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E19: Power Supplies & Regulation

WJEC Eduqas C690QS

Rectification, smoothing, voltage regulation and Zener diodes

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Power Supplies & Regulation

Rectification, smoothing, voltage regulation and Zener diodes

Key Fact: A power supply converts AC mains to a stable DC voltage through transformer, rectifier, smoothing and regulation stages.
Key Fact: A step-down transformer reduces the AC mains voltage to a lower AC voltage suitable for the circuit.
Key Fact: Half-wave rectification uses a single diode and only rectifies one half of the AC cycle, giving 50% efficiency.
Key Fact: Full-wave rectification using a bridge rectifier uses four diodes and rectifies both halves of the AC cycle.
Key Fact: The output of a rectifier is pulsating DC; a smoothing capacitor converts this to a more constant DC level.
Key Fact: Ripple voltage is the residual AC variation on the DC output; larger capacitance reduces ripple for a given load current.
Key Fact: Ripple voltage formula: V_ripple ≈ I_load / (f × C), where f is the ripple frequency (100 Hz for full-wave on 50 Hz mains).
Key Fact: A Zener diode operates in reverse breakdown to provide a stable reference voltage; it is the simplest form of voltage regulator.
Key Fact: A Zener regulator circuit uses a series resistor to limit current: R_s = (V_in − V_Z) / (I_Z + I_load).
Key Fact: A three-terminal voltage regulator (e.g. 7805 for 5 V, 7812 for 12 V) provides a fixed, stable output with built-in protection.
Key Fact: Linear regulators dissipate excess voltage as heat; efficiency is V_out / V_in, so they run hot when the difference is large.
Key Fact: Switch-mode power supplies (SMPS) use high-frequency switching for higher efficiency and smaller size than linear regulators.

📋 Key Vocabulary and Concepts

For Power Supplies & Regulation, you must know:

❓ Practice Questions

Q: What is the ripple frequency of a full-wave rectifier supplied from 50 Hz mains?

Q: A Zener regulator has V_in = 12 V, V_Z = 5.1 V, I_Z = 20 mA and I_load = 50 mA. Calculate R_s.

Q: Explain why a smoothing capacitor reduces ripple voltage.

Q: State two advantages of a bridge rectifier over a half-wave rectifier.

Q: Why does a 7805 regulator need a heat sink when the input voltage is significantly above 5 V?

✅ Answers

  1. 100 Hz, because both halves of the AC cycle are rectified, doubling the frequency.
  2. R_s = (V_in − V_Z)/(I_Z + I_load) = (12 − 5.1)/(0.02 + 0.05) = 6.9/0.07 = 98.6 Ω ≈ 100 Ω.
  3. The capacitor charges during the rectifier peak and discharges into the load during the trough, filling in the gaps between rectified pulses.
  4. A bridge rectifier uses both halves of the AC cycle (higher output) and has lower ripple because the ripple frequency is doubled.
  5. The regulator dissipates power P = (V_in − V_out) × I_load as heat; with a large voltage difference, this power can exceed the device's dissipation limit.

🎯 Exam Tips

📝 Exam Technique

GCSE Electronics Exam Tips — Power Supplies & Regulation:
1. For Power Supplies & Regulation questions, use correct electronic symbols and terminology
2. Always show your working in calculations, including units at each step
3. When analysing circuits, state which law or rule you are applying first
4. For evaluation questions on Power Supplies & Regulation, compare component choices and consider cost, reliability and tolerance
5. Draw circuit diagrams neatly with conventional symbols

⚠️ Common Errors

✗ Using 50 Hz for the ripple frequency of a full-wave rectifier. ✓ Full-wave rectification doubles the frequency; the ripple frequency is 100 Hz on a 50 Hz mains supply.

✗ Forgetting to include load current when calculating the Zener series resistor. ✓ R_s = (V_in − V_Z)/(I_Z + I_load); both Zener and load current flow through the series resistor.

✗ Assuming a smoothing capacitor removes all ripple. ✓ A capacitor reduces ripple but does not eliminate it; some AC variation always remains on the DC output.

✗ Confusing linear and switch-mode regulator efficiency. ✓ Linear regulators dissipate excess voltage as heat (low efficiency with large V_drop); SMPS achieve higher efficiency by switching at high frequency.

✍️ Model Answer

Full-Mark Response

Describe the stages of a regulated DC power supply from mains AC input to stable DC output, explaining the role of rectification, smoothing and voltage regulation.

A regulated DC power supply converts mains AC to stable DC through four stages. First, a step-down transformer reduces the 230 V AC mains to a lower AC voltage (e.g. 12 V AC). Second, a rectifier converts the AC to pulsating DC: a half-wave rectifier uses one diode and only passes one half-cycle, while a bridge rectifier uses four diodes to pass both half-cycles, producing full-wave rectified DC with a ripple frequency of 100 Hz. Third, a smoothing capacitor (typically hundreds or thousands of microfarads) charges during the rectifier peaks and discharges into the load during the troughs, reducing the ripple voltage. The ripple is approximately V_ripple = I_load/(f × C). Fourth, a voltage regulator provides a constant output voltage regardless of remaining ripple or load current changes. A Zener diode regulator uses the reverse breakdown voltage as a reference, with a series resistor to limit current. A three-terminal IC regulator (e.g. 7805) provides tighter regulation with built-in current limiting and thermal shutdown. The result is a clean, stable DC supply suitable for powering electronic circuits.

📊 AO Deep Dive

Assessment Objective Analysis

AO1 (Knowledge & Understanding): Demonstrate knowledge and understanding of power supplies & regulation, including electronic components, circuit theory and systems concepts relevant to WJEC Eduqas C690QS.

AO2 (Application): Apply knowledge and understanding of power supplies & regulation to analyse, design and construct electronic circuits and systems.

AO3 (Evaluation): Evaluate electronic circuits and systems, making reasoned judgements about design choices, performance and practical considerations, constructing supported arguments.

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