Set 30 — AC Voltage Controllers
Every converter since Set 9 has passed through DC. Rectify then invert, rectify then chop — the link was what decoupled the input from the output and made everything controllable. Some applications need none of it. A heater, a lamp, a motor being started: all they need is less voltage at the same frequency, and putting two thyristors back to back delivers exactly that.
It is the simplest converter in this book — two devices, no reactive components, no commutation circuit — and one of the worst behaved. The output is a chopped sine with harmonics at every odd order, the power factor falls with the output, and with an inductive load the firing angle cannot go below the load angle at all.
With a resistive load the output is the sine with a bite taken out:
\[ V_o = V_s\sqrt{\frac{1}{\pi}\left(\pi-\alpha+\frac{\sin2\alpha}{2}\right)}, \qquad PF = \frac{V_o}{V_s} \]With an inductive load the current continues past the zero crossing to an extinction angle:
\[ \sin(\beta-\phi) = \sin(\alpha-\phi)e^{-(\beta-\alpha)/\tan\phi}, \qquad \gamma = \beta-\alpha \]And control is only possible for \(\alpha > \phi\) — below that the controller is fully on.
The three-phase controller has three conduction modes set by the firing angle:
\[ 0\le\alpha\le60^\circ:\ \text{3 or 2 devices}; \quad 60^\circ\text{–}90^\circ:\ \text{2}; \quad 90^\circ\text{–}150^\circ:\ \text{2 or 0} \]Integral cycle control switches whole cycles and produces no harmonics at all:
\[ k = \frac{n}{n+m}, \qquad V_o = V_s\sqrt{k}, \qquad PF = \sqrt{k} \]Induction motor torque follows the square of the voltage:
\[ T \propto V^2, \qquad P_{rotor\ loss} = sP_{airgap}, \qquad \eta_{max} = 1-s \]A soft starter limits current at the cost of torque:
\[ \frac{I_{start}}{I_{DOL}} = \frac{V}{V_{rated}}, \qquad \frac{T_{start}}{T_{DOL}} = \left(\frac{V}{V_{rated}}\right)^2 \]
A single-phase AC voltage controller feeds a \(10\ \Omega\) resistive heater from 230 V, 50 Hz at \(\alpha = 60^\circ\). Find the RMS output voltage, current and power, the fundamental and THD, and the power factor. Show that the power factor equals the voltage ratio.
The RMS output. Each half cycle carries the sine from \(\alpha\) to \(\pi\), so integrating the square over that interval:
Note that removing a third of each half cycle removes only 10% of the RMS voltage. The sine is small near its zero crossing, so early conduction contributes little — which makes the control characteristic strongly nonlinear near \(\alpha = 0\).
Current and power follow directly, since the load is resistive:
The power factor is the voltage ratio, and this is exact for a resistive load:
Neat, and worth remembering: the source current is in phase with the source voltage whenever it flows, so all of the power factor degradation is distortion, none of it displacement — at least at the level of the whole waveform. The fundamental does lag, as the next step shows.
The fundamental. Fourier analysis of the chopped waveform gives:
The \(a_1\) term is non-zero, meaning the fundamental is phase-shifted from the supply — so there is a displacement component, and the load draws reactive power despite being a pure resistor. Phase control manufactures reactive power out of nothing.
The control characteristic:
| \(\alpha\) | \(V_o\) | \(P\) | PF |
|---|---|---|---|
| 0° | 230.0 V | 5.29 kW | 1.000 |
| 30° | 224.9 V | 5.06 kW | 0.978 |
| 60° | 206.3 V | 4.26 kW | 0.897 |
| 90° | 162.6 V | 2.64 kW | 0.707 |
| 120° | 105.8 V | 1.12 kW | 0.460 |
| 180° | 0 | 0 | — |
Half power at 90°, and the useful control range is compressed into the middle of the firing range — the first and last 30° barely change anything. A closed-loop controller must account for that varying gain.
Where this is used, and where it is not:
| Suitable | Unsuitable |
|---|---|
| Resistive heating | anything harmonic-sensitive |
| Incandescent lamp dimming | electronic loads |
| Motor soft starting | continuous motor speed control |
| Transformer tap-change assist | power factor correction |
The two motor entries look contradictory and are not. Soft starting uses the controller for a few seconds and then bypasses it; speed control would leave it in circuit permanently, with the consequences of Problem 5.
The same controller feeds a load of \(R = 5\ \Omega\) and \(L = 15\) mH at \(\alpha = 60^\circ\). Find the load angle, the extinction and conduction angles, and explain what happens for firing angles below the load angle.
The load parameters:
The current does not stop at the voltage zero. Inductive energy keeps it flowing past 180°, holding the thyristor on until it reaches zero at the extinction angle \(\beta\):
The same transcendental equation as the RL rectifier of Set 9 — it must be solved numerically, and the answer determines everything else.
Solving numerically:
A gap of 17.5° in every half cycle during which no current flows at all. Note also that the current continues 42.5° past the voltage zero — during which the thyristor is conducting with the supply voltage reversed, returning stored inductive energy.
The critical constraint. Consider what happens as \(\alpha\) is reduced towards \(\phi\):
If the current from one half cycle has not reached zero when the other thyristor is fired, the gate pulse is wasted — the device cannot turn on because it is reverse biased by its still-conducting partner. The controller loses control.
What actually happens below the load angle:
| \(\alpha\) | Behaviour | Output |
|---|---|---|
| \(\alpha < \phi\) | no control | full voltage |
| \(\alpha = \phi\) | boundary, \(\gamma = 180^\circ\) | full voltage |
| \(\phi < \alpha < 180^\circ\) | control range | reduced |
| \(\alpha \to 180^\circ\) | — | zero |
So the usable range is only \(\phi\) to 180° — here 43.3° to 180°. For a highly inductive load with \(\phi = 80^\circ\), the range shrinks to 80–180°, and the output cannot be trimmed at all near full power.
The practical consequence for gate drive. A short gate pulse can fail even inside the control range:
The standard remedies are the ones Set 5 designed: a long gate pulse, or a train of pulses extending until the device certainly conducts. An AC controller feeding a motor — whose load angle changes from nearly 90° at standstill to perhaps 30° when running — must use pulse trains, because no fixed pulse position works across that range.
A three-phase AC voltage controller feeds a star-connected resistive load of \(10\ \Omega\) per phase from a 415 V supply. Find the output at \(\alpha = 30^\circ\), identify the three conduction modes and their firing-angle ranges, and state the angle beyond which the output is zero.
Why there are modes at all. In a three-wire star connection, current can only flow if at least two phases are conducting — a single conducting phase has no return path:
That last line is what creates the modes. As the firing angle rises the conduction intervals shorten, and the pattern of how many devices overlap changes discontinuously — giving three distinct regions with different governing equations.
The three modes:
| Mode | Range | Devices conducting |
|---|---|---|
| I | \(0 \le \alpha \le 60^\circ\) | alternately 3 and 2 |
| II | \(60^\circ < \alpha \le 90^\circ\) | always 2 |
| III | \(90^\circ < \alpha \le 150^\circ\) | alternately 2 and 0 |
| — | \(\alpha > 150^\circ\) | none — zero output |
The 150° limit is geometric: beyond it, by the time a device is fired, the line-to-line voltage that would drive current through it has already reversed. Compare the single-phase case, which controls all the way to 180°.
Evaluate at \(\alpha = 30^\circ\), in mode I:
Only 2% down on the full 239.6 V — the control is very insensitive in mode I, because at small firing angles the removed conduction is near the voltage zero where little power flows.
The control characteristic across all three modes:
| \(\alpha\) | Mode | \(V_o/V_{ph}\) | Power |
|---|---|---|---|
| 0° | I | 1.000 | 100% |
| 30° | I | 0.978 | 96% |
| 60° | I/II | 0.897 | 80% |
| 90° | II/III | 0.707 | 50% |
| 120° | III | 0.400 | 16% |
| 150° | — | 0 | 0 |
The useful range is compressed into 60–150°, with the first 60° achieving almost nothing. A controller with a linear firing-angle ramp will therefore appear to do nothing for the first third of its travel and then act abruptly — which matters for the soft starter of Problem 6, where the ramp shape determines the acceleration.
The harmonic content. As with any three-wire three-phase converter, the triplens cancel:
So the three-phase controller is inherently better behaved than the single-phase one, which carried a 33% third harmonic. But the amplitudes still vary strongly with the firing angle, peaking near \(\alpha = 90^\circ\) where the waveform is most distorted — and the delta-connected variant, with a circulating path for triplens, behaves differently again.
The same 230 V, \(10\ \Omega\) heater is controlled by integral cycle control, conducting for 5 cycles out of every 10. Find the RMS output, power and power factor, describe the frequency content, and explain why the method is excellent for a heater and unusable for a lamp.
The principle. Instead of chopping each cycle, whole cycles are passed or blocked, with the devices always fired at a voltage zero crossing:
Firing at the zero crossing means the current starts from zero with no step — and that single fact is what removes every harmonic the phase controller produced.
The output:
Exactly half the power for half the cycles, which is the whole appeal — the control characteristic is perfectly linear in power, unlike the phase controller's compressed one.
The power factor:
Which looks poor and is entirely different in character from the phase controller's 0.707 at the same power. Here the displacement factor is unity — every cycle that flows is a perfect sinusoid in phase with the supply. The apparent power penalty comes only from the intermittency, and there is no reactive power at all.
The frequency content is the striking part:
| Component | Phase control | Integral cycle |
|---|---|---|
| Harmonics (3rd, 5th, 7th…) | large | none |
| Displacement factor | lags | unity |
| Reactive power | substantial | zero |
| Subharmonics | none | at \(f/(n+m)\) |
The waveform repeats every ten cycles rather than every one, so the spectrum contains components below the fundamental. Those cannot be filtered by anything of reasonable size, and they are precisely the frequencies that cause visible and audible disturbance.
Why 5 Hz is the worst possible answer for lighting. Human flicker sensitivity peaks between 8 and 10 Hz:
And not only in the controlled load. The 5 Hz current variation drops across the supply impedance, modulating the voltage at the point of common coupling — so every lamp on the same feeder flickers, not just the controlled one. Utilities enforce this through flicker limits, and a large integral-cycle load can fail them easily.
Where the trade comes out right:
| Load | Suitable? | Why |
|---|---|---|
| Furnace, oven, kiln | ideal | thermal time constant minutes — 5 Hz is invisible |
| Water heater | ideal | same |
| Lighting | never | flicker |
| Motors | never | torque pulsation, mechanical resonance |
| Small load on a stiff supply | acceptable | flicker below the limit |
The rule is the load's own time constant. Anything that averages over many seconds cannot tell the difference between continuous half power and alternating full-and-zero power — and gets a harmonic-free supply current as a bonus.
An AC voltage controller is used for continuous speed control of an induction motor by reducing the stator voltage. Analyse the torque, slip and efficiency, determine where the lost power goes, and identify the only load type for which the method is defensible.
Torque follows the square of the voltage. At any given slip the airgap flux is proportional to the voltage, and the torque to the flux times the rotor current — itself proportional to the flux:
Halving the voltage quarters the torque at every slip — so the whole speed–torque curve scales down, and the motor finds a new operating point at higher slip where the reduced curve meets the load.
Which is where the problem lies. The rotor loss is a fixed fraction of the airgap power, set by slip alone:
This is not a property of the controller — it is a property of the induction machine. Any method that slows the rotor by increasing slip dissipates the difference in the rotor bars, where there is no way to remove it except through the rotor's own limited cooling.
Put numbers to it:
| Slip | Speed | Max efficiency | Rotor loss |
|---|---|---|---|
| 0.03 | 97% | 97% | 3% of airgap power |
| 0.20 | 80% | 80% | 20% |
| 0.50 | 50% | 50% | 50% |
| 0.80 | 20% | 20% | 80% |
At half speed, half the airgap power is dissipated in the rotor. A 30 kW motor would be putting kilowatts into a rotor cage designed to dissipate a few hundred watts — it overheats within minutes.
The one defensible case. If the load torque itself falls with speed, the airgap power falls faster than the loss fraction rises:
A 20% speed reduction on a fan halves the shaft power while the rotor loss rises to only 13% of rating — survivable. That is why the method persisted for fan speed control, and why it is useless for anything with constant torque.
Against a proper drive:
| Property | Stator voltage control | VFD (Sets 26–28) |
|---|---|---|
| Speed range | narrow, load-dependent | 0 to above base |
| Torque at low speed | \(\propto V^2\) — collapses | full torque |
| Efficiency at half speed | ≤ 50% | > 95% |
| Rotor heating | severe | normal |
| Supply harmonics | large, varying | fixed, filterable |
| Cost | very low | higher |
| Suitable loads | fans and pumps only | anything |
A VFD keeps the flux constant by varying frequency and voltage together, so the machine always runs at low slip and high efficiency. That is why every row favours it except cost — and why the price collapse of power electronics eliminated stator voltage control from continuous-duty applications.
What the controller is still good for. Everything above concerns continuous operation:
During a start the motor is at high slip anyway — a direct-on-line start dissipates the same rotor energy regardless. The controller merely spreads it over a longer, gentler ramp, and once the motor is up to speed a contactor shorts out the thyristors entirely. That is the soft starter of Problem 6, and it is a legitimate and widespread application.
Design a soft starter for a 30 kW, 415 V induction motor with a direct-on-line starting current of six times full load. Find the rated and starting currents, the effect of a 50% voltage start on current and torque, the thyristor duty during a 10 second ramp, and explain the role of the bypass contactor.
The motor's rated current:
Three hundred and twenty-seven amps drawn for several seconds. On a weak supply that causes a visible voltage dip; on a generator it can stall the set; and it is the reason many utilities restrict direct-on-line starting above a certain rating.
Starting at 50% voltage. The motor at standstill is essentially a short-circuited transformer, so current is proportional to voltage:
Halving the current quarters the torque — the fundamental soft-starter trade, and the reason a soft start is not always possible. If the DOL starting torque is 1.5 times rated, a 50% start gives 0.375 times rated, which will not move a loaded conveyor or a compressor with a head of pressure.
The design constraint, stated properly:
| Load | Breakaway torque | Minimum start voltage |
|---|---|---|
| Centrifugal fan | ~10% of rated | ~30% |
| Centrifugal pump | ~20% | ~40% |
| Conveyor, loaded | ~70% | ~70% |
| Positive-displacement compressor | ~100% | not startable — unload it |
The last row is why soft starters are sold with unloading valves and why some applications need a VFD instead — only a drive can deliver full torque from zero speed.
The thyristor duty during the ramp. Over a 10 second ramp from 30% to 100% voltage, the current averages roughly the 50% figure:
Compare a semiconductor fuse's clearing \(I^2t\) from Set 4 and the difference is stark: this is a thermal duty lasting seconds, not a fault duty lasting milliseconds, so it is the junction temperature rise that matters rather than the fusing integral. The devices must be sized for the ramp, typically three to four times the motor's full-load current.
Conduction loss during the ramp:
Six hundred watts for ten seconds is 6.6 kJ into the device heatsinks — manageable as a transient, and unacceptable continuously. That figure alone explains the next step.
The bypass contactor. Once the motor is at speed, a contactor shorts out the thyristors entirely:
| Benefit | Detail |
|---|---|
| Conduction loss removed | 660 W → nearly zero |
| Heatsink sized for 10 s, not continuous | much smaller |
| Harmonics removed | the supply sees a plain motor again |
| Power factor restored | no phase-control displacement |
| Thyristors protected | not exposed to running transients |
The third row is easily overlooked and is worth as much as the first. A phase-controlled motor injects harmonics into the supply continuously; a bypassed soft starter injects them only during the few seconds of starting, so it has essentially no steady-state power quality impact at all.
The complete specification:
| Item | Value |
|---|---|
| Motor full-load current | 54.6 A |
| Thyristor rating | ~200 A rms, 1200 V |
| Start voltage / ramp | 30% rising to 100% over 10 s |
| Peak starting current | ~164 A (3× FLC) |
| Ramp \(I^2t\) | \(2.7\times10^5\ \text{A}^2\text{s}\) |
| Bypass contactor | AC-1, 60 A |
| Gate drive | pulse train — the load angle varies |
| Protection | semiconductor fuses, thermal model, phase loss |
The gate-drive row returns to Problem 2. A motor's load angle starts near 90° at standstill and falls as it accelerates, so no fixed pulse timing works throughout — the pulse must be repeated until conduction is confirmed, exactly as Set 5 required.
Key Formulas
| Quantity | Relation | Notes |
|---|---|---|
| Output, R load | \(V_s\sqrt{\frac{1}{\pi}\left(\pi-\alpha+\frac{\sin2\alpha}{2}\right)}\) | — |
| Power factor, R load | \(PF = V_o/V_s\) | Exact |
| Fundamental | \(a_1 = \frac{V_m}{2\pi}(\cos2\alpha-1)\), \(b_1 = \frac{V_m}{2\pi}(\sin2\alpha+2(\pi-\alpha))\) | \(a_1 \ne 0\) → displacement |
| Load angle | \(\phi = \tan^{-1}(\omega L/R)\) | Control needs \(\alpha > \phi\) |
| Extinction angle | \(\sin(\beta-\phi) = \sin(\alpha-\phi)e^{-(\beta-\alpha)/\tan\phi}\) | Solve numerically |
| Conduction angle | \(\gamma = \beta-\alpha\) | 180° at \(\alpha = \phi\) |
| 3-φ mode I output | \(V_{ph}\sqrt{1-\frac{3\alpha}{2\pi}+\frac{3\sin2\alpha}{4\pi}}\) | \(0 \le \alpha \le 60^\circ\) |
| 3-φ zero output | \(\alpha > 150^\circ\) | Not 180° |
| Integral cycle output | \(V_o = V_s\sqrt{k}\) | \(k = n/(n+m)\) |
| Its power factor | \(PF = \sqrt{k}\) | Unity displacement |
| Its subharmonic | \(f/(n+m)\) | Flicker risk |
| Motor torque | \(T \propto V^2\) | — |
| Rotor loss | \(sP_{airgap}\) | \(\eta \le 1-s\) |
| Soft start current | \(I \propto V\) | Linear |
| Soft start torque | \(T \propto V^2\) | Quadratic — the trade |
Common Mistakes
Assuming a resistive load draws no reactive power through a phase controller. The fundamental is displaced by the chopping — Problem 1.
Expecting the output to fall linearly with firing angle. The first 30° barely changes anything — Problems 1 and 3.
Firing below the load angle. For \(\alpha < \phi\) the controller has no effect at all — Problem 2.
Using a short gate pulse with an inductive load. The device may be reverse biased when the pulse arrives — use a pulse train — Problems 2 and 6.
Applying the single-phase formula to a three-phase controller. It has three conduction modes with different equations — Problem 3.
Assuming the three-phase controller works to 180°. The output is zero beyond 150° — Problem 3.
Calling integral cycle control harmonic-free without qualification. It has no harmonics and a substantial subharmonic — Problem 4.
Using integral cycle control on lighting. The subharmonic falls in the peak flicker band — Problem 4.
Proposing stator voltage control for constant-torque loads. Efficiency cannot exceed \(1-s\) and the rotor overheats — Problem 5.
Promising a soft start without checking the breakaway torque. At 50% voltage only 25% of the DOL torque is available — Problem 6.
Six problems on the simplest converter in this book. Two thyristors chopped the supply directly and produced 37.7% distortion, a displaced fundamental from a purely resistive load, and a control range that an inductive load truncates at the load angle. Integral cycle control removed every harmonic and introduced a 5 Hz subharmonic instead. And stator voltage control turned out to be limited not by the converter but by the machine: efficiency cannot exceed \(1-s\), whatever the electronics does.
All of it changed only the magnitude of the output. The frequency stayed at 50 Hz throughout, because two anti-parallel thyristors can only select which parts of the supply waveform reach the load — they cannot synthesise a new one. Selecting from three phases rather than one, and selecting continuously rather than once per half cycle, changes that: a converter that picks the most suitable input phase at each instant can construct an output at a genuinely different frequency, with no DC link at all.
Next: Set 31 — Cycloconverters and Matrix Converters, where thyristor groups synthesise a low-frequency output directly from the mains, the matrix converter replaces them with bidirectional switches and hits a hard 0.866 voltage ratio, and the absence of any freewheeling path makes commutation a four-step problem.