ideal diode the peak output across the load equals the peak of one half of the transformer secondary, V_max. The DC (average) value of the full wave output is 2 V_max / pi, which is about 0.637 V_max.
