Part of Lecture series on EEE-413, Electrical Drives (DC Drives) delivered by me to students of VIII Semester B.E. (Electrical), Session 2018-19. Z. H. College of Engg. & Technology, Aligarh Muslim University, Aligarh. Missing materials will be uploaded shortly. Please comment and feel free to ask anything related. Thanks!

1. EEE-413N
Electric Drives
Lec. 13 & 14
Mohd. Umar Rehman
umar.ee.amu@gmail.com
February 27, 2019
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2. Task
1 Why a DC motor drive is essentially a 4 quadrant drive?
2 Name the power electronic converters that are capable of single
quadrant, two quadrant and four quadrant operation.
?
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3. Ex. 5.3
A 220 V, 200 A, 800 RPM separately excited DC motor has an armature
resistance of 0.06 Ω. The motor armature is fed from a variable voltage
source with an internal resistance of 0.04 Ω. Calculate internal voltage of
the variable source when the motor is operating in regenerative braking at
80% of the rated motor torque and 600 RPM.
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4. Solution:
Since torque is proportional to the armature current, during regeneration
the armature current will be
Ia2 = 0.8 × 200 =
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5. Solution:
Since torque is proportional to the armature current, during regeneration
the armature current will be
Ia2 = 0.8 × 200 = 160 A
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6. Solution:
Since torque is proportional to the armature current, during regeneration
the armature current will be
Ia2 = 0.8 × 200 = 160 A
Back EMF during rated operation,
E1 = V − Ia1Ra =
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7. Solution:
Since torque is proportional to the armature current, during regeneration
the armature current will be
Ia2 = 0.8 × 200 = 160 A
Back EMF during rated operation,
E1 = V − Ia1Ra = 220 − 200 × 0.06 = 208 V
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8. Solution:
Since torque is proportional to the armature current, during regeneration
the armature current will be
Ia2 = 0.8 × 200 = 160 A
Back EMF during rated operation,
E1 = V − Ia1Ra = 220 − 200 × 0.06 = 208 V
Back EMF during regeneration,
E2 =
N2
N1
E1 =
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9. Solution:
Since torque is proportional to the armature current, during regeneration
the armature current will be
Ia2 = 0.8 × 200 = 160 A
Back EMF during rated operation,
E1 = V − Ia1Ra = 220 − 200 × 0.06 = 208 V
Back EMF during regeneration,
E2 =
N2
N1
E1 =
600
800
× 208 = 156 V
Internal voltage of the variable voltage source =
E2 − Ia2 (Ra + Rint)
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10. Solution:
Since torque is proportional to the armature current, during regeneration
the armature current will be
Ia2 = 0.8 × 200 = 160 A
Back EMF during rated operation,
E1 = V − Ia1Ra = 220 − 200 × 0.06 = 208 V
Back EMF during regeneration,
E2 =
N2
N1
E1 =
600
800
× 208 = 156 V
Internal voltage of the variable voltage source =
E2 − Ia2 (Ra + Rint) = 156 − 160(0.06 + 0.04) = 140 V
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11. Ex. 5.8
A 220 V, 970 RPM, 100 A separately excited DC motor has an armature
resistance of 0.05 Ω. It is braked by plugging from an initial speed of 1000
RPM. Calculate:
(a) resistance to be placed in the armature circuit to limit braking current
to twice the full load value
(b) braking torque, and
(c) torque when the speed has fallen to zero.
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12. Solution:
At rated operation (970 RPM),
E1 = 220 − 0.05 × 100 =
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13. Solution:
At rated operation (970 RPM),
E1 = 220 − 0.05 × 100 = 215 V
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14. Solution:
At rated operation (970 RPM),
E1 = 220 − 0.05 × 100 = 215 V
At 1000 RPM, E2 = 1000
970 × 215
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15. Solution:
At rated operation (970 RPM),
E1 = 220 − 0.05 × 100 = 215 V
At 1000 RPM, E2 = 1000
970 × 215 = 221.65 V
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16. Solution:
At rated operation (970 RPM),
E1 = 220 − 0.05 × 100 = 215 V
At 1000 RPM, E2 = 1000
970 × 215 = 221.65 V
(a) For plugging operation,
RB + Ra =
E + V
Ia
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17. Solution:
At rated operation (970 RPM),
E1 = 220 − 0.05 × 100 = 215 V
At 1000 RPM, E2 = 1000
970 × 215 = 221.65 V
(a) For plugging operation,
RB + Ra =
E + V
Ia
=
221.65 + 220
200
= 2.21 Ω
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18. Solution:
At rated operation (970 RPM),
E1 = 220 − 0.05 × 100 = 215 V
At 1000 RPM, E2 = 1000
970 × 215 = 221.65 V
(a) For plugging operation,
RB + Ra =
E + V
Ia
=
221.65 + 220
200
= 2.21 Ω
RB = 2.21 − 0.05 = 1.16 Ω
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19. Solution...Contd
(b) In steady state, Pm = Pe
T =
E × Ia
ωm
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20. Solution...Contd
(b) In steady state, Pm = Pe
T =
E × Ia
ωm
=
221.65 × 200
1000 × 2π/60
= 423.3 N-m
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21. Solution...Contd
(b) In steady state, Pm = Pe
T =
E × Ia
ωm
=
221.65 × 200
1000 × 2π/60
= 423.3 N-m
(c) At zero speed, E = 0
Ia =
V
RB + Ra
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22. Solution...Contd
(b) In steady state, Pm = Pe
T =
E × Ia
ωm
=
221.65 × 200
1000 × 2π/60
= 423.3 N-m
(c) At zero speed, E = 0
Ia =
V
RB + Ra
=
220
2.21
= 99.55 A
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23. Solution...Contd
(b) In steady state, Pm = Pe
T =
E × Ia
ωm
=
221.65 × 200
1000 × 2π/60
= 423.3 N-m
(c) At zero speed, E = 0
Ia =
V
RB + Ra
=
220
2.21
= 99.55 A
As T ∝ Ia,
T = 423.3 ×
99.55
200
= 210.7 N-m
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24. Try Yourself
Ex. 5.7
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25. Speed Control of DC Motors
Torque-Speed Equation of a DC Motor
ωm =
V
Keφ
−
Ra
(Keφ)2
T
Speed of the motor can be controlled by following methods:
1 Armature Voltage Control
2 Field Flux Control
3 Armature Resistance Control
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26. 1. Armature Voltage Control
If the voltage applied to the armature circuit is changed, the
speed-torque characteristics of a separately excited DC
motor(constant flux) and series motor change as shown
Thus, speed of the motor decreases both for separate and series
motor as the voltage applied to the armature circuit is reduced.
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27. Armature Voltage Control...Contd
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28. Armature Voltage Control...Contd
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29. Armature Voltage Control...Contd
Consider a separately excited DC motor supplying a constant torque
load.
If the armature voltage is reduced by a small amount say V1, the
operating point shifts to A, and Ia & T decreases
As T becomes less than TL, the motor decelerates and T & Ia
increase till the motor again produces a torque equal to TL
If the armature voltage is reduced by a large amount, to V2, the back
EMF becomes more than V2 resulting in reversal of Ia & T, and the
machine now starts working as a generator.
The torque is opposite to the motor speed and the braking torque
decelerates the motor.
After the operating point reaches D, the operation again changes to
motoring.
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30. Armature Voltage Control...Contd
The motor speed continues to fall till motor torque becomes equal to
TL
If the voltage is again increased to rated voltage, the motor draws a
very large current that may damage the motor.
Therefore, the voltage should be changed gradually to limit the
armature current to a safe value.
Note: As the flux is kept constant and motor may be loaded to its
rated current, the torque capability of the motor is not affected.
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31. Armature Voltage Control...Contd
Advantages:
1 The nature of speed-torque characteristics is unaffected with the
change in voltage. Thus, speed regulation remains same at all the
voltages.
2 Torque capability of the motor is not affected.
Therefore, it acts as a constant torque drive (Trated = KφIrated)
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32. Armature Voltage Control...Contd
Limitations:
1 The armature voltage cannot be increased above the rated voltage.
Therefore, this method is used to get motor operation below its
natural speed-torque characteristics.
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33. 2. Field Control
Torque-Speed Equation of a DC Motor (revisited)
ωm =
V
Keφ
−
Ra
(Keφ)2
T
=
V
Keφ
−
Ra
Keφ
Ia
For a separately excited DC motor, if the field flux is decreased
keeping the armature voltage constant, the speed-torque
characteristics of the motor change as shown in the fig. below
Both no-load speed ωm0 and the slope of the ωm-T curve increase as
given by above equation.
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34. 2. Field Control...Contd
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35. 2. Field Control...Contd
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36. 2. Field Control...Contd
As the field is weakened, to supply the same torque, the armature
current increases (T = KφIa)
Thus the load torque must be reduced to limit the armature current
to its rated value.
The power rating of this motor remains nearly constant.
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37. Precaution in Field Control
As the field is weakened, the back EMF drops initially (E = Kφωm)
as the motor speed does not change instantly. This results in sharp
increase in current Ia = V −E
Ra
leading to increased torque
Due to this heavy torque, the motor accelerates quickly and the back
EMF rises
In steady state, T = TL, and new value of armature current can be
found as follows:
Iaφ = Iaφ ⇒ Ia = Ia
φ
φ
Thus, the field should be weakened slowly and gradually to limit the
inrush of current to a safe value and to avoid sparking at the brushes.
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38. Drawbacks of Field Control
1 The field cannot be increased to any substantial amount, limited by
the the heating of field coils and saturation in the machine. Thus,
speed control cannot be achieved above the rated speed.
2 At weak fields, the armature reaction becomes more effective causing
deterioration of the machine performance. The load has to be
decreased to an extent so as to reduce the armature current much
below the rated current and to avoid sparking at the brushes.
Note: Highest speed of the motor is limited by sparking at the brushes
and the mechanical strength of the motor. With normal design, highest
speed is up to 1.5-2 times the rated speed. With special design, the speed
may be increased to up to 6 times the rated speed.
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39. Implementation of Field Control
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40. DC Motor Capability Curves
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41. DC Motor Capability Curves...Contd
Write about this in your own words
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42. 3. Rheostatic Control
By inserting an external variable resistance in the armature circuit,
the resistance Ra can be varied, thus varying the speed of the motor.
For a constant torque load, the armature current Ia remains constant,
hence the power input to the motor is not changed.
However, mechanical power developed by the motor Pm = EIa
decreases while the armature power loss I2
aRa increases as Ra is
increased.
Hence, efficiency of the motor becomes poor, and this method is
quite inefficient.
It is used only in those applications, where low speed operation is
required for short duration like traction loads.
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43. 3. Rheostatic Control...Contd
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44. Ex. 5.11
A 220 V, 500 A, 600 RPM separately excited DC motor has an armature
and field resistance of 0.02 Ω & 10 Ω, respectively. The load torque is
given by the expression TL = 2000 − 2N, N-m, where N is the speed in
RPM. Speed below the rated are obtained by armature voltage control and
speeds above the rated are obtained by field control.
(i) Calculate motor terminal voltage and armature current when the
speed is 450 RPM.
(ii) Calculate field winding voltage and armature current when the speed
is 750 RPM.
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45. Solution:
(i) At 450 RPM, TL = 2000 − 2 × 450 = 1100 N-m
At rated operation, E1 = 220 − 500 × 0.02 = 210 V
Rated torque
=
E1Ia1
ωm1
=
210 × 500
600 × 2π/60
= 1671 N-m
For a torque of 1100 N-m,
Ia2 =
1100
1671
× 500 = 329 A
At 450 RPM,
E2 =
450
600
× 210 = 157.5 V
V = E2 + Ia2Ra = 157.5 + 329 × 0.02 = 164 V
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46. Solution...contd
(ii) At 750 RPM, TL = 2000 − 2 × 750 = 500 N-m
At this operating point, let the flux and armature current be φ & Ia
respectively. Then
Keφ Ia = 500 ...(1)
From rated operation,
K1φ =
210
600 × 2π/60
= 3.342
Further at 750 RPM, ωm = 750 × 2π
60 = 78.54 rad/s &
V = Keφ ωm + IaRa
or
220 = 78.54Keφ + 0.02Ia
Substituting from (1)
220 = 78.54 ×
500
Ia
+ 0.02Ia
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47. Solution...contd
or
0.02Ia
2
− 220Ia + 39270 = 0
Solving this quadratic equation and discarding the large value (unfeasible),
we get Ia = 181.5 A
From (1),
Keφ =
500
181.5
= 2.755
Field voltage
= 220 ×
Keφ
Keφ1
= 220 ×
2.755
3.342
= 181.3V
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48. Try Yourself
Ex. 5.12
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49. Announcement
Mid-Semester Examination Syllabus:
Till Today’s Lecture
Best of Luck!
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