Chapter 6: Single-Phase Uncontrolled Rectifiers — Power Electronics

Where this sits

Part 2 · AC–DC Converters
Chapter 6 of 30 — the first converter chapter

You should already know

Average versus RMS (Ch. 1) and diode ratings, freewheeling and \(I_{FSM}\) (Ch. 2).

By the end you can

Compute \(V_{dc}\), \(V_{rms}\), FF, RF, η, TUF and PIV for any single-phase circuit, size a filter capacitor, and explain a 0.9 power factor at unity displacement.

Time

≈ 65 min reading · ≈ 55 min problems

i What you'll learn

The five performance parameters that grade every rectifier: form factor, ripple factor, efficiency, TUF, and PIV.
The half-wave rectifier — \( V_{dc} = V_m/\pi \) — and why its 40.5% efficiency and 1.21 ripple factor make it useless for real power.
What an inductive load does: the extinction angle \( \beta \), the negative output excursion, and the role of the freewheeling diode.
The full-wave bridge and centre-tapped circuits — \( V_{dc} = 2V_m/\pi \) — and the PIV and TUF trade-off that decides between them.
How a capacitor filter cuts ripple by a factor of forty and multiplies the peak diode current by twenty.
Why a diode bridge draws a distorted current with 48% THD and a power factor of 0.9 even at unity displacement.

Section 6-1

What a Rectifier Must Do — and How We Grade It

A rectifier performs the AC–DC conversion of Chapter 1's map. Because it uses only diodes, it is uncontrolled: the output depends entirely on the supply and the load, with no adjustment available. Chapters 8 and 9 add gates and get control; here we establish what an unaided rectifier can do, and the language for saying how well.

Real rectifier output is not DC — it is a DC average with a large AC ripple on top. Five parameters quantify the gap between what we got and what we wanted, and they recur in every problem in Part 2.

Parameter	Definition	What it tells you
Form factor	\( FF = V_{rms}/V_{dc} \)	How far the waveform is from pure DC (ideal = 1)
Ripple factor	\( RF = \sqrt{FF^2 - 1} \)	AC content relative to the DC value (ideal = 0)
Rectification efficiency	\( \eta = P_{dc}/P_{ac} \)	Fraction of delivered power that is useful DC
Transformer utilisation factor	\( TUF = P_{dc}/(\text{transformer VA}) \)	How well the transformer is used — sizing and cost
Peak inverse voltage	\( PIV \)	Maximum reverse voltage each diode must block

◇Intuition · what the five numbers are really asking

Five parameters looks like a list to memorise. It is really four different people asking four different questions, plus one for the diode:

Form factor and ripple factor — the load asking: "how far is this from the clean DC I wanted?" \(FF = 1\) and \(RF = 0\) would be perfect DC.
Rectification efficiency — the accountant asking: "of the power I delivered, how much was useful DC rather than wasted ripple?"
TUF — the purchasing department asking: "how big a transformer must I buy for this much DC?" A low TUF means an expensive, oversized transformer.
PIV — the diode asking: "what is the worst reverse voltage you will ever ask me to hold off?"

Once you know which question a parameter answers, you never have to remember which is "good high" and which is "good low" — you can reason it out.

Note also that \(RF\) is derived from \(FF\), not measured separately: \(RF = \sqrt{FF^2-1}\) follows from \(V_{rms}^2 = V_{dc}^2 + V_{ac}^2\), the same Pythagorean split you know from AC circuit theory. Compute \(V_{dc}\) and \(V_{rms}\), and all five numbers fall out.

Two averages, one distinction. Chapter 1 warned that a filtered load responds to the average while heating and power follow the RMS. In rectifier analysis both appear in every problem, and the form factor is precisely their ratio. If you find yourself computing \(V_{dc}^2/R\) for a resistive load, stop — the correct power is \(V_{rms}^2/R\), and the difference is the factor \(FF^2\).

!Common mistake

"The output is 100 V DC into 10 Ω, so the load takes \(100^2/10 = 1000\) W."

Only if the output is actually DC. A rectifier output is a DC average with a large ripple on it, and a resistor heats according to the RMS.

For a half-wave rectifier, \(FF = 1.57\), so the true power is \(FF^2 = 2.47\) times your answer — 2470 W, not 1000 W.
For a full-wave rectifier, \(FF = 1.11\), so it is \(1.23\) times — 1230 W.
Only after a good capacitor filter, where \(FF \to 1\), does \(V_{dc}^2/R\) become correct.

This is the Chapter 1 average-versus-RMS error wearing a different hat, and the form factor is precisely the correction factor. If you catch yourself squaring a DC value to find power in a rectifier problem, ask first: is this waveform smooth enough for that to be true?

Section recap. A rectifier's output is a DC average with ripple. Five numbers grade it: \(FF = V_{rms}/V_{dc}\) and \(RF = \sqrt{FF^2-1}\) measure the ripple, \(\eta\) measures useful power, TUF measures transformer sizing, and PIV sizes the diode. Compute \(V_{dc}\) and \(V_{rms}\) and the rest follows.

Section 6-2

The Half-Wave Rectifier with a Resistive Load

One diode, one direction, half the waveform

Half the energy discarded, and a huge ripple on what remains

The diode conducts for the positive half cycle and blocks the negative one, so the output is a train of half sines. Averaging over a full period — noting the second half contributes nothing — gives the DC value, and the RMS follows from integrating the square over the same period:

🔑

Half-wave rectifier, resistive load

\[ V_{dc} = \frac{1}{2\pi}\int_0^{\pi} V_m\sin\omega t\ d(\omega t) = \frac{V_m}{\pi} = 0.318V_m, \qquad V_{rms} = \frac{V_m}{2} = 0.5V_m \]

From these two, everything else follows mechanically: \( FF = 1.57 \), \( RF = 1.21 \), \( \eta = 40.5\% \), \( TUF = 0.286 \), and \( PIV = V_m \). A ripple factor above one means the AC content exceeds the DC content — the output is more ripple than signal.

→Why \(V_m/\pi\), and why \(V_m/2\)

Both results come from the same move — integrate over the whole period, even though the diode conducts for only half of it — and forgetting that is the most common slip in this chapter.

Average. The area under one half sine is \(\int_0^\pi V_m\sin\theta\,d\theta = 2V_m\). Spread that over a full period of \(2\pi\): \(V_{dc} = 2V_m/2\pi = V_m/\pi\).
RMS. The mean of \(\sin^2\) over its own half cycle is \(\tfrac12\), so the mean square over a full period is \(\tfrac12 \times \tfrac12 = \tfrac14\) of \(V_m^2\). Take the root: \(V_{rms} = V_m/2\).

The sanity check that catches the error: a full-wave circuit uses both half cycles, so both results must simply double — \(2V_m/\pi\) and \(V_m/\sqrt2\). If your half-wave and full-wave answers are not in a 2:1 ratio, you divided by the wrong period somewhere.

The circuit is also unkind to its transformer. Current flows in one direction only, so the secondary carries a DC component that drives the core toward saturation, and the utilisation factor of 0.286 means a transformer nearly four times the DC power rating is needed. Together with the ripple, this confines the half-wave rectifier to signal-level and trickle-charging duties. Its value here is pedagogical: it is the simplest circuit in which every parameter of Section 6-1 can be derived by hand.

Section recap. Half-wave: \(V_{dc} = V_m/\pi\), \(V_{rms} = V_m/2\), \(FF = 1.57\), \(RF = 1.21\), \(\eta = 40.5\%\), \(TUF = 0.286\), \(PIV = V_m\). A ripple factor above 1 means more AC than DC. Useful for learning the method, and for almost nothing else.

Section 6-3

The Half-Wave Rectifier with an Inductive Load

Replace the resistor with an \(RL\) load and the behaviour changes qualitatively. Inductor current cannot stop instantaneously, so when the supply voltage crosses zero the current is still flowing — and it keeps the diode conducting into the negative half cycle. Conduction ceases only at the extinction angle \(\beta\), where the current finally reaches zero, found from the transcendental condition

Extinction angle for an RL load

\[ \sin(\beta - \phi) + \sin\phi\,e^{-\beta/(\omega\tau)} = 0, \qquad \phi = \tan^{-1}\!\left(\frac{\omega L}{R}\right),\ \ \tau = \frac{L}{R} \]

!Common mistake

"An inductor smooths things out, so an inductive load should give a better, higher DC output."

Wrong Inductance smooths the current, so the output must improve.

Right It smooths the current by keeping the diode conducting past the zero crossing — into the negative half cycle. That negative voltage area subtracts from the average, so \(V_{dc}\) falls.

Trace it slowly:

Inductor current cannot change instantly, so at \(\omega t = \pi\) it is still flowing.
Current flowing means the diode is still conducting.
A conducting diode connects the load to the supply — and the supply has now gone negative.
So the load sees negative voltage from \(\pi\) until the extinction angle \(\beta\), and that area comes off the average.

The more inductive the load, the later \(\beta\) falls and the more average is lost. In the limit of very large \(L\), \(\beta \to 2\pi\) and the average approaches zero.

During the interval from \(\pi\) to \(\beta\) the output voltage is negative, because the load is connected to a supply that has already reversed. That negative area subtracts from the average, so an inductive load makes the DC output lower than the resistive case — the opposite of most students' first guess.

The cure is a freewheeling diode across the load, exactly as described in Chapter 2. When the supply reverses, the freewheeling diode picks up the inductor current and circulates it through the load, isolating the load from the negative supply. The output voltage is then clamped at zero instead of going negative, restoring \(V_{dc} = V_m/\pi\), and the load current becomes smoother because it never has to fall to zero. The main diode is also relieved of conducting during the negative half cycle.

The freewheeling diode does three things at once. It restores the DC output to its resistive-load value; it smooths the load current by giving it a continuous path; and it protects the main diode and the source from the \(L\,di/dt\) that would otherwise appear. One component, three benefits — which is why virtually every practical inductive-load rectifier has one, and why the same device reappears in every chopper in Part 3.

◇Intuition · giving the current somewhere else to go

The whole problem is that inductor current must keep flowing, and the only path available runs back through the supply.

So do not fight the current — offer it a better road. A diode across the load provides a short local loop the current would rather take than push against a reversed supply.

Water in a pipe is the right picture: you cannot stop it dead without bursting something, but you can open a bypass and let it circulate harmlessly until it slows on its own.

Trace the consequences and the three benefits are just one mechanism seen from three sides:

The load is disconnected from the reversed supply → output is clamped at 0 instead of going negative → \(V_{dc}\) is restored to \(V_m/\pi\).
The current never has to reach zero → it becomes smoother, and may become continuous.
There is no abrupt interruption of inductor current → no \(L\,di/dt\) spike to destroy the main diode.

Remember this circuit. The freewheeling diode is the single most reused component in the book — it is in every chopper in Part 3 and every inverter leg in Part 4.

Section recap. An inductive load keeps the diode conducting past the zero crossing to the extinction angle \(\beta\), so the output goes negative and \(V_{dc}\) falls. A freewheeling diode across the load fixes all of it at once: it restores \(V_{dc}\), smooths the current, and removes the \(L\,di/dt\) spike.

Section 6-4

Full-Wave Rectifiers: Bridge and Centre-Tap

Using both half cycles doubles the output and halves the ripple. Two circuits achieve it. The bridge uses four diodes and an ordinary secondary; the centre-tapped circuit uses two diodes and a secondary with a mid-point. Both produce identical output waveforms, but they load the transformer and stress the diodes very differently.

Bridge — four diodes, ordinary secondary, PIV = V_m

Centre-tap — two diodes, but PIV = 2V_m and poor TUF

Because the output now consists of every half sine, the averaging integral runs over a half period rather than a full one, and both the average and the RMS double relative to the half-wave case:

🔑

Full-wave rectifier, resistive load

\[ V_{dc} = \frac{1}{\pi}\int_0^{\pi} V_m\sin\omega t\ d(\omega t) = \frac{2V_m}{\pi} = 0.636V_m, \qquad V_{rms} = \frac{V_m}{\sqrt{2}} = 0.707V_m \]

Hence \( FF = 1.11 \), \( RF = 0.482 \), and \( \eta = 81.2\% \) — twice the voltage, a quarter of the power for the same load, and a ripple factor two and a half times better. The ripple frequency also doubles to \(2f\), which makes any subsequent filter smaller.

The choice between the two circuits is a genuine engineering trade. The bridge needs four diodes and drops two forward voltages in series — a real penalty at low output voltage — but each diode blocks only \(V_m\), the secondary is simple, and the TUF of 0.812 is the best available. The centre-tap needs only two diodes with a single drop in the path, which suits low-voltage high-current supplies, but each diode must block \(2V_m\), and because each half of the secondary works only on alternate half cycles the TUF falls to 0.573 — meaning a transformer nearly 40% larger for the same DC output.

Quantity	Half-wave	Centre-tap	Bridge
Diodes	1	2	4
\(V_{dc}\)	\(V_m/\pi\)	\(2V_m/\pi\)	\(2V_m/\pi\)
\(V_{rms}\)	\(V_m/2\)	\(V_m/\sqrt2\)	\(V_m/\sqrt2\)
Form factor	1.57	1.11	1.11
Ripple factor	1.21	0.482	0.482
Efficiency	40.5%	81.2%	81.2%
TUF	0.286	0.573	0.812
PIV	\(V_m\)	\(2V_m\)	\(V_m\)
Ripple frequency	\(f\)	\(2f\)	\(2f\)
Diode drops in path	1	1	2

?Think about it

The bridge beats the centre-tap on PIV and on TUF, and needs no special transformer. Yet a 5 V, 50 A supply will often use the centre-tap circuit anyway. Why?

A way in

Count the diode drops in the current path, and then compute what they cost as a fraction of the output.

Bridge: two diodes in series carry the load current, so about 1.4–2.0 V is lost. At a 5 V output that is up to 40% of the output voltage wasted as heat — 100 W at 50 A.
Centre-tap: one diode in the path, so roughly 0.7–1.0 V. Half the loss: 50 W.

At 5 V output, saving 50 W matters far more than a transformer 40% larger or a diode that must block \(2V_m\) instead of \(V_m\) — because at these voltages \(2V_m\) is still a low number and cheap diodes are plentiful.

Now flip it. At a 400 V output, the same 1.4 V of bridge drop is 0.35% — negligible — while a PIV of \(2V_m \approx 1130\) V would demand expensive high-voltage diodes and a much bigger transformer. The bridge wins decisively.

The transferable lesson: a fixed penalty (a diode drop) matters enormously at low output voltage and not at all at high output voltage. Always ask what a loss is a fraction of, never just how big it is. The same reasoning decided MOSFET versus IGBT in Chapter 5.

Section recap. Both full-wave circuits give \(V_{dc} = 2V_m/\pi\), \(FF = 1.11\), \(RF = 0.482\), \(\eta = 81.2\%\) and ripple at \(2f\). The bridge wins on PIV (\(V_m\)) and TUF (0.812); the centre-tap wins on having only one diode drop in the path — decisive at low output voltage.

Section 6-5

The Effect of Load Type

The output voltage waveform of a full-wave rectifier is fixed by the supply and the diodes — the load cannot change it. What the load changes is the current, and therefore everything that depends on current: the RMS ratings, the losses, and the shape of the current drawn from the supply.

With a resistive load the current is a scaled copy of the voltage, so it is a train of half sines with \(I_{rms} = 1.11\,I_{dc}\). With a highly inductive load — the standard idealisation for a DC motor armature or a filtered supply — the inductance is large enough that the current becomes essentially constant at \(I_o\). This assumption simplifies analysis enormously and is used throughout the rest of Part 2, so it is worth stating its consequences explicitly.

A constant DC load current becomes a square wave on the supply side

Trace the bridge with constant load current \(I_o\). During the positive half cycle \(D_1\) and \(D_4\) conduct and draw \(+I_o\) from the supply; during the negative half cycle \(D_2\) and \(D_3\) conduct and draw \(-I_o\). The supply current is therefore a square wave of amplitude \(I_o\), in phase with the supply voltage. Each diode carries \(I_o\) for exactly half the cycle, so \(I_{D(AV)} = I_o/2\) and \(I_{D(RMS)} = I_o/\sqrt2\) — the ratings needed for the diode selection of Chapter 2.

→Why the "highly inductive load" idealisation is worth so much

Assuming the load current is perfectly constant looks like a convenience. It is actually the assumption that makes all of Part 2 tractable, and it earns its place three times over:

The output current becomes a known constant \(I_o\) instead of an unknown function of time, so diode ratings drop straight out: \(I_o/2\) average, \(I_o/\sqrt2\) RMS.
The supply current becomes a square wave, whose Fourier series you already know: fundamental \(0.9I_o\), odd harmonics falling as \(1/n\). That is the whole of Section 6-7 handed to you.
The output voltage becomes independent of the load, so voltage and current analysis decouple — you can solve each separately instead of together.

And it is not a fiction. A DC motor armature, a well-designed LC filter, or a battery all behave this way to good accuracy. When you meet Chapter 8's controlled rectifiers you will use exactly this assumption again, which is why it is worth internalising here rather than treating it as a trick.

The one thing to stay alert to: at light load the assumption breaks down and the current becomes discontinuous. That regime behaves quite differently, and Chapter 8 treats it explicitly.

Section recap. The load cannot change the output voltage waveform, only the current. With a highly inductive load the current is constant at \(I_o\), so each bridge diode carries \(I_o/2\) average and \(I_o/\sqrt2\) RMS, and the supply current becomes a square wave in phase with the voltage.

Section 6-6

The Capacitor Filter

A ripple factor of 0.482 is far too coarse for electronic loads, so almost every practical rectifier puts a large capacitor across the output. The capacitor charges to near the peak when the supply rises, then supplies the load by discharging while the supply falls away — so the output becomes a nearly flat voltage with a small sawtooth ripple.

Small ripple, bought with very large peak charging currents

If the ripple is small, the capacitor discharges at very nearly the constant load current for very nearly the whole ripple period, so the peak-to-peak ripple follows directly from \(Q = CV\):

🔑

Ripple with a capacitor filter

\[ V_{r(pp)} \approx \frac{I_o}{f_r C}, \qquad V_{dc} \approx V_m - \frac{V_{r(pp)}}{2}, \qquad RF \approx \frac{1}{4\sqrt3\,f_r RC} \]

Here \(f_r\) is the ripple frequency — \(2f\) for a full-wave circuit, \(f\) for half-wave — which is the second reason full-wave rectification is preferred: it halves the required capacitance for a given ripple.

Interactive · capacitor filter

A full-wave bridge on a 230 V, 50 Hz supply, so \(V_m = 325\) V and the ripple frequency is 100 Hz. Increase \(C\) and watch the ripple collapse — then look at what happened to the peak diode current and the conduction angle. That is the bargain nobody mentions.

Capacitance 470 µF

Load current 1.0 A

Ripple V_r(pp)—

Output V_dc—

Peak diode current—

Conduction angle—

—

The cost is severe and often overlooked. Because the diodes now conduct only during the brief interval when the supply exceeds the capacitor voltage — typically 10–20° of each half cycle — the same average current must be delivered in roughly a tenth of the time, so the peak repetitive diode current is an order of magnitude above the DC output current. The supply current becomes a pair of narrow spikes per cycle rather than a square wave, with correspondingly worse harmonics than Section 6-7 describes. And at switch-on, a discharged capacitor is a short circuit: the inrush current is limited only by the source and wiring resistance, and can reach hundreds of amperes unless an NTC thermistor or a soft-start resistor is fitted. Chapter 2's \(I_{FSM}\) rating exists precisely for this moment.

!Common mistake

"The supply delivers 2 A DC, so I'll fit 3 A diodes to be safe."

With a capacitor filter, the average current tells you almost nothing about what the diode experiences. Three separate ratings are in play, and they differ by more than an order of magnitude:

Rating	Typical value here	Why it is that big
Average, \(I_{F(AV)}\)	1 A per diode	Half the 2 A load, as always in a bridge
RMS, \(I_{F(RMS)}\)	4–6 A	The current arrives in narrow tall pulses, so RMS ≫ average
Surge, \(I_{FSM}\)	100 A+	At switch-on a discharged capacitor is a short circuit

A "3 A" diode chosen on the average rating survives the bench and fails on the tenth power-up — an intermittent, infuriating fault.

What real designs do: fit an NTC thermistor or a soft-start resistor in series to limit the inrush (this is the small black bead you will find in every PC power supply), size the diode's RMS rating from the actual conduction angle, and check \(I_{FSM}\) against the worst-case inrush with the capacitor fully discharged and the mains at its peak.

Section recap. \(V_{r(pp)} \approx I_o/(f_rC)\), with \(f_r = 2f\) for a full-wave circuit — the second reason full-wave wins. The price is a conduction angle of 10–20° and peak diode currents an order of magnitude above the DC output, plus an inrush at switch-on limited only by wiring resistance.

Section 6-7

Harmonics and Input Power Factor

A diode rectifier draws a non-sinusoidal current from a sinusoidal supply. Since only the fundamental component of current can produce average power with a sinusoidal voltage, all the harmonic current circulates uselessly — heating cables, distorting the supply voltage, and degrading the power factor. Two separate effects combine:

Power factor of a distorting load

\[ PF = \underbrace{\frac{I_{s1}}{I_s}}_{\text{distortion factor}} \times \underbrace{\cos\phi_1}_{\text{displacement factor}}, \qquad THD = \sqrt{\left(\frac{I_s}{I_{s1}}\right)^2 - 1} \]

For the square-wave current of Section 6-5, Fourier analysis gives a fundamental RMS of \(I_{s1} = (2\sqrt2/\pi)I_o = 0.9\,I_o\), while the total RMS is \(I_s = I_o\). The harmonics are odd only — 3rd, 5th, 7th, and so on — with amplitudes falling as \(1/n\). Substituting:

🔑

The diode bridge's unavoidable power factor

\[ THD = \sqrt{\left(\tfrac{1}{0.9}\right)^2 - 1} = 48.3\%, \qquad PF = 0.9 \times 1 = 0.9 \]

The displacement factor is unity — the square wave is perfectly in phase with the voltage — yet the power factor is still only 0.9. No amount of capacitor correction can fix this, because there is no phase angle to correct; the deficit is entirely distortion. With a capacitor filter the current becomes narrow spikes instead of a square wave, and the power factor falls further, typically to 0.5–0.7.

!Common mistake

"Power factor is 0.9, so I'll add a capacitor bank to correct it."

Wrong Poor power factor means lagging current, and capacitors correct lagging current.

Right Here the displacement factor is already exactly 1 — the square wave is perfectly in phase. The entire 0.9 comes from distortion, and a capacitor cannot remove harmonics.

Power factor has two independent factors, and you must know which one is failing:

Displacement factor \(\cos\phi_1\) — is the fundamental current in phase with the voltage? Fixed by capacitors or by a synchronous condenser. This is the classical power-systems problem.
Distortion factor \(I_{s1}/I_s\) — is the current a sine wave at all? Fixed only by changing the current's shape: an input inductor, a passive LC filter, or an active PFC stage.

It is worse than useless to add capacitors here. A capacitor bank presents a low impedance to harmonic currents, so it can actually draw more harmonic current, and it may resonate with the supply inductance at a harmonic frequency — amplifying exactly what you were trying to reduce. Harmonic-induced capacitor failure is a well-known problem in installations full of rectifier loads.

The right answer is the active power-factor-correction boost stage of Chapter 12, which forces the input current to follow the voltage waveform and reaches a power factor above 0.99.

This is why standards such as IEC 61000-3-2 limit the harmonic current that equipment may inject, and why the power-factor-correction boost stage of Chapter 12 exists. It is also the first appearance of a theme that runs through the rest of the book: a converter's effect on the supply is as much a design specification as its effect on the load.

⚙Where this shows up in a real building

An office floor full of computers, LED lighting and phone chargers is an office floor full of diode bridges with capacitor filters. Three consequences that facilities engineers deal with routinely:

The neutral conductor overheats. In a three-phase supply, the fundamental currents of a balanced load cancel in the neutral — but the third harmonic of each phase is in phase with the others and adds. Neutral current can exceed phase current, in a conductor traditionally sized smaller. Modern installations serving IT loads use a double-sized neutral for exactly this reason.
Transformers derate. Harmonic currents cause extra eddy-current loss, so a transformer feeding rectifier loads must be oversized or specified with a "K-factor" rating.
Voltage distortion spreads. Harmonic current flowing through the supply impedance distorts the voltage everyone else sees — which is why IEC 61000-3-2 limits what each piece of equipment may inject, rather than leaving it to the installation.

The rectifier in Section 6-6 is one of the most-manufactured circuits in history, and its input current is a genuine, ongoing problem for the electricity supply industry.

Section recap. A diode bridge with a constant load current draws a square wave: fundamental \(0.9I_o\), odd harmonics falling as \(1/n\), \(THD = 48.3\%\) and \(PF = 0.9\) with unity displacement. Capacitors cannot fix it, because the deficit is distortion, not phase. With a capacitor filter it falls further, to 0.5–0.7.

Section 6-8

Worked Examples

1 Half-wave rectifier, full parameter set

Problem. A half-wave rectifier supplies a 20 Ω resistor from 230 V, 50 Hz. Find \(V_{dc}\), \(I_{dc}\), \(V_{rms}\), \(I_{rms}\), the powers, efficiency, form and ripple factors, and the PIV.

Solution. First the peak: \(V_m = \sqrt2(230) = 325.3\) V.

Working

\[ V_{dc} = \frac{325.3}{\pi} = 103.5\ \text{V}, \qquad I_{dc} = \frac{103.5}{20} = 5.18\ \text{A} \]

\[ V_{rms} = \frac{325.3}{2} = 162.6\ \text{V}, \qquad I_{rms} = 8.13\ \text{A} \]

\[ P_{dc} = (103.5)(5.18) = 536\ \text{W}, \qquad P_{ac} = (162.6)(8.13) = 1322\ \text{W}, \qquad \eta = 40.5\% \]

\[ FF = \frac{162.6}{103.5} = 1.57, \qquad RF = \sqrt{1.57^2-1} = 1.21, \qquad PIV = 325.3\ \text{V} \]

Note the trap: the resistor actually dissipates 1322 W, not 536 W. The DC power is only the useful part; the remaining 786 W is ripple heating.

2 The same load on a bridge

Problem. Repeat Example 1 for a full-wave bridge, and compare.

Solution.

Working

\[ V_{dc} = \frac{2(325.3)}{\pi} = 207.0\ \text{V}, \qquad I_{dc} = 10.35\ \text{A}, \qquad P_{dc} = 2142\ \text{W} \]

\[ V_{rms} = \frac{325.3}{\sqrt2} = 230\ \text{V}, \qquad I_{rms} = 11.5\ \text{A}, \qquad P_{ac} = 2645\ \text{W}, \qquad \eta = 81.0\% \]

\[ FF = 1.11, \qquad RF = 0.482, \qquad PIV = 325.3\ \text{V} \]

The DC voltage doubled, so the DC power went up four-fold — and the ripple factor improved by a factor of 2.5, with the ripple now at 100 Hz instead of 50 Hz. Three extra diodes buy a great deal.

3 Bridge versus centre-tap: sizing the transformer

Problem. A supply must deliver 100 V DC at 5 A to a resistive load. Compare the transformer secondary rating and diode PIV for a bridge and for a centre-tapped circuit.

Solution. Both need \(V_{dc} = 2V_m/\pi = 100\), so \(V_m = 157.1\) V and each active winding must be \(157.1/\sqrt2 = 111.1\) V RMS. The load draws \(I_{rms} = 1.11(5) = 5.55\) A. The DC power is 500 W.

Working — bridge

\[ \text{VA} = (111.1)(5.55) = 617\ \text{VA}, \qquad TUF = \frac{500}{617} = 0.81, \qquad PIV = 157\ \text{V} \]

Working — centre-tap

\[ \text{Total secondary} = 2(111.1) = 222.2\ \text{V}, \qquad I_{per\ half} = \frac{5.55}{\sqrt2} = 3.93\ \text{A} \]

\[ \text{VA} = (222.2)(3.93) = 873\ \text{VA}, \qquad TUF = \frac{500}{873} = 0.573, \qquad PIV = 314\ \text{V} \]

The centre-tap needs a transformer 40% larger and diodes rated for twice the voltage, to save two diodes. At 100 V that is a poor bargain; at 5 V output, where saving one forward drop recovers 10% of the output voltage, the arithmetic reverses — which is exactly why low-voltage supplies still use centre-tapped secondaries.

4 Capacitor filter design

Problem. A bridge on 230 V, 50 Hz feeds a 500 Ω load through a 470 µF capacitor. Find the ripple, the DC output, and the ripple factor. Compare with the unfiltered case.

Solution. With \(V_m = 325.3\) V and \(f_r = 100\) Hz, estimate \(I_o\), find the ripple, then correct:

Working

\[ I_o \approx \frac{325}{500} = 0.65\ \text{A} \;\Rightarrow\; V_{r(pp)} = \frac{0.65}{(100)(470\times10^{-6})} = 13.8\ \text{V} \]

\[ V_{dc} \approx 325.3 - 6.9 = 318.4\ \text{V} \;\Rightarrow\; I_o = 0.637\ \text{A} \;\Rightarrow\; V_{r(pp)} = 13.6\ \text{V} \]

\[ RF \approx \frac{V_{r(pp)}}{2\sqrt3\,V_{dc}} = \frac{13.6}{(3.46)(318.4)} = 0.0123 = 1.2\% \]

The ripple factor has fallen from 0.482 to 0.012 — a factor of forty — and the DC output has risen from 207 V to 318 V, because the output now sits near the peak rather than the average. One capacitor transforms the circuit.

5 The price of the capacitor: peak and inrush currents

Problem. For the circuit of Example 4, find the conduction angle and estimate the peak repetitive diode current. Then find the inrush current at switch-on if the total series resistance is 1 Ω, and size a soft-start resistor to hold it to 40 A.

Solution. The diodes conduct while the supply exceeds the capacitor's minimum voltage of \(325.3 - 13.6 = 311.7\) V:

Working

\[ \cos\theta = \frac{311.7}{325.3} = 0.958 \;\Rightarrow\; \theta = 16.6^\circ \;\Rightarrow\; t_c = \frac{16.6}{180}(10\ \text{ms}) = 0.92\ \text{ms} \]

\[ Q = I_o \times \tfrac{T}{2} = (0.637)(10\ \text{ms}) = 6.37\ \text{mC} \;\Rightarrow\; I_{peak} \approx \frac{2Q}{t_c} = 13.8\ \text{A} \]

\[ I_{inrush} = \frac{V_m}{R} = \frac{325.3}{1} = 325\ \text{A}, \qquad R_{soft} = \frac{325.3}{40} = 8.1\ \Omega \]

A 0.64 A DC output demands diodes that survive 14 A repetitively and 325 A once — a peak-to-average ratio above 20, and an inrush 500 times the load current. This is why the \(I_{FSM}\) and \(I^2t\) ratings of Chapter 2 matter more than the average rating in capacitor-filtered supplies.

6 Harmonics with an inductive load

Problem. A bridge on 230 V feeds a highly inductive load drawing a constant 10 A. Find the supply current RMS and fundamental, the THD, the power factor, and verify the input power against the output.

Solution. The supply current is a square wave of amplitude 10 A:

Working

\[ I_s = 10\ \text{A}, \qquad I_{s1} = \frac{2\sqrt2}{\pi}(10) = 9.0\ \text{A} \]

\[ THD = \sqrt{\left(\frac{10}{9}\right)^2 - 1} = 48.4\%, \qquad PF = \frac{9}{10}\times 1 = 0.9 \]

\[ P_{in} = V_sI_{s1}\cos\phi_1 = (230)(9)(1) = 2070\ \text{W}, \qquad P_{out} = V_{dc}I_o = (207)(10) = 2070\ \text{W}\ \checkmark \]

The two agree exactly, which confirms the key point: only the fundamental current transfers power. The other 4.4 A RMS of harmonic current does nothing but heat the wiring and distort the supply voltage for everyone else on the feeder.

Review

Summary & Formula Sheet

✓ Five parameters

\(FF = V_{rms}/V_{dc}\), \(RF = \sqrt{FF^2-1}\), \(\eta = P_{dc}/P_{ac}\), TUF, and PIV grade every rectifier.

✓ Half-wave

\(V_{dc} = V_m/\pi\), \(V_{rms} = V_m/2\). \(RF = 1.21\), \(\eta = 40.5\%\), \(TUF = 0.286\). Teaching circuit only.

✓ Inductive load

Conduction extends past \(\pi\) to the extinction angle \(\beta\), lowering \(V_{dc}\). A freewheeling diode restores it and smooths the current.

✓ Full-wave

\(V_{dc} = 2V_m/\pi\), \(RF = 0.482\), \(\eta = 81.2\%\), ripple at \(2f\). Bridge: PIV \(V_m\), TUF 0.812. Centre-tap: PIV \(2V_m\), TUF 0.573.

✓ Capacitor filter

\(V_{r(pp)} = I_o/(f_rC)\); output rises toward \(V_m\). Ripple falls 40×, peak diode current rises 20×, and inrush must be limited.

✓ Supply side

Square-wave current gives 48.3% THD and \(PF = 0.9\) at unity displacement. Distortion, not phase, is the problem.

Formula sheet · Chapter 6

Form and ripple factoreverything else follows from these two

\( FF = \dfrac{V_{rms}}{V_{dc}}, \qquad RF = \sqrt{FF^2 - 1} \)

Half-wave, R load\(FF=1.57\), \(RF=1.21\), \(\eta=40.5\%\)

\( V_{dc} = \dfrac{V_m}{\pi}, \qquad V_{rms} = \dfrac{V_m}{2} \)

Full-wave, R load\(FF=1.11\), \(RF=0.482\), \(\eta=81.2\%\)

\( V_{dc} = \dfrac{2V_m}{\pi}, \qquad V_{rms} = \dfrac{V_m}{\sqrt2} \)

Extinction anglehalf-wave, RL load — solve numerically

\( \sin(\beta-\phi) + \sin\phi\;e^{-\beta/(\omega\tau)} = 0, \quad \phi = \tan^{-1}\dfrac{\omega L}{R} \)

Bridge diode ratingsconstant load current \(I_o\)

\( I_{D(AV)} = \dfrac{I_o}{2}, \qquad I_{D(RMS)} = \dfrac{I_o}{\sqrt2} \)

Capacitor filter ripple\(f_r = 2f\) full-wave, \(f\) half-wave

\( V_{r(pp)} \approx \dfrac{I_o}{f_r C}, \qquad V_{dc} \approx V_m - \dfrac{V_{r(pp)}}{2} \)

Ripple factor with filterthe design equation for \(C\)

\( RF \approx \dfrac{1}{4\sqrt3\,f_r R C} \)

Power factortwo independent factors

\( PF = \dfrac{I_{s1}}{I_s}\cos\phi_1 \)

Total harmonic distortionsquare wave: 48.3%

\( THD = \sqrt{\left(\dfrac{I_s}{I_{s1}}\right)^2 - 1} \)

Square-wave fundamentalbridge with constant load current

\( I_{s1} = \dfrac{2\sqrt2}{\pi}I_o = 0.9\,I_o \)

Key terms

Form factor, \(FF\): \(V_{rms}/V_{dc}\). How far the output is from pure DC. Ideal is 1. Also the correction factor when computing power from \(V_{dc}\).
Ripple factor, \(RF\): AC content relative to DC, \(\sqrt{FF^2-1}\). Above 1 means more ripple than signal.
Transformer utilisation factor: DC power divided by the transformer VA rating. A low TUF means an oversized, expensive transformer for a given DC output.
Peak inverse voltage: The largest reverse voltage a diode must block. \(V_m\) for half-wave and bridge; \(2V_m\) for centre-tap.
Extinction angle, \(\beta\): The angle at which current in an RL load finally reaches zero. Beyond \(\pi\), so output goes negative and \(V_{dc}\) falls.
Freewheeling diode: A diode across an inductive load giving current a local path. Restores \(V_{dc}\), smooths current, removes the \(L\,di/dt\) spike.
Highly inductive load: The standard idealisation where load current is constant at \(I_o\). Decouples voltage and current analysis throughout Part 2.
Conduction angle: The fraction of each half cycle a diode conducts. With a capacitor filter it falls to 10–20°, which is why peak currents are so large.
Inrush current: The surge at switch-on into a discharged filter capacitor, limited only by source and wiring resistance. Sized against \(I_{FSM}\).
Distortion factor: \(I_{s1}/I_s\). The part of power factor caused by the current's shape rather than its phase. Capacitors cannot correct it.

Self-assessment

Test Yourself

Six questions on the first converter of the course. If these come easily, Chapters 7 to 10 are variations on the same method.

Chapter 6 · six questions answers hidden until you ask

A full-wave bridge on a 230 V, 50 Hz supply feeds a 20 Ω resistor with no filter. Find \(V_{dc}\), the true load power, and the power a careless \(V_{dc}^2/R\) would predict.

\(V_m = \sqrt2(230) = 325\) V.

\(V_{dc} = 2V_m/\pi = 207\) V
\(V_{rms} = V_m/\sqrt2 = 230\) V (unchanged by rectification — a useful check)
True power \(= V_{rms}^2/R = 230^2/20 = \) 2645 W
Careless answer \(= V_{dc}^2/R = 207^2/20 = 2142\) W

The ratio is \(2645/2142 = 1.235 = FF^2 = 1.11^2\) — exactly as the form factor predicts.

Note the elegant check: a full-wave rectifier does not change the RMS at all, because squaring destroys the sign. The RMS in equals the RMS out. If a full-wave problem gives you a different \(V_{rms}\), you have made an arithmetic slip.

Why does the ripple frequency of a full-wave rectifier being \(2f\) rather than \(f\) matter twice over?

First, less ripple to begin with. Two half sines per period instead of one gives \(RF = 0.482\) against 1.21 — two and a half times better before any filtering.

Second, the filter is half the size. From \(V_{r(pp)} = I_o/(f_rC)\), doubling \(f_r\) halves the capacitance needed for the same ripple. Electrolytic capacitors are among the largest, most expensive and shortest-lived components in a power supply, so halving them is a real saving in cost, volume and reliability.

And it compounds: the two effects multiply. For a given output ripple specification, a full-wave circuit needs roughly a fifth of the capacitance of a half-wave one.

This principle scales. A three-phase bridge (Chapter 7) has a ripple frequency of \(6f\) and a ripple factor of only 4%, which is why large DC supplies are three-phase whenever three phases are available.

A 5 V, 40 A supply is being designed. Would you use a bridge or a centre-tap rectifier, and why?

Centre-tap — the diode drops dominate everything at this output voltage.

Bridge: two diodes in series in the path. At roughly 0.8 V each that is 1.6 V lost, so the transformer must produce 6.6 V to deliver 5 V, and \(1.6 \times 40 = \) 64 W is dissipated in the diodes — against 200 W delivered. Efficiency of the rectification stage alone is only 76%.
Centre-tap: one diode in the path. 0.8 V lost, 32 W dissipated, 86% — and half the heatsinking.

The costs you accept: a PIV of \(2V_m\) (trivial at these voltages — a 40 V diode instead of a 20 V one) and a TUF of 0.573 instead of 0.812, meaning a transformer about 40% larger. At 5 V that transformer penalty is far cheaper than 32 W of continuous heat.

In practice a modern 5 V/40 A supply goes further still and uses synchronous rectification — MOSFETs in place of diodes, with \(I R_{DS(on)}\) of perhaps 0.1 V instead of 0.8 V. The Chapter 5 insight applies directly: the MOSFET's drop falls with current while a diode's does not.

A capacitor-filtered supply must deliver 2 A with under 5 V of ripple from a full-wave bridge. Find \(C\), then comment on the diode.

\(C = I_o/(f_r V_{r(pp)}) = 2/(100 \times 5) = \) 4000 µF — so specify 4700 µF, the nearest standard value.

Now the diode. With \(V_m = 325\) V and \(V_{r(pp)} = 5\) V, the conduction angle is \(\cos^{-1}(1 - 5/325) = 10.0°\), or 0.555 ms. Charge balance over each ripple period gives a peak diode current of roughly \(2I_o/(f_r\,\Delta t) = 2(2)/(100 \times 5.55\times10^{-4}) \approx \) 72 A.

So a supply rated "2 A" needs diodes that tolerate a 72 A repetitive peak — a crest factor of 36 — and an \(I_{FSM}\) rating well above the switch-on inrush into 4700 µF, which with a few hundred milliohms of source resistance could reach several hundred amperes.

The design conclusion: tighter ripple specifications are expensive in ways the ripple formula does not reveal. If 5 V of ripple is not truly required, relaxing it to 15 V cuts the capacitor to 1500 µF and reduces the peak current by about 40%. Ask what the ripple specification is actually for before designing to it.

A rectifier load has a measured power factor of 0.65. Your colleague proposes a capacitor bank. What do you say?

Ask first: is the deficit displacement or distortion? For a diode rectifier it is almost entirely distortion, and a capacitor bank cannot help.

A capacitor supplies leading reactive current to cancel a lagging fundamental. Here the fundamental current is already in phase — there is nothing to cancel. The 0.65 comes from the harmonic content of a spiky, non-sinusoidal current.

Worse, the capacitor bank is actively harmful here:

Its impedance falls as \(1/(n\omega C)\), so it is a low-impedance path for harmonics and will draw more of them, heating and eventually failing.
It can resonate with the supply inductance at or near a harmonic frequency, amplifying that harmonic instead of damping it. Harmonic resonance in installations full of rectifier loads is a well-documented failure mode.

What actually works, in increasing order of cost and effectiveness: an AC line inductor or DC-link choke (widens the conduction angle, PF up to about 0.85); a passive tuned filter; a 12-pulse arrangement if it is a three-phase load; or an active PFC boost stage (Chapter 12), which reaches above 0.99 and is now standard in equipment above 75 W because IEC 61000-3-2 effectively requires it.

You are told a rectifier has \(TUF = 0.286\). What does that mean in money, and which circuit is it?

It is the half-wave rectifier, and it means the transformer must be rated at \(1/0.286 = 3.5\) times the DC power it delivers.

To supply 100 W of DC you must buy and install a 350 VA transformer — three and a half times the iron, copper, weight, cost and volume.

Why it is so bad has two independent causes:

Half the time is wasted. The secondary carries current for only one half cycle, so its RMS rating is set by a duty it only half uses.
DC magnetisation. Current flows in one direction only, so the secondary carries a DC component that pushes the core toward saturation. The core must be oversized — or gapped — purely to tolerate a current component that does no useful work.

Compare the bridge at \(TUF = 0.812\): a 123 VA transformer for the same 100 W. That single number, more than the ripple, is why half-wave rectification is confined to trickle chargers and signal-level circuits.

Practice

Problems

Two questions fix every formula you will need, so ask them before anything else:

Which circuit? Half-wave, centre-tap, or bridge — this fixes \(V_{dc}\), \(V_{rms}\), PIV, TUF and the ripple frequency.
Which load? R, RL, RL with a freewheeling diode, constant current, or capacitor-filtered — this fixes the current waveform and therefore all the ratings.

Problems 1–5 are direct application; 6–10 need judgement; the remainder are design questions worth discussing in a tutorial.

State the ripple frequency and PIV for each of: half-wave, centre-tap, and bridge, on a 50 Hz supply of peak \(V_m\).
Explain why an inductive load reduces the average output of a half-wave rectifier, and what the freewheeling diode does about it.
A half-wave rectifier feeds 30 Ω from 110 V, 50 Hz. Find \(V_{dc}\), \(I_{dc}\), \(V_{rms}\), \(P_{dc}\), \(P_{ac}\), \(\eta\) and the PIV.
Repeat Problem 3 for a bridge and state the ratio of DC powers.
A bridge feeds 25 Ω from 240 V, 50 Hz. Find the form and ripple factors, and the average and RMS current in each diode.
A 400 W, 48 V DC supply is to be built. Compare the transformer VA and diode PIV for a bridge and a centre-tapped design, and recommend one with reasons.
A bridge with a capacitor filter supplies 24 V DC at 2 A with 1 V peak-to-peak ripple from a 50 Hz supply. Find the required capacitance.
For Problem 7, find the ripple factor and compare with the unfiltered value.
A capacitor-filtered bridge delivers 320 V at 1 A with 15 V of ripple. Estimate the conduction angle and the peak repetitive diode current.
For Problem 9, the total series resistance at switch-on is 0.8 Ω. Find the inrush current and size an NTC to limit it to 50 A.
A bridge feeds a highly inductive load drawing 25 A from a 415 V supply. Find the supply current RMS and fundamental, the THD, the power factor, and the input real power.
An engineer measures a power factor of 0.62 on a capacitor-filtered bridge and proposes fitting a power-factor-correction capacitor across the input. Explain why this will not work, what the measurement actually indicates, and what circuit change would genuinely improve it.

Tip: nearly every rectifier problem is solved in the same four steps. Find \(V_m\) from the RMS supply. Write \(V_{dc}\) and \(V_{rms}\) from the circuit type. Divide by \(R\) — or read \(I_o\) directly for a constant-current load — to get the currents. Then assemble whatever performance figures are asked. The only genuinely new judgement is deciding whether the load is resistive (current follows voltage) or highly inductive (current is flat), because that single choice determines the shape of every current in the circuit.