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Lecture 2: Scalar-Controlled Induction Motor Drive ELEC-E8402 Control of Electric Drives and Power Converters Marko Hinkkanen Spring 2019 1 / 14 Learning Outcomes After this lecture and exercises you will be able to: I Express the dynamic inverse-Γ model in synchronous coordinates I Calculate steady-state operating points and draw the corresponding vector diagrams I Explain the operating principle of the scalar control 2 / 14 Outline Motor Model Scalar Control 3 / 14 Model in Stator Coordinates I Voltage equations u ss = Rs i ss + dψ ss u sR = RR i sR + I dt dψ sR dt i ss − jωm ψ sR = 0 Flux linkages u ss Rs dψ ss dt RR Lσ i sM LM dψ sR dt i sR jωm ψ sR ψ ss = Lσ i ss + ψ sR ψ sR = LM (i ss + i sR ) I Steady state: d/dt = jωs 4 / 14 Model in Synchronous Coordinates I Synchronous (dq) coordinates rotate at the angular speed ωs I Coordinate transformation i ss = i s ejϑs , where no superscript is used in synchronous coordinates I Voltage equations become u s = Rs i s + dψ s + jωs ψ s dt dψ R u R = RR i R + + jωr ψ R = 0 dt I q β d ωs ϑs ωm α Angular speed of the coordinate system I I ωs with respect to the stator ωr = ωs − ωm with respect to the rotor 5 / 14 Model in Synchronous Coordinates I Voltage equations dψ s + jωs ψ s dt dψ R u R = RR i R + + jωr ψ R = 0 dt u s = Rs i s + I Flux linkages is Lσ iR iM dψ s dt LM dψ R dt ψ s = Lσ i s + ψ R ψ R = LM (i s + i R ) I Steady state: d/dt = 0 6 / 14 Power Balance 3 3 3 dWf ωm Re {u s i ∗s + u R i ∗R } = Rs |i s |2 + RR |i R |2 + + TM 2 2 2 dt p I Electromagnetic torque TM = I n o 3p Im i s ψ ∗s 2 Rate of change of the magnetic field energy dψ dψ 1 dWf 3 3 d 1 = Re i ∗s s + i ∗R R = Lσ |i s |2 + LM |i M |2 dt 2 dt dt 2 dt 2 2 is zero in the steady state 7 / 14 Vector Diagram: Currents and Flux Linkages is I Airgap and leakage flux paths are sketched I All vectors are constant in synchronous coordinates in the steady state (but the rotor slips at −ωr ) ψs ψσ ψR iR 8 / 14 Steady-State Torque I TM Torque in the steady state Tb 2Tb TM = ωr /ωrb + ωrb /ωr I Breakdown torque Tb = I 3p LM ψs2 2 LM + Lσ 2Lσ 0 ωrb ωr Breakdown slip ωrb = RR σLM where σ = Lσ LM + Lσ See the derivation in Exercise 1. 9 / 14 Outline Motor Model Scalar Control 10 / 14 Stator Voltage vs. Stator Frequency I I Steady-state stator voltage u s = Rs i s + jωs ψ s us /umax ψs /ψsN Approximate magnitude 1 ψs ≈ ψsN us = umax ψs ∝ 1/|ωs | us = |ωs |ψs 0.5 us ≈ |ωs |ψsN where us = |u s | and ψs = |ψ s | I Maximum voltage is limited us < umax 0 0 0.5 1 1.5 2 ωs /ωsN 11 / 14 Scalar Control or Constant-Volts-per-Hertz Control I Based on the steady-state equations I Supply frequency ωs,ref corresponds to the desired rotor speed I Some speed error due to the slip (can be partly compensated for) I Slow or oscillating dynamics I Torque cannot be controlled I Current cannot be limited I For simple applications Udc ψs,ref ωs,ref Scalar control u ss,ref PWM M us,ref = ωs,ref ψs,ref (+Rs is compensation) R ϑs = ωs,ref dt u ss,ref = us,ref ejϑs 12 / 14 TM /TN Breakdown torque Tb Example load torque TL 2 Steady-state operating point −2 −1 0 1 2 ωm /ωmN −1 −2 ωs = ωsN ψs = ψsN ωs = 0 ψs = ψsN 13 / 14 TM /TN Breakdown torque Tb Continuous weakened-field operating area 2 ωs = 2ωsN ψs = ψsN /2 −2 Continuous full-field operating area 0 1 2 ωm /ωmN −2 ωs = ωsN ψs = ψsN ωs = 0 ψs = ψsN 14 / 14