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

2. Conventional electric drives (variable speed)
Bulky
Inefficient
inflexible

3. Modern electric drives (With power electronic converters)
Small
Efficient
Flexible

5. Electric Drives AC Drives: DC Drives: uses AC motors as prime movers
More difficult to control  need variable AC voltage (magnitude and frequency)
DC Drives:
uses DC motors as prime movers
Easy to control  need variable DC voltage

6. 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

7. Overview of AC and DC drives
Extracted from Boldea & Nasar

8. DC DRIVES DC Machine: Can be operated as motor
Can be operated as generator
Armature
terminals
Field
In basic construction, it consist of 4 terminals:
Armature terminals
Field terminals

9. DC Motors - 2 pole DC DRIVES Field flux Rotor
Field circuit produces field flux
Stator

10. DC Motors - 2 pole DC DRIVES
Current flowing in armature circuit interact with field winding to produce torque X

11. DC Motors - 2 pole Torque Torque DC DRIVES
Current flowing in armature circuit interact with field winding to produce torque
Torque produced is proportional to the armature current
Torque X
The torque will make the rotor to rotate (clockwise)
As the rotor rotates, voltage will be induced in the rotor – which is known as the back emf
Torque
In order to look on how the speed is control on DC motor, we need to model the DC motor with electric circuit

12. dt di L i R v = DC DRIVES Lf Rf if + ea _ La Ra ia Vt Vf
Electric torque
Armature back e.m.f.
The flux per pole is proportional to if

13. 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

14. DC DRIVES
Armature circuit:
In steady state current will not change with time (dia/dt =0),
Therefore steady state speed is given by,
For the case kt = ke,

15. 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

16. Controlling Vt and if using Controlled Rectifier
DC DRIVES
Controlling Vt and if using Controlled Rectifier
Controlled Rectifier
Controlled Rectifier if + Vt − + Vf −
3-phase
supply αt αf
Current if is changed by changing Vf

17. Controlling Vt using Controlled Rectifier
DC DRIVES
Controlling Vt using Controlled Rectifier
Controlling if using resistance
Controlled Rectifier
Un-controlled Rectifier if Rf + Vt − + Vf −
3-phase
supply αt
Current if is changed by changing Rf

18. 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?

19. Torque developed is Te= ktIa = 0.5 *15 = 0.5*15 = 7.5 Nm
DC DRIVES
Example:
Torque developed is Te= ktIa = 0.5 *15
= 0.5*15 = 7.5 Nm + Vt − αt
Controlled Rectifier
From DC machine equation,
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.
Using
αt can be calculated as:

20. 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,

21. Controlling Vt using a DC-DC converter
DC DRIVES
Controlling Vt using a DC-DC converter
Un-controlled Rectifier
DC-DC converter + Vt − +
VDC −
3-phase
supply 
Vt is change by changing the duty cyle 

22. Controlling Vt using a DC-DC converter
DC DRIVES
Controlling Vt using a DC-DC converter T1 D1 + Vt - ia Ra +
VDC −
3-phase
supply

23. Controlling Vt using a DC-DC converter
DC DRIVES
Controlling Vt using a DC-DC converter
When T1 ON, vt = VDC T1 D1 + Vt - ia Ra
When T1 OFF, D1 will ON and vt = 0
ton
VDC T If
= duty cycle, then,
 The voltage Vt canbe controlled by changing the duty cycle 

24. 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.
Calculate the duty cycle of the converter.
If the period of the waveform is 20 s, draw the output voltage waveform of the converter.

25. From the motor equation, we can calculate the terminal voltage
DC DRIVES
12Nm of torque is due to Te/kt = 12/0.5 = 24 A of armature current RPM is equivalent to: T1 D1 + Vt - ia Ra
Hence the back e.m.f.,
Ea = *0.5 = 62.83V
From the motor equation, we can calculate the terminal voltage
Since

26. DC DRIVES
9.2s
200V
20 s

27. AC DRIVES
In modern AC drive systems, the speed of the AC motor is controlled by controlling:
Magnitude of the applied voltage
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:
Six-step voltage source inverter
PWM voltage source inverter

28. AC DRIVES
(i) Six-step voltage source inverter (assuming available source is 3-phase supply)
Controlled Rectifier
3-phase VSI
Variable DC a +
Vdc _
3-phase
supply b
AC motor c
Amplitude control A
Frequency control f

29. AC DRIVES
(i) Six-step voltage source inverter (assuming available source is 3-phase supply)
With the correct switching signals to the VSI, the following voltages can be obtained:
line-line voltage
phase voltage

30. AC DRIVES
(i) Six-step voltage source inverter (assuming available source is 3-phase supply) A1
Example of amplitude control
A1 < A2 < A3 A2 A3

31. Example of frequency control
AC DRIVES
(i) Six-step voltage source inverter (assuming available source is 3-phase supply)
1/f1
1/f2
1/f3
Example of frequency control
f1> f2 > f3

32. Frequency and amplitude control
AC DRIVES
(ii) PWM voltage source inverter (assuming available source is 3-phase supply)
Controlled Rectifier
3-phase VSI
Fixed DC a +
Vdc _
3-phase
supply b
AC motor c
Frequency and amplitude control
A, f

33. AC DRIVES
(ii) PWM voltage source inverter (assuming available source is 3-phase supply)
With the correct switching signals to the VSI, the following voltages can be obtained:
line-line voltage
phase voltage

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

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

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

37. Stator – 3-phase winding Rotor – squirrel cage / wound
INDUCTION MOTOR DRIVES
120o a c’
Construction of induction machine b’ c b a’
Stator – 3-phase winding
Rotor – squirrel cage / wound

38. Single N turn coil carrying current i Spans 180o elec
INDUCTION MOTOR DRIVES
Single N turn coil carrying current i
Spans 180o elec
Permeability of iron >> o
→ all MMF drop appear in airgap a  a’  
/2
-/2 -
Ni / 2
-Ni / 2

39. INDUCTION MOTOR DRIVES
Distributed winding
– coils are distributed in several slots
Nc for each slot 
(3Nci)/2
(Nci)/2 -
-/2
/2  

40. Phase a – sinusoidal distributed winding
INDUCTION MOTOR DRIVES
Phase a – sinusoidal distributed winding 
Air–gap mmf
F()   2

41. Combination of 3 standing waves resulted in MMF wave rotating at:
INDUCTION MOTOR DRIVES
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:
p – number of poles
f – supply frequency

42. INDUCTION MOTOR DRIVES

43. Rotating flux induced:
INDUCTION MOTOR DRIVES
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

44. Stator voltage equation: Vs = Rs Is + j(2f)LlsIs + Eag
INDUCTION MOTOR DRIVES
Stator voltage equation:
Vs = Rs Is + j(2f)LlsIs + Eag
Eag – airgap voltage or back emf
Eag = k f ag
Rotor voltage equation:
Er = Rr Ir + js(2f)Llr
Er – induced emf in rotor circuit
Er /s = (Rr / s) Ir + j(2f)Llr

45. Per–phase equivalent circuit
INDUCTION MOTOR DRIVES
Per–phase equivalent circuit
Llr
Lls Ir Rs + Vs – +
Eag – +
Er/s – Is Lm
Rr/s Im
Rs – stator winding resistance
Rr – rotor winding resistance
Lls – stator leakage inductance
Llr – rotor leakage inductance
Lm – mutual inductance
s – slip

46. Per–phase equivalent circuit
INDUCTION MOTOR DRIVES
Per–phase equivalent circuit
We know Eg and Er related by
Where a is the winding turn ratio
The rotor parameters referred to stator are:
rotor voltage equation becomes
Eg = (Rr’ / s) Ir’ + j(2f)Llr’ Ir’

47. Per–phase equivalent circuit
INDUCTION MOTOR DRIVES
Per–phase equivalent circuit
Rr’/s + Vs – Rs
Lls
Llr’
Eag Is
Ir’ Im Lm
Rs – stator winding resistance
Rr’ – rotor winding resistance referred to stator
Lls – stator leakage inductance
Llr’ – rotor leakage inductance referred to stator
Lm – mutual inductance
Ir’ – rotor current referred to stator

48. Converted to mechanical power = (1–s)Pag
INDUCTION MOTOR DRIVES
Power and Torque
Power is transferred from stator to rotor via air–gap, known as airgap power
Lost in rotor winding
Converted to mechanical power = (1–s)Pag

49. Mechanical power, Pm = Tem r
INDUCTION MOTOR DRIVES
Power and Torque
Mechanical power, Pm = Tem r
But, ss = s - r  r = (1-s)s
 Pag = Tem s
Therefore torque is given by:

50. Power and Torque sm Tem Pull out Torque (Tmax) Trated r 0 rated s s
INDUCTION MOTOR DRIVES
Power and Torque
Tem
Pull out Torque
(Tmax)
Trated r
rated s sm s

51. Speed Control of IM Rr’/s Lls Llr’ Is Ir’
INDUCTION MOTOR DRIVES
Speed Control of IM
Control of induction machine based on steady-state model (per phase steady-state equivalent circuit) is known as scalar control
Rr’/s + Vs – Rs
Lls
Llr’
Eag Is
Ir’ Im Lm

52. Speed Control of IM Te sm Te Pull out Torque (Tmax) r rotor s s
INDUCTION MOTOR DRIVES
Speed Control of IM Te
Pull out Torque
(Tmax)
Intersection point (Te=TL) determines the steady –state speed Te TL r sm
rotor s s

53. INDUCTION MOTOR DRIVES
Speed Control of IM
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

54. Variable voltage, fixed frequency
INDUCTION MOTOR DRIVES
Speed Control of IM
Variable voltage, fixed frequency
Low speed  high slip
Therefore, low efficiency at low speed

55. Variable voltage, variable frequency
INDUCTION MOTOR DRIVES
Speed Control of IM
Variable voltage, variable frequency
T-ω characteristic of IM when air-gap is kept constant:

56. Variable voltage, variable frequency
INDUCTION MOTOR DRIVES
Speed Control of IM
Variable voltage, variable frequency
How do we keep air-gap flux constant?
Eag = k f ag
At high speed, Eag ≈ Vs
= 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

57. Variable voltage, variable frequency
INDUCTION MOTOR DRIVES
Speed Control of IM
Variable voltage, variable frequency
Constant V/Hz – open-loop Vs
Vrated
frated f

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

59. Variable voltage, variable frequency
INDUCTION MOTOR DRIVES
Speed Control of IM
Variable voltage, variable frequency
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 = K2. 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.
What are the synchronous and motor speed at a load torque of 15Nm?

60. Variable voltage, variable frequency
INDUCTION MOTOR DRIVES
Speed Control of IM
Variable voltage, variable frequency
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= ω2 , such that the motor is operated at its rated speed, when the torque is at its rated value.
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?
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 ?