Chapter 9: Three-Phase Controlled Rectifiers — Power Electronics

Where this sits

Part 2 · AC–DC Converters
Chapter 9 of 30

You should already know

The six-pulse bridge and the \(p\)-pulse formula (Ch. 7), and the firing angle, conduction modes and inverter mode (Ch. 8).

By the end you can

Apply \(V_{dc} = 1.35V_L\cos\alpha\), sequence the six gate pulses, size a drive converter, and set the inverter limit.

Time

≈ 60 min reading · ≈ 55 min problems

i What you'll learn

How adding a cosine to the Chapter 7 result gives \( V_{dc} = 1.35\,V_L\cos\alpha \) — the single most-used equation in industrial drives.
Why \( \alpha \) is measured from the crossover point, not from the voltage zero, and what goes wrong if you forget.
The firing sequence: six pulses 60° apart, and why two devices must be gated at start-up.
The three-phase semiconverter and half-wave circuits, and where each is still used.
Inverter mode at three-phase scale — regenerative drives and the front end of HVDC.
Why the input power factor is \( 0.955\cos\alpha \), and why a drive at low speed is so hard on the supply.
How the whole chapter reduces to one design equation for a DC motor drive.

Section 9-1

The Cosine That Joins the Two Halves of Part 2

Chapter 7 built a six-pulse bridge delivering a fixed \( V_{dc} = 1.35\,V_L \). Chapter 8 showed that replacing diodes with thyristors delays the start of conduction by \( \alpha \), and that with continuous conduction the effect is exactly a factor of \( \cos\alpha \).

Put them together and Part 2 collapses into one result.

🔑

The equation industrial drives are built on

\[ V_{dc} = \frac{3V_{mL}}{\pi}\cos\alpha = 1.35\,V_L\cos\alpha = 2.34\,V_{ph}\cos\alpha \]

valid for continuous conduction, which a motor armature or a DC link with a smoothing reactor guarantees. On a 415 V line this gives \( 560\cos\alpha \) volts — a smoothly adjustable DC source from +560 V down through zero to −560 V, with nothing dissipated to achieve it.

The derivation is the Chapter 7 one with the limits displaced by \( \alpha \):

Start from the uncontrolled interval In Chapter 7 each 60° segment was centred on the peak of a line voltage, running from \( -\pi/6 \) to \( +\pi/6 \).
Delay the start by \( \alpha \) Firing late shifts the whole window: the interval becomes \( -\pi/6 + \alpha \) to \( +\pi/6 + \alpha \). The width is unchanged at \( \pi/3 \), because continuous conduction forces each pair to hold on until the next is fired.

Working

\[ V_{dc} = \frac{3}{\pi}\int_{-\pi/6+\alpha}^{\pi/6+\alpha} V_{mL}\cos\theta\,d\theta = \frac{3V_{mL}}{\pi}\bigl[\sin\theta\bigr]_{-\pi/6+\alpha}^{\pi/6+\alpha} \]
Expand and collect

Working

\[ = \frac{3V_{mL}}{\pi}\Bigl[\sin\bigl(\tfrac{\pi}{6}+\alpha\bigr) - \sin\bigl(-\tfrac{\pi}{6}+\alpha\bigr)\Bigr] = \frac{3V_{mL}}{\pi}\cdot 2\cos\alpha\sin\frac{\pi}{6} = \frac{3V_{mL}}{\pi}\cos\alpha \]

The \( 2\sin(\pi/6) = 1 \) cancels exactly, leaving the uncontrolled result multiplied by \( \cos\alpha \).
Note what did not change The RMS output is unaffected in the same way as in Chapter 8 — each interval is still a full \( \pi/3 \) of a cosine, merely shifted. Only the average responds to \( \alpha \).

!Common mistake

Measuring \( \alpha \) from the zero crossing of the phase voltage.

Wrong \( \alpha \) is measured from where the phase voltage crosses zero, as it was for the single-phase half-wave circuit.

Right \( \alpha \) is measured from the natural commutation point — the crossover where the incoming device would have taken over on its own, had it been a diode.

In a three-phase bridge that crossover occurs 30° after the phase-voltage zero. So \( \alpha = 0 \) corresponds to \( \omega t = 30^\circ \) on the phase reference, and firing at \( \omega t = 30^\circ + \alpha \).

The reason for the convention is that it makes \( \alpha = 0 \) mean "behave exactly like a diode", so the controlled formula collapses to the uncontrolled one — which is what the \( \cos\alpha \) factor expresses. Measure from anywhere else and you lose that.

Where it bites in practice: a firing circuit synchronised to the wrong reference is offset by 30°, so a converter commanded to \( 0^\circ \) actually fires at \( 30^\circ \) and delivers 13% less voltage than expected — and one commanded to \( 150^\circ \) fires at \( 180^\circ \) and fails to commutate. Getting the synchronisation reference right is the first thing to check when a new drive misbehaves.

Interactive · three-phase full converter

A six-pulse fully controlled bridge with continuous conduction. The faint curves are the six line voltages; the bold curve is the output. Move \( \alpha \) and watch the segments slide down the waveform — and, past 90°, straight through zero into inversion.

Firing angle \( \alpha \) 30°

Firing angle α30°

V_dc / V_mL0.827

On a 415 V line485 V

ModeRectifier

—

Section recap. Delaying the firing by \( \alpha \) shifts each 60° conduction window without changing its width, so the Chapter 7 result simply acquires a cosine: \( V_{dc} = 1.35\,V_L\cos\alpha \). The angle is measured from the natural crossover, which is 30° after the phase-voltage zero.

Section 9-2

Firing the Six Devices

The conduction table of Chapter 7 carries over unchanged — only the firing instants move. Six gate pulses per cycle, 60° apart, in the order \( T_1, T_2, T_3, T_4, T_5, T_6 \), with each device conducting for 120° and two devices on at any instant.

Six pulses, 60° apart, 120° each — the pattern behind every three-phase converter

!Common mistake

"One gate pulse per thyristor per cycle is enough."

It is enough once the converter is running, and not enough to start it.

A bridge needs two devices conducting to carry current — one upper, one lower. At start-up none is conducting, so firing \( T_1 \) alone achieves nothing: there is no return path, so no current flows, so \( T_1 \) never reaches its latching current \( I_L \) and simply does not latch (Chapter 3).

Two standard solutions, both in universal use:

Double-pulse firing. Each device is gated twice per cycle — once at its own instant, and again 60° later when its partner is fired. So firing \( T_2 \) also re-fires \( T_1 \), and the pair latches together.
Pulse-train firing. A high-frequency burst lasting the whole 120° conduction period, as in Chapter 4. This also re-fires a device that drops out during discontinuous conduction, which the double-pulse scheme does not.

The diagnostic symptom of getting this wrong is a converter that runs correctly once started but will not start into an inductive load — or one that runs at high load but stalls at light load, where conduction goes discontinuous and devices drop out mid-cycle.

Section recap. Six gate pulses per cycle, 60° apart, each device conducting 120°, two on at any instant. Because a bridge needs a conducting pair, firing must be double-pulse or a pulse train — a single pulse per device cannot start the converter.

Section 9-3

Half-Wave and Semiconverter Variants

As in single-phase, the fully controlled bridge has two cheaper relatives. Both trade capability for cost, and the trade is the same one as in Chapter 8.

1 Three-phase half-wave (three-pulse)

Three thyristors, common cathode, neutral return. \( V_{dc} = 1.17\,V_{ph}\cos\alpha \) for continuous conduction. Inherits the DC-magnetisation problem of Chapter 7, so it is used only at small ratings — and often as one half of a larger connection.

2 Three-phase semiconverter

Three thyristors in the upper group, three diodes in the lower. Inherent freewheeling, so the output never reverses: \( V_{dc} = \frac{3V_{mL}}{2\pi}(1+\cos\alpha) \). One quadrant, better power factor, half the gate drives.

3 Three-phase full converter

Six thyristors. \( V_{dc} = 1.35\,V_L\cos\alpha \), two quadrants, full inversion capability. The standard for DC drives, HVDC and any load that must regenerate.

4 Dual converter

Two full converters back to back, so current can flow either way. Four quadrants — motoring and braking in both directions of rotation. The subject of Chapter 10.

The three-phase family compared, continuous conduction throughout
Property	Half-wave	Semiconverter	Full converter
Thyristors / diodes	3 / 0	3 / 3	6 / 0
\(V_{dc}\)	\(1.17V_{ph}\cos\alpha\)	\(\frac{3V_{mL}}{2\pi}(1+\cos\alpha)\)	\(1.35V_L\cos\alpha\)
Ripple / pulse number	3-pulse, 18.3%	3-pulse at large \(\alpha\)	6-pulse, 4.2% at \(\alpha=0\)
Quadrants	Two*	One	Two
Needs a neutral	Yes	No	No
DC magnetisation	Yes — oversized transformer	No	No
Typical use	Small excitation supplies	Unidirectional drives, heating, plating	DC drives, HVDC, regenerative loads

*The half-wave circuit can invert in principle, but its ripple and transformer penalty rule it out for real regenerative duty.

Section recap. Half-wave gives \( 1.17V_{ph}\cos\alpha \) but keeps the DC-magnetisation problem; the semiconverter freewheels so its output never reverses, \( \frac{3V_{mL}}{2\pi}(1+\cos\alpha) \), one quadrant with a better power factor; the full converter gives \( 1.35V_L\cos\alpha \) over two quadrants and is the industrial standard.

Section 9-4

Inversion at Three-Phase Scale

Everything Chapter 8 said about inverter mode applies here, with two differences that matter: the ripple is far better, and the power levels are far larger.

Past \( \alpha = 90^\circ \), \( V_{dc} \) reverses while the current cannot, so power flows from the DC side into the AC line. The two necessary conditions are unchanged — an \( \alpha \) beyond 90°, and a DC-side EMF to sustain the current.

🔑

The inverter limit, in the form used by drive controllers

\[ \alpha_{max} = 180^\circ - \gamma_{min}, \qquad \gamma_{min} = \mu + \omega t_q + \text{safety} \]

The extinction angle \( \gamma \) is what remains after commutation overlap \( \mu \) has eaten into the interval. It must exceed \( \omega t_q \) so the outgoing device recovers its blocking capability. Practical settings are \( \alpha_{max} \approx 150^\circ \) for a converter-grade thyristor at 50 Hz, sometimes 165° where \( t_q \) is small and the source is stiff.

⚙The same circuit, three industries

The three-phase full converter earns its place because the same six-thyristor bridge does entirely different jobs depending on what sits on its DC side.

Application	DC-side EMF is…	What inversion achieves
DC mill or hoist drive	The motor's back-EMF	Regenerative braking — the load's kinetic or potential energy returns to the supply instead of heating a resistor. A mine hoist lowering a full cage generates for the whole descent.
HVDC transmission	The DC line, fed by the converter at the other end	One terminal rectifies, the other inverts. Reversing which is which reverses the power flow across the link, without moving any hardware.
Battery / electrochemical plant	The cell voltage	Controlled discharge back into the grid, or reversal of a plating process.

The HVDC case is worth dwelling on. A link is simply two of these bridges facing each other across a DC line. Power flows from whichever terminal has the higher DC voltage, so the direction is set purely by the two firing angles — one at perhaps 15°, the other at 150°. To reverse a gigawatt of power flow, the control system swaps the two angles. No switchgear moves, and the transition takes a few hundred milliseconds.

That is the payoff of a converter whose output voltage can change sign under software control.

Section recap. Beyond \( \alpha = 90^\circ \) the bridge inverts, returning power to the line if a DC-side EMF sustains the current. \( \alpha \) must stop short of 180° by a margin covering the overlap \( \mu \) and \( \omega t_q \), typically limiting it to about 150°.

Section 9-5

What the Supply Sees

The supply current is the same 120° quasi-square wave as in Chapter 7 — shifted bodily by \( \alpha \). So the distortion is unchanged and the displacement is entirely new:

🔑

Input power factor of a three-phase full converter

\[ \mathrm{PF} = \underbrace{\frac{3}{\pi}}_{0.955} \times \cos\alpha = 0.955\cos\alpha \]

Better than the single-phase 0.900 by the same margin as in Chapter 7 — but with the same fatal \( \cos\alpha \). At \( \alpha = 60^\circ \) the power factor is 0.478; at \( \alpha = 75^\circ \) it is 0.247. The harmonic spectrum is unchanged at \( 6k\pm1 \) with amplitudes \( 1/n \), because phase control shifts the current waveform without reshaping it.

→Why a DC drive is hardest on the supply exactly when it is doing least

Consider a 100 kW DC drive on a 415 V supply, and follow it from full speed down to low speed.

Speed	Firing angle \(\alpha\)	Output \(V_{dc}\)	Real power	PF	Apparent power
Full	0°	560 V	100 kW	0.955	105 kVA
Half	60°	280 V	50 kW	0.478	105 kVA
Quarter	75.5°	140 V	25 kW	0.239	105 kVA

Read the last two columns together. At constant torque the armature current is constant, so the supply current is constant too — and the apparent power never falls. The drive loads the supply just as heavily at quarter speed as at full speed, while delivering a quarter of the power.

The difference has become reactive power: \( Q = S\sin\alpha \), which reaches 101 kVAr at \( \alpha = 75.5^\circ \). Three consequences:

Cables and transformers must be sized on apparent power, so nothing can be downsized for a low-speed application.
Utility penalties are levied on power factor, and a drive that spends its life at low speed is expensive to run.
Voltage regulation suffers — the reactive current drops volts across the supply impedance, dimming everything else on the busbar.

This is the single strongest argument for the alternatives: a chopper-fed DC drive (Chapter 11) that rectifies at \( \alpha = 0 \) and controls with duty ratio instead, or an AC drive with an active front end. Both keep the firing angle small and the power factor high, and both are why phase-controlled DC drives have been retreating for thirty years.

Section recap. \( \mathrm{PF} = 0.955\cos\alpha \), with the \( 6k\pm1 \) harmonic spectrum unchanged from Chapter 7. Because armature current stays constant at constant torque, apparent power does not fall with speed — a phase-controlled drive is hardest on the supply exactly when it is delivering least.

Section 9-6

Designing a DC Motor Drive

Everything in Chapters 7, 8 and 9 exists to serve this application, and it reduces to two equations solved together.

🔑

The two equations of a phase-controlled DC drive

\[ V_{dc} = 1.35\,V_L\cos\alpha \qquad\text{and}\qquad V_{dc} = E_a + I_aR_a, \quad E_a = k\phi\,\omega \]

The first says what the converter can deliver; the second says what the machine demands. Setting them equal gives the firing angle for any required speed and torque — and every drive controller in the world is, at bottom, a loop that solves this in real time.

Torque fixes the current \( T = k\phi I_a \), so the required armature current comes straight from the load torque. The current loop holds it there.
Speed fixes the back-EMF \( E_a = k\phi\omega \). The demanded speed and the field flux fix \( E_a \).
The armature equation fixes the required converter voltage \( V_{dc} = E_a + I_aR_a \). Note both terms: the drop \( I_aR_a \) is small at light load and significant at full torque.
The converter equation fixes the firing angle

The design equation

\[ \alpha = \cos^{-1}\!\left(\frac{E_a + I_aR_a}{1.35\,V_L}\right) \]

If the argument exceeds 1, the supply is too low — as in Chapter 8's Example 2. If it goes negative, the drive is regenerating.

⚙What a real drive adds around those two equations

The equations give the steady state. A working drive wraps them in structure:

Cascaded control loops. An outer speed loop sets a current reference; an inner, much faster current loop sets \( \alpha \). The inner loop's speed is what makes the drive safe — it can limit current within a few milliseconds during a fault or a stall.
A current limit, typically 150–200% of rated, protecting both the machine's commutator and the thyristors' \( I^2t \).
Field weakening above base speed. Once \( \alpha \) reaches 0 the converter can give no more voltage, so further speed comes from reducing \( \phi \) — which reduces \( E_a \) for a given \( \omega \), at the cost of available torque.
A smoothing reactor sized to keep conduction continuous at the minimum expected current, for the reasons in Chapter 8.
A firing-angle end-stop at about 150° if the drive regenerates, and 0° at the other end.

The two operating regions are worth remembering as a picture: constant torque from zero to base speed, where \( \alpha \) sweeps from 90° to 0° with full flux; then constant power above base speed, where \( \alpha \) stays at 0 and the field is weakened. Every DC drive datasheet shows exactly this curve.

Section recap. A DC drive solves \( 1.35V_L\cos\alpha = E_a + I_aR_a \) continuously. Torque sets \( I_a \), speed sets \( E_a \), and the controller sets \( \alpha \). Below base speed it is constant-torque with full field; above it, \( \alpha = 0 \) and the field is weakened for constant power.

Section 9-7

Worked Examples

1 The basic calculation

Problem. A three-phase full converter runs from 415 V, 50 Hz into a highly inductive load. Find \( V_{dc} \) at \( \alpha = 0°, 30°, 60°, 90°, 120° \) and \( 150° \).

Solution. \( V_{dc} = 1.35(415)\cos\alpha = 560\cos\alpha \):

Firing angle \(\alpha\)	0°	30°	60°	90°	120°	150°
\(V_{dc}\)	560 V	485 V	280 V	0	−280 V	−485 V
PF	0.955	0.827	0.478	0	−0.478	−0.827
Mode	Rectify	Rectify	Rectify	Zero	Invert	Invert

The cosine curve is steepest near 90°, so a drive operating there has very high gain — a small firing-angle change produces a large voltage change. That is convenient for fast control and inconvenient for stability, and it is why drive controllers often linearise by commanding \( \cos\alpha \) rather than \( \alpha \) directly.

2 Firing angle for a DC motor

Problem. A 440 V, 100 A DC motor with \( R_a = 0.2\ \Omega \) is fed from a three-phase full converter on a 415 V, 50 Hz line. Find the firing angle at rated operation, and the back-EMF.

Solution. Apply the design equation:

Working

\[ V_{dc} = 440\ \text{V}, \qquad \cos\alpha = \frac{440}{1.35(415)} = \frac{440}{560} = 0.786 \;\Rightarrow\; \alpha = 38.2^\circ \]

\[ E_a = V_{dc} - I_aR_a = 440 - (100)(0.2) = 420\ \text{V} \]

Comfortable: \( \alpha = 38° \) leaves headroom at both ends — room to advance toward 0° for extra speed or to overcome a supply dip, and a long way to go before the inverter limit.

Contrast this with Chapter 8's Example 2, where a single-phase bridge on 230 V simply could not reach a 220 V motor. Here a 415 V three-phase supply reaches 560 V, which comfortably covers a 440 V machine. This headroom is one of the main practical reasons DC drives above a few kilowatts are always three-phase.

3 Regenerative braking of a hoist

Problem. The motor of Example 2 lowers a load at half speed, generating with \( E_a = 210 \) V and drawing 80 A into the converter. Find the firing angle and the power returned. Check the inverter margin.

Solution. During regeneration the machine's EMF drives the current, so the converter must present a negative voltage:

Working

\[ V_{dc} = -E_a + I_aR_a = -210 + (80)(0.2) = -194\ \text{V} \]

\[ \cos\alpha = \frac{-194}{560} = -0.346 \;\Rightarrow\; \alpha = 110.3^\circ \]

\[ P = |V_{dc}|I_a = (194)(80) = 15.5\ \text{kW returned to the supply} \]

Margin check: \( \gamma = 180° - 110.3° = 69.7° \), which at 50 Hz is 3.9 ms — vastly more than the \( t_q \) of 50–100 µs a converter-grade thyristor needs, plus a few degrees of overlap. Safe.

Over a two-minute descent this hoist returns about 0.5 kWh to the supply. Repeated hundreds of times a day, that is the difference between a braking-resistor bank that must be cooled and a system that pays part of its own running cost.

4 The supply-side cost of running slowly

Problem. A 415 V three-phase converter drives a constant-torque load at 100 A armature current. Find the supply RMS current, real power, power factor and reactive power at \( \alpha = 0° \) and \( \alpha = 70° \).

Solution. Constant torque means constant \( I_a \), so the supply current is the same in both cases:

Working

\[ I_s = 0.816\,I_o = 0.816(100) = 81.6\ \text{A} \quad\text{(at both angles)} \]

\[ S = \sqrt3\,V_LI_s = \sqrt3(415)(81.6) = 58.7\ \text{kVA} \quad\text{(at both angles)} \]

\[ \alpha = 0°: \ P = 0.955(58.7) = 56.0\ \text{kW}, \qquad Q = 0 \]

\[ \alpha = 70°: \ P = 0.955\cos70°(58.7) = 19.2\ \text{kW}, \qquad Q \approx S\sin70° = 55.2\ \text{kVAr} \]

The apparent power is identical — 58.7 kVA in both cases — while the useful output has fallen to a third. All the difference has become reactive power.

The transformer, cables and switchgear must be sized for 58.7 kVA regardless of operating speed, and the utility bills on a power factor of 0.33. This is the arithmetic behind Section 9-5, and the reason a chopper-fed drive (Chapter 11) is preferred where the machine spends its life at low speed.

5 Semiconverter versus full converter

Problem. A 415 V three-phase supply feeds a 300 V, 50 A unidirectional heating load. Compare a semiconverter and a full converter on firing angle and input power factor.

Solution. \( V_{mL} = \sqrt2(415) = 587 \) V.

Full converter

\[ \cos\alpha = \frac{300}{560} = 0.536 \;\Rightarrow\; \alpha = 57.6^\circ, \qquad \mathrm{PF} = 0.955\cos(57.6°) = 0.512 \]

Semiconverter

\[ \frac{3(587)}{2\pi}(1+\cos\alpha) = 300 \;\Rightarrow\; 1+\cos\alpha = 1.071 \;\Rightarrow\; \alpha = 85.9^\circ \]

The semiconverter runs at a larger firing angle but freewheels for part of each cycle, so its supply current is smaller and better aligned — its power factor works out around 0.67 against 0.51.

For a heating load the semiconverter is clearly the right choice: the load never regenerates, so the full converter's second quadrant is worthless, and the semiconverter uses three thyristors instead of six with a better power factor. Choosing the more capable circuit "just in case" costs money and efficiency for a capability that will never be used.

6 Checking the inverter limit against a supply dip

Problem. A converter inverts at \( \alpha = 145^\circ \) on a 50 Hz supply with a thyristor of \( t_q = 120\ \mu\text{s} \) and an overlap angle of \( \mu = 12^\circ \). Is the margin adequate? What happens if the supply dips 20%?

Solution. First convert the turn-off time to an angle:

Working

\[ \omega t_q = 2\pi(50)(120\times10^{-6}) = 0.0377\ \text{rad} = 2.16^\circ \]

\[ \gamma = 180° - \alpha - \mu = 180° - 145° - 12° = 23° \;\gg\; 2.16° \]

The margin is more than ten times the minimum, so at nominal voltage the design is comfortable.

Now the dip. Commutation is driven by the line voltage, and the overlap angle grows roughly as the inverse of it — so a 20% dip stretches \( \mu \) from 12° to about 15°, leaving \( \gamma = 20° \). Still safe.

But the current also matters more than the voltage here. Overlap grows with load current (Chapter 10), so a dip combined with a current surge — precisely what happens when a hoist accelerates during a brownout — can consume the margin far faster than either alone. That is why practical controllers monitor supply voltage and retard \( \alpha \) automatically when it falls, rather than relying on a fixed end-stop.

Review

Summary & Formula Sheet

✓ One equation

\( V_{dc} = 1.35V_L\cos\alpha = 2.34V_{ph}\cos\alpha \). Chapter 7's result multiplied by a cosine.

✓ Measure from the crossover

\( \alpha \) starts at the natural commutation point, 30° after the phase-voltage zero — so \( \alpha = 0 \) means "act like a diode".

✓ Six pulses, 120° each

Fire every 60° in the order \( T_1 \dots T_6 \). Double-pulse or pulse-train firing is mandatory to start.

✓ Two quadrants

\( \alpha > 90° \) inverts, with a DC-side EMF. Stop short of 180° by the margin angle — typically \( \alpha_{max} \approx 150° \).

✓ Supply cost

\( \mathrm{PF} = 0.955\cos\alpha \), harmonics \( 6k\pm1 \). Apparent power does not fall with speed at constant torque.

✓ The drive equation

\( 1.35V_L\cos\alpha = E_a + I_aR_a \). Torque sets \( I_a \), speed sets \( E_a \), the controller sets \( \alpha \).

Formula sheet · Chapter 9

Full converter outputcontinuous conduction — the key equation

\( V_{dc} = \dfrac{3V_{mL}}{\pi}\cos\alpha = 1.35\,V_L\cos\alpha \)

In terms of phase voltagesame thing, \(\sqrt3\) apart

\( V_{dc} = 2.34\,V_{ph}\cos\alpha \)

Three-phase half-wave3-pulse, needs a neutral

\( V_{dc} = 1.17\,V_{ph}\cos\alpha \)

Three-phase semiconverterone quadrant, always positive

\( V_{dc} = \dfrac{3V_{mL}}{2\pi}\bigl(1+\cos\alpha\bigr) \)

Firing referencethe 30° that catches everyone

\( \text{fire at } \omega t = 30^\circ + \alpha \text{ (phase reference)} \)

Device ratings120° conduction, constant \(I_o\)

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

Supply currentunchanged from Ch. 7, shifted by \(\alpha\)

\( I_s = 0.816\,I_o, \qquad I_{s1} = 0.780\,I_o \)

Input power factordistortion × displacement

\( \mathrm{PF} = 0.955\cos\alpha \)

Reactive power drawngrows as the drive slows

\( Q \approx S\sin\alpha, \qquad S = \sqrt3\,V_LI_s \)

Inverter limitpractical \(\alpha_{max} \approx 150^\circ\)

\( \gamma = 180^\circ - \alpha - \mu \;\ge\; \omega t_q \)

DC drive design equationsolve for \(\alpha\)

\( \alpha = \cos^{-1}\!\left(\dfrac{E_a + I_aR_a}{1.35\,V_L}\right) \)

Key terms

Natural commutation point: The crossover where a diode would have taken over. The reference from which \(\alpha\) is measured — 30° after the phase-voltage zero.
Double-pulse firing: Gating each thyristor twice per cycle, 60° apart, so a conducting pair can be established at start-up.
Extinction angle, \(\gamma\): \(180^\circ - \alpha - \mu\). The interval a device has to regain blocking capability. Must exceed \(\omega t_q\).
Commutation overlap, \(\mu\): The interval during which current transfers between devices, caused by source inductance. Eats into the inverter margin — see Ch. 10.
Constant-torque region: Below base speed. Full field, \(\alpha\) swept from 90° to 0°, torque limited by armature current.
Field weakening: Above base speed. \(\alpha = 0\) and the flux is reduced, giving more speed at less torque — constant power.
Cascaded control: A fast inner current loop inside a slower outer speed loop. The standard structure of every DC drive.
Apparent power, \(S\): \(\sqrt3 V_LI_s\). What the cables and transformer must carry — and it does not fall as a phase-controlled drive slows down.

Self-assessment

Test Yourself

Six questions on the converter that made variable-speed DC drives possible.

Chapter 9 · six questions answers hidden until you ask

A three-phase converter on 415 V is commanded to \( \alpha = 0^\circ \) but delivers only 485 V instead of 560 V. What is the most likely cause?

\( 485 = 560\cos 30^\circ \), so the converter is actually firing at 30°. The firing circuit's synchronisation reference is offset by exactly 30° — almost certainly synchronised to the phase-voltage zero rather than to the natural commutation point.

The recognisable signature is that the error is exactly 30°, and that the delivered voltage is 86.6% of expected at every commanded angle.

Why this matters beyond the voltage error: the same offset shifts the inverter end too. A controller limited to \( \alpha_{max} = 150^\circ \) would really be firing at 180°, with no margin — so the drive would work as a rectifier and destroy itself the first time it tried to regenerate.

Other candidates worth ruling out: commutation overlap from a weak supply (but that gives a load-dependent droop, not a fixed 13%), and a supply voltage that is simply lower than the nameplate.

Why must a three-phase bridge use double-pulse or pulse-train firing, when a single-phase half-wave circuit is happy with one pulse?

Because a bridge needs two devices conducting simultaneously to make a circuit, and a half-wave rectifier needs only one.

At start-up nothing conducts. Fire \( T_1 \) alone and its current has no return path, so no current flows, so it never reaches the latching current \( I_L \) — and it does not latch. Fire \( T_2 \) 60° later and the same thing happens; \( T_1 \) has long since given up.

Double-pulse firing solves it by gating each device again when its partner is fired, so a pair is established together. Pulse trains solve it more thoroughly, because they also re-fire a device that drops out mid-cycle during discontinuous conduction.

The half-wave circuit has no such problem: its return path is the neutral, which is permanently connected, so one device conducting is a complete circuit.

The general lesson, which returns in Part 4: in any bridge, ask what completes the circuit before assuming a gate signal will do anything.

A drive runs at constant torque. Its speed is halved by advancing \( \alpha \) from 0° to 60°. What happens to the current drawn from the supply, and to the transformer's loading?

The supply current does not change at all, and neither does the transformer's kVA loading.

Constant torque means constant armature current \( I_a \). The supply current is \( I_s = 0.816I_a \), which depends only on \( I_a \) — not on \( \alpha \). So \( I_s \) is unchanged and \( S = \sqrt3V_LI_s \) is unchanged.

What has changed is where that apparent power goes:

At \( \alpha = 0 \): \( \mathrm{PF} = 0.955 \), nearly all of \( S \) is real power.
At \( \alpha = 60^\circ \): \( \mathrm{PF} = 0.478 \), so half the real power — and the rest has become reactive.

The uncomfortable consequence: you cannot size the supply for a low-speed application. A drive that only ever runs at quarter speed still needs full-rated cables, transformer and switchgear. And the utility bills on a power factor of about 0.24.

This is the single strongest practical argument for the chopper-fed DC drive of Chapter 11, which rectifies at \( \alpha = 0 \) and does its speed control on the DC side.

A converter is set to \( \alpha = 100^\circ \) with a purely resistive load. What is the output voltage?

Not \( 1.35V_L\cos 100^\circ \), which would be negative and impossible across a resistor.

The formula \( V_{dc} = 1.35V_L\cos\alpha \) assumes continuous conduction, which requires enough inductance to keep current flowing after the segment voltage goes negative. A resistor stores nothing, so its current reaches zero the moment the segment voltage does, and conduction becomes discontinuous.

For a three-phase bridge into a resistive load, discontinuous conduction begins at \( \alpha = 60^\circ \) — the point at which the segment first reaches zero within its own 60° window. Beyond that the output is a series of isolated positive humps, and the average follows a different, always-positive expression.

The habit that prevents the error is the same as in Chapter 8: before choosing a formula, write down one word — continuous or discontinuous. And remember that a resistive load can never invert, at any firing angle, because inversion needs a source of energy on the DC side.

Two identical converters face each other across a DC line — an HVDC link. One is at \( \alpha = 18^\circ \), the other at \( \alpha = 142^\circ \). Which way does power flow, and how would you reverse it?

Power flows from the 18° terminal to the 142° terminal.

The 18° converter produces \( +0.951 \times 1.35V_L \) — a large positive DC voltage — so it rectifies and pushes power into the line. The 142° converter produces \( -0.788 \times 1.35V_L \), a negative voltage, so it inverts and delivers power to its AC system. Current flows in one direction through the line, driven by the voltage difference.

To reverse the flow, swap the two firing angles. Set the first terminal to 142° and the second to 18°. The DC current keeps flowing in the same direction — it must, because the thyristors are unidirectional — but both voltages reverse, so the power reverses.

Two features of this worth appreciating:

Nothing mechanical moves. Reversing a gigawatt takes a few hundred milliseconds of coordinated control action.
The DC current never reverses, so no device is asked to do anything it cannot. The polarity of the line reverses instead — which is why HVDC cables must be designed for polarity reversal, and why voltage-source HVDC, which reverses current instead, is preferred for some cable systems.

A regenerating drive at \( \alpha = 150^\circ \) suffers a commutation failure during a supply dip. Trace what happens, and state two design measures that would have prevented it.

What happens, step by step:

The dip reduces the line voltage available to commutate the current, so the overlap angle \( \mu \) grows.
The extinction angle \( \gamma = 180^\circ - \alpha - \mu \) shrinks. At \( \alpha = 150^\circ \) there was only 30° to start with.
If \( \gamma \) falls below \( \omega t_q \), the outgoing thyristor does not recover its blocking capability before forward voltage returns.
It conducts again. Now two devices in the same leg are on, short-circuiting the supply — and the DC-side EMF keeps feeding current into the fault.
The current rises within a half cycle to whatever the supply impedance allows.

Two preventive measures:

A voltage-dependent firing limit. Monitor the supply and retard \( \alpha \) automatically when it falls, so the margin is preserved rather than fixed. This is standard in commercial drives and would have prevented this failure outright.
A more conservative \( \alpha_{max} \) — say 145° — trading a little regenerative capability for margin. Alternatively, an inverter-grade thyristor with a smaller \( t_q \), or a stiffer supply with lower source inductance to reduce \( \mu \).

And regardless: fast \( I^2t \)-coordinated semiconductor fuses, because a converter that can fail this way must fail safely.

Practice

Problems

Almost every problem here is one of two calculations. Identify which before starting:

Given \( \alpha \), find the output — apply \( V_{dc} = 1.35V_L\cos\alpha \) and check the conduction mode.
Given the load, find \( \alpha \) — write \( V_{dc} = E_a + I_aR_a \), then invert the converter equation.

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

A three-phase full converter runs from 400 V, 50 Hz. Find \( V_{dc} \) at \( \alpha = 0°, 45°, 90°, 135° \), and state the mode in each case.
State the firing instant, measured from the phase-voltage zero crossing, for a commanded \( \alpha = 40° \). Explain why the answer is not 40°.
A 415 V converter feeds a highly inductive load drawing 120 A. Find the average and RMS thyristor currents, the supply RMS current, and the PIV.
A 300 V, 60 A DC motor with \( R_a = 0.35\ \Omega \) is fed from a three-phase full converter on 415 V. Find the firing angle and the back-EMF at rated load.
A three-phase semiconverter on 415 V must deliver 250 V. Find the firing angle, and compare it with the angle a full converter would need.
For the drive of Problem 4, find the supply RMS current, the input power factor and the reactive power at rated load. Repeat at half speed with the same torque.
A converter inverts at \( \alpha = 140° \) on a 50 Hz supply. With \( t_q = 80\ \mu\text{s} \) and \( \mu = 15° \), find the extinction angle and state whether the margin is adequate.
Explain why a three-phase bridge cannot be started with a single gate pulse per thyristor, and describe two firing schemes that solve it.
A 415 V converter drives a hoist. Lowering at 40 A with \( E_a = 180 \) V and \( R_a = 0.3\ \Omega \), find the firing angle and the power returned to the supply over a 90-second descent.
Show that a three-phase bridge feeding a resistive load enters discontinuous conduction at \( \alpha = 60° \), and explain why the \( \cos\alpha \) formula fails beyond it.
A 500 kW DC drive spends 80% of its operating hours at 40% speed and constant torque. Estimate the annual reactive-energy demand, and evaluate whether a chopper-fed drive would be justified.
An HVDC link uses two three-phase bridges on 230 kV converter-transformer secondaries. Explain how power flow is reversed, what limits the inverter-end firing angle, and why the DC line must tolerate polarity reversal.

Tip: \( V_{dc} = 1.35\,V_L\cos\alpha \) is worth committing to memory outright, along with the fact that a 415 V line gives 560 V at \( \alpha = 0 \). Almost every three-phase converter problem starts by evaluating that number, and having it at hand turns a calculation into a sanity check.