Solved Problems · Set 35

Vector Control and Field Orientation

Part 6 · Electric Drives — give the induction machine a virtual commutator. Six problems on the control that made it a servo.

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

Set 35 — Vector Control and Field Orientation

Everything V/f could not do traces to one omission: the drive never knows where the rotor flux is. It commands a stator frequency and hopes.

The DC machine of Set 33 had no such problem, because its commutator holds the armature MMF perpendicular to the field mechanically — so armature current commands torque and nothing else. Vector control is the observation that if the flux position can be computed rather than mechanically enforced, the stator current can be resolved into a component along it and a component across it — one setting flux, the other setting torque — and the induction machine becomes the DC drive that was already designed.

Part 6 · Chapter 26 · 6 solved problems

i Method Recap
  • The rotor flux responds with the rotor time constant:

    \[ \tau_r = \frac{L_r}{R_r}, \qquad \lambda_r(s) = \frac{L_mi_d}{1+s\tau_r} \]
  • In rotor-flux coordinates the torque equation separates:

    \[ T = \frac32\cdot\frac{P}{2}\cdot\frac{L_m}{L_r}\lambda_ri_q, \qquad \lambda_r = L_mi_d\ \text{(steady state)} \]
  • Indirect field orientation computes the slip frequency:

    \[ \omega_{sl} = \frac{R_r}{L_r}\cdot\frac{i_q}{i_d}, \qquad \theta_e = \int\left(\omega_r+\omega_{sl}\right)dt \]
  • Direct torque control selects a vector from a table instead of modulating:

    \[ \text{6 sectors}\times\text{flux band}\times\text{torque band} \;\to\; \vec V_k \]
  • Sensorless estimation fails when the back EMF falls below the resistive drop:

    \[ f_{crit} = f_{rated}\frac{I_1R_1}{V_{rated}} \]
  • Loop bandwidths separate by a decade at each level:

    \[ f_{current} \approx 1\ \text{kHz} \;\gg\; f_{speed} \approx 100\ \text{Hz} \;\gg\; f_{flux} \approx 10\ \text{Hz} \]
Problem 1CoreWhy V/f Cannot Be Fast

For the machine of Set 34 with \(L_m = 0.13\) H, \(L_{lr} = 0.005\) H and \(R_r = 0.8\ \Omega\), find the rotor time constant and the settling time of a flux change. Explain why this makes open-loop V/f unsuitable for dynamic applications, and what vector control does about it.

Solution

The rotor time constant:

\[ L_r = L_m+L_{lr} = 0.130+0.005 = 0.135\ \text{H} \]
\[ \tau_r = \frac{L_r}{R_r} = \frac{0.135}{0.8} = 169\ \text{ms} \]

A sixth of a second. The rotor cage is a short-circuited winding of substantial inductance and low resistance, so any change in its flux linkage decays slowly — and there is no terminal at which to force it faster.

What that means for a torque command. In a V/f drive, torque is commanded by changing the frequency, which changes the slip:

\[ \Delta f \to \Delta f_{sl} \to \Delta\lambda_r \to \Delta T \]
\[ t_{settle} \approx 3\tau_r = 500\ \text{ms} \]

Half a second to establish a new torque. For a pump that is irrelevant — the hydraulic system is slower still. For a servo, a spindle, a traction drive or anything that must reject a load disturbance, it is useless.

Compare the timescales in play:

QuantityTime constantSet by
Stator current~2 msleakage inductance / stator resistance
Rotor flux169 ms\(L_r/R_r\)
Mechanical~1 s\(J\omega/T\)

Nearly two orders of magnitude separate the current from the flux. That separation is the opportunity: if the flux could be held constant while only the current changes, torque would respond at the current's timescale rather than the flux's.

Which is precisely what vector control does. Resolve the stator current into two components relative to the rotor flux:

\[ i_d\ \text{along the flux} \;\Longrightarrow\; \text{sets } \lambda_r,\ \text{slowly } (\tau_r) \]
\[ i_q\ \text{across the flux} \;\Longrightarrow\; \text{sets } T,\ \text{fast (current loop)} \]

Hold \(i_d\) constant and the flux never changes, so its slow dynamics are simply never excited. Torque is then commanded through \(i_q\) alone, which responds in milliseconds.

The resulting improvement:

ControlTorque responseLimited by
Open-loop V/f~500 msrotor time constant
Slip-frequency V/f~100 mspartial flux disturbance
Vector control (FOC)~5 mscurrent loop bandwidth
Direct torque control~1 msswitching frequency

A hundredfold improvement, and no change to the machine or the inverter — the hardware of Set 26, Problem 6 serves either. The entire difference is in what the controller computes.

The one thing that stays slow. Establishing the flux from zero still takes \(3\tau_r\):

\[ \text{magnetising at start-up: } \approx 500\ \text{ms} \]

So a vector drive cannot produce torque instantly from a cold start — it must first build the rotor flux, which takes half a second. Drives handle this by pre-magnetising: applying \(i_d\) with \(i_q = 0\) before the brake releases. It is also why field weakening is a slow manoeuvre, and why a drive commanded rapidly through base speed can briefly lose orientation.

The rotor flux and the stator current live two orders of magnitude apart in time. Vector control exploits that gap by never disturbing the flux — holding \(i_d\) fixed so the 169 ms dynamics are simply not excited — and commanding torque entirely through a current that settles in milliseconds.
Answera\(\tau_r = 169\ \text{ms}\)   b V/f torque settles in ~500 ms   c FOC holds the flux constant and commands torque via \(i_q\) — ~5 ms
Problem 2CoreTwo Currents, Two Jobs

For the same 4-pole machine at a rotor flux of 0.9 Wb, find the torque constant, the \(i_d\) and \(i_q\) needed for a rated torque of 47.7 N·m, and the resulting stator current. Verify against the machine's rated current.

id = 6.9 A iq = 18.3 A is = 19.6 A rotor flux axis (d) torque = 2.6 × iq flux = 0.13 × id q
Perpendicular by construction — the commutator's job, done in software
Solution

The torque equation in rotor-flux coordinates:

\[ T = \frac32\cdot\frac{P}{2}\cdot\frac{L_m}{L_r}\lambda_ri_q \]
\[ k_T = \frac32(2)\left(\frac{0.130}{0.135}\right)(0.9) = 2.60\ \text{N}\!\cdot\!\text{m/A} \]

Compare Set 33's DC machine: \(T = K_a\Phi I_a\) with \(K_a\Phi = 1.321\). The form is identical — a constant times a current — and that identity is the entire point of the transformation.

The torque-producing current:

\[ i_q = \frac{T}{k_T} = \frac{47.7}{2.60} = 18.3\ \text{A} \]

The flux-producing current, from the steady-state relation:

\[ \lambda_r = L_mi_d \;\Longrightarrow\; i_d = \frac{0.9}{0.130} = 6.9\ \text{A} \]

Note that this is the magnetising current, and in a V/f drive it is drawn automatically as reactive current. Here it is commanded explicitly, which means it can also be reduced — the basis of field weakening and of efficiency optimisation at light load.

The stator current is their vector sum, because they are perpendicular by construction:

\[ \left|\vec i_s\right| = \sqrt{i_d^2+i_q^2} = \sqrt{6.9^2+18.3^2} = 19.6\ \text{A peak} \]
\[ I_{s,rms} = \frac{19.6}{\sqrt2} = 13.9\ \text{A} \]
\[ \text{against the rated } 13.6\ \text{A}\ \checkmark \]

Within 2% of the nameplate current computed in Set 34 by an entirely different route — power, voltage, power factor and efficiency. The dq model and the equivalent circuit are describing the same machine.

What the two components mean physically:

\(i_d\)\(i_q\)
Setsrotor fluxtorque
DC machine analoguefield currentarmature current
Time constant169 ms~2 ms
Normallyheld constantvaried freely
In field weakeningreducedreduced to stay within \(I_{max}\)
Reactive/activereactiveactive

The second row is why the technique works at all. An induction machine has no field winding, so \(i_d\) is a synthetic field current — carried by the same three wires as \(i_q\) and separable only because the controller knows the flux angle.

The transformations that make it possible:

\[ \text{Clarke: } (i_a,i_b,i_c) \to (i_\alpha,i_\beta) \quad\text{— three phases to two, stationary} \]
\[ \text{Park: } (i_\alpha,i_\beta) \to (i_d,i_q) \quad\text{— rotating with } \theta_e \]

Both are pure algebra; the difficulty is entirely in \(\theta_e\), the rotor flux angle. It cannot be measured — there is no terminal at the rotor flux — so it must be computed, which is Problem 3 and the whole practical content of vector control.

The dq transformation gives the induction machine a virtual commutator. Once the current is resolved relative to the rotor flux, torque is a constant times one current and flux is a constant times the other — exactly the DC machine's equations, with the perpendicularity enforced by software rather than by brushes.
Answera\(k_T = 2.60\ \text{N}\!\cdot\!\text{m/A}\)   b\(i_q = 18.3\ \text{A},\ i_d = 6.9\ \text{A}\)   c\(\left|\vec i_s\right| = 19.6\ \text{A}\) peak = 13.9 A rms
Problem 3Exam levelFinding the Flux Without Looking

Compute the slip frequency required for field orientation at the operating point of Problem 2, verify it against Set 34's rated slip frequency, and determine the effect of a 50% rise in rotor resistance as the machine heats up.

Solution

The problem. The Park transformation needs \(\theta_e\), the angle of the rotor flux — which is not the rotor's mechanical angle, and cannot be measured:

\[ \theta_e = \theta_r+\theta_{sl} \]

The rotor position \(\theta_r\) is available from an encoder. The slip angle is not — but it can be computed from the currents the controller is itself commanding. That is indirect field orientation.

The slip relation. Requiring the rotor flux to lie entirely on the d-axis forces a specific slip:

\[ \omega_{sl} = \frac{R_r}{L_r}\cdot\frac{i_q}{i_d} = \frac{1}{\tau_r}\cdot\frac{i_q}{i_d} \]
\[ = \frac{0.8}{0.135}\cdot\frac{18.3}{6.9} = (5.93)(2.65) = 15.7\ \text{rad/s} \]
\[ f_{sl} = \frac{15.7}{2\pi} = 2.50\ \text{Hz} \]

Against Set 34's rated slip frequency of 2.0 Hz, computed from the equivalent circuit — agreement to within the difference between the two models' assumptions. The machine's rated slip frequency and its field-orientation slip are the same physical quantity.

The complete IFOC scheme:

\[ \omega_e = \omega_r+\omega_{sl}, \qquad \theta_e = \int\omega_e\,dt \]
StepOperation
1Measure \(\omega_r\) from the encoder
2Compute \(\omega_{sl}\) from the commanded \(i_q/i_d\)
3Integrate the sum to get \(\theta_e\)
4Park-transform the measured currents using \(\theta_e\)
5Regulate \(i_d\) and \(i_q\) with PI loops
6Inverse Park and hand the voltage vector to SVM — Set 28

Note that step 6 hands a voltage vector straight to the space vector modulator — no conversion to three-phase references and back. That fit is why SVM and vector control are always paired.

Now the weakness. The slip computation depends on \(R_r\), which is a rotor temperature measurement in disguise:

\[ R_r\ \text{rises} \approx 50\%\ \text{from cold to hot (20}^\circ\text{C to 150}^\circ\text{C)} \]
\[ \omega_{sl,correct} = 23.6\ \text{rad/s},\quad \text{but the drive still commands } 15.7 \]

A 50% error in the commanded slip frequency — and the error is in the integrated angle, so it does not merely scale the torque. It rotates the entire dq frame away from the actual flux.

What detuning does:

ConsequenceDetail
Orientation lostthe d-axis no longer lies on the flux
Torque errorcommanded and actual diverge, typically 10–20%
Coupling returns\(i_q\) now disturbs the flux
Efficiency fallsflux is wrong for the operating point
Dynamic response degradesthe fast/slow separation is broken
Instability at high torquein severe cases

The third row is the subtle one. Field orientation's whole benefit was that \(i_q\) does not affect the flux; once the frame is misaligned that guarantee fails, and a torque step drags the flux with it — reintroducing the 169 ms dynamics the scheme was built to avoid.

The remedies:

ApproachNote
Thermal model of the rotorestimate temperature from load history — cheap, approximate
Online \(R_r\) adaptationestimate from a model mismatch — standard in good drives
Direct field orientationmeasure or observe the flux instead — avoids \(R_r\) at speed
Accept the erroradequate for many industrial duties

Direct field orientation observes the flux from the stator voltage and current rather than computing the slip — which removes the \(R_r\) dependence but introduces an integration that drifts at low speed, and a dependence on \(R_s\) instead. Most drives use indirect orientation with adaptation, and Problem 5 explains why.

Indirect field orientation trades a measurement for a model. The flux angle is computed from the commanded currents and one machine parameter, so no flux sensor is needed — but \(R_r\) rises 50% as the rotor heats, and a wrong slip frequency rotates the whole reference frame off the flux, destroying the decoupling that justified the scheme.
Answera\(\omega_{sl} = 15.7\ \text{rad/s} = 2.50\ \text{Hz}\)   b matches Set 34's rated slip frequency   c 50% \(R_r\) rise misorients the frame and reintroduces coupling
Problem 4Exam levelSkipping the Modulator

Describe direct torque control, explain how it achieves a torque response several times faster than field-oriented control, and identify what it gives up in exchange.

Solution

The premise. The stator flux moves in the direction of the applied voltage vector:

\[ \frac{d\vec\lambda_s}{dt} = \vec v_s-R_s\vec i_s \approx \vec v_s \]

So applying one of the six active vectors of Set 28 pushes the stator flux directly, and the torque — which depends on the angle between stator and rotor flux — changes immediately. No current loop, no transformation to a rotating frame, no modulator.

The control law is a lookup table:

InputValues
Flux error band2 (increase / decrease)
Torque error band3 (increase / hold / decrease)
Flux sector6 — the hexagon of Set 28
Outputone of the 8 voltage vectors
\[ 2\times3\times6 = 36\ \text{table entries} \]

Two hysteresis comparators and a 36-entry table — that is the entire controller. There is no PI regulator anywhere in the torque path, and no PWM: the selected vector is applied directly until a comparator changes state.

Why it is faster. Count what stands between a torque command and a change in the inverter's output:

StageFOCDTC
Coordinate transformClarke + ParkClarke only
Current regulatorstwo PI loopsnone
Inverse transforminverse Parknone
ModulatorSVM — one period of delaynone
Response~5 ms~1 ms

The two rows in bold are the difference. A current loop necessarily takes several switching periods to change a current; DTC changes the applied voltage on the very next sample, which is the fastest anything can happen in a switching converter.

What it costs:

DrawbackCause
Variable switching frequencyhysteresis, not a fixed carrier
Broadband acoustic noiseno discrete carrier tone
Harder EMI filteringspread spectrum — no line to notch
Torque and flux rippleset by the hysteresis bands
Poor at very low speedstator flux estimated by integration
High sampling rate neededtypically 40 kHz for a 4 kHz effective rate

The first row propagates into the next two and is DTC's main practical objection: an EMI filter designed against a fixed carrier, as Set 32 did, has nothing fixed to work against. The fifth row is shared with sensorless FOC and is the subject of Problem 5.

The three schemes side by side:

V/fFOCDTC
Torque response500 ms5 ms1 ms
Torque controlnoneexcellentexcellent
Switching frequencyfixedfixedvariable
Parameter sensitivitylow\(R_r\)\(R_s\)
Position sensornousuallyno
Complexitylowesthighmoderate
Zero-speed torquenoyes (with sensor)marginal

Modern drives blur these lines: space-vector-modulated DTC adds a modulator back to fix the variable frequency, and predictive control generalises the vector selection over a longer horizon. The three columns are landmarks rather than a complete taxonomy.

DTC is fast because it removes the modulator, and awkward for the same reason. Selecting a voltage vector directly from a hysteresis table changes the output on the next sample rather than the next PWM period — a fivefold improvement in response, bought with a switching frequency that is no longer a design parameter at all.
Answera hysteresis bands select from a 36-entry vector table   b ~1 ms against FOC's 5 — no current loops, no modulator   c variable switching frequency and torque ripple
Problem 5ChallengeRunning Without a Sensor

Vector control needs the rotor position. Determine the frequency below which voltage-model estimation fails for the machine of Set 34, explain why, and describe the technique that works to zero speed.

Solution

The voltage model. The stator flux can be found by integrating the terminal quantities:

\[ \vec\lambda_s = \int\left(\vec v_s-R_s\vec i_s\right)dt \]

Elegant, and it needs only one parameter. From the stator flux and the currents, the rotor flux angle follows — and hence the rotor position, without any sensor. It works well at speed and fails completely at low speed, for a reason that can be quantified exactly.

Why it fails. The integrand is a difference of two terms, and their relative size depends entirely on frequency:

\[ \text{back EMF} \propto f; \qquad I_1R_1\ \text{independent of } f \]
\[ \text{at } 1\ \text{Hz: back EMF} = \frac{415}{50} = 8.3\ \text{V} \]
\[ I_1R_1 = (13.6)(1.0) = 13.6\ \text{V} \]

The resistive drop is larger than the signal being estimated. Subtracting a 13.6 V correction from a 13.6 V measurement to find an 8.3 V quantity leaves nothing but the error in \(R_s\).

The crossover frequency:

\[ f_{crit} = f_{rated}\frac{I_1R_1}{V_{rated}} = (50)\frac{13.6}{415} = 1.64\ \text{Hz} \]
\[ = 3.3\%\ \text{of base speed} \]

Below about 1.6 Hz the resistive drop dominates and estimation degrades rapidly; practical drives quote a minimum of 1–3 Hz for sensorless operation, which agrees. Note that this frequency is a property of the machine — specifically of its per-unit stator resistance — not of the algorithm.

Three compounding problems at low speed:

ProblemEffect
\(R_s\) uncertainty±50% with temperature — and it is the dominant term
Integrator driftany DC offset integrates without bound
Inverter nonlinearitydead-time error is 12.7% at low modulation — Set 25
Measurement resolutionsmall signals against full-scale sensors

The third row connects directly to Set 25, Problem 5: the drive does not actually apply the voltage it commanded, and the discrepancy is worst at exactly the low modulation index where the estimator is already struggling. Sensorless drives therefore need dead-time compensation before they need a better observer.

The technique that works to zero speed:

\[ \text{inject a high-frequency carrier (500 Hz–2 kHz) and detect the response} \]

If the machine has any magnetic saliency — a difference between \(L_d\) and \(L_q\) — the injected carrier produces a current response whose magnitude varies with rotor position. Demodulating that response gives position without depending on back EMF at all, so it works perfectly at standstill.

Its own limitations:

RequirementDifficulty
Sufficient saliencya cage induction motor has very little
Audible noisethe carrier is in the audible band
Extra losscarrier current does no work
Reduced current capabilitythe carrier occupies headroom
Handover to the voltage modelmust blend smoothly around 2–5 Hz

The first row explains where the technique is actually used. An interior permanent magnet machine, with its buried magnets, has strong saliency and responds beautifully; a cage induction motor has only the small saturation-induced saliency, so injection works but marginally. That is one reason Set 36's IPMSM has displaced the induction machine in traction.

When to fit an encoder:

ApplicationSensor
Pumps, fans, conveyorssensorless — low speed never needed
Extruders, mixerssensorless with injection
Hoists, cranes, liftsencoder — full torque at zero speed
Machine tools, servosencoder or resolver — position accuracy
Tractionresolver — environment and reliability

The third row is the elevator of Set 33 again, now with an induction machine. A hoist must produce full torque at exactly zero speed to hold a load while the brake releases — the one condition under which every sensorless method fails — so an encoder is not a refinement but a safety requirement.

The sensorless limit is a property of the machine, not the algorithm. Estimation fails when the back EMF drops below the stator resistive drop, at \(f_{rated}I_1R_1/V_{rated}\) — 1.64 Hz here, or 3.3% of base speed. No observer can recover a signal that is smaller than the correction being subtracted from it.
Answera\(f_{crit} = 1.64\ \text{Hz}\) — 3.3% of base speed   b the \(I_1R_1\) drop exceeds the back EMF   c high-frequency injection works to zero speed — if the machine is salient
Problem 6ChallengeWriting It Down

Specify a complete field-oriented control implementation for the 7.5 kW machine: the computation sequence per sampling period, the loop bandwidths, the cross-coupling terms that must be cancelled, and the limits that must be enforced.

Solution

The per-period sequence, at a 5 kHz sampling rate synchronised to the PWM centre — the sampling instant Set 28 established:

StepOperation
1Sample \(i_a,i_b\) and \(V_{dc}\) at the period centre
2Read \(\theta_r\) from the encoder
3Clarke: \((i_a,i_b) \to (i_\alpha,i_\beta)\)
4Compute \(\omega_{sl}\) and integrate to \(\theta_e\)
5Park: \((i_\alpha,i_\beta,\theta_e) \to (i_d,i_q)\)
6Speed PI → \(i_q^*\); flux controller → \(i_d^*\)
7Two current PIs → \(v_d,v_q\)
8Add decoupling terms
9Inverse Park → \((v_\alpha,v_\beta)\)
10SVM → three compare registers — Set 28

The bandwidths, each a decade below the one inside it — the cascade principle of Set 33:

LoopBandwidthLimited by
Current (\(i_d,i_q\))1 kHzPWM at 5 kHz — a fifth of it
Speed100 Hza decade below current
Flux10 Hz\(\tau_r = 169\ \text{ms}\)
Position (if fitted)10 Hza decade below speed

The flux loop is deliberately slow because the plant is slow — there is no point demanding a faster response than \(1/\tau_r\) can deliver, and trying only saturates the voltage.

The decoupling terms. The d and q axes are not actually independent in the voltage equations — rotation couples them:

\[ v_d = R_si_d+\sigma L_s\frac{di_d}{dt}-\omega_e\sigma L_si_q \]
\[ v_q = R_si_q+\sigma L_s\frac{di_q}{dt}+\omega_e\left(\sigma L_si_d+\frac{L_m}{L_r}\lambda_r\right) \]

The \(\omega_e\) terms are speed-dependent disturbances that each axis injects into the other, and they grow with speed. Adding their computed values to the PI outputs — feedforward decoupling — leaves each current loop seeing a clean first-order plant, which is what the pole-cancellation design of Set 33 assumed.

Without decoupling, the picture at rated speed:

\[ \omega_e\frac{L_m}{L_r}\lambda_r = (157)(0.963)(0.9) = 136\ \text{V} \]

A 136 V disturbance in the q-axis voltage equation — more than half the phase voltage. The PI would eventually reject it, but only within its bandwidth, so a speed transient produces a large transient current error. Feedforward removes it before the loop ever sees it.

The limits that must be enforced:

LimitValueHandling
Current magnitude\(\sqrt{i_d^2+i_q^2} \le I_{max}\)prioritise \(i_d\), clip \(i_q\)
Voltage magnitude\(\left|\vec v\right| \le V_{dc}/\sqrt3\)Set 28's inscribed circle
Field weakeningreduce \(i_d\) when the voltage saturateskeeps the vector inside the circle
Integrator anti-windupon every PIback-calculation
Minimum pulse widthSet 28, Problem 6drop short pulses
Dead-time compensationSet 25, Problem 5essential at low speed

The first row's priority matters. When the current limit binds, reducing \(i_q\) costs torque but reducing \(i_d\) costs flux — and losing flux takes 169 ms to recover while losing torque is instantly reversible. So \(i_d\) is protected and \(i_q\) absorbs the limit.

What the whole structure achieves:

CapabilityV/f (Set 34)FOC
Torque response500 ms5 ms
Speed accuracy4–40%< 0.1% with encoder
Full torque at zero speednoyes
Torque limitingnoexact
Four-quadrantcrudeseamless
Computationtrivial~20 µs per period

Every entry in the last column matches or beats the DC drive of Set 33 — in a machine with no brushes, no commutator and no maintenance. Twenty microseconds of arithmetic per period is what replaced a mechanical commutator, and it is why the DC drive has essentially disappeared.

Vector control is the DC drive's control structure applied through a computed reference frame. The cascade of Set 33, the space vector modulator of Set 28 and the dead-time compensation of Set 25 all reappear unchanged — with two coordinate transforms and a slip calculation added. That addition replaced the commutator.
Answera 10 steps at 5 kHz, sampled at the PWM centre   b 1 kHz / 100 Hz / 10 Hz   c decouple the 136 V \(\omega_e\) term   d prioritise \(i_d\) at the current limit
Formulas

Key Formulas

QuantityRelationNotes
Rotor time constant\(\tau_r = L_r/R_r\)169 ms here
Flux dynamics\(\lambda_r = \dfrac{L_mi_d}{1+s\tau_r}\)Settles in \(3\tau_r\)
Torque\(T = \dfrac32\dfrac{P}{2}\dfrac{L_m}{L_r}\lambda_ri_q\)DC-machine form
Torque constant\(k_T = \dfrac32\dfrac{P}{2}\dfrac{L_m}{L_r}\lambda_r\)2.60 N·m/A
Flux current\(i_d = \lambda_r/L_m\)6.9 A
Stator current\(\sqrt{i_d^2+i_q^2}\)Perpendicular by construction
Slip frequency\(\omega_{sl} = \dfrac{R_r}{L_r}\dfrac{i_q}{i_d}\)The IFOC relation
Flux angle\(\theta_e = \int(\omega_r+\omega_{sl})dt\)Computed, not measured
Detuning sensitivity\(\omega_{sl} \propto R_r\)+50% hot
Voltage model\(\vec\lambda_s = \int(\vec v_s-R_s\vec i_s)dt\)Fails at low \(f\)
Sensorless limit\(f_{crit} = f_{rated}\dfrac{I_1R_1}{V_{rated}}\)1.64 Hz here
d-axis decoupling\(-\omega_e\sigma L_si_q\)Feedforward
q-axis decoupling\(+\omega_e\left(\sigma L_si_d+\frac{L_m}{L_r}\lambda_r\right)\)136 V at rated speed
Bandwidth ladder1 kHz / 100 Hz / 10 Hzcurrent / speed / flux
Torque response500 / 5 / 1 msV/f / FOC / DTC
Pitfalls

Common Mistakes

  1. Expecting instant torque from a cold start. The flux takes \(3\tau_r \approx 500\) ms to establish — pre-magnetise — Problem 1.

  2. Using \(L_m\) instead of \(L_r\) in the torque equation. The ratio \(L_m/L_r\) appears, not either alone — Problem 2.

  3. Confusing the rotor flux angle with the rotor mechanical angle. They differ by the slip angle — Problem 3.

  4. Ignoring \(R_r\) variation in IFOC. A 50% rise misorients the frame and reintroduces cross-coupling — Problem 3.

  5. Assuming detuning only scales the torque. It rotates the reference frame, so \(i_q\) begins disturbing the flux — Problem 3.

  6. Specifying DTC where EMI is filtered against a fixed carrier. Its switching frequency varies — Problem 4.

  7. Expecting a sensorless drive to hold zero speed. Estimation fails below 1.64 Hz for this machine — Problem 5.

  8. Applying high-frequency injection to a cage induction motor. It has very little saliency — Problem 5.

  9. Omitting the \(\omega_e\) decoupling terms. The q-axis disturbance reaches 136 V at rated speed — Problem 6.

  10. Clipping \(i_d\) rather than \(i_q\) at the current limit. Lost flux takes 169 ms to recover — Problem 6.

Looking Ahead

Six problems and the induction machine has become a servo. The rotor time constant of 169 ms made V/f dynamics hopeless and was then simply never excited; the torque equation collapsed to a constant times a current, identical in form to the DC machine's; indirect field orientation computed the flux angle from a slip relation and paid for it with sensitivity to a rotor resistance that rises 50% when hot; and sensorless operation was shown to fail below 1.64 Hz for a reason arising from the machine rather than the algorithm.

Every difficulty in this set came from the same source: the rotor's flux is induced, so it must be created by stator current, maintained against a slow time constant, and located by computation. A machine whose rotor carries permanent magnets has none of those problems. Its flux is simply there — constant, free of excitation current, and mechanically locked to a rotor position that an encoder measures directly.

That removes \(i_d\) from the torque equation, removes the magnetising current from the stator, removes \(R_r\) from the control, and removes the rotor copper loss entirely. It also removes the ability to switch the flux off — which becomes the central problem above base speed.

Next: Set 36 — Synchronous, BLDC and PMSM Drives, where a permanent-magnet machine needs 357 V of DC link at base speed and would need 533 at twice it, field weakening injects negative \(i_d\) to fight the magnets, an interior machine's reluctance torque contributes 11% and reduces the current by 7%, and the technologies are compared for a traction application.