PhysicsISC Class 12 · Electronic Devices

Rectifiers: Half-Wave and Full-Wave Rectification, and the Zener Diode Regulator

Updated 15 October 2026 · Fact-checked

A rectifier converts alternating current into unidirectional (direct) current using diodes, which conduct only when forward biased. A half-wave rectifier uses one diode and passes only alternate half cycles; a centre-tapped full-wave rectifier uses two diodes that conduct in turn, so both halves reach the load and the output pulses at twice the supply frequency.

Half-wave blocks the negative half (red dashed), so gaps appear; full-wave uses D1 and D2 in turn, so pulses double to 100 Hz.

Illustration drawn for QuestionBuddy.

Understand Rectifiers and the Zener Voltage Regulator

Why rectify? Mains supply is a.c., but electronic circuits (phone chargers, LED lamps, computers) need steady d.c. A diode conducts in one direction only, so it can let through one polarity and block the other.

Half-wave rectifier. Components: a transformer (steps the mains down to the required voltage and isolates the circuit), one diode, and a load resistor $R_{L}$ .

Positive half cycle: the diode is forward biased and conducts; current flows through $R_{L}$ and the output follows the input (less about 0.7 V for silicon).
Negative half cycle: the diode is reverse biased and blocks; output is zero.
Output: positive half-sine pulses with gaps; output frequency equals the input frequency (50 Hz for 50 Hz mains). Half the input wave is wasted.

Full-wave rectifier (centre-tap, two diodes). The transformer secondary has a centre tap; the two ends A and B are always in antiphase relative to the centre C. Diodes $D_{1}$ and $D_{2}$ connect A and B to one end of $R_{L}$ ; the other end of $R_{L}$ goes to C.

First half cycle: A positive relative to C: $D_{1}$ conducts, $D_{2}$ is reverse biased; current flows through $R_{L}$ in one direction.
Second half cycle: B positive: $D_{2}$ conducts, $D_{1}$ blocks; current through $R_{L}$ is in the same direction.
Output: a continuous series of positive half-sine pulses; the output frequency is twice the input frequency (100 Hz for 50 Hz). It is more efficient and smoother than half-wave. (The four-diode bridge rectifier is not in the syllabus.)
Each half of the secondary supplies the peak voltage, so the peak output is $V_{m} = 2 V_{r m s (ha l f)}$ , less one diode drop.

Function of each component (qualitative). Transformer: change voltage level and isolate. Diodes: allow current one way only. Load resistor: where the d.c. output is taken. A capacitor across the load (filter) charges at each peak and discharges slowly between them, smoothing the output.

Zener diode. A heavily doped p-n junction designed to work in reverse breakdown. Its V-I characteristic in reverse shows a very sharp breakdown at the Zener voltage $V_{Z}$ : beyond it, the voltage stays almost constant while the current can vary over a wide range. Forward, it behaves like an ordinary diode.

Zener diode as a voltage regulator. The Zener is connected in reverse bias across the load, with a series resistor $R_{s}$ between the unregulated supply and the Zener-load combination.

Output voltage across the load = $V_{Z}$ , nearly constant.
If the input rises, the current through $R_{s}$ rises, but the extra current goes through the Zener; the extra voltage is dropped across $R_{s}$ , and the output stays at $V_{Z}$ . If the input falls, the Zener current falls, again keeping $V_{Z}$ constant.
Equations: $I_{s} = \frac{V _{in} - V _{Z}}{R _{s}}$ , $I_{L} = \frac{V _{Z}}{R _{L}}$ , $I_{Z} = I_{s} - I_{L}$ (must stay positive).

A complete power supply is: transformer → rectifier → filter → Zener regulator → load.

Rectifiers and the Zener Voltage Regulator: formulas and key facts

Peak output (Si): $V_{o u t, p e ak} = 2 V_{r m s} - 0.7 V$; One diode conducting.
Output frequency: $f_{ha l f} = f, f_{f u l l} = 2 f$; 50 Hz → 50 or 100 Hz.
Zener series current: $I_{s} = \frac{V _{in} - V _{Z}}{R _{s}}$; Through the series resistor.
Load current: $I_{L} = \frac{V _{Z}}{R _{L}}$; Load sees V_Z.
Zener current: $I_{Z} = I_{s} - I_{L}$; Must be positive.

How to solve Rectifiers and the Zener Voltage Regulator questions

Rectifier questions are mostly diagrams and waveforms; Zener questions are Kirchhoff's current law.

1Draw: transformer, diode(s), load resistor; label input and output terminals.
2Sketch input (sine) and output waveforms one below the other on the same time axis.
3Peak output: $2 V_{r m s}$ (of the relevant secondary half) minus 0.7 V per conducting Si diode, or no drop if ideal.
4Output frequency: f for half-wave, 2f for full-wave.
5Zener: $I_{s} = \frac{V _{in} - V _{Z}}{R _{s}}$ , $I_{L} = \frac{V _{Z}}{R _{L}}$ , $I_{Z} = I_{s} - I_{L}$ ; check $I_{Z} > 0$ for regulation.

Rectifiers and the Zener Voltage Regulator in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Inside a phone charger

Suppose a simple charger turns 230 V a.c. mains into about 5 V d.c. for a phone.

List the stages and the job of each.

Try it first, then show the solutionSolution

Transformer steps 230 V down to a few volts (and isolates you from the mains).
Rectifier (diodes) makes the current flow one way only; a capacitor filter smooths the pulses.
A regulator (such as a Zener-based circuit) holds the output steady at about 5 V despite mains fluctuations.

Answer: Step down, rectify, smooth, regulate.

Memory hook: Transform, rectify, filter, regulate: "To Really Feel Right".

Keeping an LED panel steady

Imagine a battery-powered LED panel whose battery voltage sags from 13 V to 10 V as it discharges, while the panel needs a steady 5.1 V at 5.1 mA. A 5.1 V Zener with a 150 Ω series resistor is used.

Find the Zener current at 13 V and at 10 V.

Try it first, then show the solutionSolution

At 13 V: $I_{s} = \frac{13 - 5.1}{150} \approx 52.7$ mA; $I_{Z} \approx 52.7 - 5.1 = 47.6$ mA.
At 10 V: $I_{s} = \frac{10 - 5.1}{150} \approx 32.7$ mA; $I_{Z} \approx 27.6$ mA.

Answer: ≈ 47.6 mA and ≈ 27.6 mA; the output stays at 5.1 V because the Zener absorbs the change.

Memory hook: The Zener soaks up the surplus.

Tricks to remember Rectifiers and the Zener Voltage Regulator

Short memory aids. Say each one aloud once, then use it on a question the same day.

Half wastes half
Half-wave uses one half cycle; full-wave uses both.
Use it when: Comparison questions.
Full doubles frequency
50 Hz in, 100 Hz out for full-wave.
Use it when: Output frequency.
Zener sits reversed
Regulator Zener is always reverse biased, in breakdown, across the load.
Use it when: Circuit diagrams.
Series takes the difference
$R_{s}$ drops $V_{in} - V_{Z}$ .
Use it when: Zener numericals.

Rectifiers and the Zener Voltage Regulator: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Output frequency · 1 mark · MCQBoard level

A full-wave rectifier is fed from 50 Hz mains. The frequency of the output pulses is:
(a) 25 Hz
(b) 50 Hz
(c) 100 Hz
(d) 200 Hz

Try it first, then show the solutionSolution

Each half cycle gives one pulse: 2 × 50 = 100 per second.

Answer: (c).

Example 2: Half-wave peak output · 2 marks · Short answerBoard level

A transformer secondary gives 9 V (r.m.s.), 50 Hz to a half-wave rectifier with a silicon diode. Find the peak output voltage and the output frequency.

Try it first, then show the solutionSolution

Peak input $= 92 \approx 12.73$ V; peak output ≈ 12.73 − 0.7 ≈ 12.0 V. (1)
Half-wave: output frequency = 50 Hz. (1)

Answer: ≈ 12.0 V; 50 Hz.

Example 3: Full-wave rectifier · 4 marks · Long answerHarder board level

Draw a labelled circuit of a full-wave rectifier using two junction diodes. Explain its working and show the input and output waveforms.

Try it first, then show the solutionSolution

Circuit: transformer with centre-tapped secondary (ends A, B, centre C); $D_{1}$ from A and $D_{2}$ from B joined to one end of load $R_{L}$ ; other end of $R_{L}$ to C; output across $R_{L}$ . (1)
First half: A positive w.r.t. C, $D_{1}$ forward biased, conducts; $D_{2}$ reverse biased. Current through $R_{L}$ in one direction. (1)
Second half: B positive, $D_{2}$ conducts, $D_{1}$ blocks; current through $R_{L}$ in the same direction. (1)
Waveforms: input sine; output a continuous train of positive half-sine pulses at twice the input frequency, with input and output axes labelled. (1)

Answer: Both halves of the a.c. give current in the same direction through the load.

Example 4: Zener regulator currents · 3 marks · Short answerHarder board level

A 5.1 V Zener diode with a 220 Ω series resistor regulates a 12 V supply for a 1 kΩ load. Find the current through the series resistor, the load and the Zener. What is the Zener current if the supply rises to 14 V?

Try it first, then show the solutionSolution

$I_{s} = \frac{12 - 5.1}{220} \approx 31.4$ mA; $I_{L} = \frac{5.1}{1000} = 5.1$ mA. (1)
$I_{Z} = 31.4 - 5.1 \approx 26.3$ mA. (1)
At 14 V: $I_{s} \approx 40.5$ mA, $I_{Z} \approx 35.4$ mA; load still gets 5.1 V. (1)

Answer: ≈ 31.4 mA, 5.1 mA, ≈ 26.3 mA; ≈ 35.4 mA at 14 V.

Example 5: Smallest load the regulator can supply · 2 marks · Short answerHOTS / competency

For the regulator above (5.1 V Zener, 220 Ω, 12 V input), the Zener needs at least 5 mA to stay in breakdown. Find the smallest load resistance that can be regulated.

Try it first, then show the solutionSolution

$I_{s} \approx 31.36$ mA; maximum load current $= 31.36 - 5 = 26.36$ mA. (1)
$R_{L, min} = \frac{5.1}{0.02636} \approx 193$ Ω. (1)

Answer: ≈ 193 Ω.

Example 6: Building a d.c. supply · 4 marks · Case-basedHOTS / competency

Suppose you build a power supply from 50 Hz mains using a transformer, two diodes, a capacitor and a Zener diode.
(i) What is the function of the transformer? (1 mark)
(ii) Why are two diodes and a centre-tapped secondary used? (1 mark)
(iii) What is the frequency of the ripple before the capacitor? (1 mark)
(iv) What do the capacitor and the Zener each do? (1 mark)

Try it first, then show the solutionSolution

(i) Steps the mains voltage down to the level needed (and isolates the circuit).
(ii) For full-wave rectification: each diode conducts in alternate half cycles, so both halves give current in the same direction.
(iii) 100 Hz.
(iv) Capacitor: smooths the pulsating output (filter). Zener: holds the output at a constant voltage (regulator).

Answer: (i) step down; (ii) full-wave; (iii) 100 Hz; (iv) filter and regulator.

Where students lose marks in Rectifiers and the Zener Voltage Regulator

The slips examiners see most often, and what to write instead.

Mistake: Confusing a rectifier with a transformer
The ISC 2024 examiners (Q1(B)(vii)) saw the transformer's job given as the rectifier's, or waveforms drawn instead of a statement.
Do this: Rectifier: converts a.c. to d.c. (unidirectional).
Usually costs: 1 mark.
Mistake: Drawing half-wave when full-wave is asked
The 2025 examiners (Q20(ii)) saw half-wave circuits, missing load resistors and incomplete circuits.
Do this: Two diodes, centre-tapped secondary, load from the diode junction to the centre tap.
Usually costs: 1 to 2 marks.
Mistake: Unlabelled waveforms
Also 2025: input and output waveforms not labelled or wrongly drawn.
Do this: Draw input above output with a common time axis; label both.
Usually costs: 1 mark.
Mistake: Naming the wrong regulator diode
2025 Q20(iii): some named another diode type.
Do this: Voltage regulator: Zener diode.
Usually costs: 1 mark.

Exam tips

Mark the direction of current through the load in each half cycle.
Show output pulses all on the same side of the time axis.
Draw the Zener reverse biased (cathode to positive) in the regulator circuit.
Remember the bridge rectifier is not required for ISC 2027.

Practice questions on Rectifiers and the Zener Voltage Regulator

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The function of a rectifier is to:
(a) step up a.c. voltage
(b) convert a.c. into d.c.
(c) convert d.c. into a.c.
(d) amplify signals
Hint
Not a transformer.
Answer
(b).
Q2 · 1 mark · Short answerBoard level
What happens in a half-wave rectifier during the negative half cycle? Why?
Hint
Bias of the diode.
Answer
The diode is reverse biased and does not conduct, so the output is zero.
Q3 · 1 mark · Assertion-reasonHarder board level
Assertion (A): A full-wave rectifier gives a smoother output than a half-wave rectifier.
Reason (R): It uses both half cycles, so the gaps between output pulses are removed.
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.
Hint
Pulses per second.
Answer
(a).
Q4 · 2 marks · Short answerHarder board level
A centre-tapped transformer gives 6 V (r.m.s.) on each half to a full-wave rectifier with silicon diodes on 50 Hz mains. Find the peak output and output frequency.
Hint
$62 - 0.7$ .
Answer
≈ 7.8 V; 100 Hz.
Q5 · 3 marks · Long answerHarder board level
Sketch the V-I characteristic of a Zener diode, mark $V_{Z}$ , and explain how it acts as a voltage regulator.
Hint
Sharp reverse breakdown.
Answer
Reverse current is tiny until $V_{Z}$ , then rises almost vertically at nearly constant voltage. Used reverse biased across the load with a series resistor: changes in input change the Zener current and the drop across $R_{s}$ , while the output stays at $V_{Z}$ .
Q6 · 2 marks · Short answerHOTS / competency
A 9 V Zener must supply a 900 Ω load from a 16 V supply with a Zener current of 10 mA. Find the series resistance.
Hint
$I_{L}$ = 10 mA, $I_{s}$ = 20 mA.
Answer
$R_{s} = \frac{16 - 9}{0.020} = 350$ Ω.
Q7 · 2 marks · Short answerHOTS / competency
In a two-diode full-wave rectifier, one diode burns out and becomes an open circuit. Describe the new output and its frequency.
Hint
Only one diode works.
Answer
It works as a half-wave rectifier: pulses only in alternate half cycles, at 50 Hz for 50 Hz input (and of the same peak).
Q8 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a student views rectifier outputs on an oscilloscope with an input of 10 V peak, 50 Hz, using ideal diodes.
(i) Sketch the half-wave output and state its peak. (1 mark)
(ii) How many pulses per second does it give? (1 mark)
(iii) How many pulses per second does the full-wave circuit give? (1 mark)
(iv) Why is the full-wave circuit preferred for a charger? (1 mark)
Hint
Ideal: no diode drop.
Answer
(i) Positive half-sine pulses with gaps, peak 10 V. (ii) 50. (iii) 100. (iv) It uses both halves, so it is more efficient and the output is easier to smooth.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Function of a rectifier?Show answer
Convert a.c. into d.c.
2.Output frequency of half-wave for 50 Hz?Show answer
50 Hz.
3.Diodes in a centre-tap full-wave rectifier?Show answer
Two.
4.Diode used as a voltage regulator?Show answer
Zener diode.
5.Bias of a Zener in a regulator?Show answer
Reverse (in breakdown).

Rectifiers and the Zener Voltage Regulator: frequently asked questions

Is the output of a rectifier pure d.c.?

No. It is unidirectional but pulsating; a filter capacitor smooths it, and a regulator steadies it.

Why does the Zener not get damaged in breakdown?

It is designed to operate there, and the series resistor limits the current to a safe value.

Why is the centre tap needed?

It provides a common return point so that each diode sees an opposite-phase half of the secondary and the load current flows the same way in both half cycles.

What does the ISC 2027 syllabus need here?

Diode as a rectifier: half-wave and two-diode full-wave circuits and graphs, with the function of each component (qualitative), and the Zener diode, its V-I characteristic and its working as a voltage regulator. The bridge rectifier is excluded.

How was this asked recently?

ISC 2025 Q20 (5 marks) asked the difference between p and n types, a labelled full-wave rectifier with output graph, and the diode used as a voltage regulator; ISC 2024 Q1(B)(vii) asked the function of a rectifier.