Chapter 7: Three-Phase Uncontrolled Rectifiers — Power Electronics

Where this sits

Part 2 · AC–DC Converters
Chapter 7 of 30

You should already know

The five performance parameters, the constant-load-current idealisation and input harmonics from Chapter 6, and diode ratings from Chapter 2.

By the end you can

Derive \(V_{dc}\) for any pulse number, size the diodes and transformer for a six-pulse bridge, and explain why 12-pulse exists.

Time

≈ 60 min reading · ≈ 55 min problems

i What you'll learn

Why three-phase supplies deliver constant instantaneous power, and why that single fact makes every performance figure better.
The three-pulse (half-wave) rectifier — \( V_{dc} = 1.17\,V_{ph} \) — and the DC magnetisation problem that limits it.
The six-pulse bridge — \( V_{dc} = 1.35\,V_{L} \) — the most-built converter circuit in industry.
The single \(p\)-pulse formula that reproduces every result in Chapters 6 and 7 from one line of algebra.
Diode and transformer ratings for 120° conduction: \( I_{D(AV)} = I_o/3 \), \( I_{D(RMS)} = I_o/\sqrt3 \).
Why the supply current has no triplen harmonics, a THD of 31%, and a power factor of 0.955.
How 12-pulse connections cancel the 5th and 7th harmonics, and where that matters.

Section 7-1

Why Three Phase Changes Everything

Chapter 6 ended with a single-phase bridge that was 81.2% efficient, had a 48% ripple factor, and pushed 48% current distortion back into the supply. Those numbers are not a failure of the circuit — they are a consequence of the source.

Here is the reason, and it is worth stating before any circuit appears.

→Single-phase power pulsates; three-phase power does not

Take a single-phase supply feeding a resistive load. The instantaneous power is

Single-phase instantaneous power

\[ p(t) = v i = V_m I_m \sin^2\omega t = \frac{V_m I_m}{2}\bigl(1 - \cos 2\omega t\bigr) \]

It has an average, and a component at twice the supply frequency that swings all the way down to zero a hundred times a second. There is no arrangement of diodes that can extract steady DC from a source that keeps stopping.

Now take three balanced phases. Add their instantaneous powers:

Three-phase instantaneous power

\[ p(t) = \frac{V_mI_m}{2}\sum_{k=0}^{2}\Bigl[1 - \cos\bigl(2\omega t - \tfrac{4\pi k}{3}\bigr)\Bigr] = \frac{3V_mI_m}{2} = \text{constant} \]

The three double-frequency terms are spaced 120° apart in \(2\omega t\), so they sum to zero exactly. A balanced three-phase supply delivers perfectly constant instantaneous power.

That is the whole chapter in one sentence. Every improvement below — smaller ripple, a better transformer utilisation factor, lower harmonics, no DC magnetisation — is a consequence of feeding the rectifier from a source that never lets go.

What changes when you move from one phase to three
Property	1-φ bridge (Ch. 6)	3-φ bridge (this chapter)
Pulse number \(p\)	2	6
Ripple frequency	\(2f\) = 100 Hz	\(6f\) = 300 Hz
Ripple factor	48.2%	4.2%
Rectification efficiency	81.2%	99.8%
Transformer utilisation	0.812	0.954
Input current THD	48.3%	31.1%
Input power factor	0.900	0.955
Practical power range	Watts to a few kW	Kilowatts to megawatts

Section recap. A balanced three-phase supply delivers constant instantaneous power, while a single-phase supply's power falls to zero twice per cycle. Every advantage in this chapter — an order of magnitude less ripple, a far better transformer utilisation factor, lower distortion — follows from that one fact.

Section 7-2

The Three-Pulse (Half-Wave) Rectifier

Three diodes, their cathodes joined to form the positive output terminal, each anode connected to one phase of a star-connected secondary. The load returns to the neutral.

The rule that governs it is simple enough to state in one line, and it is the rule behind every circuit in this chapter:

🔑

The common-cathode rule

In a common-cathode group, the diode connected to the most positive phase conducts; the others are reverse biased by that same voltage.

Conduction therefore transfers from one diode to the next at each crossover of the phase voltages — at 30°, 150° and 270° on the usual reference — so each diode conducts for exactly 120°, and the output follows the upper envelope of the three phase voltages.

Three diodes, common cathode — the most positive phase wins

The output is the upper envelope — and it never reaches zero

Deriving the average, in three steps

Pick the interval and centre it on the peak Each diode conducts for \(2\pi/3\). Take the interval symmetric about the peak of its phase, so the output is \(V_m\cos\theta\) for \(-\pi/3 \le \theta \le \pi/3\). Centring on the peak turns an awkward integral into a symmetric one.
Average over that interval

Working

\[ V_{dc} = \frac{1}{2\pi/3}\int_{-\pi/3}^{\pi/3} V_m\cos\theta\,d\theta = \frac{3V_m}{2\pi}\bigl[\sin\theta\bigr]_{-\pi/3}^{\pi/3} = \frac{3V_m}{2\pi}\bigl(2\sin\tfrac{\pi}{3}\bigr) \]
Collect

Three-pulse average output

\[ V_{dc} = \frac{3\sqrt3}{2\pi}V_m = 0.827\,V_m = 1.17\,V_{ph} \]

where \(V_{ph}\) is the RMS phase voltage. The same integral with \(\cos^2\theta\) gives \( V_{rms} = 0.8407\,V_m \).

From those two numbers everything else follows exactly as in Chapter 6: \( FF = 1.017 \), \( RF = 18.3\% \), \( \eta = 96.8\% \), ripple frequency \(3f\), and \( \mathrm{PIV} = \sqrt3\,V_m \) — because a blocking diode sees the full line voltage, not the phase voltage.

!Common mistake

"Each diode blocks the phase voltage, so PIV = \(V_m\)."

Wrong The diode is connected to one phase, so the worst reverse voltage is that phase's peak.

Right Its cathode is tied to whichever phase is currently conducting, so it sees the difference between two phases — a line voltage. \(\mathrm{PIV} = \sqrt3\,V_m = 1.73\,V_m\).

Check it concretely. When phase a is at its peak \(+V_m\), \(D_1\) conducts and pulls the cathode bus to \(+V_m\). At that instant phase b sits at \(V_m\cos 120° = -0.5V_m\), so \(D_2\) has \(-0.5V_m\) on its anode and \(+V_m\) on its cathode — a reverse bias of \(1.5V_m\). Sweep through the cycle and the worst case reaches \(\sqrt3\,V_m\).

Under-rating the diode by a factor of 1.73 destroys it on the first cycle. Always ask what is on both terminals, not just the one you connected.

!Common mistake

"The three-pulse rectifier is efficient, so it should be the standard industrial circuit."

It is not, and the reason is in the transformer rather than the diodes. Each secondary winding carries current in one direction only, for 120° of every cycle.

That unidirectional current has a DC component of \(I_o/3\).
A DC component in a winding produces a steady magnetising force that pushes the core toward saturation.
The core must therefore be oversized, or gapped, purely to tolerate a current component that transfers no power.

The result is a transformer utilisation factor of only 0.664 — you buy 1.5 kVA of transformer for every kW of DC. It also needs a neutral, which many industrial supplies do not bring out.

So the three-pulse circuit survives mainly as a teaching step and as the building block of larger connections. The six-pulse bridge of the next section has no DC magnetisation at all, and that is the circuit industry actually builds.

Section recap. Common cathode means the most positive phase conducts, so each diode carries 120° and the output is the upper envelope: \(V_{dc} = 3\sqrt3 V_m/2\pi = 1.17\,V_{ph}\), \(RF = 18.3\%\), ripple at \(3f\), \(\mathrm{PIV} = \sqrt3 V_m\). Its weakness is DC magnetisation of the transformer, giving a TUF of only 0.664.

Section 7-3

The Six-Pulse Bridge

Add a second group of three diodes with their anodes joined, forming the negative terminal, and you have the three-phase bridge — six diodes, no neutral required, and the most-manufactured power converter circuit in the world.

Two independent rules now operate at once:

The upper (common-cathode) group connects the load's positive terminal to the most positive phase.
The lower (common-anode) group connects the load's negative terminal to the most negative phase.

So the load always sees the largest line-to-line voltage available at that instant. That single sentence is the whole analysis.

Six diodes, two groups — no neutral, no DC magnetisation

Six shallow humps — the output barely moves

The derivation is the three-pulse one with two changes: the interval is now \(\pi/3\) wide instead of \(2\pi/3\), and the quantity being averaged is the line voltage of peak \(V_{mL} = \sqrt3\,V_m\).

🔑

Six-pulse bridge — the number to memorise

\[ V_{dc} = \frac{1}{\pi/3}\int_{-\pi/6}^{\pi/6}V_{mL}\cos\theta\,d\theta = \frac{3}{\pi}V_{mL} = 1.35\,V_{L} = 2.34\,V_{ph} \]

With \(V_{rms} = 0.9558\,V_{mL}\), the performance figures are \( FF = 1.0009 \), \( RF = 4.2\% \), \( \eta = 99.8\% \), \( TUF = 0.954 \), ripple frequency \(6f\), and \( \mathrm{PIV} = V_{mL} \). On a 415 V, 50 Hz line that is 560 V DC with 300 Hz ripple of about 24 V peak-to-peak — usable as a DC link with no filtering at all.

!Common mistake

Mixing up phase and line voltage in \(V_{dc} = 1.35\,V_L\).

This costs more marks than anything else in the chapter, because both forms of the formula are correct and they differ by \(\sqrt3\):

If you are given…	Use	On a 415 V system
RMS line voltage \(V_L\)	\(V_{dc} = 1.35\,V_L\)	\(1.35 \times 415 = 560\) V
RMS phase voltage \(V_{ph}\)	\(V_{dc} = 2.34\,V_{ph}\)	\(2.34 \times 240 = 560\) V
Peak line voltage \(V_{mL}\)	\(V_{dc} = 0.955\,V_{mL}\)	\(0.955 \times 587 = 560\) V

The habit that prevents it: write down which voltage you have before choosing the coefficient, and sanity-check the answer. A three-phase bridge on a 415 V line gives about 560 V — if your answer is near 325 V you have used phase voltage with the line coefficient.

A useful anchor: \(V_{dc}\) is always a little below the peak line voltage, because the output is the line-voltage envelope dipping only to \(\cos 30°\) of its peak.

Which diodes conduct, and when

Conduction transfers in the upper group every 120°, and in the lower group every 120°, but the two groups are offset by 60°. So something changes every 60°, which is why there are six pulses per cycle. The standard numbering \(D_1 \dots D_6\) is chosen so that the devices fire in numerical order:

Conduction sequence of a three-phase bridge
Interval (ωt)	Upper device	Lower device	Output voltage
30°–90°	D₁ (phase a)	D₆ (phase b)	\(v_{ab}\)
90°–150°	D₁ (a)	D₂ (c)	\(v_{ac}\)
150°–210°	D₃ (b)	D₂ (c)	\(v_{bc}\)
210°–270°	D₃ (b)	D₄ (a)	\(v_{ba}\)
270°–330°	D₅ (c)	D₄ (a)	\(v_{ca}\)
330°–30°	D₅ (c)	D₆ (b)	\(v_{cb}\)

Read the two device columns downward and you will see each diode appear in two consecutive rows — that is its 120° of conduction. Read across and exactly two devices conduct at any instant, one from each group, never two from the same leg.

!Common mistake

"Six pulses per cycle, so each diode conducts 60°."

Wrong 360° ÷ 6 pulses = 60° each.

Right Each pair combination lasts 60°, but each individual diode stays on through two consecutive combinations — 120°.

The distinction is not academic; it changes every current rating by \(\sqrt2\). With constant load current \(I_o\):

Correct (120°): \(I_{D(AV)} = I_o/3\), \(I_{D(RMS)} = I_o/\sqrt3 = 0.577I_o\).
Wrong (60°): \(I_o/6\) and \(0.408I_o\) — an under-estimate that halves the calculated loss.

The check that settles it: six diodes, each on for 120°, gives \(6 \times 120° = 720°\) of device-conduction per 360° cycle — exactly two devices on at all times, which is what a bridge requires. Sixty degrees each would give only one, and the current would have nowhere to go.

Section recap. The upper group picks the most positive phase and the lower group the most negative, so the load sees the largest available line voltage: \(V_{dc} = 3V_{mL}/\pi = 1.35\,V_L = 2.34\,V_{ph}\). Ripple is 4.2% at \(6f\), \(\mathrm{PIV} = V_{mL}\), and each diode conducts 120° — two devices on at every instant.

Section 7-4

One Formula for Every Rectifier

Chapters 6 and 7 have now produced five sets of numbers — half-wave, single-phase full-wave, three-pulse, six-pulse — and they look like four separate results to memorise. They are not. Every one of them is the same integral with a different pulse number \(p\).

The output of any \(p\)-pulse rectifier is the envelope of \(p\) sinusoids spaced \(2\pi/p\) apart. Average over one \(2\pi/p\) interval centred on a peak, exactly as in Section 7-2, and you get:

🔑

The p-pulse rectifier — one line that contains the whole of Part 2 so far

\[ V_{dc} = \frac{p}{\pi}V_m\sin\frac{\pi}{p}, \qquad V_{rms} = V_m\sqrt{\frac{1}{2} + \frac{p}{4\pi}\sin\frac{2\pi}{p}} \]

where \(V_m\) is the peak of the constituent sinusoid — the phase peak for a half-wave connection, the line peak for a bridge. Ripple frequency is \(pf\). As \(p\) grows, \(\sin(\pi/p) \to \pi/p\) and \(V_{dc} \to V_m\): the output approaches the peak, and the ripple vanishes.

Every rectifier in Part 2, from one formula
Pulses \(p\)	Circuit	Ratio \(V_{dc}/V_m\)	Ratio \(V_{rms}/V_m\)	Ripple factor	Ripple freq.
2	Single-phase bridge or centre-tap	0.637	0.707	48.2%	\(2f\)
3	Three-phase half-wave	0.827	0.841	18.3%	\(3f\)
6	Three-phase bridge	0.955	0.956	4.2%	\(6f\)
12	Twelve-pulse (Section 7-6)	0.989	0.989	1.03%	\(12f\)
24	Twenty-four-pulse (HVDC)	0.997	0.997	0.26%	\(24f\)

Interactive · pulse number

Move the pulse number and watch the output climb toward the peak while the ripple collapses. The faint curves are the constituent sinusoids; the bold curve is their envelope, which is what the load actually sees.

Pulse number \(p\) 6

Pulse number6

V_dc / V_m0.955

Ripple factor4.2%

Ripple at 50 Hz300 Hz

—

?Think about it

Going from \(p=2\) to \(p=6\) cuts the ripple by a factor of eleven. Going from \(p=12\) to \(p=24\) cuts it by only a factor of four, and almost nobody builds a 24-pulse rectifier. If more pulses are strictly better, why does the industry stop at six or twelve?

A way in

Look at what has to be built to raise \(p\), rather than at the ripple figure.

\(p = 6\) needs an ordinary three-phase supply and six diodes. No transformer at all if the line voltage suits the DC requirement — which is why every small drive is six-pulse.
\(p = 12\) needs a transformer with two secondaries, one star and one delta, to create the 30° phase shift — plus twelve diodes. The transformer is now a significant part of the cost.
\(p = 24\) needs four secondaries at 15° intervals, which requires zig-zag or extended-delta windings, precise phase-angle accuracy, and careful impedance matching between the groups. Any imbalance between them lets the harmonics you paid to cancel reappear.

Meanwhile the benefit is shrinking. From \(p=6\) to \(p=12\) the ripple falls by 3.2 percentage points; from 12 to 24 it falls by only 0.77.

The engineering conclusion: raise \(p\) while it is cheap and the gains are large, then switch to a different tool. Above 12-pulse it is almost always better to fit a filter, or to use an active front end (Chapter 12), than to keep adding transformer windings. Recognising when to stop pushing a technique and change approach is a large part of design judgement.

Section recap. Every rectifier in Part 2 obeys \(V_{dc} = (p/\pi)V_m\sin(\pi/p)\) with ripple at \(pf\). Learn the formula, not the four special cases — and remember that \(V_m\) means the peak of whatever sinusoid the circuit is taking the envelope of.

Section 7-5

Diode, Transformer and Load Ratings

Use the constant-load-current idealisation from Chapter 6 — a highly inductive load or a well-filtered DC link, so \(i_o = I_o\). Every rating then follows from a single question: what fraction of the cycle does this component carry that current?

Ratings for a three-phase bridge with constant load current \(I_o\)
Quantity	Value	Where it comes from
Diode average current	\(I_{D(AV)} = I_o/3\)	120° of 360°
Diode RMS current	\(I_{D(RMS)} = I_o/\sqrt3 = 0.577I_o\)	\(I_o\sqrt{1/3}\)
Diode PIV	\(V_{mL} = \sqrt3\,V_m\)	Full line voltage across the off device
Supply line RMS current	\(I_s = I_o\sqrt{2/3} = 0.816I_o\)	\(+I_o\) for 120°, \(-I_o\) for 120°, zero for 120°
Supply fundamental RMS	\(I_{s1} = \dfrac{\sqrt6}{\pi}I_o = 0.780I_o\)	Fourier series of the 120° block
Transformer VA	\(\sqrt3\,V_L I_s\)	Three-phase apparent power
On-state voltage loss	\(2V_F\)	Two diodes always in series

!Common mistake

"The output is 560 V, so I'll ignore the diode drops."

Usually right — but check the fraction before you decide, exactly as in the Chapter 6 bridge-versus-centre-tap question.

At 560 V output, two diode drops of 1.0 V each cost \(2/560 = 0.36\%\). Ignore them.
At 24 V output (a battery charger from a step-down transformer), the same 2 V is 8.3% of the output — and at 200 A that is 400 W of heat needing real heatsinking.

Two more things the ideal analysis hides at high current: the transformer's winding resistance and leakage reactance also drop volts, and the leakage reactance causes commutation overlap, which reduces \(V_{dc}\) further. That effect is important enough to have its own chapter — it is the first topic of Chapter 10.

Section recap. With constant \(I_o\): each diode carries \(I_o/3\) average and \(I_o/\sqrt3\) RMS for its 120°, blocks \(V_{mL}\), and two are always in series so the path loses \(2V_F\). The supply line carries \(0.816I_o\) RMS in 120° blocks.

Section 7-6

The Supply Current and Its Harmonics

Trace one supply line through a full cycle. Phase a is connected to the positive rail while \(D_1\) conducts (120°), to the negative rail while \(D_4\) conducts (120°), and to nothing at all for the remaining 120°. The line current is therefore a quasi-square wave: \(+I_o\), zero, \(-I_o\), zero.

The line current — no phase shift, but plenty of distortion

Fourier analysis of that block gives the numbers that matter:

🔑

Supply-side performance of a six-pulse bridge

\[ I_s = \sqrt{\tfrac23}\,I_o, \qquad I_{s1} = \frac{\sqrt6}{\pi}I_o, \qquad \mathrm{PF} = \frac{I_{s1}}{I_s} = \frac{3}{\pi} = 0.955 \]

The displacement factor is again unity — the current block is centred on the voltage — so the entire deficit is distortion, giving \( \mathrm{THD} = 31.1\% \). Compare the single-phase bridge of Chapter 6 at 48.3% THD and 0.900 power factor: three phases improve matters substantially, but do not fix them.

→Why the triplen harmonics vanish

The single-phase bridge's biggest harmonic was the third. In the three-phase line current it is completely absent — the spectrum starts at the fifth. That is not an accident of the waveform shape; it is a consequence of three-phase symmetry, and it is worth understanding because it explains an entire class of transformer connections.

The three line currents are identical waveforms displaced by 120° in the fundamental.
A harmonic of order \(n\) is therefore displaced by \(120n\) degrees between phases.
For \(n = 3\) (or 9, 15, …), \(120 \times 3 = 360°\) — so the third harmonics of all three phases are exactly in phase with each other.
Three identical in-phase currents cannot flow in a three-wire system: they would have to return through a neutral, and a bridge has none.

So the triplens are excluded by the circuit topology. What remains is the family \(n = 6k \pm 1\): the 5th, 7th, 11th, 13th, 17th, 19th, with amplitudes falling as \(1/n\).

The general rule, worth carrying into Section 7-7: a \(p\)-pulse converter draws harmonics of order \(n = kp \pm 1\). Six-pulse gives 5, 7, 11, 13. Twelve-pulse gives 11, 13, 23, 25 — the 5th and 7th are gone. That is the entire motivation for multipulse connections.

Note also the contrast with a four-wire installation full of single-phase rectifiers, where the triplens have a neutral to flow in, add arithmetically, and overheat it — the problem described at the end of Chapter 6. Here, the same harmonics simply cannot exist.

Harmonic content of the ideal 120° line current
Harmonic \(n\)	Amplitude \(I_n/I_{s1}\)	Present?	Why
3, 9, 15	—	No	Triplen — in phase in all three lines, no return path
5	1/5 = 20%	Yes	\(6k-1\), negative sequence
7	1/7 = 14.3%	Yes	\(6k+1\), positive sequence
11	1/11 = 9.1%	Yes	\(6k-1\)
13	1/13 = 7.7%	Yes	\(6k+1\)
even	—	No	Half-wave symmetry

⚙Why the 5th harmonic gets its own budget line

Of the family, the fifth matters most — it is the largest, and it is negative sequence.

A positive-sequence set of currents produces a magnetic field rotating with the machine; a negative-sequence set produces one rotating against it. In a motor connected to the same busbar as a large rectifier, fifth-harmonic current therefore produces a braking torque at six times line frequency and extra rotor heating, for no useful output.

Three consequences you will meet in practice:

Tuned filters in industrial installations are almost always tuned to the 5th first, then the 7th.
Motors near rectifier loads are derated, or fed from a separate transformer.
Standards such as IEEE 519 set limits per harmonic order, not just on total THD, precisely because the orders differ in what they do.

And it is why the 12-pulse connection — which cancels the 5th and 7th outright — is worth a second transformer secondary at large ratings.

Section recap. The line current is a 120° quasi-square wave: \(I_s = 0.816I_o\), \(I_{s1} = 0.780I_o\), \(\mathrm{PF} = 3/\pi = 0.955\) at unity displacement, \(\mathrm{THD} = 31.1\%\). Triplens cannot flow in a three-wire circuit, so the spectrum is \(n = 6k\pm1\) — and generally \(kp \pm 1\) for a \(p\)-pulse converter.

Section 7-7

Twelve-Pulse and Multipulse Connections

The harmonic rule \(n = kp \pm 1\) points straight at the remedy. Raise the pulse number and the low-order harmonics disappear — not because they are filtered away, but because they cancel.

The standard arrangement uses one transformer with two secondaries:

A star secondary feeding one six-pulse bridge.
A delta secondary feeding a second six-pulse bridge.

A delta winding is inherently 30° displaced from a star winding on the same core. Two six-pulse bridges 30° apart produce output ripples that interleave, and supply currents whose 5th and 7th harmonics are in antiphase.

Two bridges, 30° apart — the harmonics cancel rather than being filtered

◇Intuition · why 30° cancels the fifth

Shift a fundamental by 30°, and its \(n\)th harmonic shifts by \(30n\) degrees.

Fundamental (\(n=1\)): shifted 30°. The transformer connection undoes this, so the two bridges' fundamentals add. Good — that is the useful current.
5th: shifted \(30 \times 5 = 150°\). Combined with the 180° inherent in the negative-sequence 5th, the two contributions arrive in antiphase and cancel.
7th: shifted \(30 \times 7 = 210°\), which is \(180° + 30°\) — again antiphase, again cancelled.
11th and 13th: shifted 330° and 390°, both equivalent to \(\pm 30°\) — in phase, so they survive and add.

Two identical noise sources, deliberately offset so their strongest components arrive out of step, is the same principle as noise-cancelling headphones. And the surviving harmonics are pushed to a higher order where they are both smaller (amplitude \(1/n\)) and much easier to filter.

The result: THD falls from 31% to about 15%, output ripple from 4.2% to 1.03%, and ripple frequency doubles to \(12f\) = 600 Hz.

⚙Where each connection is actually used

Connection	Typical rating	Where you find it
Single-phase bridge	< 3 kW	Chargers, appliances, small SMPS front ends
Six-pulse bridge	1 kW – 1 MW	The front end of essentially every industrial variable-frequency drive, DC drives, welding sets, plating rectifiers, UPS
Twelve-pulse	> 500 kW	Large mill drives, mine hoists, electrochemical plant, HVDC terminals, traction substations
24-pulse and above	Very large	HVDC where harmonic limits are severe and filters alone are insufficient

Notice that the choice is driven almost entirely by harmonic obligations at the point of common coupling, not by ripple. A six-pulse bridge's 4.2% ripple is already fine for a DC link; it is the 31% current distortion the utility objects to. That is why the alternative to going multipulse is not a bigger output capacitor but an active front end — a fully controlled bridge with PWM (Chapter 12 and Chapter 17), which draws near-sinusoidal current and reaches a power factor above 0.99.

Section recap. Two six-pulse bridges fed 30° apart by star and delta secondaries make a 12-pulse converter: ripple 1.03% at \(12f\), and the 5th and 7th harmonics cancel, leaving \(12k\pm1\). Multipulse is chosen for harmonic compliance, not for ripple.

Section 7-8

Worked Examples

1 Sizing a six-pulse bridge from the nameplate

Problem. A three-phase bridge on a 415 V, 50 Hz supply feeds a highly inductive load drawing a constant 60 A. Find \(V_{dc}\), the ripple frequency, the diode ratings, and the DC power delivered.

Solution. "415 V" is the RMS line voltage, so use the \(1.35\) coefficient:

Working

\[ V_{dc} = 1.35\,V_L = 1.35(415) = 560\ \text{V}, \qquad f_{ripple} = 6f = 300\ \text{Hz} \]

\[ I_{D(AV)} = \frac{I_o}{3} = 20\ \text{A}, \qquad I_{D(RMS)} = \frac{I_o}{\sqrt3} = 34.6\ \text{A} \]

\[ \mathrm{PIV} = V_{mL} = \sqrt2(415) = 587\ \text{V} \;\Rightarrow\; \text{specify } 1200\ \text{V} \]

\[ P_{dc} = V_{dc}I_o = (560)(60) = 33.6\ \text{kW} \]

Applying the Chapter 2 margins — twice the PIV and 1.5 times the average current — gives a 1200 V, 30 A diode, which is a standard three-phase bridge module. Note that 1200 V is the next standard grade above \(2 \times 587 = 1174\) V, which is exactly why 1200 V devices exist for 415 V systems.

2 Ripple without a filter

Problem. For the same bridge, find the peak-to-peak output ripple and the RMS ripple voltage. Comment on whether a filter is needed for a DC link.

Solution. The output swings between the peak line voltage and its value 30° away:

Working

\[ V_{max} = V_{mL} = 587\ \text{V}, \qquad V_{min} = V_{mL}\cos 30° = 587(0.866) = 508\ \text{V} \]

\[ V_{r(pp)} = 587 - 508 = 79\ \text{V} \quad (14\%\ \text{of } V_{dc}\ \text{peak-to-peak}) \]

\[ V_{r(rms)} = RF \times V_{dc} = 0.042(560) = 23.5\ \text{V} \]

Two different-looking numbers that do not contradict each other: the waveform swings 79 V peak-to-peak, but because it spends most of its time near the top, its RMS ripple content is only 23.5 V.

Is a filter needed? For a DC link feeding an inverter, no — 4.2% at 300 Hz is easily handled by the link capacitor that has to be there anyway for other reasons. For a DC motor armature, also no, since the armature inductance smooths the current. For a precision electronic supply, yes — but that would be a switching converter downstream, not a passive filter.

3 Working backwards from a required DC voltage

Problem. A 240 V DC, 150 A electroplating bath is to be supplied from a 415 V, 50 Hz line through a transformer and a six-pulse bridge. Find the required transformer secondary line voltage and its VA rating.

Solution. Invert the bridge relation to get the secondary voltage:

Working

\[ V_{L2} = \frac{V_{dc}}{1.35} = \frac{240}{1.35} = 178\ \text{V} \]

\[ I_s = \sqrt{\tfrac23}\,I_o = 0.816(150) = 122.5\ \text{A} \]

\[ S = \sqrt3\,V_{L2}I_s = \sqrt3(178)(122.5) = 37.8\ \text{kVA} \]

The DC power is \(240 \times 150 = 36\) kW, so the transformer VA is \(37.8/36 = 1.05\) times the DC power — the reciprocal of the 0.954 TUF, as it must be.

Add practical margin before ordering. Allow for the \(2V_F\) diode drop (2 V on 240 V — about 1%), transformer regulation of perhaps 4%, commutation overlap (Chapter 10), and supply voltage tolerance. Specify around 190 V secondary with taps, and a 45 kVA transformer. Compare this with a single-phase design at TUF 0.812, which would need over 44 kVA for the same load — and would put 100 Hz ripple into a bath that prefers steady current.

4 Three-pulse versus six-pulse on the same transformer

Problem. A 240 V (phase), 50 Hz three-phase secondary is available. Compare the DC output and ripple of a three-pulse and a six-pulse rectifier built from it.

Solution. Both take the envelope, but of different waveforms:

Working

\[ \text{3-pulse:}\quad V_{dc} = 1.17\,V_{ph} = 1.17(240) = 281\ \text{V}, \qquad RF = 18.3\%,\ f_r = 150\ \text{Hz} \]

\[ \text{6-pulse:}\quad V_{dc} = 2.34\,V_{ph} = 2.34(240) = 562\ \text{V}, \qquad RF = 4.2\%,\ f_r = 300\ \text{Hz} \]

The bridge gives exactly twice the DC voltage — because \(2.34 = 2 \times 1.17\) — with a quarter of the ripple, from the same transformer, at the cost of three more diodes.

It also removes the DC magnetisation, raising the TUF from 0.664 to 0.954, and needs no neutral. This is why the three-pulse circuit is essentially never built on its own: three cheap diodes buy an improvement in every single figure of merit.

5 Harmonic current at the point of common coupling

Problem. The 33.6 kW bridge of Example 1 draws 60 A DC. Find the RMS line current, the fundamental, the 5th and 7th harmonic currents, and the THD.

Solution.

Working

\[ I_s = 0.816(60) = 49.0\ \text{A}, \qquad I_{s1} = \frac{\sqrt6}{\pi}(60) = 0.780(60) = 46.8\ \text{A} \]

\[ I_5 = \frac{I_{s1}}{5} = 9.36\ \text{A}, \qquad I_7 = \frac{I_{s1}}{7} = 6.69\ \text{A} \]

\[ \mathrm{THD} = \sqrt{\left(\frac{I_s}{I_{s1}}\right)^2 - 1} = \sqrt{\left(\frac{49.0}{46.8}\right)^2 - 1} = 0.311 = 31.1\% \]

Nearly 9.4 A of fifth-harmonic current is being injected into the supply from a single 33.6 kW drive. Put twenty such drives on one busbar and that is 187 A of fifth harmonic — enough to distort the voltage waveform for every other customer on the transformer, which is precisely what IEEE 519 exists to limit.

The available remedies, in increasing order of cost: a line reactor (3–5% impedance, widens the conduction angle and cuts THD to roughly 25–30%), a DC-link choke, a tuned passive filter at the 5th, a 12-pulse connection, or an active front end.

6 Checking the p-pulse formula against 12-pulse

Problem. A 12-pulse converter is fed so that the peak of each constituent segment is 587 V. Find \(V_{dc}\), \(V_{rms}\), the ripple factor and the ripple frequency, and verify against the six-pulse result.

Solution. Apply the formula of Section 7-4 directly with \(p = 12\):

Working

\[ V_{dc} = \frac{12}{\pi}(587)\sin\frac{\pi}{12} = 3.8197(587)(0.2588) = 580.3\ \text{V} \]

\[ V_{rms} = 587\sqrt{\tfrac12 + \tfrac{12}{4\pi}\sin\tfrac{\pi}{6}} = 587\sqrt{0.5 + 0.4775} = 580.4\ \text{V} \]

\[ FF = \frac{580.4}{580.3} = 1.00005 \;\Rightarrow\; RF = \sqrt{FF^2-1} = 1.03\%, \qquad f_r = 600\ \text{Hz} \]

Compare the six-pulse result from the same peak: \(V_{dc} = 0.955(587) = 560\) V at 4.2% ripple. So 12-pulse gives 3.6% more DC voltage and a quarter of the ripple.

Notice how close \(V_{dc}\) and \(V_{rms}\) have become — they differ in the fifth significant figure. Carrying enough precision matters here: rounding to three figures would make \(FF = 1.000\) and the ripple factor would come out as zero. When two large numbers are subtracted, keep the digits.

Review

Summary & Formula Sheet

✓ Constant power

Balanced three-phase power is constant in time; single-phase power falls to zero twice a cycle. Every advantage follows.

✓ Envelope rule

Common cathode picks the most positive phase, common anode the most negative. The load sees the largest available voltage.

✓ The two numbers

3-pulse: \(V_{dc} = 1.17V_{ph}\), \(RF = 18.3\%\). 6-pulse bridge: \(V_{dc} = 1.35V_L = 2.34V_{ph}\), \(RF = 4.2\%\).

✓ One formula

\(V_{dc} = (p/\pi)V_m\sin(\pi/p)\), ripple at \(pf\). Reproduces every result in Chapters 6 and 7.

✓ 120°, not 60°

Each bridge diode conducts 120°: \(I_{D(AV)} = I_o/3\), \(I_{D(RMS)} = I_o/\sqrt3\). Two devices always on.

✓ Harmonics \(kp\pm1\)

Six-pulse: 5, 7, 11, 13; no triplens; THD 31%, PF 0.955. Twelve-pulse cancels the 5th and 7th.

Formula sheet · Chapter 7

The \(p\)-pulse average\(V_m\) = peak of the constituent sinusoid

\( V_{dc} = \dfrac{p}{\pi}V_m\sin\dfrac{\pi}{p} \)

The \(p\)-pulse RMSthen \(FF\) and \(RF\) as in Ch. 6

\( V_{rms} = V_m\sqrt{\dfrac{1}{2} + \dfrac{p}{4\pi}\sin\dfrac{2\pi}{p}} \)

Three-pulse output\(RF=18.3\%\), \(\mathrm{PIV}=\sqrt3 V_m\)

\( V_{dc} = \dfrac{3\sqrt3}{2\pi}V_m = 1.17\,V_{ph} \)

Six-pulse bridge output\(RF=4.2\%\), \(\mathrm{PIV}=V_{mL}\)

\( V_{dc} = \dfrac{3}{\pi}V_{mL} = 1.35\,V_{L} = 2.34\,V_{ph} \)

Line and phase peaksthe \(\sqrt3\) that causes most errors

\( V_{mL} = \sqrt3\,V_m, \qquad V_m = \sqrt2\,V_{ph} \)

Bridge diode currents120° conduction, constant \(I_o\)

\( I_{D(AV)} = \dfrac{I_o}{3}, \qquad I_{D(RMS)} = \dfrac{I_o}{\sqrt3} \)

Supply line currentquasi-square, 120° blocks

\( I_s = \sqrt{\dfrac{2}{3}}\,I_o = 0.816\,I_o \)

Supply fundamentalin phase with the voltage

\( I_{s1} = \dfrac{\sqrt6}{\pi}I_o = 0.780\,I_o \)

Input power factorall distortion, no displacement

\( \mathrm{PF} = \dfrac{3}{\pi} = 0.955, \qquad \mathrm{THD} = 31.1\% \)

Harmonic orders present\(p\)-pulse converter

\( n = kp \pm 1, \qquad I_n \approx \dfrac{I_{s1}}{n} \)

Output ripple boundsenvelope maximum and minimum

\( V_{max} = V_m, \qquad V_{min} = V_m\cos\dfrac{\pi}{p} \)

Key terms

Pulse number, \(p\): Output pulses per supply cycle. Sets the ripple, the ripple frequency \(pf\), and the harmonic orders \(kp\pm1\).
Common cathode / common anode: Diode groups whose cathodes (or anodes) are joined. The most positive (or most negative) phase conducts.
Crossover point: The instant two phase voltages are equal, where conduction transfers from one diode to the next. At 30°, 150°, 270° for a three-pulse circuit.
DC magnetisation: A unidirectional current component in a transformer winding, pushing the core toward saturation. The three-pulse circuit's main defect.
Line and phase voltage: \(V_L = \sqrt3\,V_{ph}\) in a star system. Choosing the wrong one is the most common error in this chapter.
Quasi-square wave: The 120°-on, 60°-off, 120°-reverse, 60°-off supply current of a three-phase bridge with constant load current.
Triplen harmonics: Orders 3, 9, 15… They are in phase in all three lines, so they cannot flow in a three-wire circuit.
Negative sequence: A harmonic set rotating against the fundamental — the 5th, 11th… It produces braking torque and heating in nearby motors.
Multipulse connection: Two or more bridges fed at deliberately displaced phase angles so their low-order harmonics cancel. 12-pulse uses a 30° shift.
Point of common coupling: The busbar where an installation meets the utility supply, and where harmonic limits such as IEEE 519 are assessed.

Self-assessment

Test Yourself

Six questions on the workhorse converter of industry. Try each aloud before revealing the answer.

Chapter 7 · six questions answers hidden until you ask

A three-phase bridge on a 400 V line is measured at 540 V DC. A colleague says the circuit is faulty because \(400 \times \sqrt2 = 566\) V. Are they right?

No — 540 V is exactly what the circuit should produce.

\(566\) V is the peak line voltage, which the output only touches momentarily six times per cycle. The average is lower, because the envelope dips to \(\cos 30°\) of the peak between crossovers:

\(V_{dc} = 1.35(400) = 540\) V, which is \(0.955 \times 566\) V.

Two useful sanity anchors: \(V_{dc}\) is always a few percent below the peak line voltage for a six-pulse bridge, and the ratio \(V_{dc}/V_{L} = 1.35\) is worth memorising outright — it converts a nameplate straight into a DC-link voltage. On a 415 V supply, 560 V; on a 400 V supply, 540 V; on a 480 V supply, 648 V.

Why does a three-phase bridge need no neutral, when the three-pulse rectifier cannot work without one?

Because of where the load current returns.

Three-pulse: only the positive terminal is switched between phases. The load's other terminal must connect to something, and the only available reference is the star point — so the neutral carries the entire load current, and it carries it in one direction.

Bridge: both terminals are switched. The lower group connects the load's negative terminal to the most negative phase, so the return path is through another phase, never through a neutral.

Two consequences follow immediately:

The bridge works on any three-wire supply — no transformer needed if the line voltage suits, which is why it is so cheap to deploy.
Each secondary winding now carries current in both directions (once through an upper diode, once through a lower one), so the DC magnetisation problem disappears and the TUF rises from 0.664 to 0.954.

A six-pulse bridge feeds 100 A. What average and RMS current must each diode carry, and what happens if you size on the average alone?

Each diode conducts for 120°, so \(I_{D(AV)} = 100/3 = 33.3\) A and \(I_{D(RMS)} = 100/\sqrt3 = 57.7\) A.

Sizing on the average alone under-predicts the loss. Using the Chapter 2 model with \(V_{T0} = 0.9\) V and \(r_D = 5\) mΩ:

Correct: \(P = (0.9)(33.3) + (0.005)(57.7)^2 = 30.0 + 16.6 = 46.6\) W per diode.
Average only: \((0.9)(33.3) + (0.005)(33.3)^2 = 30.0 + 5.5 = 35.5\) W — a 24% under-estimate.

Across six diodes that is 67 W of heat the heatsink was never designed to remove. The resistive term always needs the RMS, exactly as in Chapters 2 and 6 — the shorter the conduction fraction, the worse the error becomes.

The third harmonic is the largest in a single-phase rectifier but absent from a three-phase bridge's line current. Where has it gone?

It has not gone anywhere — it was never able to flow.

Each phase's third harmonic is displaced by \(3 \times 120° = 360°\) from its neighbours, which is to say not displaced at all. All three third harmonics are in phase, so they behave like a single-phase current that needs a common return path. A three-wire bridge has none, so the current simply cannot exist.

But be careful with the general claim. Triplens are excluded from the line currents of a three-wire circuit. In other places they are very much present:

Inside a delta winding they circulate happily, heating it — which is one reason delta windings are used deliberately, as a triplen trap.
In a four-wire installation of single-phase rectifiers, they add arithmetically in the neutral, as Chapter 6 described.

The topology, not the waveform, decides.

Twelve-pulse halves the low-order harmonics without any filter. Why is a 415 V, 30 kW drive still built as six-pulse?

Because the cost is in the transformer, and a six-pulse drive at that rating does not need one at all.

A 415 V six-pulse bridge connects straight to the line and produces a 560 V DC link — which is precisely the voltage a 415 V inverter output needs. The bill of materials is six diodes in one module.

Going 12-pulse means adding a transformer with two secondaries, one star and one delta, sized for the full 30 kW. That transformer costs more than the entire rest of the rectifier, adds mass and losses, and needs floor space.

What actually gets done at this rating: fit a 3–5% line reactor or a DC-link choke. It costs a fraction of a transformer, cuts THD from 31% to roughly 25–30%, and is usually enough for compliance at 30 kW.

Twelve-pulse earns its keep above roughly 500 kW, where the harmonic current is large enough in absolute terms that the utility objects — and where the installation needs a dedicated transformer anyway, so the incremental cost is only the second secondary.

A 6-pulse bridge and a 12-pulse converter both feed 500 A. Which draws more RMS supply current, and which distorts the supply more?

Supply current: essentially the same. Both deliver the same DC power from the same voltage, so both must draw the same fundamental current. The RMS differs only by the small harmonic contribution.

Distortion: the six-pulse is far worse, and the difference is concentrated where it hurts.

Property	6-pulse	12-pulse
THD	31%	~15%
Lowest harmonic	5th (20%)	11th (9%)
Output ripple	4.2% at 300 Hz	1.03% at 600 Hz

Why the harmonic order matters as much as the amplitude: a filter tuned to the 11th is physically smaller than one tuned to the 5th, because filter component size scales inversely with frequency. And the 5th is negative sequence, so it actively brakes nearby motors — the 11th does too, but at less than half the amplitude.

So 12-pulse improves the situation twice over: it removes the worst harmonics entirely, and it makes whatever remains cheaper to filter.

Practice

Problems

Three questions fix almost every problem in this chapter, so ask them first:

What is the pulse number? It gives \(V_{dc}\), the ripple factor, the ripple frequency and the harmonic orders.
Line or phase voltage? Write down which one you were given before choosing a coefficient.
What fraction of the cycle does each device conduct? 120° in a bridge — from which every current rating follows.

Problems 1–5 are direct application; 6–10 need judgement; 11–12 are design questions worth discussing in a tutorial.

State the pulse number, ripple frequency on a 50 Hz supply, ripple factor and PIV for: (a) a single-phase bridge, (b) a three-pulse rectifier, (c) a three-phase bridge, (d) a 12-pulse converter.
A three-phase bridge operates from a 400 V, 50 Hz line. Find \(V_{dc}\), the peak and minimum instantaneous output, and the peak-to-peak ripple.
The bridge of Problem 2 supplies a constant 80 A. Find the average and RMS diode currents, the PIV, and specify a device with sensible margins.
A three-pulse rectifier is fed from a 230 V phase voltage. Find \(V_{dc}\), \(V_{rms}\), the form and ripple factors, and the PIV.
Using the \(p\)-pulse formula, compute \(V_{dc}/V_m\) and the ripple factor for \(p = 2, 3, 6, 12\). Verify each against the values quoted in Section 7-4.
A six-pulse bridge feeds 45 kW at 540 V. Find the RMS supply line current, the fundamental component, and the 5th and 7th harmonic currents.
Explain why the three-pulse rectifier's transformer must be oversized, and estimate the kVA needed for a 20 kW DC output. Repeat for a six-pulse bridge and compare.
A 415 V three-phase bridge feeds a 240 V DC motor through a transformer. Find the required secondary line voltage. If the motor draws 90 A, find the transformer kVA.
Two diodes are always in series in a bridge. For a 24 V, 400 A supply with \(V_F = 0.85\) V each, find the total diode loss and express it as a percentage of the DC output power. Comment on the device choice.
A 12-pulse converter replaces a six-pulse one on a 2 MW load. State which harmonic orders disappear, which remain, and what happens to the output ripple frequency and magnitude.
An installation has six identical 415 V, 50 kW six-pulse drives on one busbar. Estimate the total fifth-harmonic current. Discuss two remedies and the circumstances under which you would choose each.
A three-phase bridge feeds a DC link that supplies an inverter. Explain why a DC-link choke improves the input power factor even though it is on the output side of the rectifier, and state what it costs you.

Tip: when a three-phase problem looks unfamiliar, draw the three phase voltages and mark the crossover points. Everything else — which device conducts, for how long, what the output is, what the off devices block — can be read directly off that sketch. The algebra is only there to turn the picture into a number.