1.12 Rectifiers, PIV and Power Supply Smoothing

1.12 Rectifiers, PIV and Power Supply Smoothing

ACMA Exam Focus: Syllabus items 4.2 and 4.3 — understand the operation of half-wave, full-wave centre-tapped and bridge rectifier circuits and the ripple each one produces, and understand the smoothing and regulation that turn pulsating DC into a supply a transceiver can use, including the peak inverse voltage a rectifier diode must survive.

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From 230 V AC to 13.8 V DC

Australian mains is 230 V RMS at 50 Hz, and a typical HF transceiver wants 13.8 V DC. The chain that gets you from one to the other has four stages, and the syllabus expects you to name all four in order:

1. Transformer — steps the mains down and provides isolation.
2. Rectifier — converts alternating current into unidirectional, pulsating DC.
3. Filter, or smoothing — reduces that pulsation to a small ripple.
4. Regulator — holds the output steady as the load current and the mains voltage vary.

Rectification is the diode's job. A silicon diode conducts in one direction only, once about 0.7 V of forward voltage is present across it.

The three rectifier circuits (item 4.2)

Half-wave. A single diode passes only the half-cycles of one polarity and throws the other half of the waveform away. The output is one pulse per input cycle, so from 50 Hz mains the ripple frequency is 50 Hz. It is cheap and rarely used for anything that matters: the transformer works for only half of each cycle, and the long gaps between pulses make the smoothing capacitor's job much harder.

Full-wave centre-tapped. Two diodes work from a secondary with an earthed centre tap. Each diode handles one half-cycle and both halves reach the load with the same polarity, giving two pulses per cycle and a ripple frequency of 100 Hz. It needs a more expensive transformer and only half the winding conducts at any instant, but just one diode drop sits in the current path — genuinely useful in low-voltage, high-current supplies.

Full-wave bridge. Four diodes arranged in a bridge use both halves of the cycle from an ordinary secondary, again giving 100 Hz ripple from 50 Hz mains. It makes the fullest use of the transformer and is the standard arrangement in amateur linear supplies. The price is two diode drops in series, about 1.4 V, in the conducting path.

That 50 Hz against 100 Hz distinction is examined directly, and it has a practical use as well: 100 Hz ripple is twice as easy to filter as 50 Hz for the same size capacitor, and mains hum heard in a receiver at 100 Hz points at the power supply, while hum at 50 Hz points at a half-wave circuit or at stray pickup.

Smoothing (item 4.3)

Raw rectifier output is pulsating DC — the right polarity, but nowhere near steady. A reservoir or smoothing capacitor is connected across the output. It charges to the peak while a diode conducts and discharges into the load during the gap, so the output sags a little instead of dropping to zero. What is left of the sawtooth is the ripple voltage:

$V_{ripple(p\text{-}p)} \approx \frac{I_{load}}{f_{ripple} \times C}$

Worked example 1. A bridge supply delivers 2.0 A into its load, smoothed by a 10 000 µF capacitor, with 100 Hz ripple. Note that 10 000 µF is 0.01 F.

$V_{ripple} = \frac{2.0}{100 \times 0.01} = 2.0\ \text{V peak-to-peak}$

Doubling the capacitor to 20 000 µF halves the ripple to 1.0 V, and halving the load current does the same. Run the identical load from a half-wave circuit, where the ripple frequency is only 50 Hz, and the ripple doubles to 4.0 V — the practical penalty for throwing half the waveform away.

Worked example 2. A 15.0 V RMS secondary feeds a bridge and a reservoir capacitor. The capacitor charges towards the peak, less two diode drops:

$V_{DC} \approx (15.0 \times 1.414) - 1.4 = 21.2 - 1.4 = 19.8\ \text{V}$

That is the figure a regulator has to work down from, and it explains why a nominally "15 volt" transformer is the right choice behind a 13.8 V supply.

Chokes. Adding a series choke — a large iron-cored inductor — makes an L-section or pi-section filter. The choke's reactance rises with frequency, so it opposes the ripple while passing the DC almost unhindered, and the capacitor after it shunts what remains to earth. Choke-input filters give better regulation and gentler diode current pulses. They are heavy and are now seen mostly in valve equipment, but the syllabus expects you to recognise the arrangement.

Peak inverse voltage (item 4.2)

While a diode is not conducting it has to hold off the reverse voltage applied across it. The peak inverse voltage (PIV), also called peak reverse voltage, is the largest reverse voltage a diode can survive. Exceed it and the junction breaks down, normally short circuit, which puts raw AC onto the reservoir capacitor and the load and takes the fuse, the capacitor and often the transformer with it.

• In a bridge, each diode blocks the peak secondary voltage, $V_{pk}$.
• In a full-wave centre-tapped circuit, the non-conducting diode has the peak of its own half-winding plus the peak of the other half across it, so it must withstand twice the peak of one half-winding.
• With a capacitor-input filter, the diode in a half-wave circuit sees the secondary peak plus the charged capacitor voltage — again about twice the peak.

Worked example 3. A 15.0 V RMS secondary feeds a bridge. The peak is $15.0 \times 1.414 = 21.2$ V, so each diode must block at least 21.2 V. Never buy to the calculated figure: allow two or three times over for mains surges and switching transients, which makes a 100 V device such as a 1N4002 the sensible choice. The same reasoning applies to the current rating, remembering that a capacitor-input filter draws its current in short, tall pulses rather than smoothly.

Regulation basics (item 4.3)

Smoothing removes ripple, but it does nothing to stop the output sagging when the load increases or drifting when the mains voltage wanders. That is regulation, expressed as a percentage:

$\text{Regulation} = \frac{V_{no\ load} - V_{full\ load}}{V_{full\ load}} \times 100$

A supply measuring 15.0 V off load and 13.8 V at full load has a regulation of $(15.0 - 13.8) / 13.8 \times 100 \approx 8.7%$. The lower the figure, the stiffer the supply.

• A Zener diode run in reverse breakdown holds a fixed voltage and makes the simplest reference or low-current regulator.
• A linear regulator, such as a three-terminal 7812 or a series pass transistor, is quiet and free of radio-frequency interference, but it dissipates the surplus voltage as heat and so needs a heatsink.
• A switch-mode power supply chops the DC at tens or hundreds of kilohertz through a small ferrite transformer. It is light and 85 to 95 per cent efficient, but it can radiate hash right across the HF bands unless it is well shielded and filtered — a common and genuine nuisance in an amateur shack.