Control of AC servo motors

Transcription

1 Advanced Control and Automation L1:1 Control of AC servo motors -phase permanent magnet synchronous motors brushless DC motor trapezoidal back-emf profile rectangular pulse current profile requires only Hall-effect position sensors for electronic commutation AC servo motor sinusoidal back-emf profile balanced sinusoidal current profile requires precise motor position measurement (resolver or encoder)

2 L1: Recall DC servo motor

3 Permanent magnet AC servo motor L1:

4 L1:4 Control of brushless DC motor Source: P. Krause, O. Wasynczuk, S. Sudhoff, Analysis of Electric Machinery and Drive Systems, Wiley (00)

5 Electrical torque production Tmω m = eaia + ebib + ecic Back-EMF e = k ( θ ) ω, e = k ( θ ) ω, e = k ( θ ) ω, a a e m b b e m c c e m L1:5 For balanced windings and -phase supply, current and back-emf waveform shapes are identical, but displaced by 10ºE (electrical degrees) Hence, motor torque is T ( θ ) = k ( θ ) i ( θ ) + k ( θ π ) i ( θ π ) + k ( θ + π ) i ( θ + π ) m e a e a e a e a e a e a e Two standard ways of producing constant torque: Trapezoidal back-emf and square wave current Sinusoidal back-emf and sinusoidal current Shape of back-emf profile depends on geometry of magnets and windings

6 Brushless DC waveforms and torque production L1:6 I p K p Tω m = eaia + ebib + ecic T( θ ) = K I e p p Three Hall-effect sensors provide commutation switching points six-step drive Source: D. Hanselman, Brushless Permanent Magnet Motor Design, nd ed, 00

7 L1:7

8 L1:8 AC servomotor waveforms and torque production T ( θ ) = k ( θ ) i ( θ ) + k ( θ ) i ( θ ) + k ( θ + ) i ( θ + ) m e a e a e a e a e a e a e Sinusoidal back-emf: Sinusoidal phase current: Hence i.e., k i ( θ ) = K cos( θ ) a e p e ( θ ) = I cos( θ ) a e p e π π Tm( θe) = KpI p cos θe + cos ( θe ) + cos ( θe + ) Tm( θ e) = KpIp Need precise measurement of rotor position to generate phase current with correct phase; i.e., encoder or resolver

9 Mechanical and electrical position L1:9 Coil of phase a Coil 1 of phase a 4 pole motor i ( θ ) = I cos( ω t) a e p e Balanced sinusoidally-distributed phase windings supplied with balanced -phase currents generate a spatially-sinusoidal MMF which rotates at angular speed ω m = (/P)ω e radm/s (mechanical radians), where ω e is frequency of phase currents, P is number of motor poles: N s P MMFs = I p cos ωet φ s P where φ s = stator angular coordinate (radm) 40 Ref: P. Krause, O. Wasynczuk, S. Sudhoff, Analysis of Electric Machinery and Drive Systems, Wiley (00) P = ω = t e o 10 E ω = t e o 60 E ω = 0 t e

10 Dynamics of AC servo motor L1:10 KVL for phases: d iabc vabc = Rsiabc + Ls + eabc dt T v = v v v, i = i i i [ ] [ ] abc a b c abc a b c T R L s = R I ph L + L L L 1 1 l ph ph ph 1 1 s = Lph Ll + Lph Lph 1 1 Lph Lph Ll + Lph L l, L ph e = leakage and magnetising inductances of coils cosθr π = ω K cos( θ ) P π cos( θr + ) abc r p r (sinusoidal back-emf) All this results in a very complex, nonlinear expression for the motor torque: T m = T m (i a, i b, i c, θ r )

11 Park transformation to qd0 variables The equations are simplified by a transformation of variables from the machine frame abc to a quadrature-direct-zero qd0 reference frame rotating with the rotor, at speed ω r = (P/)ω m rade/s Transformation for voltage, current, flux linkage or charge variables: ( π) ( π ) + ( π) ( π) vq cosθr cos θr cos θr va v d = sin θr sin θr sin θr + v v b 0 = K ( θ ) v v 0 vc va cosθr sinθr 1 vq ( ) ( v π π cos ) b θr sin θr 1 v 1 = d vabc = Ks ( θr ) vqd 0 ( π) ( π v cos ) c θr + sin θr + 1 v0 Then, total power: P = vi + vi + vi abc a a b b c c = P = vi + vi + vi ( ) qd 0 q q d d 0 0 L1:11 qd s r abc

12 Equations of motion in transformed variables KVL di ( ) q P v = R i + L + L + ω ( L + L ) i + K dt di ( ) d v ( ) d = Rphid + Ll + Lph ωr Ll + Lph iq dt di0 v0 = Rphi0 + Ll dt Motor torque Tm = Kpiq q ph q l ph r l ph d p r Mechanical dynamics J ωm = Tm Bωm Tl i.e., J ω = T B ω T P P r m r l ω K ω L1:1 P = K ω p m p r i q is the torque producing component of the stator currents

13 Balanced -phase set Phase voltages are controlled to have a frequency equal to the rotor speed (in rade/s): ω e = ω r va cos( ωrt+ φv) π vabc = v b = V s cos( ωrt + φv ) π vc cos( ωrt+ + φv) L1:1 Quadrature and direct voltages and currents vq cosφv vqd 0 = Ks ( θr ) vabc = v d = V s sinφ v v 0 0 To maximise torque production and minimise losses, φ v = 0 v = V, v = 0 q s d

14 Electromechanical model of AC servo motor L1:14 Source: S. Lyshevski, Electromechanica Systems, Electirc Machines, and Applied Mechatronics, CRC Press (000)

15 Control based on quadrature and direct currents L1:15 Source: SIEMENS ducumentation for SIMODRIVE 611U