INDUCTION MOTOR Scalar Control (squirrel cage) MEP 1523 ELECTRIC DRIVES INDUCTION MOTOR Scalar Control (squirrel cage)

Scalar control of induction machine: Control of induction machine based on steady-state model (per phase SS equivalent circuit): Is Lls Ir’ Llr’ Rs + Vs – + Eag – Lm Rr’/s Im

Scalar control of induction machine Te Pull out Torque (Tmax) rotor TL Te Intersection point (Te=TL) determines the steady –state speed sm rated Trated r s s

Scalar control of induction machine Given a load T– characteristic, the steady-state speed can be changed by altering the T– of the motor: Pole changing Synchronous speed change with no. of poles Discrete step change in speed Variable voltage (amplitude), frequency fixed E.g. using transformer or triac Slip becomes high as voltage reduced – low efficiency Variable voltage (amplitude), variable frequency Using power electronics converter Operated at low slip frequency

Variable voltage, fixed frequency e.g. 3–phase squirrel cage IM V = 460 V Rs= 0.25  Rr=0.2  Lr = Ls = 0.5/(2*pi*50) Lm=30/(2*pi*50) f = 50Hz p = 4 Lower speed  slip higher Low efficiency at low speed

Variable voltage, variable frequency Constant V/f operation At low slip

If Φag is constant  Te α slip frequency Variable voltage, variable frequency – Constant V/f If Φag is constant  Te α slip frequency (regardless of the synchronous frequency)

Approximates constant air-gap flux when Eag is large Variable voltage, variable frequency – Constant V/f How do we make constant ? Approximates constant air-gap flux when Eag is large Eag = k f ag = constant Speed is adjusted by varying f - maintaining V/f to approximate constant air-gap flux

Characteristic with constant Variable voltage, variable frequency – Constant V/f Characteristic with constant

Constant  constant V/f Variable voltage, variable frequency Constant  constant V/f Vs Vrated Constant slope frated f

Constant V/f – open-loop Variable voltage, variable frequency Constant V/f – open-loop Rectifier 3-phase supply VSI IM C f Ramp Pulse Width Modulator V s* + rate limiter is needed to ensure the slip change within allowable range (e.g. rated value)

Constant V/f – open-loop : Power converters Variable voltage, variable frequency Constant V/f – open-loop : Power converters + Vdc  IM 3-phase supply Uncontrolled Diode rectifier Crow bar circuit 3-phase VSI with PWM control Power flow only in one direction Transistor will be switched on when capacitor voltage increases beyond allowable during dynamic braking Power flow in both direction

Constant V/f – open-loop : Power converters Variable voltage, variable frequency Constant V/f – open-loop : Power converters + Vdc  IM Let’s assume we have 415 V (rms) Assuming output of rectifier is In actual the value is lower if LC filter is used to remove the AC components. Peak fundamental phase voltage for ma =1: Therefore RMS of fundamental line-line voltage is With overmodulation, this can be increased to (0.78)587 = 458V

Constant V/f – open-loop Variable voltage, variable frequency Constant V/f – open-loop Simulation example: 415V, 50Hz, 4 pole, Rs = 0.25, Rr = 0.2, Lr=Ls= 0.0971 H, Lm = 0.0955, J = 0.046 kgm2 , Load: k2

Constant V/f – open-loop Variable voltage, variable frequency Constant V/f – open-loop Simulation example: 415V, 50Hz, 4 pole, Rs = 0.25, Rr = 0.2, Lr=Ls= 0.0971 H, Lm = 0.0955, J = 0.046 kgm2 , Load: k2 constant_vhz_withoutBoost/Signal Builder : Group 1 Signal 1 50 40 30 20 10 0.5 1 1.5 2 2.5 3 3.5 Time (sec) 15

Constant V/f – open-loop Variable voltage, variable frequency Constant V/f – open-loop Simulation example: 415V, 50Hz, 4 pole, Rs = 0.25, Rr = 0.2, Lr=Ls= 0.0971 H, Lm = 0.0955, J = 0.046 kgm2 , Load: k2 16

Constant V/f – open-loop Variable voltage, variable frequency Constant V/f – open-loop Simulation example: 415V, 50Hz, 4 pole, Rs = 0.25, Rr = 0.2, Lr=Ls= 0.0971 H, Lm = 0.0955, J = 0.046 kgm2 , Load: k2 With almost no rate limiter 17

Constant V/f – open-loop Variable voltage, variable frequency Constant V/f – open-loop Simulation example: 415V, 50Hz, 4 pole, Rs = 0.25, Rr = 0.2, Lr=Ls= 0.0971 H, Lm = 0.0955, J = 0.046 kgm2 , Load: k2 With 628 rad/s2 18

Constant V/f – open-loop low speed problems Variable voltage, variable frequency Constant V/f – open-loop low speed problems Problems with open-loop constant V/f At low speed, voltage drop across stator impedance is significant compared to airgap voltage - poor torque capability at low speed Solution: (i) Voltage boost at low frequency (ii) Maintain Im constant  stator current control

Constant V/f – open-loop low speed problems (i) voltage boost Variable voltage, variable frequency Constant V/f – open-loop low speed problems (i) voltage boost Torque deteriorate at low frequency – hence compensation commonly performed at low frequency In order to truly compensate need to measure stator current – seldom performed

Constant V/f – open-loop low speed problems (i) voltage boost Variable voltage, variable frequency Constant V/f – open-loop low speed problems (i) voltage boost With voltage boost of Irated*Rs Torque deteriorate at low frequency – hence compensation commonly performed at low frequency In order to truly compensate need to measure stator current – seldom performed

Constant V/f – open-loop low speed problems (i) voltage boost Variable voltage, variable frequency Constant V/f – open-loop low speed problems (i) voltage boost Voltage boost at low frequency Vrated Linear offset Boost Non-linear offset – varies with Is frated

Constant V/f – open-loop low speed problems (i) voltage boost Variable voltage, variable frequency Constant V/f – open-loop low speed problems (i) voltage boost Idc Rectifier + Vdc - 3-phase supply VSI IM C f Ramp Pulse Width Modulator s* + V + Vboost

Constant V/f – open-loop low speed problems (ii) Constant Im Variable voltage, variable frequency Constant V/f – open-loop low speed problems (ii) Constant Im ag, constant → Eag/f , constant → Im, constant (rated) Controlled to maintain Im at rated Is Lls Llr’ Ir’ Rs + Vs – Lm + Eag – Rr’/s maintain at rated Im

Constant V/f – open-loop low speed problems (ii) Constant Im Variable voltage, variable frequency Constant V/f – open-loop low speed problems (ii) Constant Im From per-phase equivalent circuit, Current is controlled using current-controlled VSI The problem of stator impedance drop is solved Dependent on rotor parameters – sensitive to parameter variation

Current reference generator Variable voltage, variable frequency Constant V/f – open-loop low speed problems (ii) Constant Im 3-phase supply VSI Rectifier IM C Current controller Tacho slip * + |Is| PI - + s r Current reference generator +

Problems with open-loop constant V/f Variable voltage, variable frequency Constant V/f Problems with open-loop constant V/f Poor speed regulation Solution: (i) Slip compensation (ii) Closed-loop control

Constant V/f – poor speed regulation: (i) slip compensation Variable voltage, variable frequency Constant V/f – poor speed regulation: (i) slip compensation Motor characteristic AFTER slip compensation T ωr (rad/s) ωs1* Tload ωs2*=ωs1*+ωslip1 ωslip1 Motor characteristic BEFORE slip compensation ωr2≈ωs1* T2 ωr1 T1 ωslip1

Constant V/f – poor speed regulation: (i) slip compensation Variable voltage, variable frequency Constant V/f – poor speed regulation: (i) slip compensation Idc Rectifier + Vdc - 3-phase supply VSI IM C f Ramp Pulse Width Modulator s* + + V + + Vboost Slip speed calculator Vdc Idc

Constant V/f – poor speed regulation: (i) slip compensation Variable voltage, variable frequency Constant V/f – poor speed regulation: (i) slip compensation How is the slip frequency calculated ? + Vdc  Idc INV Pdc= VdcIdc Pmotor,in= Pdc – Pinv,losses Pmotor,in Stator Copper lossess Stator Core losses ROTOR STATOR Pair-gap 30

Constant V/f – poor speed regulation: (i) slip compensation Variable voltage, variable frequency Constant V/f – poor speed regulation: (i) slip compensation How is the slip frequency calculated ? Pair-gapc = Tesyn Te =  Pair-gap/syn For constant V/f control, 31

Constant V/f – poor speed regulation: (ii) closed-loop speed Variable voltage, variable frequency Constant V/f – poor speed regulation: (ii) closed-loop speed Require speed encoder Increase complexity 32

INDUCTION MOTOR DRIVES Speed control : Constant V/Hz Example A 4–pole, 3-phase, 415V, 50 Hz IM has a rated torque and speed of 20 Nm and 1450 rpm respectively. The motor is supplied by a 3-phase PWM-VSI using a constant V/Hz control method. It is used to drive a load with TL–ω characteristic given by TL = Kω2. The load torque demand is such that it equals the rated torque of the motor at the rated motor speed.   i) Find the constant K in the TL–ω characteristic of the load. What are the synchronous and motor speeds at point A and point B of the TL-ω characteristic (See figure)? If it is required that the starting torque equals 1.2 x rated torque of the motor, what should be the voltage and frequency applied at start–up? If Vdc of the inverter is 630V and the switching frequency is 5 kHz, what should be the values of the amplitude modulation index and frequency modulation index of the VSI for point B?

INDUCTION MOTOR DRIVES Speed control : Constant V/Hz Trated= 20 Nm Rated synchronous speed 1500 rpm = 157.1 rad/s Rated speed 1450 rpm = 151.8 rad/s Rated slip frequency= 5.3 rad/s

INDUCTION MOTOR DRIVES Speed control : Constant V/Hz ωsyn1 ωm1 Slip frequency, ωslip1= ?

INDUCTION MOTOR DRIVES Speed control : Constant V/Hz ωsyn2 ωm2 Slip frequency, ωslip2= ?

INDUCTION MOTOR DRIVES Speed control : Constant V/Hz 24Nm Slip frequency, ωstart= ?

At 43 Hz, the L-L rms voltage should be INDUCTION MOTOR DRIVES Speed control : Constant V/Hz (iv) At B, ωsyn = 135.475 rad/s  43 Hz Hence mf =5000/43 = 116 At 43 Hz, the L-L rms voltage should be Since we know that