The document discusses various armature voltage control methods for electric drives, focusing on the Ward-Leonard method and rectifier control techniques. It highlights the advantages and drawbacks of each method, including regenerative braking capabilities, efficiency concerns, and system size. Additionally, the document details operational principles and configurations involving transformers and rectifiers for different power ratings.

1. EEE-413N
Electric Drives
Lec. 15 & 16
Mohd. Umar Rehman
umar.ee.amu@gmail.com
March 13, 2019
EEE-413N L-15 & 16 March 13, 2019 1 / 30

2. Armature Voltage Control Methods
1 Ward-Leonard Method
2 Rectifier control
3 Chopper control
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3. 1. Ward Leonard Method
−
+
Variable
DCSupply
−
V = E−IaRa
+
IF
3 ϕ IM
3 ϕ
Supply Load
Constant
DC Supply
Pin
G
w
Pout
M

4. 1. Ward Leonard Method
In this method, the motor to be controlled is fed from a separately
excited DC generator.
The DC generator is run by a prime-mover which is usually an
induction motor, and the field winding of the generator is fed by a
variable DC power supply.
As the power required by the field winding is < 1% of the machine
power rating, a low power rating variable power supply is required.
For low power machines, a potential divider may also be used. For
high power rating, a controlled rectifier may be used.
By controlling the field of the generator, the armature voltage is
controlled to control the speed of the DC motor.
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5. Ward Leonard...contd
For regenerative braking, the field of the DC generator is reduced
to get V < E causing reversal of current and torque.
The machine M now works as a generator and G works as a
motor.
The AC motor works as a generator and feeds the electrical power
back to the supply.
This is II quadrant operation. For III & IV quadrant operation, field
winding current is reversed.
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6. Advantages of WL Scheme
1 Regenerative braking can be employed.
2 Four quadrant operation is possible.
3 Short duration power supply disturbances do not affect the
operation of the motor due to inertia.
4 Enough time is available for the standby power supply to take the
load.
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7. Drawbacks
1 High initial cost.
2 Poor efficiency due to the use of two additional machines of the
same rating.
3 Large size system.
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8. 2. Rectifier Control
(a) Transformer & uncontrolled rectifier
AC
Supply
Tap Changing t/f or
Auto-t/f
Filter
Uncontrolled
Rectifier or Diode
Bridge
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9. 2. Rectifier Control...contd
(a) Transformer & uncontrolled rectifier
Filter is used to reduce current ripples.
The scheme is employed in 25 kV, 1-phase AC traction.
Advantages
1 Rectifier output voltage waveform is not changed as the voltage is
reduced leading to a good PF at low voltages(speeds).
2 Source current distortion is less at low voltages.
Drawbacks
Regeneration is not possible.
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10. Rectifier Control...contd
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11. Rectifier Control...contd
(b) Fully Controlled Rectifier
For low and medium power applications (up to 10 kW), single
phase controlled rectifier is used.
For higher power rating, a three phase rectifier may be used.
In fully controlled rectifier, the output voltage may be negative
(for α > 90◦), while the current can only be positive.
Therefore, rectifier can work in two quadrants of the V-I plane and
regeneration is possible.
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12. Rectifier Control...contd
(c) Half Controlled Rectifier
If regeneration is not required, half controlled rectifiers are used.
Here, both V & I are positive, hence only single quadrant
operation is possible.
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13. Single Phase Fully Controlled Rectifier fed S. E.
Motor
T1 T2
T4 T3
AC
vs = Vm sinwt
Ra
La
E
ia
va
EEE-413N L-15 & 16 March 13, 2019 13 / 30

14. Contd...
In +ve half cycle, T1 & T3 are given firing pulses from ωt = α to π
In −ve half cycle, T2 & T4 are given firing pulses from
ωt = π +α to 2π
When T1 & T3 conduct,va = vS, vT2 = vT4 = −vS
When T2 & T4 conduct,va = −vS, vT1 = vT3 = vS
When none of the thyristors conduct, va = E, ia = 0
Motor speed will vary as the instantaneous armature voltage
varies, but the ripple will be negligible due to large mechanical
torque.
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15. Contd...
Figure: Circuit Diagram during +ve half cycle
ACvs
R
L
E
ia
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16. Contd...
Figure: Circuit Diagram during -ve half cycle
ACvs
R
L
E
ia
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17. Waveforms for CCM

18. Waveforms for DCM

19. Waveforms at the boundary of CCM & DCM

20. Contd...
In steady state,
ω = constant ⇒ E = constant = V −IaRa
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21. Contd...
In steady state,
ω = constant ⇒ E = constant = V −IaRa
(Almost ripple free)
KVL equation
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22. Contd...
In steady state,
ω = constant ⇒ E = constant = V −IaRa
(Almost ripple free)
KVL equation
va = Raia +La
dia
dt
+E
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23. Contd...
In steady state,
ω = constant ⇒ E = constant = V −IaRa
(Almost ripple free)
KVL equation
va = Raia +La
dia
dt
+E
If iaRa drop is neglected, then
va = La
dia
dt
+E
⇒
dia
dt
=
va −E
La

24. Contd...
In steady state,
ω = constant ⇒ E = constant = V −IaRa
(Almost ripple free)
KVL equation
va = Raia +La
dia
dt
+E
If iaRa drop is neglected, then
va = La
dia
dt
+E
⇒
dia
dt
=
va −E
La
EEE-413N L-15 & 16 March 13, 2019 20 / 30

25. Contd...
Case (i)
va > E ⇒
dia
dt
> 0 ⇒ ia increases
Energy supplied by the source is:
(i) stored in armature inductance La
(ii) converted into mechanical energy
(iii) dissipated in the armature resistance Ra
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26. Contd...
Case (ii)
0 < va < E ⇒
dia
dt
< 0 ⇒ ia decreases
Energy from the source as well as the energy released by the
inductance is:
(i) converted into mechanical energy Eia
(ii) dissipated in the armature resistance i2
aRa
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27. Contd...
Case (iii)
va < 0 ⇒
dia
dt
< 0 ⇒ ia decreases
Energy released by the inductance La is:
(i) absorbed by the source vaia
(ii) converted into mechanical energy Eia
(iii) dissipated in the armature resistance i2
aRa
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28. Contd...
Note:
1 In steady state, the energy stored in La becomes constant.
In other words, storage of energy in one part of the cycle is equal
to the release of energy in the other part of the cycle. Thus,
current at the beginning of the cycle is equal to that at the end of
the cycle.
2 ia is minimum at the instant when a rising va becomes equal to E
(ωt = γ). ia is maximum when a decreasing va = E (ωt = π −γ)
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29. Waveforms to depict γ
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30. Contd...
Flow of Energy:
Period Source Sink
α < ωt < γ vs, La E, Ra
γ < ωt < π −γ vs E, Ra, La
π −γ < ωt < π vs, La E, Ra
π < ωt < π +α La E, Ra, vs
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31. Steady State Analysis
1. Continuous Conduction Mode (CCM)
Assumptions:
1 Power semiconductor switches are ideal.
2 Motor speed is constant.
Average value of armature voltage:
Va =
1
π
π+α
α
sin(ωt) d(ωt) =
2Vm
π
cosα ...(1)
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32. Steady State Analysis
1. Continuous Conduction Mode (CCM)
Assumptions:
1 Power semiconductor switches are ideal.
2 Motor speed is constant.
Average value of armature voltage:
Va =
1
π
π+α
α
sin(ωt) d(ωt) =
2Vm
π
cosα ...(1)
Va =



2Vm
π (Max), at α = 0
0, at α = π
2
−2Vm
π (Min), at α = π
Vm is the rated machine voltage.
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33. Steady State Analysis...contd
The motor may be operated in I or IV quadrant
The motor cannot be operated in II & III quadrant, as Ia cannot be
reversed.
For I quadrant operation (Forward Motoring), α ≤ 90◦. The
polarities of Va & E are shown below in the figure.
The average power supplied by the rectifier is consumed by the
motor, and it runs in the forward direction.
For IV quadrant operation (Reverse generating or braking
mode), the polarity of E should reverse. This is necessary for
regeneration, as Ia cannot reverse.
Reversal of speed may take place if active load is connected
which drives the motor in reverse direction.
The polarity of Va should also be reversed which is done by
making α > 90◦ such that |Va| < |E|
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34. Contd...
The polarities of Va & E are shown below in the figure.
The machine works as a generator and supplies the generated
power to the rectifier.
Rectifier now works as an inverter and feeds back energy to the
source.
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35. Contd...
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