Chapter 27 of 30 — the last of Part 6
- Why a synchronous machine has no slip at all, and what that costs as well as what it buys.
- The one distinction that separates BLDC from PMSM: trapezoidal versus sinusoidal back-EMF — and why it changes the sensor, the modulation and the torque ripple.
- Why six-step commutation gives \( \pi/6 \) of unavoidable torque ripple, and where the 15% figure comes from.
- How Hall sensors give sixty-degree resolution — enough for BLDC, useless for PMSM.
- Saliency: why an interior-magnet rotor produces torque by a second, entirely different mechanism.
- The MTPA current angle — how to get the most torque per amp out of a salient machine, and why it is not simply \( \beta = 0 \).
- Why PM machines are hard to field-weaken, and the uncomfortable failure mode that limits how far you can go.
- How synchronous reluctance and PM-assisted machines fit between the induction motor and the PMSM.
The Machine With No Slip
The induction machine of Chapter 26 makes its rotor field by induction, which is why it must slip. A synchronous machine does not: its rotor carries its own field, from a DC winding or from permanent magnets. The rotor field and the stator field lock together and turn at exactly the same speed.
What varies with load is not the speed but the load angle \( \delta \) — the angular displacement between the rotor field and the stator field. The rotor lags the stator field by an angle that grows with torque, like a spring stretching, while the average speed stays locked.
| Machine | Rotor field source | Distinctive property | Where used |
|---|---|---|---|
| Wound-field synchronous | DC field winding, via slip rings or a brushless exciter | Field is adjustable, so power factor can be commanded — including leading | Large generators; very large drives above roughly 1 MW |
| Surface PM (SPM / BLDC) | Magnets bonded to the rotor surface | \( L_d \approx L_q \) — magnetically round, no reluctance torque | Fans, pumps, drones, appliances, small servos |
| Interior PM (IPM / PMSM) | Magnets buried inside the rotor iron | \( L_q > L_d \) — salient, so reluctance torque adds to magnet torque | Traction, EVs, high-efficiency industrial servos |
| Synchronous reluctance | None — no magnets, no winding | Torque comes entirely from saliency | Efficient industrial pumps and fans, where magnet cost matters |
| PM-assisted SynRel | Small ferrite magnets in a reluctance rotor | Most of the performance of an IPM without rare-earth magnets | Cost-sensitive traction and industrial drives |
BLDC and PMSM: One Distinction
Both machines have permanent magnets on the rotor, three-phase stator windings, and no brushes. They are frequently confused, and drives are frequently mismatched to them. The difference is in one waveform: the shape of the back-EMF, which is decided by how the stator is wound and how the magnets are shaped.
Instantaneous power in a phase is \( p = e\,i \), and torque is total power divided by speed. If \( i \) has the same shape as \( e \), the three phases sum to a constant and the torque is smooth.
- Trapezoidal EMF → rectangular current, injected only during the 120° flat top. Two phases conduct at a time; the third is open.
- Sinusoidal EMF → sinusoidal current, all three phases conducting continuously.
Everything else — the sensors, the modulation, the controller, the torque ripple — follows from this one choice.
| Property | BLDC (trapezoidal) | PMSM (sinusoidal) |
|---|---|---|
| Back-EMF | Trapezoidal, 120° flat top | Sinusoidal |
| Stator winding | Concentrated | Distributed or fractional-slot, shaped for a sinusoidal MMF |
| Current | Rectangular, two phases at a time | Sinusoidal, three phases continuously |
| Position feedback | Three Hall sensors, 60° resolution | Resolver or encoder, typically 12 bits or better |
| Control | Six-step commutation with a current loop | Field-oriented control with SVM |
| Torque ripple | About 15% inherent | Under 2%, limited by cogging and winding imperfection |
| Torque per ampere | Higher for the same peak current | Slightly lower, but smoother |
| Acoustic noise | Higher, with commutation tones | Low |
| Controller cost | Low | Higher — needs transforms and a good position signal |
| Typical use | Fans, pumps, drones, tools, appliances | Servos, machine tools, traction, EV drives |
- Work out the BLDC torque ripple, since the 15% figure is often quoted without justification In six-step commutation, two phases conduct and the current vector is held at one of six fixed positions, 60° apart. The rotor turns continuously between commutations, so the angle between the current vector and the ideal position sweeps from \( -30° \) to \( +30° \).
- Torque follows the cosine of that angle
Maximum at \( \theta = 0 \) — mid-sector — and minimum at the commutation instants.Instantaneous torque\[ T(\theta) = T_{pk}\cos\theta, \qquad -30° \le \theta \le +30° \]
- Find the extremes
Working\[ T_{max} = T_{pk}, \qquad T_{min} = T_{pk}\cos 30° = 0.866\,T_{pk} \]
- Average over the sector
Result\[ T_{avg} = \frac{1}{\pi/3}\int_{-\pi/6}^{\pi/6} T_{pk}\cos\theta\,d\theta = \frac{3}{\pi}T_{pk}\left[\sin\theta\right]_{-\pi/6}^{\pi/6} = \frac{3}{\pi}T_{pk} = 0.955\,T_{pk} \]
- Express the ripple
That is where the "about 15%" comes from, and it is inherent — a consequence of holding the current vector still while the rotor moves. No amount of current-loop bandwidth removes it. It appears at six times the electrical frequency, and it is the reason a BLDC fan hums.Peak-to-peak, as a fraction of the mean\[ \frac{T_{max}-T_{min}}{T_{avg}} = \frac{1 - 0.866}{0.955} = 14.0\% \]
Saliency and the Second Torque
Bury the magnets inside the rotor iron instead of gluing them to its surface and something useful happens. Magnets have a relative permeability of about 1.05 — magnetically, they are almost air. So a rotor with buried magnets presents a different reluctance depending on which way the stator field points.
- Along the d axis — the magnet axis — the flux path passes through the magnets, so the reluctance is high and \( L_d \) is small.
- Along the q axis — between the magnets — the flux path is all iron, so the reluctance is low and \( L_q \) is large.
For an interior-magnet machine, \( L_q/L_d \) is typically 1.5 to 3. And any machine with \( L_d \ne L_q \) produces reluctance torque — the rotor tries to align its low-reluctance axis with the stator field, exactly as a compass needle aligns with a field.
Write the current as a magnitude \( I_s \) at an angle \( \beta \) ahead of the q axis, so that \( i_q = I_s\cos\beta \) and \( i_d = -I_s\sin\beta \). Substituting:
Two terms, with two different behaviours. The magnet term peaks at \( \beta = 0 \) and is proportional to current. The reluctance term peaks at \( \beta = 45° \) and is proportional to current squared. The optimum is somewhere between, and it moves with current.
An eight-pole machine with \( \psi_m = 0.15 \) Wb, \( L_d = 0.30 \) mH, at a fixed stator current of 120 A peak. Vary the current angle and see the magnet and reluctance contributions separate. Then change the saliency ratio and watch the optimum angle move.
—
- Differentiate the torque expression with respect to \( \beta \)
Working\[ \frac{dT}{d\beta} \propto -\psi_m I_s\sin\beta + (L_q-L_d)I_s^2\cos 2\beta = 0 \]
- Substitute \( \cos 2\beta = 1 - 2\sin^2\beta \) and let \( x = \sin\beta \)
A quadratic in \(\sin\beta\)\[ 2(L_q-L_d)I_s\,x^2 + \psi_m x - (L_q-L_d)I_s = 0 \]
- Solve, taking the physically meaningful root
The MTPA angle\[ \sin\beta_{MTPA} = \frac{-\psi_m + \sqrt{\psi_m^2 + 8(L_q-L_d)^2I_s^2}}{4(L_q-L_d)I_s} \]
- Check the two limits, which is the useful part
- No saliency \( (L_q = L_d) \): the equation degenerates and \( \beta_{MTPA} = 0 \). All current on the q axis, none on the d axis. This is the surface-magnet case.
- Very large current: the magnet term becomes negligible against the reluctance term, and \( \sin\beta \to 1/\sqrt2 \), so \( \beta \to 45° \). This is the pure-reluctance limit.
Field Weakening, and Why It Is Uncomfortable
A DC machine weakens its field by reducing the field current. An induction machine weakens its field by running out of volts. A permanent-magnet machine has a problem neither of them has: you cannot turn the magnets off.
The magnet flux \( \psi_m \) is there whenever the rotor is turning, so the back-EMF rises linearly with speed whether the drive wants it to or not:
At some speed \( E \) equals the maximum voltage the inverter can produce, and above that the drive can no longer force the current it wants. The only way to go faster is to inject a negative \( i_d \) — a current component that produces flux opposing the magnets.
- Write the steady-state voltage limit
neglecting the stator resistance, which is small at high speed.Voltage constraint\[ V_s^2 = \left(\omega_e L_d i_d + \omega_e\psi_m\right)^2 + \left(\omega_e L_q i_q\right)^2 \le V_{max}^2 \]
- Note that the d-axis term can be made to cancel The bracket \( \left(L_d i_d + \psi_m\right) \) is the net d-axis flux. Making \( i_d \) negative reduces it — which is exactly field weakening, achieved with stator current rather than a field winding.
- Find the current that would cancel the magnets completely
This one number decides the machine's field-weakening capability.The characteristic current\[ i_{ch} = \frac{\psi_m}{L_d} \]
- Compare it with the machine's rated current
- \( i_{ch} \gg I_{rated} \) — the magnets are strong relative to \( L_d \), and the machine cannot be weakened much. Constant-power range under 1.5:1. Typical of surface-magnet machines.
- \( i_{ch} \approx I_{rated} \) — the ideal case. Infinite theoretical constant-power range, and the machine is described as optimally designed for field weakening. Traction machines are designed to land here.
- \( i_{ch} \ll I_{rated} \) — easy to weaken, but the magnets are contributing little and most of the torque is reluctance torque. This is heading towards a synchronous reluctance machine.
Choosing a Machine
Part 6 has now covered three machine families. It is worth putting them side by side, because in practice the choice is made on four or five criteria and almost never on efficiency alone.
| Criterion | DC | Induction | PMSM / IPM | SynRel |
|---|---|---|---|---|
| Efficiency at rated point | 85–90% | 90–95% | 94–97% | 92–95% |
| Efficiency at part load | poor | poor | good | good |
| Power density | low | medium | high | medium |
| Maintenance | brushes | bearings only | bearings only | bearings only |
| Position sensor | not required | optional | required | required |
| Control complexity | low | medium to high | high | high |
| Field weakening range | 3–4:1 | 2–3:1 | 1.5–4:1 by design | 2–3:1 |
| Cost of active materials | medium | low | high (rare earths) | low |
| Fails safe when de-energised | yes | yes | no | yes |
| Typical use today | legacy, small niches | most industrial power | traction, servos, appliances | efficient pumps and fans |
Worked Examples
Problem. An eight-pole BLDC machine has a back-EMF constant of 0.42 V·s/rad (line-to-line, per mechanical rad/s) and is driven with 30 A rectangular current. Find the average torque, the torque ripple, and the ripple frequency at 3000 rev/min.
Why the frequency matters more than the amplitude. 1200 Hz sits squarely in the range where the human ear is most sensitive and where machine structures have their first bending modes. A 14% torque ripple at 20 Hz is felt; at 1200 Hz it is heard, and it is the characteristic whine of a BLDC fan.
What can and cannot be done about it. The ripple is inherent to holding the current vector still for 60°, so no current-loop improvement removes it. What does help is sinusoidal commutation — driving the same machine with sinusoidal currents from an interpolated position estimate. Torque per ampere drops a few percent because the current no longer sits on the flat top, but the ripple falls to a few percent and the noise largely disappears. Many appliance drives do exactly this, on machines that are physically BLDC.
Problem. An eight-pole IPM has \( \psi_m = 0.15 \) Wb, \( L_d = 0.30 \) mH, \( L_q = 0.75 \) mH. At a stator current of 120 A peak, find the MTPA angle, the magnet and reluctance contributions, and the gain over operating at \( \beta = 0 \).
Compare with putting all the current on the q axis: \( \beta = 0 \) gives \( T = 108.0 + 0 = 108.0 \) N·m. Advancing to 17.3° is worth 5.7% more torque for the same current — and therefore the same copper loss.
Now see how it scales, because this is the part that determines whether an MTPA table is worth implementing. At 240 A:
against 216.0 N·m at \( \beta = 0 \) — a gain of 18.2%. The reluctance term grows as \( I^2 \) while the magnet term grows only as \( I \), so MTPA matters most at peak torque, which is exactly where a traction machine spends its most demanding moments.
And this is why \( \beta_{MTPA} \) is stored as a table. It moved from 17.3° to 26.1° for a doubling of current, and in a real machine \( L_d \), \( L_q \) and even \( \psi_m \) all vary with current through saturation — so the table is populated from measurement or finite-element analysis rather than from the closed-form expression, which is used mainly to understand the trend.
Problem. The machine of Example 2 runs on a 400 V DC link with SVM, giving a peak phase voltage of 231 V. Its rated current is 120 A and its maximum is 240 A. Find the base speed, the characteristic current, and assess the machine's behaviour if the inverter trips at 12,000 rev/min.
The design is not safe, and the number tells you immediately. The characteristic current is 500 A against a machine maximum of 240 A. The drive can never inject enough negative \( i_d \) to cancel the magnet flux, so:
- Field weakening is limited. At 240 A the very best the d axis can do is \( L_di_d = -0.072 \) Wb against \( \psi_m = 0.15 \) Wb — a 48% reduction, and only if all the current goes to the d axis, leaving none for torque.
- The fault case is dangerous. At 12,000 rev/min the open-circuit back-EMF is \( (12000/3677)\times231 = 754 \) V peak per phase. If the inverter trips, the freewheel diodes rectify that into the 400 V link, and the link sees a rectified line-to-line voltage well above 1000 V.
What the design should have been. For inherently safe uncontrolled-generator behaviour, \( i_{ch} \le I_{max} \):
So \( L_d \) must be roughly doubled, or \( \psi_m \) roughly halved with the lost magnet torque made up by more saliency. Both routes point at the same machine — a strongly salient IPM or a PM-assisted reluctance machine with weaker magnets and a larger inductance. This is not a coincidence; it is why traction machines look the way they do.
And with that design in place, the active short circuit becomes available. On a fault, all three low-side devices are turned on deliberately. The machine sees a three-phase short, the current settles towards \( i_{ch} = 240 \) A — within its rating — the braking torque is modest and bounded, and nothing reaches the DC link at all. That is the standard protection in every modern EV drive, and it only works because of the inequality above.
Summary and Formula Sheet
Chapter 27, and Part 6, in five sentences:
- A synchronous machine has no slip — load changes the load angle, and exceeding pull-out loses synchronism rather than merely stalling.
- BLDC and PMSM differ in back-EMF shape, and everything else — sensor, modulation, ripple — follows from matching the current shape to it.
- Six-step commutation carries an inherent 14% torque ripple at six times the electrical frequency, because the current vector is held still for 60°.
- Saliency adds a reluctance torque proportional to \( I^2\sin2\beta \), so the best current angle grows from 0° towards 45° as current rises.
- A PM machine cannot switch its magnets off, so its field-weakening capability and its fault behaviour are both decided by \( i_{ch} = \psi_m/L_d \).
Key terms
- Load angle
- The angle between rotor and stator fields in a synchronous machine. It carries the torque; the speed does not change.
- Pull-out
- Loss of synchronism when the load exceeds the maximum torque. Unlike an induction stall, it is not self-recovering.
- BLDC
- A PM machine with trapezoidal back-EMF, driven with rectangular current in six steps from Hall sensor feedback.
- PMSM
- A PM machine with sinusoidal back-EMF, driven with sinusoidal current under field-oriented control.
- Saliency
- Unequal d- and q-axis inductances. Produces reluctance torque, and gives an interior-magnet machine its extra capability.
- Current angle β
- The angle of the stator current vector ahead of the q axis. Zero puts all current into magnet torque.
- MTPA
- Maximum torque per ampere: the current angle that extracts the most torque from a given current magnitude.
- Characteristic current
- \( \psi_m/L_d \) — the d-axis current that would exactly cancel the magnet flux. It sets both field-weakening range and fault safety.
- Uncontrolled generator operation
- A PM machine feeding the DC link through the freewheel diodes after the inverter turns off at high speed.
- Active short circuit
- Deliberately shorting the machine through the low-side devices on a fault, to bound the current and protect the DC link.
Test Yourself
A PMSM will not start when the drive is powered up with no position feedback. Why is this a fundamental problem rather than a commissioning error?
Because a synchronous machine has no self-starting mechanism at all. Its torque depends on the angle between the rotor and the stator field, and if the drive does not know that angle it cannot choose which way to push.
Contrast the two families squarely:
- Induction machine. Energise the stator at any frequency and the rotor is dragged along by induction. The slip provides the torque automatically; no position information is needed to start.
- Synchronous machine. Apply a rotating field to a stationary rotor and the torque alternates in direction as the field sweeps past. The average is zero, and the rotor merely vibrates.
So the drive must know where the rotor is before it applies anything. The methods, in order of cost:
- Fit an absolute encoder or resolver. Position is known at power-up, and this is the only fully reliable answer for a machine that must produce full torque from standstill under load.
- Hall sensors give the sector, which is enough to start in the right direction and to run six-step commutation. This is why BLDC machines nearly always have them.
- Alignment. Apply a DC current along an arbitrary axis and wait; the rotor pulls into alignment with it, and the drive then knows the angle. Simple, but it moves the shaft — unacceptable for a hoist or a positioning axis — and it needs the load to be light enough to allow the movement.
- Signal injection. Inject a high-frequency voltage and measure the resulting current. In a salient machine the current response depends on the angle between the injection and the d axis, so the position can be extracted at zero speed without moving the shaft. This works only because \( L_d \ne L_q \), so it is available on an IPM and not on a surface-magnet machine.
- Back-EMF observers give excellent position above a few percent of rated speed and nothing at all at standstill, because the back-EMF they observe is proportional to speed. Hence the standard architecture: start with one of the methods above, then hand over to the observer.
The general statement: a PM machine and its drive are a single system. The machine cannot run without the drive, and the drive must be matched to that specific machine's parameters. An induction machine and its drive are far more loosely coupled, which is a large part of why induction drives dominate general industry.
Why does a surface-magnet machine have no reluctance torque, when a magnet is clearly not the same material as iron?
Because the magnet layer is present in every direction, so the flux path looks the same whichever way the stator field points.
Work through the geometry. Magnets have \( \mu_r \approx 1.05 \), so magnetically they behave almost exactly like air.
- Surface magnets form a continuous shell of magnet material over the rotor. Flux entering along the d axis crosses the magnet thickness; flux entering along the q axis crosses the same magnet thickness. \( L_d \approx L_q \), and the machine is magnetically round.
- Interior magnets are buried in slots inside the iron. Along the d axis the flux must cross the magnets — a large effective air gap, so \( L_d \) is small. Along the q axis it can travel entirely through iron between the magnet pockets — a small effective gap, so \( L_q \) is large.
The counter-intuitive part is the direction of the inequality. In a wound-field synchronous machine \( L_d > L_q \), because the d axis lies along the pole where the iron is. In an interior-magnet machine \( L_q > L_d \), because the magnets sit on the d axis and block it. The saliency is inverted relative to the classical machine, and the sign of \( (L_d - L_q) \) in the torque equation follows.
The consequence for control: reluctance torque requires \( i_di_q(L_d-L_q) > 0 \). With \( L_q > L_d \) this needs \( i_d \) negative — that is, the current vector advanced ahead of the q axis. Conveniently, negative \( i_d \) is also what field weakening needs, so in an IPM the MTPA trajectory and the field-weakening trajectory lie in the same quadrant and blend into one another smoothly. In a classical wound-field machine they would oppose.
And a practical note: surface-magnet machines are not inferior — they are simpler to build, have lower cogging and are entirely adequate where the constant-power range is narrow. They are the right choice for fans, pumps and low-speed direct drives. The interior magnet earns its complexity only where field weakening or peak torque density is genuinely needed.
Where does the 14% figure for six-step torque ripple actually come from?
From the fact that the current vector can only occupy six positions, so it is held still for 60° while the rotor keeps turning.
The geometry. With two phases conducting, the stator current vector points in one of six directions, 60° apart. Ideally it should lead the rotor flux by exactly 90° at all times; in practice it is placed at 90° in the middle of each sector and drifts \( \pm30° \) either side.
The extremes:
- Maximum \( T_{pk} \) at mid-sector, where the alignment is ideal.
- Minimum \( T_{pk}\cos30^\circ = 0.866\,T_{pk} \) at each commutation instant.
The average:
So the peak-to-peak ripple as a fraction of the mean is \( (1-0.866)/0.955 = 14.0\% \), and it repeats six times per electrical cycle.
Three things follow that are worth remembering:
- It is inherent. It comes from the quantisation of the current vector, not from any imperfection. A faster current loop, a better sensor or a higher switching frequency change nothing.
- Real machines are worse. The figure above assumes ideal trapezoidal EMF and instantaneous commutation. Finite phase inductance means the outgoing current cannot fall instantly, producing a commutation notch on top of this — 20% total ripple is common in practice.
- The cure is to stop using six steps. Sinusoidal commutation, with the position interpolated between Hall edges, reduces the ripple to a few percent at the cost of a small drop in torque per ampere and a more capable controller. This is now standard in appliance drives.
A traction motor is specified with a characteristic current equal to its maximum current. What is being bought, and what is being given up?
What is bought is an unbounded field-weakening range and inherent fault safety. What is given up is magnet strength, and therefore torque per ampere at low speed.
Take the field-weakening argument first. The d-axis flux is \( \psi_m + L_di_d \). Setting \( i_d = -i_{ch} = -\psi_m/L_d \) makes it exactly zero. With the flux cancelled, the back-EMF no longer rises with speed, and the voltage limit is never reached — so in theory the machine can run at any speed, and the constant-power region has no upper bound. In practice mechanical and loss limits intervene, but ratios of 4:1 or more become achievable.
Now the fault argument, which is the one that actually gets written into the specification. If the inverter trips at high speed, the freewheel diodes turn the machine into an uncontrolled generator. The steady-state short-circuit current of a PM machine converges to \( i_{ch} \). So:
- If \( i_{ch} \le I_{max} \), the fault current is within what the machine and the devices can survive, and the braking torque is bounded.
- This makes the active short circuit a legitimate protection strategy: on detecting a fault, turn all three low-side transistors on. The machine is shorted, the current settles at \( i_{ch} \), and the DC link is completely isolated from the back-EMF.
- If instead \( i_{ch} \gg I_{max} \), neither option is safe — the short circuit would produce a destructive current, and leaving the diodes to rectify would destroy the link.
The price. \( i_{ch} = \psi_m/L_d \) is made small by weakening the magnets or by increasing \( L_d \). Weaker magnets mean less magnet torque for the same current, so the design must recover it elsewhere:
- More saliency, so the reluctance term contributes a larger share. This pushes the design towards a strongly salient IPM or a PM-assisted reluctance machine.
- More current, which means more copper loss and more cooling.
- More poles or a longer stack, which is a size and cost decision.
A further honest cost: holding \( i_d = -i_{ch} \) at high speed dissipates copper loss continuously while producing no torque at all. Efficiency at high speed and light load — motorway cruising, in a vehicle — is measurably worse than it would be with a machine that did not need so much weakening current. Real designs therefore compromise: \( i_{ch} \) close to \( I_{max} \) but not below it, tuned against the drive cycle the vehicle is actually expected to see.
Why is the MTPA angle stored as a lookup table rather than computed from the closed-form expression?
Because the closed-form expression assumes \( \psi_m \), \( L_d \) and \( L_q \) are constants, and in a real machine none of them is.
What actually varies, and by how much:
- \( L_q \) saturates heavily. The q-axis flux path is all iron, so at high \( i_q \) that iron approaches saturation and \( L_q \) can fall by 30–40% from its small-signal value. The saliency \( \Delta L \) therefore shrinks exactly where the reluctance torque is largest.
- \( L_d \) varies much less, because its path already crosses the magnets — a large effective air gap that dominates the reluctance and keeps the path linear.
- Cross-saturation. \( L_d \) depends on \( i_q \) and \( L_q \) depends on \( i_d \), because both fluxes share the same iron. There is no way to represent this with two independent inductances.
- \( \psi_m \) falls with temperature — about 0.12% per kelvin for NdFeB — so a hot machine has measurably less magnet flux and a slightly different optimum angle.
What that does to the formula. Substituting small-signal inductances typically gives an MTPA angle several degrees away from the true optimum at high current. The torque error is second-order near a maximum — a 5° error costs well under 1% — so it is not catastrophic, but it is measurable, and in a traction application 1% of peak torque is worth having.
How real drives handle it:
- A two-dimensional table of \( i_d \) and \( i_q \) against torque demand and speed, populated from finite-element analysis and then corrected on a dynamometer.
- Temperature correction of \( \psi_m \) from a rotor thermal model or a magnet temperature estimate.
- Online search in some high-end drives — perturb the angle slightly and observe whether the current needed for the commanded torque rises or falls, then walk towards the minimum. Slow, but it tracks everything automatically.
The closed-form expression is still worth knowing, because it tells you the trend: the angle starts at zero, grows with current, and approaches 45°. That is what lets you sanity-check a table and recognise when one is wrong.
Looking back over Part 6 — DC, induction and permanent-magnet drives — what is the single idea that connects all three?
Torque is always flux times current at right angles, and every drive is an argument about how to arrange that.
How each family achieves the orthogonality:
- DC machine (Chapter 25). Mechanically, by the commutator. The brushes physically switch the armature current so its MMF stays 90° from the field, whatever the rotor does. Simple, effective, and it wears out.
- Induction machine (Chapter 26). Not at all, in scalar control — which is precisely why V/f is slow and cannot command torque. Field-oriented control supplies it computationally, by finding the rotor flux angle and resolving the current about it.
- PM machine (Chapter 27). The flux is fixed to the rotor by the magnets, so the drive must place the current relative to a measured rotor position. And because the flux is no longer adjustable, the current angle acquires a second job: trading magnet torque for reluctance torque, and later for field weakening.
The second connecting idea is the two-region picture, which appeared in all three chapters in the same shape:
- Constant torque below base speed, where flux is held at its maximum and torque is limited by current — that is, by heating.
- Constant power above base speed, where voltage is at its ceiling and flux must be reduced.
- And in every case a limit on how far the second region can extend: commutation in the DC machine, pull-out torque falling as \( 1/f^2 \) in the induction machine, and the characteristic current in the PM machine.
The third is the cascade. Every drive in this part is built the same way: an inner current loop, tuned fast, that protects the semiconductors and makes the machine look like a torque source; a speed loop outside it; and often a position loop outside that. The machine changes, the physics changes, the transformations change — the control architecture does not.
And the practical thread: in all three chapters the decisive question for choosing a drive was rarely efficiency. It was what happens at zero speed, what happens when the load drives the machine, and what happens when the electronics stop.
Problems
Three habits for synchronous and PM drive work:
- Identify the back-EMF shape first. It decides the sensor, the modulation and the achievable ripple.
- Compute \( i_{ch} = \psi_m/L_d \) early. One number tells you the field-weakening range and the fault behaviour.
- Ask what happens when the inverter stops. For a PM machine this is a design requirement, not an afterthought.
Problems 1–5 are direct application; 6–9 need judgement; 10–12 are design questions worth discussing in a tutorial.
- A six-pole PMSM runs at 2400 rev/min. Find the electrical frequency, and the frequency of the torque ripple if it were driven by six-step commutation.
- A BLDC machine has \( k_e = 0.085 \) V·s/rad and is driven with 18 A. Find the average torque, the peak and minimum instantaneous torque, and the percentage ripple.
- An eight-pole IPM has \( \psi_m = 0.11 \) Wb, \( L_d = 0.42 \) mH and \( L_q = 1.05 \) mH. At 150 A peak, find the MTPA angle and the two torque components.
- For the machine of Problem 3, find the characteristic current and comment on whether an active short circuit would be a safe protection strategy if \( I_{max} = 300 \) A.
- A surface-magnet machine has \( \psi_m = 0.20 \) Wb and \( L_d = L_q = 1.8 \) mH, with \( I_{max} = 60 \) A. Find \( i_{ch} \), and estimate the maximum field-weakening speed ratio.
- Explain why three Hall sensors are sufficient for BLDC commutation but not for field-oriented control, and quantify the worst-case torque error if FOC were attempted with raw Hall states.
- A 4 kW PMSM drive is retrofitted to a machine that was previously run six-step, and the measured torque per ampere falls by 4% while the acoustic noise drops sharply. Explain both observations.
- A PM traction drive trips at 9000 rev/min on a machine with \( \psi_m = 0.09 \) Wb, four pole pairs and a 350 V link. Estimate the open-circuit back-EMF and state what the drive should do in the first millisecond.
- Compare an IPM and an induction machine for a 2 MW marine propulsion drive. Address efficiency, fault behaviour, redundancy and maintenance, and state which you would choose and why.
- Design the drive for a 30 kW electric vehicle traction motor: choose the machine type, specify the position sensor, decide the switching frequency with reference to Chapter 18, size the field-weakening range for a 4:1 speed range, verify the characteristic current condition, and specify the fault response.
- A domestic washing machine must spin at 1200 rev/min and tumble at 50 rev/min with high torque. Compare a BLDC with Hall sensors, a PMSM with sinusoidal commutation, and an induction machine with a V/f drive, and recommend one with reasons.
- Part 6 has covered three machine families. Construct a decision procedure — a short ordered list of questions — that leads an engineer from an application description to a machine and control choice, and test it against three applications of your own choosing.