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Q.Differentiate between half-wave and full-wave rectification. With the help of a circuit diagram, explain the working of a full-wave rectifier.

CBSECBSE Class XII Board 2025Subjective· 3mImportance★★★★★
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A full-wave rectifier uses two diodes and a centre-tapped transformer to convert both halves of an AC input into a unidirectional output, giving higher efficiency and less ripple than a half-wave rectifier.

The Core Idea: Why Rectification?

A rectifier is the simplest power supply stage — it turns alternating current (AC) into pulsating direct current (DC). The key difference between half-wave and full-wave lies in how much of the AC cycle they use.

Half-wave rectification uses only one half (say, the positive half) of each AC cycle. The negative half is simply blocked. This wastes half the input power and produces a large gap between pulses, making the output very "ripply" — hard to smooth into clean DC.

Full-wave rectification uses both halves of the AC cycle. It cleverly inverts the negative half so that it also contributes to the output. The result is a stream of pulses at twice the input frequency, with no dead time between them. This gives:

  • Higher average output voltage
  • Lower ripple (easier to filter)
  • Better transformer utilisation
Watch out

A common mistake is to think a full-wave rectifier uses four diodes (that's a bridge rectifier). The centre-tapped full-wave rectifier uses exactly two diodes — a different topology.

Step-by-Step: Full-Wave Rectifier with Centre-Tapped Transformer

1. The Circuit Setup

The full-wave rectifier requires:

  • A centre-tapped transformer — the secondary winding has a tap exactly at its midpoint
  • Two diodes (D1D_1 and D2D_2)
  • A load resistor (RLR_L)

Draw the circuit like this. The AC mains connects to the transformer primary. The secondary winding has three terminals: end A at the top, end B at the bottom, and the centre tap (CT) at the midpoint. Diode D1D_1's anode connects to A; diode D2D_2's anode connects to B. The cathodes of both diodes are joined together at a common point, and the load resistor RLR_L is connected from this common cathode point back to the centre tap. The centre tap serves as the return (reference) terminal.

The voltage from A to CT is VAC=Vmsin⁡(ωt)V_{AC} = V_m \sin(\omega t), and from B to CT is VBC=−Vmsin⁡(ωt)V_{BC} = -V_m \sin(\omega t) — they are 180∘180^\circ out of phase.

2. How It Works: The Two Half-Cycles

During the positive half-cycle (when A is positive with respect to CT):

  • Diode D1D_1 is forward-biased (anode > cathode) and conducts
  • Diode D2D_2 is reverse-biased (its anode at B is negative relative to CT) and does not conduct
  • Current flows: A →\to D1D_1 →\to RLR_L →\to CT
  • Voltage across RLR_L is positive: Vout=Vmsin⁡(ωt)V_{out} = V_m \sin(\omega t) (ignoring the 0.7 V diode drop)

During the negative half-cycle (when A is negative, B is positive with respect to CT):

  • Now D2D_2 is forward-biased and conducts
  • D1D_1 is reverse-biased and blocks
  • Current flows: B →\to D2D_2 →\to RLR_L →\to CT
  • Notice: current through RLR_L flows in the same direction as before (from the joined cathodes toward the centre tap)
  • Voltage across RLR_L is again positive: Vout=Vmsin⁡(ωt)V_{out} = V_m \sin(\omega t) (but this time from the B-side)
Tip

The trick is that the centre tap gives you two voltages that are mirror images. Each diode "handles" one half-cycle, and both push current through the load in the same direction. The output frequency doubles to 2finput2f_{input}.

3. The Output Waveform

The output is a series of positive half-sine pulses, one from each half-cycle of the input. There is no gap — the next pulse starts immediately after the previous one ends.

The average (DC) output voltage is:

Vdc=2VmπV_{dc} = \frac{2V_m}{\pi}

Compare this to a half-wave rectifier, where Vdc=VmπV_{dc} = \frac{V_m}{\pi} — exactly half.

The ripple frequency is 2f2f, where ff is the input AC frequency (e.g., 100 Hz for a 50 Hz input). This higher frequency makes filtering much easier.

For a full-wave rectifier with centre-tapped transformer:

Vdc=2Vmπ,Idc=2VmπRLV_{dc} = \frac{2V_m}{\pi}, \quad I_{dc} = \frac{2V_m}{\pi R_L}

Ripple factor: γ=0.48\gamma = 0.48 (compared to 1.21 for half-wave)

4. Key Differences: Half-Wave vs Full-Wave

| Parameter | Half-Wave | Full-Wave (Centre-Tap) | …

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