Motor Drives Dr. Nik Rumzi Nik Idris Dept. of Energy Conversion, UTM 2013

Motor Drives

Dr. Nik Rumzi Nik IdrisDept. of Energy Conversion, UTM

2013

Conventional electric drives (variable speed)

• Bulky

• Inefficient

• inflexible

Modern electric drives (With power electronic converters)

• Small

• Efficient

• Flexible

DC Drives:• uses DC motors as prime movers• Easy to control need variable DC

voltage

AC Drives: • uses AC motors as prime movers• More difficult to control need variable

AC voltage (magnitude and frequency)

Electric Drives

DC motors and AC motors

DC motors: Regular maintenance, heavy, expensive, speed limit

Easy control, decouple control of torque and flux

AC motors: Less maintenance, light, less expensive, high speed

Coupling between torque and flux – variable spatial angle between rotor and stator flux

Overview of AC and DC drives

Extracted from Boldea & Nasar

Armatureterminals

Fieldterminals

DC Machine:• Can be operated as motor• Can be operated as generator

In basic construction, it consist of 4 terminals:• Armature terminals• Field terminals

DC DRIVES

DC Motors - 2 pole

Stator

Rotor

Field circuit produces field flux

Field flux

DC DRIVES

DC Motors - 2 pole

X

X

X

X

X

• Current flowing in armature circuit interact with field winding to produce torque

DC DRIVES

DC Motors - 2 pole

X

X

X

X

X

Torque

Torque

• Torque produced is proportional to the armature current

• The torque will make the rotor to rotate (clockwise)

• Current flowing in armature circuit interact with field winding to produce torque

• As the rotor rotates, voltage will be induced in the rotor – which is known as the back emf

In order to look on how the speed is control on DC motor, we need to model the DC motor with electric circuit

DC DRIVES

Electric torque

Armature back e.m.f.

Lf Rf

if+

ea

_

LaRa

ia+

Vt

_

+

Vf

_

dtdi

LiRv ffff +=

The flux per pole is proportional to if

DC DRIVES

• If the field flux comes from permanent magnet, the motor is called the permanent magnet DC motor

• For permanent magnet DC motor, the field current CANNOT be controlled, therefore the torque and back e.m.f. can be written as

kt is the torque constant ke is the back e.m.f. constant

• Most of the time, kt = ke

DC DRIVES

aaat EIRV • In steady state current will not change with time (dia/dt =0),

2

et

ea

e

t

kk

TR

k

V

• Therefore steady state speed is given by,

• Armature circuit:

2

t

ea

t

t

k

TR

k

V

• For the case kt = ke,

DC DRIVES

• Three possible methods of speed control:

• Armature voltage Vt

• Field flux• Armature resistance Ra

• Control using Ra is seldom used because it is inefficient due to the losses in the external armature resistance

2

t

ea

t

t

k

TR

k

V

DC DRIVES

Controlling Vt and if using Controlled Rectifier

3-phasesupply

+

Vt

−

+Vf

−

αt αf

if

Current if is changed by changing Vf

Controlled Rectifier Controlled Rectifier

DC DRIVES

Controlling Vt using Controlled RectifierControlling if using resistance

3-phasesupply

+

Vt

−

if

+Vf

−

αt

Current if is changed by changing Rf

Rf

Controlled Rectifier Un-controlled Rectifier

DC DRIVES

Example:

A wound stator DC motor which is fed by a 3-phase controlled rectifier has the following parameters:

Ra = 1.2 Ω, ktΦ = 0.5 Nm/amp, keΦ = 0.5 V/rad/s

The amplitude of the line-line input voltage of the rectifier is 200V. If the motor runs at 1500 RPM, and the armature current is 15 A, calculate the delay angle of the rectifier.

What is the torque developed?

If the field is reversed and what would happen to the motor? If the current is to be maintained at 15A, what should be the delay angle of the rectifier?

DC DRIVES

Example:

+

Vt

−

αt

Controlled Rectifier

aaat EIRV

Using

where Ra = 1.2 Ω, Ia = 15 A and Ea = keω .

The given speed is in rotation per minute (RPM) and has to be converted to rad/s.

αt can be calculated as:

Torque developed is Te= ktIa = 0.5 *15 = 0.5*15 = 7.5 Nm

From DC machine equation,

DC DRIVES

+

Vt

−

αt

Controlled Rectifier

+

-78.5V

−

Ra

At the instant the field is reversed, the speed will remain at 1500rpm. The back emf be -78.5V. The DC machine will be operated as a generator. Hence to maintain the current at 15A,

DC DRIVES

3-phasesupply

+

Vt

−

ll

DCV

V 3

Vt is change by changing the duty cyle

DC-DC converterUn-controlled Rectifier

Controlling Vt using a DC-DC converter

+

VDC

−

DC DRIVES


3-phasesupply

+

VDC

−

ll

DCV

V 3

T1

D1

+

Vt

-

iaRa

DC DRIVES


When T1 ON, vt = VDC

When T1 OFF, D1 will ON and vt = 0

0

VDC

T1

D1

+

Vt

-

iaRa

T

ton

T

tonIf = duty cycle, then, DCt VV

The voltage Vt canbe controlled by changing the duty cycle

DC DRIVES

Example:

A permanent magnet DC motor which is fed by a DC-DC converter (BUCK converter) has the following parameters:

Ra = 1.2 Ω, kt = 0.5 Nm/amp, ke = 0.5 V/rad/s

When the motor runs at a speed of 1200 RPM, the torque developed by the motor is 12Nm. The DC input voltage of the converter is 200V.

(i) Calculate the duty cycle of the converter.

(ii) If the period of the waveform is 20 s, draw the output voltage waveform of the converter.

DC DRIVES

T1

D1

+

Vt

-

iaRa

VEIRV aaat 639183622421 ..).(

12Nm of torque is due to Te/kt = 12/0.5 = 24 A of armature current. 1200 RPM is equivalent to:

sradN

/.)()(

6712560

21200

60

2

Hence the back e.m.f.,

Ea = 125.67 *0.5 = 62.83V

From the motor equation, we can calculate the terminal voltage

Since DCt VV 460200

6391.

.

DC

t

V

V

DC DRIVES

0

200V

20 s

9.2s

stT

ton

on 291020460460 6 .)(..

DC DRIVES

AC DRIVES

• In modern AC drive systems, the speed of the AC motor is controlled by controlling:

(i) Magnitude of the applied voltage

(ii) Frequency of the applied voltage

• Magnitude and frequency of the AC motor can be controlled by using power electronic converters

• There are two most widely used inverter control technique:

(i) Six-step voltage source inverter

(ii) PWM voltage source inverter

AC DRIVES

(i) Six-step voltage source inverter (assuming available source is 3-phase supply)

3-phasesupply

3-phase VSIControlled Rectifier

Variable DC

AC motor

Frequency control

f

Amplitude control

A

a

b

c

+

Vdc

_


With the correct switching signals to the VSI, the following voltages can be obtained:

line-line voltage

phase voltage

AC DRIVES


Example of amplitude control

A1 < A2 < A3

A1

A2

A3

AC DRIVES


Example of frequency control

f1> f2 > f3

1/f1 1/f2 1/f3

AC DRIVES

3-phasesupply

3-phase VSIControlled Rectifier

Fixed DC

AC motor

Frequency and

amplitude control

A, f

a

b

c

+

Vdc

_

(ii) PWM voltage source inverter (assuming available source is 3-phase supply)

AC DRIVES

With the correct switching signals to the VSI, the following voltages can be obtained:

line-line voltage

phase voltage

(ii) PWM voltage source inverter (assuming available source is 3-phase supply)

AC DRIVES

(ii) PWM voltage source inverter (assuming available source is 3-phase supply) Variable amplitude

AC DRIVES

(ii) PWM voltage source inverter (assuming available source is 3-phase supply) Variable frequency

AC DRIVES

(ii) PWM voltage source inverter (assuming available source is 3-phase supply) Variable frequency

AC DRIVES

Construction of induction machine

a

b

b’

c’

c

a’

Stator – 3-phase windingRotor – squirrel cage / wound

120o120o

120o

INDUCTION MOTOR DRIVES

a

a’

Single N turn coil carrying current iSpans 180o elec

Permeability of iron >> o

→ all MMF drop appear in airgap

/2-/2-

Ni / 2

-Ni / 2


Distributed winding – coils are distributed in several slots

Nc for each slot

/2-/2-

(3Nci)/2

(Nci)/2


Phase a – sinusoidal distributed winding

Air–gap mmf

F()

2


• Sinusoidal winding for each phase produces space sinusoidal MMF and flux

• Sinusoidal current excitation (with frequency s) in a phase produces space sinusoidal standing wave MMF

• Combination of 3 standing waves resulted in MMF wave rotating at:

f2p2

s

p – number of polesf – supply frequency



• Rotating flux induced:emf in stator winding (known as back emf)Emf in rotor winding Rotor flux rotating at synchronous frequency

• Rotor current interact with flux producing torque

• Rotor ALWAYS rotate at frequency less than synchronous, i.e. at slip speed (or slip frequency):

sl = s – r

• Ratio between slip speed and synchronous speed known as slip

s

rss


Stator voltage equation:

Vs = Rs Is + j(2f)LlsIs + Eag

Eag – airgap voltage or back emf

Eag = k f ag

Rotor voltage equation: Er = Rr Ir + js(2f)Llr

Er – induced emf in rotor circuit

Er /s = (Rr / s) Ir + j(2f)Llr


Per–phase equivalent circuit

Rr/s

+

Vs

–

RsLls

Llr

+

Eag

–

Is

Ir

Im

Lm

Rs – stator winding resistanceRr – rotor winding resistanceLls – stator leakage inductanceLlr – rotor leakage inductanceLm – mutual inductances – slip

+

Er/s

–


We know Eg and Er related by

\ rotor voltage equation becomes

Eg = (Rr’ / s) Ir’ + j(2f)Llr’ Ir’

as

EE

g

r Where a is the winding turn ratio

The rotor parameters referred to stator are:

2lr

lr2r

rrr a'L

L,a

'RR,)'I(aI

INDUCTION MOTOR DRIVES Per–phase equivalent circuit

Rr’/s+

Vs

–

RsLls Llr’

+

Eag

–

Is Ir’

Im

Lm

Rs – stator winding resistanceRr’ – rotor winding resistance referred to statorLls – stator leakage inductanceLlr’ – rotor leakage inductance referred to statorLm – mutual inductanceIr’ – rotor current referred to stator

INDUCTION MOTOR DRIVES Per–phase equivalent circuit

Power and Torque

Power is transferred from stator to rotor via air–gap, known as airgap power

s1s

'RI3'RI3

s'R

I3P r2'rr

2'r

r2'rag

Lost in rotor winding

Converted to mechanical power = (1–s)Pag


Power and Torque

Mechanical power, Pm = Tem r

But, ss = s - r r = (1-s)s

Pag = Tem s

s

r2'r

s

agem s

'RI3PT

Therefore torque is given by:

2lrls

2r

s

2s

s

rem

'XXs

'RR

Vs

'R3T


Power and Torque

1 0

rs

Trated

Pull out Torque(Tmax)

Tem

0 rated s

2lrls2

s

rm

XXR

Rs

2lrls

2ss

2s

smax

XXRR

Vs3

T

sm


Control of induction machine based on steady-state model (per phase steady-state equivalent circuit) is known as scalar control

Rr’/s+

Vs

–

RsLls Llr’

+

Eag

–

Is Ir’

Im

Lm

Speed Control of IMINDUCTION MOTOR DRIVES

rs

Pull out Torque(Tmax)

Te

ssmrotor

TL

Te

Intersection point (Te=TL) determines the steady –state speed


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 polesDiscrete step change in speed

Variable voltage (amplitude), frequency fixedE.g. using transformer or triacSlip becomes high as voltage reduced – low efficiency

Variable voltage (amplitude), variable frequencyUsing power electronics converter Operated at low slip frequency


0 20 40 60 80 100 120 140 1600

100

200

300

400

500

600

Tor

que

w (rad/s)

Low speed high slip

Therefore, low efficiency at low speed

Speed Control of IMVariable voltage, fixed frequency


T-ω characteristic of IM when air-gap is kept constant:

0 20 40 60 80 100 120 140 1600

100

200

300

400

500

600

700

800

900

Tor

que

50Hz

30Hz

10Hz

Speed Control of IMVariable voltage, variable frequency


How do we keep air-gap flux constant?

Eag = k f ag

f

V

f

Eag = constant

• Speed is adjusted by varying f - maintaining V/f constant to avoid flux saturation

• This method is known as Constant V/f (or V/Hz) method


At high speed, Eag ≈ Vs


Vrated

frated

Vs

f


Constant V/Hz – open-loop


VSIRectifier

3-phase supply IM

Pulse Width

Modulators* +

Rampf

C

V

rate limiter is needed to ensure the slip change within allowable range (e.g. rated value)

Constant V/Hz – open-loop





A 4–pole, 3-phase, 50 Hz IM has a rated torque and speed of 20 Nm and 1450 rpm respectively. The motor is supplied by a 3-phase inverter using a constant V/f control method. It is used to drive a load with TL– characteristic given by TL = K2. The load torque demand is such that it equals the rated torque of the motor at the rated motor speed and frequency. i) Find the constant K in the TL– characteristic of the load. ii) What are the synchronous and motor speed at a load torque of

15Nm?



A 4-pole 3-phase induction motor has the following ratings: 330V, 50Hz, 1450 rpm

The motor is fed by a 3-phase VSI with constant V/f control strategy. The input 3-phase voltage to the VSI is 415V. The load which is coupled to the induction motor has a T- ω characteristic given by TL= 0.0015ω2 , such that the motor is operated at its rated speed, when the torque is at its rated value. i) If the motor is operated at 1000 rpm, what should be the

applied phase voltage (fundamental amplitude and frequency) fed to the IM? If the speed to be increased to 1600 rpm, what should be the amplitude of the fundamental phase voltage?

ii) If the starting torque of 60 Nm is required, what should be the amplitude and frequency of the fundamental component initially applied to the induction motor during start-up ?

Documents

Motor Drives Dr. Nik Rumzi Nik Idris Dept. of Energy Conversion, UTM 2013