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. 19 & 20
Mohd. Umar Rehman
umar.ee.amu@gmail.com
March 27, 2019
EEE-413N L-19 & 20 March 27, 2019 1 / 20

2. ωm-T Characteristics of a 1-Ph Full Converter fed
Separately Excited DC Motor
KVL equation for average values,
Va = RaIa +E
E = Va −RaIa = Kωm
ωm =
Va
K
−
IaRa
K
...(1)
ωm =
Va
K
−
Ra
K2
T ...(2)
ωm =
2Vm
Kπ
cosα −
Ra
K2
T ...(3)(For CCM)
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3. Contd...
For a constant α, the characteristics is a straight line with -ve slope same
as that with a normal DC supply.
As α changes, the characteristics shifts downward parallel to itself.
For low values of torque, the conduction becomes discontinuous.
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4. Characteristics
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5. Contd...
For DCM,
Va =
1
π
[Vm(cosα −cosβ)+E(π +α −β)]
But E = Kωm, hence
Va =
Vm
π
(cosα −cosβ)+
Kωm
π
(π +α −β) ...(4)
Using eq.(4) in eq. (2), we get
ωm =
Vm
Kπ
(cosα −cosβ)+
Kωm
Kπ
(π +α −β)−
Ra
K2
T
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6. Contd...
For DCM,
Va =
1
π
[Vm(cosα −cosβ)+E(π +α −β)]
But E = Kωm, hence
Va =
Vm
π
(cosα −cosβ)+
Kωm
π
(π +α −β) ...(4)
Using eq.(4) in eq. (2), we get
ωm =
Vm
Kπ
(cosα −cosβ)+
Kωm
Kπ
(π +α −β)−
Ra
K2
T
ωm
β −α
π
=
Vm
Kπ
(cosα −cosβ)−
Ra
K2
T
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7. Contd...
ωm = Vm
(cosα −cosβ)
K(β −α)
−
πRa
K2(β −α)
T ...(5)
Remarks:
Here, the slope depends upon α
When motor is loaded, va does not remain same, rather it depends upon
β
As T ↑, Ia ↑, β ↑, thus the slope of the ωm-T characteristics decreases
as the load is increased.
In CCM, drop in speed is only due to increase in IaRa drop. The average
voltage Va = 2Vm
π cosα is independent of the load
In DCM, the drop in speed is due to ↑ in IaRa drop as well as ↓ in average
voltage (Va ↓ with ↑ in β ). Thus speed regulation is poor in DCM.
EEE-413N L-19 & 20 March 27, 2019 7 / 20

8. No Load Speed
At no load, Ia = 0. For zero average current, ia will be zero at all the time
instants.
This is possible only if EMF at no-load is equal to the max. value of
armature voltage, i. e. E0 = va−max
No-load speed is given by:
ωm0 =
Vm
K , For α ≤ 90◦, E0 = Kωm0 = Vm
Vm
K sinα, For α > 90◦, E0 = Kωm0 = Vm sinα
Thus, the no-load speed decreases as α is increased beyond 90◦.
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9. No-load Speed...Contd
Remarks:
Supply is taken in such a manner so that the max. average voltage is
equal to the rated voltage i. e. Va(max) = 2Vm
π = Vr
Hence, no-load speed is given by ωm0 = 2Vm
Kπ
No-load speed with rectifier is ωm0 = Vm
K
It can be noted that the ratio of above two no-load speeds is π
2
EEE-413N L-19 & 20 March 27, 2019 9 / 20

10. Three-Phase Full Converter fed S. E. Motor Drive
For higher power rating and better performance, 3-Ph full wave rectifiers
(6 pulse converters) are used.
With this converter, ripples in the armature current are reduced and the
motor operates predominantly in the CCM.
Average armature voltage is given by:
Va =
3Vm
π
cosα
Vm: Peak value of line-to-line voltage
All the equations of 1-Ph rectifier fed drive for CCM are valid with the
above value of Va.
EEE-413N L-19 & 20 March 27, 2019 10 / 20

11. Waveforms

12. Home Assignment
Simulate the above three phase full converter fed S. E. motor drive using
MATLAB/Simulink or any other software of your choice and observe the
waveforms. Attach the printouts of model and waveforms in your assignment.
Firing Angles to be taken as follows:
R. No. α
1 - 17 30◦
18 - 34 60◦
35 - end 90◦
EEE-413N L-19 & 20 March 27, 2019 12 / 20

13. Effect of Current Ripples on Motor Performance
The armature current of a motor fed by a rectifier contains ripples and
varies between two extreme values ia−min & ia−max
The average value of armature current is Ia
The frequency of variation is twice the supply frequency for 1-φ rectifier.
The ripples in armature current have the following adverse effects:
1. The rms value Irms which is responsible for heat loss in armature (I2
rmsRa) is
more than the average value Ia. This results in increased Cu losses in the
armature. The motor has to be de-rated to limit the temperature rise within
safe value.
2. The peak value of ia is more than the average value resulting in
commutation failure and consequent brush sparking. Again the motor
needs to be de-rated to limit the peak value of the current to a safe limit.
EEE-413N L-19 & 20 March 27, 2019 13 / 20

14. Remedies
To reduce the armature current ripples, filter inductance may be added in the
armature circuit. This has two fold benefits: reduction in current ripples and
extension in range of CCM. However, additional inductance has the following
drawbacks:
(i) Armature time constant increases, resulting in sluggish transient response
(ii) Increased cost & weight of the drive system
EEE-413N L-19 & 20 March 27, 2019 14 / 20

15. Ex. 5.17
A 220 V, 1500 rpm, 50 A separately excited motor with armature resistance of
0.5 Ω, is fed from a 3-phase fully controlled rectifier. Available ac source has a
line voltage of 440 V, 50 Hz. A Y-∆ connected transformer is used to feed the
armature so that the motor terminal voltage equals rated voltage when
converter firing angle is zero.
(i) Calculate transformer turns ratio.
(ii) Determine the value of firing angle when: (a) motor is running at 1200 rpm
and rated torque; (b) when motor is running at − 800 rpm and twice the rated
torque.
Assume continuous conduction.
EEE-413N L-19 & 20 March 27, 2019 15 / 20

16. Solution
Armature voltage for a 3-φ fully-controlled rectifier is:
Va =
3Vm
π
cosα
⇒ Vm =
π
3
·
Va
cosα
For rated motor terminal voltage α = 0◦
Vm =
π
3
·
220
cos0◦
= 230.4 V
rms converter input line voltage (t/f secondary voltage)
= 230.4/
√
2 = 162.9 V
(i) For Y-∆ transformer connection, ratio of turns between phase windings of
primary and secondary
440/
√
3
162.9
= 1.559
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17. Solution...contd
(ii) (a) At 1500 rpm, E = 220−50×0.5 = 195 V
At 1200 rpm, E = 1200
1500 ×195 = 156 V
Va = E +IaRa = 156+50×0.5 = 181 V
Since
Va =
3
π
Vm cosα
cosα =
π
3
·
Va
Vm
=
π
3
×
181
230.4
= 0.8227
α = 34.65◦
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18. Solution...contd
(b) Try yourself, answer is
α = 104.2◦
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19. Unit-2 Ends!
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20. Home Assignment
Problems 5.51 to 5.54
Q. Describe Microprocessor controlled electric drives in detail.
EEE-413N L-19 & 20 March 27, 2019 20 / 20

21. Home Assignment
Problems 5.51 to 5.54
Q. Describe Microprocessor controlled electric drives in detail.
Submission Date: 24th
April 2019
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