Chapter 21 of 30 — the first of Part 5
- Why controlling AC without changing its frequency is the simplest converter in this book — and the one with the worst power factor.
- Phase control with a resistive load: \( V_o = V_s\sqrt{\dfrac{\pi - \alpha + \sin2\alpha/2}{\pi}} \), and why the power factor equals \( V_o/V_s \).
- What an inductive load changes: the extinction angle \( \beta \), the transcendental equation that locates it, and why control is lost for \( \alpha < \varphi \).
- Why a single gate pulse fails on an inductive load, and what a pulse train fixes.
- Integral-cycle control: no harmonics at all, but flicker — and exactly when that trade is right.
- Three-phase controllers, their conduction modes, and the 150° limit.
- Why this topology dominates soft starters and heater control while being unacceptable for lighting on a modern grid.
Changing AC Without Changing Its Frequency
Part 2 turned AC into DC. Part 4 turned DC into AC. Part 5 asks the remaining question: can you turn AC into different AC directly, without a DC link in between?
The answer splits by what you want to change. Change the frequency and you need a cycloconverter or a matrix converter — Chapter 22. Change only the voltage, keeping the frequency, and the converter is almost embarrassingly simple: two thyristors back to back in series with the load.
Phase Control with a Resistive Load
With a purely resistive load the current follows the voltage exactly, so the thyristor turns off at the supply's zero crossing. Conduction runs from \( \alpha \) to \( \pi \) in each half cycle, and the analysis needs one integral.
- Integrate the square of what actually reaches the load
The average is taken over \( \pi \), not \( 2\pi \), because both half cycles are identical.Working\[ V_o^2 = \frac{1}{\pi}\int_{\alpha}^{\pi} \bigl(\sqrt2\,V_s\sin\theta\bigr)^2 d\theta = \frac{2V_s^2}{\pi}\int_{\alpha}^{\pi}\frac{1 - \cos2\theta}{2}\,d\theta \]
- Evaluate
Working\[ V_o^2 = \frac{V_s^2}{\pi}\left[\theta - \frac{\sin2\theta}{2}\right]_{\alpha}^{\pi} = \frac{V_s^2}{\pi}\left(\pi - \alpha + \frac{\sin2\alpha}{2}\right) \]
- Check the limits before trusting it At \( \alpha = 0 \): \( V_o = V_s \) — full output, as it must be. At \( \alpha = \pi \): \( V_o = 0 \). Both correct.
- Get the power factor free
The source current is \( V_o/R \) in RMS, so the apparent power is \( V_sV_o/R \) and the real power is \( V_o^2/R \):
The power factor is simply the per-unit output voltage. Half voltage means a power factor of 0.5 — with a load that has no reactance at all.Result\[ \text{PF} = \frac{P}{S} = \frac{V_o^2/R}{V_sV_o/R} = \frac{V_o}{V_s} \]
Note the relationship between \( \alpha \) and output is far from linear. Half voltage occurs at \( \alpha = 90^\circ \); half power occurs at about \( \alpha = 90^\circ \) too, since \( P \propto V_o^2 \) — a coincidence worth checking rather than assuming.
An AC voltage controller on a load of impedance angle \( \varphi \). Watch the output voltage and load current as you delay the firing — then raise \( \varphi \) and find the setting below which the controller stops controlling anything.
—
What an Inductive Load Changes
Real loads — fans, pumps, transformers, motors — are inductive, and the analysis changes in a way that has practical consequences every user of a fan controller has met.
The current now lags the voltage, so it does not reach zero at the supply's zero crossing. The thyristor keeps conducting past \( \pi \), into a half cycle where the supply voltage has already reversed.
- Solve the circuit equation during conduction
With \( v = \sqrt2\,V_s\sin\theta \) applied to \( R + j\omega L \) from \( \theta = \alpha \) with zero initial current:
with \( Z = \sqrt{R^2 + (\omega L)^2} \) and \( \varphi = \tan^{-1}(\omega L/R) \). The first term is the steady state; the second is the transient that makes \( i(\alpha) = 0 \).Working\[ i(\theta) = \frac{\sqrt2\,V_s}{Z}\left[\sin(\theta - \varphi) - \sin(\alpha - \varphi)\,e^{-(\theta-\alpha)/\tan\varphi}\right] \]
- Find where the current stops
The extinction angle \( \beta \) is where \( i = 0 \) again:
Transcendental — no closed form. It is solved numerically, and the conduction angle is \( \gamma = \beta - \alpha \).Result\[ \sin(\beta - \varphi) = \sin(\alpha - \varphi)\,e^{-(\beta-\alpha)/\tan\varphi} \]
- Find the RMS output over the actual conduction window
which reduces to the resistive result when \( \beta = \pi \).Result\[ V_o = V_s\sqrt{\frac{1}{\pi}\left[(\beta - \alpha) - \frac{\sin2\beta - \sin2\alpha}{2}\right]} \]
- Now examine the case \(\alpha \le \varphi\) Setting \( \alpha = \varphi \) makes the transient term vanish, so \( i = (\sqrt2V_s/Z)\sin(\theta-\varphi) \) — a pure sinusoid that reaches zero exactly at \( \theta = \pi + \varphi \), by which time the other thyristor has already been fired. Conduction never breaks.
The useful control range is \( \pi - \varphi \), not \( \pi \). For a fan motor with \( \varphi = 60^\circ \), a third of the control knob does nothing at all.
Integral-Cycle Control
Phase control's harmonics come from switching within a cycle. There is an alternative that never does: switch only at zero crossings, and control the output by passing whole cycles and blocking whole cycles.
Pass \( n \) cycles out of every \( n + m \), and since power is proportional to time:
Every cycle that is passed is a complete, undistorted sine, so there are no harmonics of the supply frequency at all — a striking advantage over phase control.
| Criterion | Phase control | Integral cycle |
|---|---|---|
| Switching instant | Mid-cycle, at \(\alpha\) | Zero crossing only |
| Harmonics of \(f_s\) | Large — 3rd, 5th, 7th | None |
| Sub-harmonics | None | Present — at \(f_s/(n+m)\) |
| Switching \(dv/dt\), EMI | Severe — full voltage step | Negligible |
| Power factor | \(V_o/V_s\), and distorted | \(\sqrt{n/(n+m)}\), undistorted |
| Resolution | Continuous | Quantised to \(1/(n+m)\) |
| Response time | Half a cycle | \((n+m)\) cycles |
| Suits | Lighting, motors, anything fast | Heaters, ovens, anything thermal |
Three-Phase Controllers
Above a few kilowatts the load is three-phase, and the controller becomes three pairs of thyristors — one pair per line. The analysis is harder than three times the single-phase case, because current can only flow if at least two phases have a device conducting.
| Firing angle | Devices conducting | What happens |
|---|---|---|
| \(0 \le \alpha < 60^\circ\) | alternately 3 and 2 | Mostly continuous; output close to full |
| \(60^\circ \le \alpha < 90^\circ\) | always 2 | Current flows between two lines at a time |
| \(90^\circ \le \alpha < 150^\circ\) | alternately 2 and 0 | Discontinuous — intervals with no current at all |
| \(\alpha \ge 150^\circ\) | none | Output is zero |
Worked Examples
Problem. A 230 V, 50 Hz supply feeds a 10 Ω heater through an AC voltage controller at \( \alpha = 60^\circ \). Find the RMS output voltage, the load power, the RMS supply current, the apparent power and the power factor.
Compare with full output. Unchopped, the heater would take \( 230^2/10 = 5290 \) W. So 60° of delay has reduced the power to 80.4% — a much smaller reduction than the angle suggests, because the first 60° of a sine carries little area.
The design point worth noticing. The relationship between \( \alpha \) and power is strongly non-linear: 60° costs 20% of the power, but 120° costs 80%. A controller with a linear knob feels wrong at both ends, which is why good controllers linearise the scale in firmware.
Problem. A controller on 230 V, 50 Hz feeds \( R = 5\ \Omega \), \( L = 20 \) mH at \( \alpha = 90^\circ \). Find \( \varphi \), the extinction angle, the conduction angle and the RMS output.
Check the control range first. \( \alpha = 90^\circ > \varphi = 51.5^\circ \), so control exists. Had \( \alpha \) been below 51.5°, the output would have been fixed at 230 V.
RMS output over the actual conduction window:
Compare with the resistive case at the same \( \alpha \). A resistor at 90° would give \( 0.707 \times 230 = 163 \) V. The inductive load gives 178 V — 9% more, because conduction continues past the zero crossing into the reversed half cycle.
The practical consequence. A controller calibrated on a resistive load will over-deliver on an inductive one. This is why soft starters are set up on the actual motor rather than on a test resistor.
Problem. A 15 kW oven element on 400 V is to run at 40% power using integral-cycle control on a 50 Hz supply. Choose a pattern, find the RMS voltage and power factor, and check the flicker frequency.
Now the flicker check, which is the part that decides the design.
10 Hz is squarely in the worst band for human flicker perception. A 15 kW load switching at 10 Hz across the supply impedance will visibly modulate the lighting on the same feeder.
Two ways to improve it:
- Distributed firing. Instead of 2 on, 3 off, use a pattern such as on-off-on-off-off repeating — same average, but the dominant modulation moves to 25 Hz, well above the sensitive band. This is a firmware change with no hardware cost.
- A shorter period. \( n = 2 \), \( m = 3 \) at every opportunity rather than blocks of ten gives the same result more smoothly.
And note what has not improved: the power factor is still 0.632, and no firing pattern changes that. Integral-cycle control avoids harmonics but not the poor power factor, because the current is still absent for 60% of the time. The supply must be sized for 9.5 kVA to deliver 6 kW.
Problem. A 30 kW, 400 V induction motor draws 55 A at full load and 380 A direct-on-line. A soft starter must limit the starting current to 200 A. Find the required starting voltage, the resulting starting torque as a fraction of direct-on-line, and the initial firing angle.
So the current is halved but the torque falls to 28%. This is the fundamental trade of every soft starter, and it must be checked against the load before the design is accepted.
The firing angle. A motor at standstill has a low power factor — locked-rotor \( \varphi \approx 70^\circ \) is typical. From the extinction-angle relation, producing 0.526 per-unit voltage on such a load needs approximately:
Now the essential check. Will 28% of direct-on-line torque actually accelerate the load?
- A centrifugal pump or fan: yes comfortably. Their load torque goes as speed squared, so at standstill they need almost nothing.
- A loaded conveyor or a screw compressor: probably not. Breakaway torque may exceed 50% of full-load torque, and the motor will stall — drawing 200 A indefinitely until the overload trips.
If the torque is insufficient, the options are a higher current limit (accepting more supply disturbance), a kick-start pulse of full voltage for a few hundred milliseconds, or a variable-frequency drive — which produces full torque at any speed and is the correct answer for genuinely hard-starting loads.
Summary & Formula Sheet
Chapter 21 in five sentences:
- Two thyristors in inverse parallel, fired at \( \alpha \) and commutated naturally, vary the RMS output at fixed frequency — the cheapest converter in the book.
- With a resistive load, \( V_o = V_s\sqrt{(\pi-\alpha+\sin2\alpha/2)/\pi} \) and the power factor equals \( V_o/V_s \), degraded by distortion as much as by displacement.
- An inductive load extends conduction past the zero crossing to an extinction angle \( \beta \), and destroys all control for \( \alpha < \varphi \).
- Integral-cycle control removes every harmonic of the supply and replaces them with sub-harmonics, so it suits thermal loads and nothing else.
- Three-phase controllers need two devices conducting at once, limiting the star-connected range to 150°, and their dominant use is soft starting.
Key terms
- AC voltage controller
- A converter that varies the RMS value of an AC supply without changing its frequency.
- Firing angle, \(\alpha\)
- Delay from the supply's zero crossing to the gate pulse. The only control variable.
- Extinction angle, \(\beta\)
- The angle at which the load current reaches zero and the device commutates naturally.
- Conduction angle, \(\gamma\)
- \(\beta - \alpha\). Equals \(180^\circ - \alpha\) for a resistive load; longer for an inductive one.
- Load angle, \(\varphi\)
- \(\tan^{-1}(\omega L/R)\). Sets the lower limit of the useful control range.
- Distortion factor
- \(I_1/I_{rms}\). The part of the power factor lost to harmonic current.
- Displacement factor
- \(\cos\varphi_1\). The part lost to the fundamental current lagging the voltage.
- Pulse train
- A burst of gate pulses rather than a single one, ensuring the incoming device fires once the outgoing one stops.
- Integral-cycle control
- Passing whole cycles and blocking whole cycles. No harmonics; sub-harmonics instead.
- Sub-harmonic
- A component below the supply frequency, produced when the switching pattern repeats over several cycles.
- Soft starter
- A three-phase AC controller that ramps a motor's voltage at start, then is bypassed by a contactor.
Test Yourself
A phase-controlled heater has a purely resistive load, yet its power factor at half output is 0.5. Where does the reactive power come from?
There is no reactive power in the classical sense. The power factor is poor because of distortion, which the simple \( \cos\varphi \) picture does not capture at all.
The definition that always holds is \( \text{PF} = P/S \), where \( S = V_{rms}I_{rms} \). With a chopped waveform, \( I_{rms} \) contains a great deal of harmonic current that carries no real power — because the supply voltage is a pure sinusoid, and a harmonic current at a frequency the voltage does not contain transfers nothing.
The correct decomposition:
\[ \text{PF} = \underbrace{\frac{I_1}{I_{rms}}}_{\text{distortion}} \times \underbrace{\cos\varphi_1}_{\text{displacement}} \]
Both terms are degraded here, by the same mechanism:
- Chopping produces 3rd, 5th and 7th harmonic current, so \( I_1 < I_{rms} \) — the distortion factor falls.
- Delaying the firing shifts the fundamental component later in the cycle, so even into a resistor the fundamental current lags — the displacement factor falls too.
Why the distinction has practical teeth. Power-factor correction capacitors correct displacement only. Fit one here and you improve the second factor slightly while doing nothing about the first — and you create a resonant circuit with the supply inductance that the 5th and 7th harmonics may excite, producing overvoltages and capacitor failure.
Correcting distortion requires different equipment: a passive harmonic filter tuned to the offending orders, an active filter that injects cancelling current, or — best — a converter that does not distort in the first place. This is the argument that pushes larger installations away from phase control entirely.
Why does a fan speed controller do nothing over the first third of its travel?
Because for \( \alpha < \varphi \) the load current never reaches zero, so the thyristors conduct continuously and the load sees the full supply.
Trace it. The load current is
\[ i(\theta) \propto \sin(\theta - \varphi) - \sin(\alpha - \varphi)e^{-(\theta-\alpha)/\tan\varphi} \]
At \( \alpha = \varphi \) the second term vanishes, leaving a pure sinusoid \( \sin(\theta - \varphi) \) that reaches zero at \( \theta = \pi + \varphi \). But the opposite thyristor is fired at \( \pi + \alpha = \pi + \varphi \) — the same instant. Conduction passes seamlessly from one device to the other and never breaks.
For any \( \alpha < \varphi \) the situation is worse still: the outgoing device is still conducting when the incoming one is fired, so the transition is seamless with margin to spare. The output is a complete, undistorted sine at full amplitude.
Put numbers on a real fan. A shaded-pole motor has \( \varphi \approx 65^\circ \). So:
- \( \alpha = 0 \) to \( 65^\circ \): no effect at all — 36% of the control's travel.
- \( \alpha = 65^\circ \) to \( 180^\circ \): output falls from 100% to 0.
And the remaining range is not linear either, so the useful settings are compressed into a narrow band. This is why such controllers feel coarse and often have detented steps rather than a continuous dial — the manufacturer has picked a handful of angles that give distinguishable speeds.
The proper fix is not a better controller. Reducing an induction motor's voltage increases its slip and its rotor loss, so voltage control of a running motor is thermally poor as well as coarse. Modern fans use an electronically commutated (BLDC) motor with a small inverter, giving genuine speed control at high efficiency — Chapter 27.
Integral-cycle control produces no harmonics. Why is it not used for everything?
Because it produces sub-harmonics instead, and for most loads those are worse than the harmonics it avoided.
The mechanism. With \( n \) cycles on and \( m \) off, the pattern repeats every \( n+m \) supply cycles. The waveform's true fundamental is therefore \( f_s/(n+m) \), and Fourier places energy there and at its multiples.
With \( n = 3 \), \( m = 7 \) on 50 Hz, that is 5 Hz.
Why 5 Hz is a serious problem:
- Flicker. Human vision is most sensitive to luminance modulation between about 5 and 15 Hz. Every lamp on the same feeder will visibly pulse, and standards such as EN 61000-3-3 limit this explicitly.
- Network disturbance. The load switches on and off at 5 Hz across the supply impedance, so the voltage dips and recovers at 5 Hz — affecting other customers, not just the equipment concerned.
- Torque pulsation. On a rotating load, full torque then no torque is a violent excitation. Integral-cycle control is never used for motors for this reason alone.
- Mechanical resonance. Transformers and machines have resonances in the single-digit hertz range, and exciting one produces audible drumming and fatigue.
Where it is right: loads whose thermal time constant far exceeds \( n+m \) cycles — ovens, furnaces, process and barrel heaters. There the modulation is invisible in the controlled variable, and avoiding harmonics and switching EMI is a real gain.
The modern improvement is distributed firing: spread the on-cycles evenly rather than in a block. The average is identical but the lowest sub-harmonic moves up several fold, often enough to pass the flicker limits. It costs nothing but firmware, and is worth specifying.
One thing it does not fix. The power factor is still \( \sqrt{n/(n+m)} \) — 0.55 at 30% power — because the current is simply absent most of the time. No firing pattern changes that.
Why does a three-phase controller stop working at \(\alpha = 150^\circ\) when a single-phase one works to 180°?
Because a three-phase star load with no neutral needs two devices conducting at once to complete a circuit, and beyond 150° no two can ever overlap.
The single-phase case. One device conducting completes the circuit, because the return path — neutral or the other line — is permanently connected. Current can flow from \( \alpha \) right up to \( 180^\circ \).
The three-phase star case. Current entering one line must leave through another, so two devices in different lines must conduct simultaneously. Each is fired at \( \alpha \) after its own phase's zero crossing, and the phases are 120° apart.
What the driving voltage is. When two devices conduct, the voltage available is the line-to-line voltage between those two phases, which leads the phase voltage by 30°. Beyond \( \alpha = 150^\circ \), every firing instant falls after the relevant line voltage has already passed through zero and reversed — so no pair of devices is ever forward-biased together. Nothing conducts, regardless of the gate signal.
The delta-connected exception. If the controller sits inside a delta-connected load, each device sees one load branch directly with a complete circuit of its own. The single-phase analysis applies and the full 0°–180° range is available.
This is why large soft starters offer an "inside-delta" connection where the motor's six terminals are brought out: the extra 30° of range genuinely matters when starting a stiff load, and the device currents are lower by \( \sqrt3 \) as well.
A related benefit worth noting. With an isolated neutral, triplen harmonics cannot flow — the same cancellation as Chapters 17 and 18. So a three-phase controller's line current contains 5th, 7th, 11th and 13th but no 3rd, which is considerably better than three independent single-phase controllers would produce.
An AC controller feeding a transformer produces a large DC current and the transformer overheats. What is the likely cause?
Almost certainly a single gate pulse rather than a pulse train, leaving one of the two thyristors unfired.
The failure sequence, on an inductive load:
- \( T_1 \) is fired at \( \alpha \) and, because the current lags, keeps conducting past \( 180^\circ \).
- At \( 180^\circ + \alpha \) the controller issues \( T_2 \)'s pulse — but \( T_1 \) is still conducting, so \( T_2 \) is held reverse-biased. The pulse achieves nothing.
- \( T_1 \) stops at \( \beta \), by which time \( T_2 \)'s pulse has ended.
- \( T_2 \) never conducts. Only positive half cycles reach the load.
Why a transformer suffers so badly. Its winding resistance is a fraction of an ohm, so even a few volts of DC drives a large unidirectional current. That current biases the core towards saturation on one half cycle, the magnetising current rises sharply, and the winding heats with no useful output. Transformers fed this way fail in hours to days.
The fix: a pulse train. Apply gate pulses continuously (or as a burst at a few kilohertz) from \( \alpha \) until the end of the half cycle. Whenever \( T_1 \) finally stops, a pulse is present to turn \( T_2 \) on. Every competent phase-control circuit does this.
Two related causes worth checking:
- Asymmetric firing. If the two devices are fired at slightly different angles — a component tolerance in an analogue trigger — a smaller but persistent DC component appears. Measure both firing instants on a scope.
- One device failed open. Rare, but it produces the same symptom exactly. Check the voltage across each device.
And a design note. Feeding a transformer from a phase-controlled supply is difficult even when done correctly: the chopped waveform has a DC-free average only if perfectly symmetric, and the inrush at each firing is severe. Where a transformer must be voltage-controlled, integral-cycle control or a tap changer is usually the better choice.
Why fit a soft starter when a variable-frequency drive is better in every technical respect?
Because for a load that starts occasionally and then runs at full speed, the VFD's advantages are never used and its disadvantages run continuously.
What the soft starter actually costs and does:
- About a fifth the price of a VFD of the same rating — six thyristors against six IGBTs, a rectifier, a DC link and a control platform.
- About a third the size, with no DC link capacitors and much less heatsinking.
- Bypassed once running. A contactor shorts out the thyristors after the ramp, so the motor runs directly on the mains with zero converter loss and no harmonics whatever. A VFD's 2–3% conversion loss runs 8,760 hours a year.
- No harmonics in normal operation, so no supply pollution and no input filter.
- No motor derating. A VFD's PWM output causes extra iron loss and insulation stress; a bypassed soft starter causes neither.
Do the arithmetic for a 30 kW pump running 6,000 hours a year. A VFD losing 2.5% wastes 750 W continuously — 4,500 kWh a year. Over a ten-year life that is a substantial figure, and it buys speed control the pump never uses.
When the VFD is nevertheless right:
- Variable speed is genuinely needed. On a pump or fan, throttling by speed rather than by valve saves energy as the cube of the flow — an enormous gain that dwarfs the conversion loss.
- The load starts hard. A soft starter's torque falls as \( V^2 \), so 50% voltage gives 25% torque. A conveyor or compressor may simply not move. A VFD produces full torque at any speed.
- Frequent starting. Many starts per hour heat the motor; a VFD's controlled acceleration is gentler.
- Precise control of speed, position or torque.
The engineering lesson, which recurs throughout this book: the technically superior converter is not automatically the correct one. Match the capability to the requirement, and count the losses over the whole operating life rather than at the rated point.
Problems
Three habits for AC controller problems:
- Check \( \alpha \) against \( \varphi \) first. Below the load angle there is no control and no calculation to do.
- Remember the power factor is \( P/S \), and that distortion degrades it even into a resistor.
- For an inductive load, find \( \beta \) before anything else. Every other quantity depends on the conduction window.
Problems 1–5 are direct application; 6–9 need judgement; 10–12 are design questions worth discussing in a tutorial.
- A 230 V, 50 Hz controller feeds a 15 Ω resistive heater at \( \alpha = 45^\circ \). Find the RMS output voltage, the power, the supply RMS current and the power factor.
- For the heater of Problem 1, find the firing angle needed for exactly half the maximum power, and comment on whether it equals the angle for half voltage.
- A controller on 400 V, 50 Hz feeds \( R = 8\ \Omega \), \( L = 30 \) mH. Find \( \varphi \), state the useful control range, and find the extinction angle at \( \alpha = 100^\circ \).
- An oven is controlled by integral cycles at 25% power on a 50 Hz supply with a 12-cycle period. Find \( n \) and \( m \), the RMS voltage from a 400 V supply, the power factor and the lowest sub-harmonic frequency.
- A 400 V three-phase controller feeds a star-connected 20 Ω per-phase resistive load at \( \alpha = 60^\circ \). State which conduction mode applies and estimate the output power.
- Show that for a resistive load the power factor of a phase-controlled AC controller equals the per-unit RMS output voltage, and explain why this is not true for an inductive load.
- Explain why a single gate pulse produces a DC component when an AC controller feeds an inductive load, and state two other faults that produce the same symptom.
- Compare phase control and integral-cycle control for (a) a 3 kW immersion heater, (b) a theatre lighting dimmer, and (c) a 2 kW fan. Recommend one for each with reasons.
- A 45 kW motor draws 600 A direct-on-line. A soft starter limits the start to 250 A. Find the starting voltage and torque as fractions of direct-on-line, and state one load type for which this would be inadequate.
- Design the control for a 20 kW, 400 V industrial oven requiring ±2 °C control with negligible flicker. Choose a control method, specify the firing pattern, compute the power factor at 60% output, and state what you would check on commissioning.
- An engineer proposes fitting power-factor correction capacitors to a bank of phase-controlled heaters running at 50% output. Analyse the proposal: how much of the poor power factor would be corrected, what risk is created, and what would you recommend instead?
- A 6-pole, 30 kW induction motor drives a loaded conveyor with a breakaway torque of 60% of full-load torque. Assess whether a soft starter can start it, and if not, propose two alternatives with their respective costs and consequences.