Notes · ElectronicsUK · A-Levels

Mains power supply systems

A regulated DC supply converts the mains alternating voltage into the steady direct voltage electronic circuits need. This chapter follows the power-supply chain block by block: transformation, half-wave and full-wave rectification, reservoir-capacitor smoothing and ripple, and voltage regulation using the zener shunt regulator and the emitter-follower series regulator.

4 sections~14 min reading time3 competenciesLevel Foundation 1 · Standard 2 · Advanced 1

T·0999 / 16

Exam profile

AO1 · Describe the transformer, rectifier, smoothing and regulator stages of a DC power supply.AO2 · Calculate rectified voltages, ripple voltage and zener and regulator component values.AO3 · Design a regulated supply to a specification and evaluate its ripple and regulation.

Operators:describecalculatedeterminedesignsketchanalyseevaluate

basic level

At AS the focus is the block chain, half-wave and full-wave rectification and simple capacitor smoothing.

higher level

The full A-Level develops ripple calculation, load and line regulation, and the zener and emitter-follower regulators quantitatively.

Depth

Text

Contents · 4 sections

Mains power supply systems

The power-supply systems chain#

FoundationWJEC/Eduqas A Level Electronics — Unit 3 Analogue Design (power supplies)

Regulated power supply block chain

Fig. 1The four sub-systems of a regulated supply: transformer, rectifier, smoothing and regulator, turning AC mains into steady DC.

A mains power supply is a textbook input-process-output system, built as a chain of four sub-systems that each do one job. The transformer steps the mains voltage down (and isolates the circuit from the mains); the rectifier converts the alternating voltage into a one-directional, pulsing voltage; the smoothing capacitor fills in the gaps to make it nearly steady; and the regulator holds the output constant against changes in load and supply. Following a signal through this chain is the way to understand and to design a supply.

The transformer exploits electromagnetic induction: an alternating current in the primary winding induces an alternating voltage in the secondary, in the ratio of the numbers of turns,

V_s/V_p = N_s/N_p

. A step-down transformer has fewer secondary turns, giving a lower, safer secondary voltage. Just as important, the transformer provides galvanic isolation — there is no direct electrical connection between the mains and the low-voltage circuit — which is an essential safety feature of a mains supply.

Each stage improves the signal at the cost of some complexity, and the quality of the final output depends on all of them. A supply with a transformer and rectifier but no smoothing delivers a badly pulsing voltage; adding a reservoir capacitor makes it usably steady; adding a regulator makes it rock-solid regardless of load. The designer includes only the stages the application needs — a simple battery charger may need no regulator, while a sensitive analogue circuit needs a well-regulated, low-ripple supply.

The two figures of merit for a supply are its ripple (the small residual AC variation left on the DC output) and its regulation (how much the output sags when the load draws more current or the mains voltage changes). Good smoothing reduces ripple; good regulation keeps the output constant. These are the quantities the later sections calculate, and they are the criteria against which a supply design is evaluated in an exam.

Seeing the supply as a chain of blocks, each transforming the signal one step closer to steady DC, is the systems insight that makes the whole topic coherent. It also mirrors the signal-processing chains elsewhere in the course: just as a sensor system is input-process-output, so is a power supply, and the same block-diagram thinking lets you reason about, fault-find and design each stage independently before combining them.

\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p}

Transformer turns ratio

Secondary voltage is set by the ratio of secondary to primary turns.

Worked example

Transformer secondary turns

A transformer steps $230\,\text{V}$ rms down to $12\,\text{V}$ rms with a $2000$ -turn primary. Find the secondary turns.

Rearrange the turns ratio
$N_s = N_p \times V_s/V_p = 2000 \times 12/230$ .
Evaluate
$N_s = 2000 \times 0.0522 = 104$ turns.
$N_s = 2000 \times \dfrac{12}{230} \approx 104\ \text{turns}$

Result: The secondary needs about $104$ turns; the transformer also isolates the low-voltage circuit from the mains for safety.

Typical mistakes

Confusing the order of the stages, or omitting smoothing between rectification and regulation.
Inverting the turns-ratio equation, giving a step-up where a step-down is intended.

Active revision

A transformer steps $230\,\text{V}$ rms mains down to $12\,\text{V}$ rms. If the primary has $2000$ turns, find the number of secondary turns and state one safety benefit of the transformer.

Active recall

Recall the key points — then reveal.

Sources: WJEC/Eduqas GCE Electronics specification (WJEC / Eduqas)

Rectification#

StandardWJEC/Eduqas A Level Electronics — Unit 3 Analogue Design (rectification)

Bridge rectifier

Fig. 2The four-diode bridge: whichever way the AC input swings, two diodes conduct to keep the load current in the same direction, giving full-wave rectification.

Rectification converts alternating voltage, which reverses direction, into a voltage of one polarity. The simplest is half-wave rectification: a single diode passes the positive half-cycles and blocks the negative ones, so the output is a series of positive humps with gaps where the negative half-cycles have been removed. It works, but it wastes half the input and leaves large gaps that are hard to smooth, so it is used only in undemanding circuits.

Full-wave rectification is far better. A bridge of four diodes routes the current so that both halves of the input cycle produce a positive output: during the positive half-cycle two diodes conduct one way through the load, and during the negative half-cycle the other two diodes conduct, keeping the load current in the same direction. The output is a continuous series of positive humps at twice the input frequency, with no gaps — twice as much energy delivered and much easier to smooth.

Every diode drops about

0.7\,\text{V}

when conducting, and this must be accounted for in the output voltage. In a bridge rectifier two diodes are in series with the load at any instant, so the peak output is the peak secondary voltage minus

1.4\,\text{V}

(

2 \times 0.7\,\text{V}

); a half-wave rectifier loses only one diode drop. Because the transformer is rated in rms but the capacitor charges to the peak, the peak is found from the rms by multiplying by

\sqrt{2}

— a step candidates frequently forget.

The doubling of the ripple frequency in full-wave rectification is a real advantage that carries into smoothing. Because the humps come twice as often, the smoothing capacitor has only half as long to discharge between them, so for the same capacitor the ripple is half that of a half-wave circuit. This is the main reason the bridge rectifier is almost always chosen over the half-wave design despite needing four diodes instead of one.

The rectified output, though one-directional, is still far from steady — it swings from the peak down to zero (half-wave) or nearly so between humps. On its own it is unusable as a DC supply; it is the smoothing stage that turns this pulsing waveform into a nearly constant voltage. Rectification is therefore best understood as the stage that fixes the polarity, leaving the smoothing stage to fix the steadiness, in the division of labour that defines the supply chain.

V_{peak} = \sqrt{2}\,V_{rms} - 1.4\,\text{V} \ \text{(bridge)}

Bridge peak output

Peak of the secondary minus two diode drops.

f_{ripple} = 2 f_{mains} \ \text{(full-wave)}

Ripple frequency

Full-wave rectification doubles the ripple frequency.

Full-wave rectified output

Fig. 3Full-wave rectification flips the negative half-cycles up, giving positive humps at twice the input frequency (marker at a peak).

Worked example

Peak output of a bridge rectifier

A bridge rectifier is fed from a $12\,\text{V}$ rms secondary of a $50\,\text{Hz}$ transformer. Find the peak DC output and the ripple frequency.

Peak of the secondary
$V_{peak(ac)} = \sqrt{2} \times V_{rms} = 1.414 \times 12 = 16.97\,\text{V}$ .
Subtract two diode drops
In a bridge, two diodes conduct in series, so $V_{peak(dc)} = 16.97 - 1.4 = 15.6\,\text{V}$ .
$V_{peak} = \sqrt{2}(12) - 1.4 = 15.6\,\text{V}$
Ripple frequency
Full-wave rectification doubles the frequency: $f_{ripple} = 2 \times 50 = 100\,\text{Hz}$ .

Result: The peak DC output is about $15.6\,\text{V}$ and the ripple frequency is $100\,\text{Hz}$ .

Typical mistakes

Forgetting the two diode drops ( $1.4\,\text{V}$ ) in a bridge, or the one drop in a half-wave circuit.
Using the rms secondary voltage as the peak; the capacitor charges to the peak, which is $\sqrt{2}$ times the rms.

Active revision

A bridge rectifier is fed from a $12\,\text{V}$ rms transformer secondary. Find the peak DC output voltage after the diode drops, and state the ripple frequency for a $50\,\text{Hz}$ mains.

Active recall

Recall the key points — then reveal.

Sources: WJEC/Eduqas GCE Electronics specification (WJEC / Eduqas)

Smoothing and ripple#

StandardWJEC/Eduqas A Level Electronics — Unit 3 Analogue Design (smoothing)

Smoothed output with ripple

Fig. 4The reservoir capacitor charges to the peak then discharges slowly into the load, leaving a small sawtooth ripple

V_r

on the DC output.

Smoothing turns the pulsing rectified output into a nearly steady voltage using a large reservoir capacitor connected across the load. The capacitor charges up almost to the peak of each rectified hump, and then, in the gaps between humps, it supplies the load current itself, discharging only a little before the next hump tops it back up. The output is therefore held near the peak, with only a small residual variation — the ripple.

The ripple voltage depends on how much the capacitor discharges between humps, which is governed by the load current, the time between humps and the capacitance. From

Q = It

and

Q = CV

, the ripple is

V_r = It/C

, where

t

is the discharge time — one half-cycle for full-wave rectification. Writing

t = 1/(2f)

gives the compact form

V_r = I/(2fC)

for a full-wave supply. A bigger capacitor, a smaller load current or a higher ripple frequency all reduce the ripple.

This formula makes the design trade-offs explicit. To halve the ripple you must double the capacitance, so low-ripple supplies use large reservoir capacitors — thousands of microfarads. Full-wave rectification helps here too: because its ripple frequency is double that of half-wave, the discharge time

t

is halved and the ripple is half as large for the same capacitor. The choice of capacitor is thus the central smoothing calculation, made against a target ripple.

There is a hidden cost to a large reservoir capacitor: it charges only during the brief interval near each peak, so it draws a short, high current pulse from the transformer and rectifier on every hump. This means the diodes and transformer must be rated for a peak current well above the average load current — a practical point that examiners raise as an evaluation of the design. Bigger is not automatically better; the capacitor must be matched to the load and the components' ratings.

Even a well-smoothed supply still has some ripple, and for sensitive circuits this residual variation must be removed by the regulator that follows. The smoothing stage reduces the ripple to a manageable level; the regulator then holds the output truly constant. Understanding smoothing quantitatively — predicting the ripple from the load and the capacitor, and choosing the capacitor for a target ripple — is a core power-supply skill and a frequent exam calculation.

V_r = \dfrac{I\,t}{C} = \dfrac{I}{2 f C} \ \text{(full-wave)}

Ripple voltage

The capacitor discharges by It/C between humps; t is one half-cycle for full-wave.

Worked example

Ripple of a smoothed supply

A full-wave supply from a $50\,\text{Hz}$ mains delivers $100\,\text{mA}$ to a load and is smoothed by a $4700\,\mu\text{F}$ capacitor. Find the ripple voltage, and the capacitor needed to halve it.

Ripple frequency
Full-wave: $f_{ripple} = 2 \times 50 = 100\,\text{Hz}$ , so the discharge time is $t = 1/100 = 10\,\text{ms}$ .
Ripple voltage
$V_r = I/(2fC) = 0.100/(2 \times 50 \times 4700\times 10^{-6}) = 0.100/0.470 = 0.213\,\text{V}$ .
$V_r = \dfrac{0.100}{2(50)(4700\times 10^{-6})} = 0.21\,\text{V}$
Halving the ripple
Ripple is inversely proportional to capacitance, so halving it needs double the capacitor: $9400\,\mu\text{F}$ (a $10\,000\,\mu\text{F}$ part).

Result: The ripple is about $0.21\,\text{V}$ ; doubling the capacitor to roughly $10\,000\,\mu\text{F}$ halves it.

Typical mistakes

Using the mains frequency instead of the ripple frequency ( $2f$ for full-wave) in the ripple formula.
Muddling microfarads and farads, giving a ripple out by a factor of a million.

Active revision

A full-wave supply from a $50\,\text{Hz}$ mains delivers $100\,\text{mA}$ to a load, smoothed by a $4700\,\mu\text{F}$ reservoir capacitor. Calculate the ripple voltage.

Active recall

Recall the key points — then reveal.

Sources: WJEC/Eduqas GCE Electronics specification (WJEC / Eduqas)

Regulation: zener and emitter-follower#

AdvancedWJEC/Eduqas A Level Electronics — Unit 3 Analogue Design (regulation)

Zener shunt regulator

Fig. 5The series resistor feeds the output node; the reverse-biased zener holds it at

V_Z

, absorbing changes as the load current varies.

A regulator holds the output voltage constant despite changes in the load current (load regulation) and in the input voltage (line regulation). The simplest is the zener shunt regulator: a zener diode is connected in reverse across the load, fed through a series resistor from the smoothed supply. Operated in reverse breakdown, the zener maintains an almost constant voltage

V_Z

across itself over a wide range of current, so the output is pinned at

V_Z

and the ripple and variation are absorbed by the series resistor.

The series resistor is the key component. It must drop the difference between the input and the zener voltage while passing both the load current and enough current to keep the zener in breakdown, so

R_S = (V_{in} - V_Z)/(I_Z + I_L)

. If the load current rises, the zener current falls to compensate (and vice versa), keeping the total through the resistor — and hence the output — roughly constant. The zener must be chosen with a high enough current rating and power rating to handle the worst case, when the load takes no current and the zener carries it all.

The zener shunt regulator is simple and cheap but wasteful and limited to small currents, because the series resistor and the zener continuously dissipate power. For larger currents an emitter-follower (series) regulator is used: a transistor is connected as an emitter follower with its base held at a reference (often a zener), so the output follows the reference less the

0.7\,\text{V}

base-emitter drop,

V_{out} = V_Z - 0.7\,\text{V}

. The transistor supplies the heavy load current from the supply, while the reference and its zener carry only the small base current.

The emitter-follower regulator therefore multiplies the current capability of a zener reference by the transistor's current gain: a zener passing a few milliamps into the base can control hundreds of milliamps in the load. It also gives better regulation, because the transistor's feedback holds the output firmly at the reference. Adding an op-amp to compare the output against the reference and drive the transistor produces the feedback-controlled regulator that underlies the three-terminal regulator chips used in practice.

Regulation is the final stage that turns a merely smoothed supply into a precise, steady source. The two circuits illustrate a recurring design trade-off: the zener shunt regulator is simple but wasteful and low-current, while the emitter-follower is more complex but efficient and high-current. Choosing between them for a stated load, and calculating the series resistor, zener rating or output voltage, is exactly the analyse-and-design reasoning that completes the power-supply topic.

R_S = \dfrac{V_{in} - V_Z}{I_Z + I_L}

Zener series resistor

Drops the input-to-zener difference while passing the zener and load currents.

V_{out} = V_Z - 0.7\,\text{V} \ \text{(emitter follower)}

Emitter-follower output

The output follows the zener reference less the base-emitter drop, at a much higher current.

Worked example

Designing a zener shunt regulator

A zener shunt regulator must give $5.1\,\text{V}$ from a $15\,\text{V}$ smoothed input, feeding a $20\,\text{mA}$ load with at least $5\,\text{mA}$ through the zener. Find the series resistor and the worst-case zener power.

Series resistor
$R_S = (V_{in} - V_Z)/(I_Z + I_L) = (15 - 5.1)/(5 + 20)\,\text{mA} = 9.9/0.025 = 396\,\Omega$ . Use $390\,\Omega$ .
$R_S = \dfrac{15 - 5.1}{(5 + 20)\,\text{mA}} = 396\,\Omega$
Worst-case zener current
If the load is disconnected, all the resistor current flows through the zener: $I = (15 - 5.1)/390 = 25.4\,\text{mA}$ .
Worst-case power
$P_Z = V_Z I = 5.1 \times 0.0254 = 0.13\,\text{W}$ , so a $0.5\,\text{W}$ zener is comfortable.

Result: A $390\,\Omega$ series resistor holds the output at $5.1\,\text{V}$ ; the zener must handle about $25\,\text{mA}$ and $0.13\,\text{W}$ at no load.

Typical mistakes

Sizing the series resistor for only the load current, forgetting the zener needs current too.
Omitting the $0.7\,\text{V}$ base-emitter drop when finding the emitter-follower output.

Active revision

Design a zener shunt regulator to give $5.1\,\text{V}$ from a $15\,\text{V}$ smoothed input, supplying a $20\,\text{mA}$ load with at least $5\,\text{mA}$ through the zener. Find the series resistor and the peak zener power.

Active recall

Recall the key points — then reveal.

Sources: WJEC/Eduqas GCE Electronics specification (WJEC / Eduqas)

References & sources

Sources

WJEC / Eduqas

WJEC/Eduqas GCE Electronics specification

Mains power supply systems

4 sections~14 min reading time3 competenciesLevel Foundation 1 · Standard 2 · Advanced 1

Mains power supply systems

Key points

Transformer secondary turns

Key points

Peak output of a bridge rectifier

Key points

Ripple of a smoothed supply

Key points

Designing a zener shunt regulator

Sources

Mains power supply systems

Key points

Transformer secondary turns

Key points

Peak output of a bridge rectifier

Key points

Ripple of a smoothed supply

Key points

Designing a zener shunt regulator

Sources