Diode as Rectifier: Half-Wave, Full-Wave & Filters

Rectification: One-Way Conduction Put to Work

The diode conducts only when forward biased. Apply an alternating voltage across it and current flows only during the half-cycles that forward-bias it — the ac is rectified into one-directional (if still pulsating) output. The circuit doing this is a rectifier.

The Half-Wave Rectifier

A transformer secondary supplies ac across terminals A and B; a single diode sits in series with the load $R_L$ (NCERT Fig. 14.18):

• A positive: diode forward biased → conducts → output voltage across $R_L$.
• A negative: diode reverse biased → no conduction (reverse saturation ≈ 0) → no output.

Output appears only for half of each input cycle — hence half-wave. Output frequency = input frequency (50 Hz in → 50 Hz pulses out: one pulse per cycle).

Key Point: The diode's reverse breakdown voltage must exceed the peak ac voltage of the secondary, or the negative half-cycles will destroy it.

The Full-Wave Rectifier (Centre-Tap)

Two diodes, their p-sides connected to the two ends of the secondary, n-sides joined; output taken between that common point and the secondary's centre tap (NCERT Fig. 14.19). Each diode sees only half the total secondary voltage.

• A positive (B negative): $D_1$ conducts, $D_2$ blocks → output across $R_L$.
• A negative (B positive): $D_2$ conducts, $D_1$ blocks → output again.

Both half-cycles deliver output — full-wave, and clearly more efficient. Output frequency = 2 × input frequency (50 Hz in → 100 Hz pulses: two per cycle). The centre-tap transformer is essential to this circuit; an alternative bridge rectifier uses four diodes and no centre tap.

Rectifier frequency summary (NCERT Exercise 14.6)

Circuit	50 Hz input → output ripple
Half-wave	50 Hz
Full-wave	100 Hz

Filters: From Pulses to DC

The rectified output is unidirectional but pulsating (half-sinusoid humps). To smooth it, connect a capacitor across the load (or an inductor in series) — a filter, so called because it 'filters out the ac ripple'.

How the capacitor filter works

• Voltage rising → capacitor charges (to the peak, if unloaded).
• With a load, it discharges through $R_L$ as the rectified voltage falls, holding the output up.
• Next hump recharges it to the peak.
• Discharge rate is set by the time constant $R_LC$: rate of fall $\propto \dfrac{1}{R_LC}$ — so a large C gives a slow droop and near-peak output.

Hence capacitor-input filters use large capacitors; the output approaches the peak rectified voltage, and this arrangement is the standard in power supplies.

[NEET Important] The 50/100 Hz pairing is NCERT Exercise 14.6 verbatim and among the most-asked facts of the chapter. [JEE Tip] Circuit-spotting: one diode + load = half-wave; two diodes + centre tap = full-wave; four diodes = bridge (no centre tap needed). Diode count and transformer type identify the circuit instantly.

Solved Examples

Example 1: Output frequencies (NCERT Exercise 14.6)

A 50 Hz ac input feeds (a) a half-wave rectifier and (b) a full-wave rectifier. Find each output frequency.

Solution:

1. (a) Half-wave: one output pulse per input cycle → 50 Hz.
2. (b) Full-wave: both half-cycles produce pulses → two per cycle → 100 Hz.
3. Takeaway: full-wave doubles the ripple frequency, never the amplitude — NCERT's exact exercise, endlessly recycled.

Example 2: Trace the conduction [Board Waveform]

In the centre-tap full-wave rectifier, the voltage at A is negative with respect to the centre tap. Which diode conducts, and why is there still output?

Solution:

1. A negative → $D_1$ reverse biased: blocks.
2. B, being out of phase, is positive → $D_2$ forward biased: conducts.
3. Current flows through $R_L$ in the same direction as before (both diodes feed the common terminal).
4. Takeaway: alternate diodes, same load direction — that unidirectional handover is the whole trick.

Example 3: Why the centre tap? [Board Conceptual]

What role does the centre tap play in the two-diode full-wave rectifier?

Solution:

1. It provides the common return terminal for the load — output is taken between the joined n-sides and the tap.
2. It splits the secondary so the two diode circuits work on opposite half-cycles, each using half the total secondary voltage.
3. Without a tap, two diodes cannot make a full-wave rectifier — you'd need the four-diode bridge instead.
4. Takeaway: centre-tap = 2 diodes + half voltage each; bridge = 4 diodes + full voltage. Cost trade-offs, same output shape.

Example 4: Peak output of a half-wave rectifier [JEE Numerical]

A half-wave rectifier's secondary supplies 20 V rms at 50 Hz to an ideal diode and load. Find the peak output voltage and output ripple frequency.

Solution:

1. Peak: $V_{peak} = \sqrt{2} \times 20 \approx 28.3$ V (ideal diode: no drop).
2. Frequency: half-wave → 50 Hz.
3. (With a real Si diode: peak ≈ 28.3 - 0.7 = 27.6 V.)
4. Takeaway: rms → peak via $\sqrt{2}$ first; then subtract the cut-in only if the question treats the diode as real.

Example 5: The breakdown-voltage safety rule [Board Conceptual]

Why must a rectifier diode's reverse breakdown voltage comfortably exceed the transformer's peak secondary voltage?

Solution:

1. During its blocking half-cycle the diode sits in reverse bias carrying the full peak voltage.
2. If that peak reached $V_{br}$, breakdown current would flow — uncontrolled, it destroys the diode by overheating.
3. NCERT: the breakdown voltage 'must be sufficiently higher than the peak ac voltage at the secondary … to protect the diode'.
4. Takeaway: rectifier design begins with a voltage-rating check, not with the forward direction at all.

Example 6: Filter action, step by step [Board Conceptual]

Describe how a capacitor across $R_L$ converts pulsating dc into near-steady dc.

Solution:

1. Rising hump: capacitor charges toward the peak.
2. Falling hump: capacitor discharges through $R_L$, holding the output voltage up; the discharge rate $\propto 1/(R_LC)$.
3. Next hump: recharges to the peak; the output ripples gently about a value near the peak voltage.
4. Takeaway: large $R_LC$ = slow droop = smooth dc — hence the physically large capacitors in power supplies.

Example 7: Choosing C [JEE Numerical]

A full-wave rectifier at 50 Hz feeds $R_L$ = 1 kohm. Roughly how large should C be so the capacitor barely discharges between humps?

Solution:

1. Time between humps (full-wave): $\dfrac{1}{100}$ s = 10 ms.
2. Condition: time constant $R_LC \gg$ 10 ms → $C \gg \dfrac{10^{-2}}{10^3} = 10\ \mu$F.
3. Practical choice: hundreds of $\mu$F — hence electrolytic capacitors.
4. Takeaway: compare $R_LC$ with the ripple period; 'much larger' is the design rule.

Example 8: Which circuit is this? [NEET Spotting]

A rectifier uses four diodes and an ordinary (untapped) transformer. Identify it and its output frequency for 50 Hz input.

Solution:

1. Four diodes, no centre tap = the bridge rectifier — NCERT's noted alternative full-wave circuit.
2. It rectifies both half-cycles: output ripple at 100 Hz.
3. Takeaway: count diodes: 1 → half-wave (50 Hz); 2 + tap → full-wave (100 Hz); 4 → bridge full-wave (100 Hz).

Example 9: Why full-wave is 'more efficient' [Conceptual]

Justify NCERT's claim that the full-wave circuit is more efficient than the half-wave.

Solution:

1. The half-wave rectifier wastes every negative half-cycle — no output half the time.
2. The full-wave circuit converts both halves into output pulses: double the pulses, double the average dc output for the same input.
3. Its 100 Hz ripple is also easier to filter (shorter gaps for the capacitor to bridge).
4. Takeaway: more output AND smoother output — efficiency in both senses.

Example 10: Inductor as filter [JEE Extra]

NCERT mentions an inductor in series with $R_L$ as an alternative filter. Why does that work?

Solution:

1. An inductor opposes changes in current ($V = L\,dI/dt$) — it passes steady dc easily but impedes the ac ripple.
2. In series with the load it smooths the current the way the parallel capacitor smooths the voltage.
3. Both 'appear to filter out the ac ripple and give a pure dc voltage — so they are called filters.'
4. Takeaway: capacitor parallel (voltage smoothing), inductor series (current smoothing) — complementary filter philosophies.