Solved Problems · Set 36

Synchronous, BLDC and PMSM Drives

Part 6 · Electric Drives — put the flux on the rotor and stop paying for it. Six problems closing Part 6.

Prof. Mithun Mondal 6 solved problems GATE · ESE · University

Set 36 — Synchronous, BLDC and PMSM Drives

Every difficulty in Set 35 came from one source: the induction machine's rotor flux is induced. It must be created by stator current, maintained against a 169 ms time constant, and located by computation using a resistance that changes 50% with temperature.

Put permanent magnets on the rotor and all of it disappears. The flux is simply there — constant, at no excitation cost, mechanically locked to a rotor position an encoder measures directly. That removes \(i_d\) from the torque equation, removes the magnetising current from the stator, removes the rotor copper loss entirely, and raises the efficiency by several points. It also removes the ability to switch the flux off — which becomes the central problem the moment the machine turns faster than its base speed.

Part 6 · Chapter 27 · 6 solved problems

i Method Recap
  • A surface-magnet PMSM has one current in its torque equation:

    \[ T = \frac32\cdot\frac{P}{2}\lambda_mi_q, \qquad E = \omega_e\lambda_m \]
  • The terminal voltage is the vector sum of EMF and reactance drop:

    \[ \hat V = \sqrt{\left(\omega_e\lambda_m\right)^2+\left(\omega_eL_qi_q\right)^2} \]
  • Field weakening opposes the magnet flux with negative \(i_d\):

    \[ \lambda_m+L_di_d \le \frac{\hat V_{max}}{\omega_e} \;\Longrightarrow\; i_d \le \frac{\hat V_{max}/\omega_e-\lambda_m}{L_d} \]
  • A salient machine gains reluctance torque:

    \[ T = \frac32\cdot\frac{P}{2}\left[\lambda_mi_q+\left(L_d-L_q\right)i_di_q\right] \]
  • MTPA chooses the current split that minimises magnitude:

    \[ i_d = \frac{\lambda_m-\sqrt{\lambda_m^2+8\left(L_q-L_d\right)^2i_q^2}}{4\left(L_q-L_d\right)} \]
  • A load-commutated inverter uses the machine's own EMF to turn the thyristors off:

    \[ \text{over-excited synchronous machine} \;\Longrightarrow\; \text{leading PF} \;\Longrightarrow\; \text{natural commutation} \]
Problem 1CoreFlux You Do Not Pay For

An 8-pole surface-mount PMSM has \(\lambda_m = 0.15\) Wb and \(L_d = L_q = 3\) mH, with a base speed of 3000 rpm. Find the torque constant, the current for 20 N·m, the back EMF and terminal voltage at base speed, and the DC link required.

Solution

The torque equation loses a term. With the flux fixed by the magnets, only the quadrature current matters:

\[ T = \frac32\cdot\frac{P}{2}\lambda_mi_q = \frac32(4)(0.15)i_q = 0.90i_q \]
\[ i_q = \frac{20}{0.90} = 22.2\ \text{A} \]

One current, one constant — simpler even than the induction machine's vector control, because \(i_d\) is set to zero rather than to a magnetising value. Below base speed a PMSM drive commands only \(i_q\).

The electrical frequency at base speed:

\[ \omega_e = \frac{2\pi(3000)}{60}\cdot\frac{P}{2} = (314.2)(4) = 1257\ \text{rad/s} = 200\ \text{Hz} \]

Two hundred hertz — four times the mains frequency, from a machine turning at 3000 rpm. High pole counts give high torque density and demand a fast inverter, which is why PMSM drives switch at 10–20 kHz where induction drives manage at 4.

The back EMF:

\[ \hat E = \omega_e\lambda_m = (1257)(0.15) = 188.5\ \text{V peak per phase} \]
\[ V_{LL} = (188.5)\sqrt{\frac32} = 230.9\ \text{V rms} \]

And this exists whenever the rotor turns, whether the drive is running or not — a safety consideration with no counterpart in an induction machine. A PMSM spun by its load with the inverter disabled produces voltage at its terminals and, if the diodes conduct, charges the DC link.

The terminal voltage adds the reactance drop, in quadrature:

\[ \omega_eL_qi_q = (1257)(0.003)(22.2) = 83.8\ \text{V} \]
\[ \hat V = \sqrt{188.5^2+83.8^2} = 206.3\ \text{V peak} \;\Longrightarrow\; V_{LL} = 252.6\ \text{V rms} \]

Hence the DC link, using the 0.707 utilisation of Set 26:

\[ V_{dc} \ge \frac{252.6}{0.707} = 357\ \text{V} \quad\to\quad \text{use } 400\ \text{V} \]

Note how directly the machine's design determines the converter's. Choosing a higher \(\lambda_m\) for more torque per amp raises the back EMF and therefore the link voltage — the two are inseparable, and machine and inverter must be specified together.

Compare with the induction machine of Set 35:

PropertyInductionPMSM
Flux source\(i_d\) from the statormagnets — free
Magnetising current6.9 A of 19.6zero
Rotor loss\(sP_{airgap}\)essentially zero
Efficiency~90%~95%
Torque densitybaseline1.5–2×
Control parameters\(R_r,L_m,L_r\)\(\lambda_m,L_d,L_q\) — stable
Position sensingflux angle computedrotor angle = flux angle
Flux switch-offyesno — Problem 2
Costlowmagnets are expensive

The seventh row removes Set 35's hardest problem outright. In a PMSM the flux is locked to the rotor, so an encoder reading the rotor angle gives the flux angle directly — no slip calculation, no \(R_r\), no detuning.

Magnets remove the magnetising current, the rotor loss and the flux angle problem all at once. Thirty-five per cent of the induction machine's stator current was doing nothing but making flux; a PMSM spends all of it on torque. What it cannot do is turn the flux off, and Problem 2 shows what that costs.
Answera\(k_T = 0.90\ \text{N}\!\cdot\!\text{m/A},\ i_q = 22.2\ \text{A}\)   b\(\hat E = 188.5\ \text{V}\), terminal 252.6 V rms   c\(V_{dc} \ge 357\ \text{V}\)
Problem 2CoreFighting the Magnets

The same machine must run to 6000 rpm from a 400 V link. Show that the back EMF alone exceeds what the inverter can supply, find the negative \(i_d\) required, and identify the hazard that permanent magnets create at high speed.

3000 rpm iq only 6000 rpm id = −6.7 A +d −d (weakening) q current limit
Above base speed the current vector must rotate into the negative d region
Solution

The problem at double speed:

\[ \omega_e = 2513\ \text{rad/s} \;\Longrightarrow\; \hat E = (2513)(0.15) = 377\ \text{V peak} \]
\[ \hat V_{max} = \sqrt2\left(0.707\right)(400)\sqrt{\tfrac23} = 326.6\ \text{V peak} \]
\[ 377 > 327 \quad\Longrightarrow\quad \textbf{the inverter cannot even reach the back EMF} \]

Without intervention no current can be forced into the machine at all — the machine is generating more voltage than the inverter can produce. Torque is not merely reduced; it is zero.

The remedy: negative \(i_d\). Current along the negative d-axis produces flux opposing the magnets:

\[ \lambda_{effective} = \lambda_m+L_di_d \]
\[ \text{require } \omega_e\lambda_{effective} \le \hat V_{max} \;\Longrightarrow\; \lambda_{effective} \le \frac{326.6}{2513} = 0.130\ \text{Wb} \]
\[ i_d \le \frac{0.130-0.150}{0.003} = -6.7\ \text{A} \]

Nearly seven amps of demagnetising current, producing no torque whatsoever — pure overhead, dissipating \(i_d^2R_s\) in the stator for the sole purpose of cancelling flux the magnets insist on providing.

The contrast is stark:

MachineField weakening methodCost
DC (Set 33)reduce field currentsaves power
Induction (Set 34)reduce \(i_d\)saves current
PMSMinject negative \(i_d\)costs current and loss

The first two machines weaken their field by supplying less; the PMSM must supply more. That inversion is the fundamental penalty of permanent excitation, and it grows with speed — at four times base speed the demagnetising current would dominate the machine's rating entirely.

The current budget at 6000 rpm. With a 25 A limit:

\[ i_q = \sqrt{I_{max}^2-i_d^2} = \sqrt{625-44.6} = 24.1\ \text{A} \]

Only 4% of the torque capability lost at double speed, because \(i_d\) is still small compared with the limit. The penalty is modest here and becomes severe only when \(i_d\) approaches the current limit — which defines the machine's maximum usable speed.

The hazard: uncontrolled generation. If the inverter shuts down — a fault, a protection trip, a loss of gate supply — while the machine is spinning fast:

\[ \hat E = 377\ \text{V} > V_{dc} = 400\ \text{V}?\ \text{At } 6000\ \text{rpm, } V_{LL} = 462\ \text{V rms} \]

The line-to-line EMF exceeds the link voltage, so the inverter's freewheeling diodes form an uncontrolled rectifier and pump power into the link. With no load to absorb it the capacitor overvolts, and there is no way to stop it short of stopping the rotor. Every magnet drive above base speed lives with this.

Mitigations, none of them free:

ApproachNote
Active short-circuitturn all three lower devices on — the standard response
Design so \(\hat E < V_{dc}\) at max speedlimits the speed range severely
Crowbar or dump resistorextra hardware, must be reliable
Link capacitor rated for itexpensive at high speed
Mechanical disconnecta clutch — heavy and slow

The first row is what production drives do: shorting the three phases together forces the machine into a controlled short circuit, where the current is limited by the synchronous reactance and the torque is a modest braking torque. It requires that the gate drives keep working during the fault, which is why they have their own energy reserve.

A permanent magnet cannot be switched off, and above base speed that becomes an active hazard. Field weakening costs current instead of saving it, and an inverter shutdown at speed leaves the machine feeding its own EMF through the freewheeling diodes into a link that cannot absorb it. Active short-circuit is the standard answer, and it must survive the fault that caused it.
Answera\(\hat E = 377\ \text{V} > \hat V_{max} = 327\ \text{V}\)   b\(i_d = -6.7\ \text{A}\) required   c uncontrolled generation on shutdown — active short-circuit
Problem 3Exam levelSix Steps and a Hall Sensor

Compare the BLDC drive — trapezoidal back EMF, 120° conduction, Hall-sensor commutation — with sinusoidal PMSM control. Explain where the BLDC's torque ripple comes from and why the topology persists.

Solution

The BLDC principle. The machine is wound so that its back EMF is trapezoidal — flat over 120° of each half cycle rather than sinusoidal:

\[ \text{drive a constant current through the flat portion} \;\Longrightarrow\; T = k_tI\ \text{constant} \]

Two phases conduct at a time and the third floats, exactly the 120° conduction mode of Set 26 — which was dismissed there for losing 13.4% of the output. Here it is the whole point, because the machine has been designed to match it.

What the controller needs. Only six commutation instants per electrical cycle:

\[ \text{3 Hall sensors} \;\Longrightarrow\; 6\ \text{states per cycle} \;\Longrightarrow\; 60^\circ\ \text{resolution} \]
RequirementBLDCPMSM (FOC)
Position resolution60° — three Hall sensorscontinuous — encoder or resolver
Current sensingone, in the DC linktwo phase currents
TransformsnoneClarke and Park
Control loopa single current looptwo current loops in dq
Modulatorsimple PWM on one pairSVM
Processoran 8-bit microcontroller sufficesDSP or capable MCU

The cost difference is real. A BLDC controller for a fan or a pump can be built around a few pounds of silicon; a PMSM drive needs a resolver or a good encoder plus the processing of Set 35.

Where the torque ripple comes from. Two mechanisms, both at the commutation instants:

\[ \text{(i) the back EMF is not perfectly flat — the corners are rounded} \]
\[ \text{(ii) the current cannot transfer instantly between phases — } L\frac{di}{dt} \]

During commutation the outgoing phase's current decays while the incoming phase's rises, and the two do not match — so the total torque dips or peaks. The result is a ripple at six times the electrical frequency, typically 10–15% of the mean.

The comparison:

PropertyBLDCPMSM
Back EMFtrapezoidalsinusoidal
Currentrectangular, 120°sinusoidal
Torque density~15% higherbaseline
Torque ripple10–15% at \(6f_e\)< 2%
Acoustic noisehigherlower
Field weakeningpoorgood
Controller costlowesthigher
Efficiencyslightly lowerslightly higher

The third row is worth explaining. A trapezoidal machine puts all its conductors under full flux for 120° instead of distributing them sinusoidally, so it extracts more torque from the same copper and iron — about 15% more. That advantage is why BLDC persists despite its ripple.

Where each belongs:

ApplicationChoiceDeciding factor
Fans, pumps, appliancesBLDCcost — ripple is irrelevant
Power tools, e-bikesBLDCcost and torque density
Hard drives, small motorsBLDCsimplicity
Traction, servosPMSMsmoothness, field weakening
Machine toolsPMSMsurface finish depends on ripple
Compressors, HVACPMSMacoustic noise

The distinction has blurred: the same magnet machine can be driven either way, and many controllers run BLDC commutation at low speed and switch to sinusoidal control above it — taking the simplicity where it helps and the smoothness where it matters.

Sensorless BLDC is unusually easy, which reinforces the cost argument:

\[ \text{the floating phase's back EMF crosses zero } 30^\circ\ \text{before each commutation} \]

One phase is always open, so its back EMF can be measured directly — no observer, no model, no parameters. Detect the zero crossing, wait 30 electrical degrees, commutate. It fails at standstill, where there is no EMF, so the machine is started open-loop and handed over once it is moving — which is why some fans hesitate briefly on start-up.

The BLDC trades 12% torque ripple for a controller that fits on an 8-bit micro. Three Hall sensors, one current sensor, no transforms and no modulator — and 15% more torque from the same iron. For a fan or a power tool that is decisively the right trade; for a machine tool the ripple appears in the workpiece.
Answera BLDC: trapezoidal EMF, 120° blocks, 60° position resolution   b ripple 10–15% at \(6f_e\) from finite \(di/dt\)   c 15% more torque density, far cheaper control
Problem 4Exam levelTorque From Saliency

An interior PMSM has \(\lambda_m = 0.15\) Wb, \(L_d = 3\) mH and \(L_q = 6\) mH, 8 poles. For \(i_q = 20\) A find the MTPA operating point, the torque produced, the reluctance contribution, and the current saving against pure quadrature operation.

Solution

Why the interior machine is different. Burying the magnets inside the rotor makes the magnetic path direction-dependent:

\[ L_q > L_d \quad\text{(the magnet's permeability is close to air's)} \]
\[ T = \frac32\cdot\frac{P}{2}\left[\lambda_mi_q+\left(L_d-L_q\right)i_di_q\right] \]

The second term is reluctance torque — the rotor's tendency to align its low-reluctance axis with the stator field. Since \(L_d-L_q\) is negative, it contributes positively when \(i_d\) is negative.

Which is a happy coincidence. Negative \(i_d\) is exactly what field weakening required in Problem 2:

\[ i_d < 0 \;\Longrightarrow\; \text{weakens the field } \textbf{and} \text{ adds torque} \]

The surface-mount machine's demagnetising current was pure loss; here it does useful work. That is why traction drives use interior magnets almost universally — the machine that most needs field weakening is the one that profits from it.

The MTPA condition. Minimising \(\left|\vec i_s\right|\) for a given torque gives:

\[ i_d = \frac{\lambda_m-\sqrt{\lambda_m^2+8\left(L_q-L_d\right)^2i_q^2}}{4\left(L_q-L_d\right)} \]
\[ = \frac{0.15-\sqrt{0.0225+8\left(0.003\right)^2(400)}}{4(0.003)} = \frac{0.15-0.2265}{0.012} = -6.37\ \text{A} \]

The torque produced:

\[ T = \frac32(4)\left[(0.15)(20)+(-0.003)(-6.37)(20)\right] \]
\[ = 6\left[3.000+0.382\right] = 20.3\ \text{N}\!\cdot\!\text{m} \]
\[ \text{reluctance share} = \frac{0.382}{3.382} = 11.3\% \]

Eleven per cent of the torque from saliency alone, with no magnet material involved. Machines designed for higher saliency — more flux barriers in the rotor — push this to 30–50%, reducing the magnet content and with it the cost and the dependence on rare-earth supply.

The current saving:

\[ \left|\vec i_s\right|_{MTPA} = \sqrt{6.37^2+20^2} = 21.0\ \text{A} \]

Against producing the same torque with \(i_d = 0\):

\[ i_q = \frac{20.3}{(6)(0.15)} = 22.6\ \text{A} \]
\[ \text{saving} = 1-\frac{21.0}{22.6} = 6.9\% \]
\[ \text{copper loss saving} = 1-\left(\frac{21.0}{22.6}\right)^2 = 13.3\% \]

Seven per cent less current and thirteen per cent less copper loss, from choosing the current's direction rather than merely its magnitude. That is free efficiency, requiring only a lookup table in the controller.

The complete operating strategy:

RegionStrategyDetermines
Below base speedMTPAminimum current for the demanded torque
Field weakening Ivoltage-limited, current-limitedfollow the voltage ellipse
Field weakening IIMTPV (max torque per volt)when the ellipse centre is reachable

All three regions are usually precomputed into a two-dimensional lookup table indexed by torque demand and speed, because solving the constrained optimisation online is expensive. Calibrating that table is a substantial part of commissioning a traction drive.

Saliency turns field weakening from a cost into a contribution. The surface machine's negative \(i_d\) produced nothing but loss; the interior machine's produces 11% of the torque and reduces the total current by 7%. That is why every automotive traction motor buries its magnets.
Answera\(i_d = -6.37\ \text{A}\)   b\(T = 20.3\ \text{N}\!\cdot\!\text{m}\), 11.3% from reluctance   c 21.0 A against 22.6 — 6.9% less current, 13.3% less loss
Problem 5ChallengeVery Large Machines

Describe the load-commutated inverter used for very large wound-field synchronous drives, explain how the machine commutates the thyristors, identify what happens below about 10% speed, and compare with the cycloconverter of Set 31.

Solution

The topology. Two six-pulse thyristor bridges back to back with a DC link inductor — a current source inverter, as in Set 29:

\[ \text{supply} \to \text{controlled rectifier} \to L_{dc} \to \text{inverter} \to \text{machine} \]

Twelve thyristors instead of the cycloconverter's thirty-six, with a DC link inductor holding the current constant. The supply-side bridge controls the current; the machine-side bridge steers it into the right phases.

How commutation happens without any commutation circuit. Set 6 established that a thyristor needs reverse voltage for its turn-off time:

\[ \text{over-excite the field} \;\Longrightarrow\; \text{machine draws leading current} \;\Longrightarrow\; \text{reverse voltage available} \]

An over-excited synchronous machine behaves capacitively at its terminals — it supplies reactive power rather than consuming it. That leading power factor provides exactly the reverse voltage the outgoing thyristor needs, so the machine's own EMF performs the commutation. No forced commutation circuit, no self-commutating devices, no snubber energy.

Which is why LCIs reach ratings nothing else approaches:

PropertyValue
Devices12 thyristors (or 24 for 12-pulse)
Power range1–100 MW
Efficiency> 98% — no switching loss
Fault behaviourcurrent-source — inherently limited
Four-quadrantyes — both bridges reverse
Applicationsgas compressors, pumped storage, ship propulsion, wind tunnels

The fourth row echoes Set 29: the DC link inductor makes a short circuit harmless. For a 40 MW compressor drive that robustness is worth far more than waveform quality.

The problem at low speed. The commutating voltage is the machine's EMF, which is proportional to speed:

\[ E \propto \omega \;\Longrightarrow\; \text{below } \approx 10\%\ \text{speed, insufficient to commutate} \]

At standstill there is no EMF at all, so a machine that must be started from rest cannot commutate its own inverter. Something else must do it.

The solution: DC link current pulsing. Force the link current to zero at each commutation instant:

\[ \text{supply bridge inverts briefly} \;\Longrightarrow\; i_{dc} \to 0 \;\Longrightarrow\; \text{machine bridge commutates} \]

The supply-side bridge is driven into inversion for a few milliseconds, collapsing the link current so the machine-side thyristors turn off naturally. Torque is then pulsating and the acceleration is rough — but it only has to work up to about 10% speed, after which the machine's own EMF takes over. It is why an LCI-driven compressor sounds distinctly unhappy for the first few seconds of a start.

Against the cycloconverter:

PropertyLCICycloconverter (Set 31)
Thyristors1236
Output frequencyany — above or below supply\(\le f_i/3\)
Input power factorpoor, but betterpoor
Low-speed torquepulsating — needs pulsingexcellent
Input harmonicscharacteristic 6-pulseinterharmonics
Machine requiredsynchronous, over-excitedsynchronous or induction
Typical usecompressors, high-speed pumpsgearless mills, rolling mills

The fourth row divides the applications cleanly. A gas compressor spends its life near rated speed and starts unloaded, so pulsed starting is acceptable; a gearless mill must be inched into position and produce full torque at 0.2 Hz, which only the cycloconverter does well. Neither displaces the other.

An over-excited synchronous machine commutates its own inverter. That single property gives twelve thyristors a hundred-megawatt drive with no forced commutation and no switching loss — and it fails completely at standstill, where the EMF that does the commutating does not exist.
Answera over-excitation gives leading PF, providing the reverse voltage   b 12 thyristors to 100 MW   c below ~10% speed the link current must be pulsed to zero
Problem 6ChallengeChoosing For a Vehicle

Compare induction, interior PMSM, synchronous reluctance and switched reluctance machines for an electric vehicle traction drive, on efficiency, torque density, field weakening range, fault behaviour and cost, and state what actually decides the choice.

Solution

What traction demands is an unusually broad set of requirements:

RequirementWhy
Full torque from zero speedhill starts, no clutch
Wide constant-power range4–5× base speed — single-ratio gearbox
High efficiency over the drive cyclerange, not peak efficiency
High torque and power densitypackaging and mass
Safe under fault at speedProblem 2's hazard, at motorway speed
Cost and supply-chain securityrare-earth magnets

The second row is what makes traction distinctive. A single-speed transmission means the machine must cover the whole speed range electrically — up to five times base speed — where a conveyor or a pump needs one.

The four candidates:

InductionIPMSMSynRMSRM
Peak efficiency93%96%94%92%
Torque density1.01.61.11.2
Field weakeningexcellent — freegood, costs currentexcellentexcellent
Zero-speed torqueyes (with sensor)yesyesyes
Fault at speedsafe — flux collapseshazard — Problem 2safesafe
Magnet costnonehighnonenone
Torque ripplelowlowmoderatehigh
Acoustic noiselowlowmoderatehigh
Converterstandardstandardstandardspecial — asymmetric bridge
Control complexityhigh (\(R_r\))moderatehighhigh

The fifth row deserves emphasis. An induction machine that loses its inverter simply stops exciting:

\[ \text{no stator current} \;\Longrightarrow\; \text{flux decays in } \tau_r \;\Longrightarrow\; \text{no EMF, no torque} \]

Safe by construction. A permanent-magnet machine at 12,000 rpm keeps generating and keeps braking, so a fault produces both an overvoltage and an uncommanded retarding torque on a driven wheel — which is why an active short-circuit strategy, and gate drives that survive the fault, are safety items in every EV.

Why the SRM keeps not winning. On paper it should:

\[ T = \tfrac12i^2\frac{dL}{d\theta} \quad\text{— no magnets, no rotor windings, no rotor loss} \]

The simplest, cheapest and most robust rotor imaginable — a lump of laminated steel — tolerant of high temperature and safe under any fault. And it is defeated by rows seven, eight and nine together: the doubly salient geometry produces large torque ripple and radial forces that make the stator ring audibly, and it needs a non-standard converter that no one makes in volume. Decades of development have narrowed but not closed those gaps.

What actually decides it:

PriorityChoiceExample
Maximum range and efficiencyIPMSMmost current EVs
Cost and magnet independenceinductionsome models' front motors
BothIPMSM rear + induction frontdual-motor vehicles
EmergingPM-assisted SynRMless magnet, most of the performance

The third row is a genuinely elegant engineering answer. Using a magnet machine on one axle for efficiency at cruise and an induction machine on the other for cost and for its zero standby drag — it can be de-energised completely, whereas a PMSM always drags its magnets past the stator — captures the strengths of both.

And the requirement that dominates everything. Traction efficiency is judged over a drive cycle, not at one point:

\[ \text{a vehicle spends most of its life at} \approx 10\text{–}30\%\ \text{of peak torque} \]

So the map of efficiency across the whole torque–speed plane matters far more than the peak figure, and the machine that wins is the one whose high-efficiency island sits where the drive cycle actually lives. That is where permanent magnets are strongest — they have no excitation loss at light load, exactly where an induction machine is still paying for its magnetising current.

Traction is decided by the efficiency map, not the peak number. A vehicle lives at 10–30% of peak torque, where the induction machine is still paying for magnetising current and the magnet machine is not — which is worth more than any single-point comparison, and is why the IPMSM dominates despite its cost and its fault hazard.
Answera IPMSM: best efficiency and density, but magnet cost and fault hazard   b induction: safe under fault, no magnets, wide field weakening   c the drive-cycle efficiency map decides
Formulas

Key Formulas

QuantityRelationNotes
PMSM torque\(T = \dfrac32\dfrac{P}{2}\lambda_mi_q\)Surface magnet
Torque constant\(k_T = \dfrac32\dfrac{P}{2}\lambda_m\)0.90 N·m/A
Electrical frequency\(\omega_e = \omega_m\dfrac{P}{2}\)200 Hz at 3000 rpm, 8-pole
Back EMF\(\hat E = \omega_e\lambda_m\)Exists whenever the rotor turns
Terminal voltage\(\sqrt{(\omega_e\lambda_m)^2+(\omega_eL_qi_q)^2}\)206 V peak here
DC link needed\(V_{LL}/0.707\)357 V
Field weakening\(i_d \le \dfrac{\hat V_{max}/\omega_e-\lambda_m}{L_d}\)\(-6.7\) A at 6000 rpm
Fault at speed\(\hat E > V_{dc}\) → uncontrolled generationActive short-circuit
IPMSM torque\(\dfrac32\dfrac{P}{2}\left[\lambda_mi_q+(L_d-L_q)i_di_q\right]\)Reluctance term
MTPA\(i_d = \dfrac{\lambda_m-\sqrt{\lambda_m^2+8(L_q-L_d)^2i_q^2}}{4(L_q-L_d)}\)\(-6.37\) A
Reluctance share11.3% here30–50% in high-saliency designs
MTPA saving6.9% current, 13.3% lossFree — a lookup table
BLDC torque ripple10–15% at \(6f_e\)Commutation \(di/dt\)
BLDC torque density~15% above sinusoidalFull flux over 120°
LCI commutationover-excited machine → leading PFFails below ~10% speed
SRM torque\(T = \tfrac12i^2\dfrac{dL}{d\theta}\)No magnets, high ripple
Pitfalls

Common Mistakes

  1. Forgetting the pole-pair factor in \(\omega_e\). An 8-pole machine at 3000 rpm runs at 200 Hz — Problem 1.

  2. Sizing the DC link from the back EMF alone. The reactance drop adds in quadrature — Problem 1.

  3. Assuming PMSM field weakening saves current. It costs current, unlike every other machine — Problem 2.

  4. Ignoring uncontrolled generation on inverter shutdown. Above base speed the machine feeds its own EMF into the link — Problem 2.

  5. Driving a trapezoidal machine sinusoidally, or the reverse. The waveform and the winding must match or torque ripple results — Problem 3.

  6. Expecting sensorless BLDC to start under load. There is no back EMF at standstill — Problem 3.

  7. Getting the sign of \((L_d-L_q)\) wrong. For an interior machine \(L_q > L_d\), so reluctance torque needs negative \(i_d\) — Problem 4.

  8. Operating an IPMSM at \(i_d = 0\). MTPA gives the same torque for 7% less current — Problem 4.

  9. Specifying an LCI for a drive that must start under load. Below 10% speed the torque pulsates badly — Problem 5.

  10. Comparing traction machines on peak efficiency. The drive cycle lives at 10–30% of peak torque — Problem 6.

Looking Ahead

That completes Part 6. Four sets took the drive from a brushed DC machine to a vector controlled magnet machine, and the control structure never changed — a fast current loop inside a slow speed loop, exactly as Set 33 built it. What changed was how the reference frame was found: mechanically by a commutator, then by computing a slip frequency, then simply by reading a rotor encoder.

The recurring lesson was that each machine's convenience is another's difficulty. The DC machine's commutator gave free decoupling and needed maintenance; the induction machine's induced flux gave a maintenance-free rotor and a 169 ms time constant; the magnet machine's permanent flux gave both, and took away the ability to switch it off.

Every converter so far has drawn power from a stiff source — a mains supply, a DC link, a battery — and delivered it to a load that took whatever was offered. Some sources are not like that. A solar panel has one operating point at which it delivers maximum power and the converter must find it, continuously, as the sun and temperature change. The load no longer sets the operating point; the converter does, and getting it wrong wastes energy that cannot be recovered.

Next: Part 7 begins with Set 37 — Solar PV Systems and MPPT, where the panel's I–V curve is worked through, the maximum power point is located and shown to move with irradiance and temperature, perturb-and-observe is compared with incremental conductance, partial shading is shown to create multiple local maxima that trap a naive tracker, and a grid-tied string inverter is designed.