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. 17 & 18
Mohd. Umar Rehman
umar.ee.amu@gmail.com
March 15, 2019
EEE-413N L-17 & 18 March 15, 2019 1 / 26

2. Speed-Torque Relations
Applying KVL,
Va = Raia +La
dia
dt
+E
Average value of inductor current is zero in steady state, hence,
Va = Raia +E ...(2)
We know that, E = K1ωm & T = K1ia, hence equation (2) can be written as:
Va = Ra
T
K1
+K1ωm ...(3)
ωm =
Va
K1
−
Ra
K2
1
T =
2Vm
K1π
cosα −
Ra
K2
1
T ...(4)
EEE-413N L-17 & 18 March 15, 2019 2 / 26

3. Ex. 5.13
A 200 V, 875 RPM, 150 A separately excited DC motor has an armature
resistance of 0.06 Ω. It is fed from a single phase fully controlled rectifier with
an AC source voltage of 220 V, 50 Hz. Assuming continuous conduction,
calculate:
(a) firing angle for rated motor torque and 750 RPM.
(b) firing angle for rated motor torque and −500 RPM.
(c) motor speed for α = 160◦ and rated torque.
EEE-413N L-17 & 18 March 15, 2019 3 / 26

4. Ex. 5.13
A 200 V, 875 RPM, 150 A separately excited DC motor has an armature
resistance of 0.06 Ω. It is fed from a single phase fully controlled rectifier with
an AC source voltage of 220 V, 50 Hz. Assuming continuous conduction,
calculate:
(a) firing angle for rated motor torque and 750 RPM.
(b) firing angle for rated motor torque and −500 RPM.
(c) motor speed for α = 160◦ and rated torque.
Solution on Board
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5. Home Assignment
Problems 5.36 to 5.39
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6. 2. Discontinuous Conduction Mode
When thyristors T1 & T3 are fired in +ve half cycle then current starts
decreasing after ωt = π −γ and continues to decay after ωt = π
During this interval, the energy is released by the inductance La.
If all the energy stored in the La is released before firing of the the
thyristors T2 & T4, the current falls to zero and armature terminal voltage
becomes equal to E.
This is called discontinuous conduction. DCM occurs due to:
(i) low value of La.
(ii) small value of average armature current Ia i.e. due to low value of load
torque.
EEE-413N L-17 & 18 March 15, 2019 5 / 26

7. Waveforms for DCM

8. Contd...
One cycle of the output voltage may be divided into two intervals:
(i) Duty interval (α < ωt < β) va = vs = Vm sin(ωt)
(ii) Zero current interval (β < ωt < π +α) ia = 0,va = E
EEE-413N L-17 & 18 March 15, 2019 7 / 26

9. Analysis during duty interval
Applying KVL, we get
Vm sin(ωt) = Raia +La
dia
dt
+E
or −E +Vm sin(ωt) = Raia +La
dia
dt
...(1)
The above equation can be solved by applying principle of superposition.
EEE-413N L-17 & 18 March 15, 2019 8 / 26

10. Analysis...contd
The steady state current due to ac voltage source alone is given by:
ia(ac)(ωt) =
Vm
Z
sin(ωt −φ) ...(2)
where, Z = R2
a +(ωLa)2
, φ = tan−1 ωLa
Ra
The steady state current due to dc source alone is given by:
ia(dc) = −
E
Ra
...(3)
EEE-413N L-17 & 18 March 15, 2019 9 / 26

11. Analysis...contd
Both ac and dc sources cause flow of a transient current which may be
combined together as:
ia(tr) = K1e− t
τa
ia(tr) = K1e−ω cotφ
...(4)
EEE-413N L-17 & 18 March 15, 2019 10 / 26

12. Analysis...contd
Both ac and dc sources cause flow of a transient current which may be
combined together as:
ia(tr) = K1e− t
τa
ia(tr) = K1e−ω cotφ
...(4)
τa =
La
Ra
⇒
1
τa
=
Ra
La
EEE-413N L-17 & 18 March 15, 2019 10 / 26

13. Analysis...contd
Both ac and dc sources cause flow of a transient current which may be
combined together as:
ia(tr) = K1e− t
τa
ia(tr) = K1e−ω cotφ
...(4)
τa =
La
Ra
⇒
1
τa
=
Ra
La
×
ω
ω
= ω cotφ
EEE-413N L-17 & 18 March 15, 2019 10 / 26

14. Analysis...contd
Combining equations (2), (3) & (4), we get
ia(ωt) =
Vm
Z
sin(ωt −φ)−
E
Ra
+K1e−ωt cotφ
...(5)
To determine K1, apply the boundary condition ia = 0 at ωt = α
0 =
Vm
Z
sin(α −φ)−
E
Ra
+K1e−α cotφ
⇒ K1 =
E
Ra
−
Vm
Z
sin(α −φ) eα cotφ
...(6)
EEE-413N L-17 & 18 March 15, 2019 11 / 26

15. Analysis...contd
Substitute the value of K1 thus obtained in equation (5)
ia(ωt) =
Vm
Z
sin(ωt −φ)−
E
Ra
+
E
Ra
−
Vm
Z
sin(α −φ) e(α−ωt)cotφ
=
Vm
Z
sin(ωt −φ)−sin(α −φ)e−(ωt−α)cotφ
− E
Ra
1−e−(ωt−α)cotφ ...(7)
EEE-413N L-17 & 18 March 15, 2019 12 / 26

16. Analysis...contd
To find the value of β, substitute ia(β) = 0 in equation (7)
ia(β) =
Vm
Z
sin(β −φ)−sin(α −φ)e−(β−α)cotφ
− E
Ra
1−e−(β−α)cotφ = 0 ...(8)
The above equation can be solved numerically to find β. Once β is known,
average voltage can be found as:
Va =
1
π
β
α
Vm sin(ωt) d(ωt)+
π+α
β
E d(ωt)
Va =
1
π
[Vm(cosα −cosβ)+E(π +α −β)] ...(9)
EEE-413N L-17 & 18 March 15, 2019 13 / 26

17. Critical Speed
If torque is high, conduction is continuous. For a given α if torque
decreases, the average value of armature current decreases and the
speed of the motor increases.
If the torque is decreased below a certain value, called critical torque, the
conduction changes from continuous to discontinuous.
The speed at the boundary of continuous & discontinuous conduction is
called critical speed ωmc.
At speeds above ωmc, conduction is discontinuous.
EEE-413N L-17 & 18 March 15, 2019 14 / 26

18. Expression for ωmc
At the boundary of continuous & discontinuous conduction,
β = π +α, ia(β) = 0
Substitute β = π +α in equation (8), then we get,
ia(β) =
Vm
Z
sin(π +α −φ)−sin(α −φ)e−(π+α−α)cotφ
− E
Ra
1−e−(π+α−α)cotφ = 0
EEE-413N L-17 & 18 March 15, 2019 15 / 26

19. Expression for ωmc
At the boundary of continuous & discontinuous conduction,
β = π +α, ia(β) = 0
Substitute β = π +α in equation (8), then we get,
ia(β) =
Vm
Z
sin(π +α −φ)−sin(α −φ)e−(π+α−α)cotφ
− E
Ra
1−e−(π+α−α)cotφ = 0
But E = Kωmc
EEE-413N L-17 18 March 15, 2019 16 / 26

20. Expression for ωmc...contd
Hence,
Kωmc
Ra
1−e−π cotφ
= −
Vm
Z
sin(α −φ) 1+e−π cotφ
⇒ ωmc =
VmRa
KZ
sin(α −φ)
e−π cotφ +1
e−π cotφ −1
...(10)
Everything in the above equation is known, hence ωmc can be found.
EEE-413N L-17 18 March 15, 2019 17 / 26

21. Types of Problems
1. To find the torque for a given α ωm
Step 1: Find ωmc using eq. (10)
If ωm ωmc → Discontinuous conduction
If ωm ωmc → Continuous conduction
Step 2: For discontinuous conduction, find β using eq. (8) and find Va using
eq. (9)
For continuous conduction, Va = 2Vm
π cosα
Step 3: Find armature current Ia = Va−E
Ra
, and torque T = KIa
EEE-413N L-17 18 March 15, 2019 18 / 26

22. Types of Problems...contd
2. To find ωm for a given α T
Step 1: Find ωmc using eq. (10), EMF at critical speed, Ec = Kωmc
Step 2: Torque at critical speed Tc = KIac, where Iac = Va−Ec
Ra
, Va = 2Vm
π cosα
Step 3: If T Tc → Continuous conduction
If T Tc → Discontinuous conduction
EEE-413N L-17 18 March 15, 2019 19 / 26

23. Ex. 5.14
If the armature circuit inductance of motor of the drive of Ex. 5.13 be 0.85 mH,
calculate the motor torque for
(i) α = 60◦ and speed = 400 rpm.
Now external inductance of 2 mH is added to the armature circuit to reduce the
region of discontinuous conduction. Calculate the torque for for
(ii) α = 120◦ and speed = −400 rpm
(iii) α = 120◦ and speed = −600 rpm
EEE-413N L-17 18 March 15, 2019 20 / 26

24. Solution
(i) First Calculate critical speed
ωmc =
VmRa
KZ
sin(α −φ)
e−π cotφ +1
e−π cotφ −1
φ = tan−1 ωLa
Ra
= tan−1 100π ×0.85×10−3
0.06
= 77.34◦
cotφ = 0.2247
Z = R2
a +(ωLa)2
= 0.062 +(100π ×0.85×10−3)2
= 0.2737 Ω
K =
E
ωm
=
191
875×2π/60
= 2.084
ωmc =
220
√
2×0.06
2.084×0.2737
sin(60◦
−77.34◦
)
e−0.2247π +1
e−0.2247π −1
= +28.8 rad/s
= +275 rpm
EEE-413N L-17 18 March 15, 2019 21 / 26

25. Solution...Contd
Since ωm ωmc, drive is operating under discontinuous conduction.
At 400 rpm E = 400
875 ×191 = 87.3 V
From Eq. (8),
Vm
Z sin(β −φ)−sin(α −φ)e−(β−α)cotφ − E
Ra
1−e−(β−α)cotφ = 0
Substituting the known values, we get a nonlinear transcedental equation in β
sin(β −77.34◦
)+2.09e−0.2247β
= 0
Trial solution of this equation gives β = 230◦
EEE-413N L-17 18 March 15, 2019 22 / 26

26. Solution...Contd
Find armature voltage using Eq. (9),
Va =
1
π
[Vm(cosα −cosβ)+E(π +α −β)]
=
1
π
220
√
2(cos60◦
−cos230◦
)+87.3 π +(60◦
−230◦
)
π
180◦
= 118 V
Ia =
Va −E
Ra
=
118−87.3
0.06
= 512 A
T = KIa = 2.084×512 = 1067 N-m
EEE-413N L-17 18 March 15, 2019 23 / 26

27. Solution...contd
(ii) See yourself in the book.
(iii) Critical Speed
ωmc =
220
√
2
2.084×0.8974
sin(120◦
−86.17◦
)
e−0.067π +1
e−0.067π −1
= −52.94 rad/s = −505.5 rpm
Since. the motor speed (−600 rpm) is less than critical speed, hence the drive
operates in continuous conduction.
Va = 2Vm
π cosα = 220
√
2
π cos120◦ = −99 V E = −600
875 ×191 = −131 V
Ia = −99−(−131)
0.06 = 533.3 A T = KIa = 2.084×533.3 = 1111.5 N-m
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28. Try Yourself
Ex. 5.15
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29. Home Assignment Submission Date
24th April, 2019
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