By the end of this chapter you should be able to:
Describe surface-mounted and interior PMSM rotor constructions and their saliency.
Write the d–q voltage and torque equations of a PMSM.
Separate magnet torque and reluctance torque, and compute each.
Explain field-oriented control, maximum-torque-per-ampere operation and flux weakening.
Construction
A PMSM has a conventional three-phase distributed stator winding, like an induction motor or alternator, and a rotor carrying permanent magnets in place of a field winding. The magnets are shaped to produce a sinusoidal back-EMF.
Magnets glued on the rotor surface. Magnets have permeability close to air, so the effective air gap is the same on both axes: \(L_d = L_q\) — non-salient. Simple, low ripple; magnets need a retaining sleeve at high speed.
Magnets buried in slots inside the rotor iron. The d-axis passes through the magnets (high reluctance), the q-axis through iron: \(L_q > L_d\) — inverse saliency. Robust at high speed and produces extra reluctance torque.
The d–q Model
Transforming the three-phase quantities to a frame rotating with the rotor (Park's transformation), with the d-axis aligned to the magnet flux \(\lambda_f\) and \(\omega_e = p\,\omega_m\) (\(p\) = pole pairs):
For an SPMSM \(L_d = L_q\) and torque is simply proportional to \(i_q\). For an IPMSM, \(L_d - L_q < 0\), so a negative \(i_d\) adds reluctance torque.
Amplitude-invariant transformation is assumed (hence the factor 3/2); \(i_d\), \(i_q\) are peak phase-current components and \(\lambda_f\) is the peak flux linkage per phase.
Problem. An 8-pole PMSM (\(p = 4\)) has \(\lambda_f = 0.1\) Wb. Find the torque for \(i_q = 10\) A when (a) it is an SPMSM with \(i_d = 0\); (b) it is an IPMSM with \(L_d = 2\) mH, \(L_q = 5\) mH and \(i_d = -5\) A. (c) Find the peak phase back-EMF at 3000 rev/min.
The IPMSM produces 15 % more torque for only a modest increase in current magnitude (\(\sqrt{10^2 + 5^2} = 11.2\) A).
Field-Oriented Control
Because \(i_d\) and \(i_q\) control flux and torque independently, a PMSM can be controlled exactly like a separately excited DC motor. An encoder or resolver measures rotor angle; Park's transformation converts measured phase currents to \(i_d\), \(i_q\); two PI regulators set \(v_d\), \(v_q\); the inverse transformation and space-vector PWM generate the inverter switching.
\(i_d = 0\) control: optimal for an SPMSM — all current produces torque.
Maximum torque per ampere (MTPA): for an IPMSM, choose the negative \(i_d\) that maximises torque for a given current magnitude.
Flux weakening: above base speed the back-EMF \(\omega_e\lambda_f\) approaches the inverter's voltage limit. Injecting negative \(i_d\) opposes the magnet flux, reducing the stator flux linkage \(L_di_d + \lambda_f\) and extending the speed range at constant power.
Merits and Applications
Highest efficiency and power density of common industrial motors (no rotor copper loss, no magnetising current); smooth torque; operation at unity power factor possible; excellent dynamic response.
Cost and supply risk of rare-earth magnets; demagnetisation risk at high temperature; cannot be started directly on the grid without a cage (line-start PMSM); requires an inverter and position sensing for variable speed.
Applications: electric-vehicle traction (mostly IPMSM), servo drives and robotics, CNC spindles, washing-machine direct drives, compressors, direct-drive wind generators.
Summary and Common Mistakes
SPMSM: \(L_d = L_q\); IPMSM: \(L_q > L_d\).
\(T_e = \tfrac32p[\lambda_fi_q + (L_d - L_q)i_di_q]\).
FOC decouples flux (\(i_d\)) and torque (\(i_q\)); negative \(i_d\) gives MTPA and flux weakening in an IPMSM.
Using the number of poles instead of pole pairs for \(p\).
Assuming \(L_d > L_q\) for an IPMSM, by analogy with wound-field machines.
Applying positive \(i_d\) in an IPMSM — it reduces torque and strengthens the flux.
Chapter Review
P90.1 A 6-pole SPMSM has \(\lambda_f = 0.08\) Wb. What \(i_q\) gives 9 N·m?
Show answer
\(p = 3\); \(i_q = 9/(1.5 \times 3 \times 0.08) = 25\) A.P90.2 For the IPMSM of Example 90.1, find the total torque with \(i_q = 10\) A and \(i_d = -10\) A.
Show answer
\(T_\text{rel} = 1.5(4)(-0.003)(-10)(10) = 1.8\) N·m; \(T_e = 7.8\) N·m.
MCQ 1. In a surface-mounted PMSM:
(a) \(L_d > L_q\) (b) \(L_d < L_q\) (c) \(L_d = L_q\) (d) \(L_d = 0\)Show answer
(c)MCQ 2. Flux weakening in a PMSM is achieved with:
(a) positive \(i_d\) (b) negative \(i_d\) (c) zero \(i_q\) (d) higher frequency onlyShow answer
(b)
Why is the PMSM more efficient than an induction motor of the same rating?
Explain why buried magnets give inverse saliency.
Why is flux weakening necessary above base speed?
Remove the magnets altogether and rely only on reluctance torque, and you arrive at the simplest and most rugged rotor of all — the switched reluctance motor of Chapter 91.