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Half-Wave Rectification

Key idea: Explain half-wave rectification using a single diode and sketch the ideal sinusoidal input and pulsating output waveforms.

  • GCE A-Level H2 Physics 2027
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Learning objectives

  • Analyse mean power in resistive a.c. loads and half-wave rectification.
  • Represent sinusoidal a.c. and use peak and r.m.s. values.

1. Definitions (Must Know)

A. Rectification

Rectification converts alternating current (a.c.) into a unidirectional output. Without smoothing, the output is usually pulsating d.c., not a steady value.

B. Diode

A diode is a component that allows current to flow mainly in one direction:

  • forward bias: diode conducts (current flows)
  • reverse bias: diode blocks (current is approximately zero)

2. Key Ideas (What Earns Marks)

  • A single diode in series with a load resistor produces half-wave rectification.
  • The output is not steady d.c.; it is pulsating d.c. (zero during half the cycle).
  • A half-wave rectifier produces one output pulse per input cycle, so its pulse repetition rate is f.
  • For an ideal diode:
    • during one half-cycle, the diode conducts and the load gets a voltage of one polarity
    • during the other half-cycle, the diode blocks and the load voltage is (approximately) zero
Syllabus focus

The syllabus specifically highlights using a single diode for half-wave rectification.

3. Detailed Explanations

A. Why a.c. must sometimes be rectified

Alternating current is convenient for power transmission, but many devices need current in one direction (e.g. charging a battery, some electronic circuits). Rectification produces a one-direction current.

B. Half-wave rectification (single diode)

The circuit:

Ideal single-diode half-wave rectifierA sinusoidal alternating-voltage source is connected in series with an ideal diode and load resistor. The diode anode is on the source side and its cathode is on the load side. An output voltmeter is connected across the load with positive polarity at the top. Arrows show clockwise conventional current during the conducting half-cycle; notes state that the opposite half-cycle is blocked.Single diode selects one half-cyclesinusoidal inputvᵢₙ changes polarityideal diodeanodecathodeconventional IloadRV+−Vₒᵤₜ across R+−conducting half-cycle:top terminal positiveForward biased: diode conductsI flows clockwise; Vₒᵤₜ is a positive pulse.Reverse biased: diode blocksIdeal model: I = 0 and Vₒᵤₜ = 0.
Scroll diagram horizontally to read all labels.
When the source makes the diode's anode positive relative to its cathode, conventional current passes through the load and Vout is positive. On the opposite half-cycle the ideal diode blocks, so I = 0 and Vout = 0.

Assume an ideal diode and a sinusoidal input.

  • When the diode is forward biased, it conducts, so current flows through the load.
  • When the diode is reverse biased, it blocks, so the load current is approximately zero.

So the load voltage has only one polarity (a “one-sided” waveform), but it is not constant.

Half-wave rectification (one cycle, scaled)

One cycle of a sinusoidal input and the half-wave rectified output across the load (ideal diode).

Scroll across the graph to read all labels.

One cycle of a sinusoidal input and the half-wave rectified output across the load (ideal diode).One cycle of a sinusoidal input and the half-wave rectified output across the load (ideal diode).
In the ideal-diode model, the negative half-cycle is blocked so the output is zero for half the cycle (pulsating d.c.).
Open full-size graph
View figure data
Values for Half-wave rectification (one cycle, scaled)
Time, t (ms)Input (AC): sin(ωt)Output (load): half-wave rectified
000
2.50.7070.707
511
7.50.7070.707
1000
12.5-0.7070
15-10
17.5-0.7070
2000

4. Common Mistakes

  • Mixing up forward bias and reverse bias.
  • Saying “it becomes d.c.” without stating “pulsating d.c.”.
  • Forgetting the load voltage is near zero during the blocked half-cycle (ideal diode model).

5. Exam Tips

  • Use the ideal diode model unless the question gives a diode drop.
  • State clearly which half-cycle the diode conducts (based on diode orientation).
  • Use r.m.s. values only when the question explicitly asks for mean power or r.m.s.; rectified waveforms are not sinusoidal.
  • Do not apply Vᵣₘₛ = V₀/square root of 2 to a half-wave output; that shortcut describes a complete sinusoid.
Compare the waveforms

Use the AC Waveform & Power Explorer to move a phase cursor across the input and ideal half-wave output.

Trace the diode current path

Use the Semiconductor Devices Lab to connect forward and reverse bias to half-wave and full-wave current paths.

6. Worked Examples

Modelled example 1

Which half-cycle conducts?

Core

Problem

In a half-wave rectifier, the diode is oriented so that it conducts when the input makes it forward biased. Describe the load current over one full cycle.
Study the worked solution

  1. Trace the forward-biased half-cycle

    Method

    The diode conducts and current flows through the load in one direction.

    Reason

    Forward bias permits conventional current through the diode.

    Working

    Forward bias: I_load follows the conducting half-cycle.

  2. Trace the reverse-biased half-cycle

    Method

    The load current is approximately zero.

    Reason

    The reverse-biased diode blocks the series path.

    Working

    Reverse bias: I_load ≈ 0.

  3. Classify the output

    Method

    Call the load current pulsating d.c.

    Reason

    It is unidirectional but falls to zero for half of every cycle.

    Working

    Output: one unidirectional current pulse per cycle.

Guided practice 2

Peak value at the load (ideal diode)

About 4 min

Problem

An a.c. input of peak value V₀ = 12 V is applied to a half-wave rectifier with a resistive load. Find the peak load voltage for an ideal diode.

Try this before viewing the solution

Unit: V

Hints

Hint 1: conducting model
At the positive peak the diode is forward biased.
Hint 2: ideal voltage drop
Use V_diode = 0 while it conducts.
View solution step by step

  1. Apply the ideal-diode model

    Method

    Assign zero voltage across the conducting diode.

    Reason

    An ideal diode has no forward voltage drop.

    Working

    V_(load,peak) = V₀-0

  2. State the peak

    Method

    Obtain 12 V.

    Reason

    The entire input peak appears across the resistive load.

    Working

    V_(load,peak) = 12 V.

Common misconception 3

Peak load voltage with a diode drop

Find and correct the mistake

Learner claim

An input has peak voltage 10 V and the diode’s forward drop is 0.70 V. A learner says the peak load voltage remains 10 V because rectification only removes a half-cycle. Diagnose the claim.

Try this before viewing the solution

Correct conducting-loop relation

View solution step by step

  1. Use the stated diode model

    Method

    Subtract the forward drop from the source peak.

    Reason

    During conduction, Kirchhoff’s voltage law gives V₀ = V_diode + V_load.

    Working

    V_(load,peak) = 10-0.70

  2. Evaluate

    Method

    Obtain 9.3 V.

    Reason

    The diode is explicitly non-ideal in this question.

    Working

    V_(load,peak) = 9.3 V.

Examiner practice 4

Peak current in the load

3 marks

Examination question

A half-wave rectifier produces a peak load voltage of 12 V across a 200 Ω resistive load. Assuming an ideal diode during conduction, find the peak load current and state the current during the blocked half-cycle. [3 marks]

Try this before viewing the solution

View solution step by step

  1. Apply Ohm's law at the peak

    1 mark

    Method

    Use Iₚₑₐₖ = Vₚₑₐₖ/R.

    Reason

    The conducting diode places the resistive load across the rectified source.

    Working

    Iₚₑₐₖ = 12/200

  2. Evaluate

    1 mark

    Method

    Obtain 0.060 A.

    Reason

    The voltage and resistance are in SI units.

    Working

    Iₚₑₐₖ = 0.060 A.

  3. Describe the blocked half-cycle

    1 mark

    Method

    State I ≈ 0.

    Reason

    The reverse-biased diode opens the series current path.

    Working

    Blocked half-cycle: I_load ≈ 0.

Challenge 5

Peak voltage from an r.m.s. input

Minimal support

Independent transfer

A sinusoidal supply of Vᵣₘₛ = 6.0 V is applied to an ideal half-wave rectifier. Find the peak load voltage during conduction and sketch or describe the output over a full cycle.

Try this before viewing the solution

Hints

Hint 1: convert the input representation
The input is a complete sinusoid before it reaches the diode, so V₀ = square root of 2 Vᵣₘₛ applies to the input.
View solution step by step

  1. Convert the sinusoidal input to peak

    Method

    Multiply its r.m.s. value by square root of 2.

    Reason

    The supplied input is a complete sinusoid.

    Working

    V₀ = square root of 2 (6.0) = 8.49 V

  2. Apply ideal rectification

    Method

    Assign 8.49 V as the conducting load peak.

    Reason

    The ideal diode has zero forward drop.

    Working

    V_(load,peak) = 8.49 V.

  3. Describe the output

    Method

    Retain one sinusoidal half-cycle and set the opposite half-cycle to zero.

    Reason

    The diode alternates between forward and reverse bias.

    Working

    Conducting half-cycle: sinusoidal pulse; blocked half-cycle: V_load = 0.

7. Mind Stretchers

Mind stretcher 1: Pulse repetition rate (half-wave)Extension

If the input frequency is 50 Hz, what is the repetition rate of the pulses in half-wave rectification? Explain.

Show Answer

Half-wave rectification produces one pulse per input cycle, so the pulse repetition rate is 50 Hz.

Mind stretcher 2: Reversing the diodeExtension

Describe what happens to the output if the diode is reversed.

Show Answer

The diode would conduct on the opposite half-cycle, so the output pulses would flip polarity (still pulsating d.c., but of the opposite sign across the load).

Continue with the next resource in this course.

Course and syllabus information
Course
GCE A-Level H2 Physics
Edition
GCE A-Level H2 Physics 2027