The document covers chopper-controlled DC drives, focusing on bidirectional control for motoring and regenerative braking operations. It details the operational principles, equations for analyzing armature current during different modes, and the advantages of using choppers over rectifiers for motor control. Additionally, it includes examples for calculating duty ratios and discusses control strategies for electric drives.

1. EEE-413N
Electric Drives
Lec. 21 & 22
Mohd. Umar Rehman
umar.ee.amu@gmail.com
April 3, 2019
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2. Unit-3
DC Drives-II
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3. Chopper Controlled DC Drives
F
a
Fig. Class-A/Step-down/Buck Chopper
S: Self-commutated power electronic switch
DF: freewheeling diode

4. Chopper Modes
a a
F
Mode 1: S is ON, Mode 2: S is OFF
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5. Chopper Waveforms
a

6. Motoring Control
Motoring Control is achieved using a step-down chopper as shown in the
above fig.
The fully controlled switch S (MOSFET/BJT/IGBT) is operated at high
frequencies (~kHZ)
If T is the time period of the chopper & TON is the ON period of the
switch, then duty ratio is defined as: δ = TON/T
When S is closed, armature current ia increases and when S is open, ia
decreases. In steady state, ia fluctuates between two extremes I1 & I2
Average output voltage is given by: V0 = δVs
Motor attains a constant speed ωm with back EMF E = Kωm
Speed-torque equation:
ωm =
δV
K
−
R
K2
T
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7. Analysis during Duty interval (0 < t < TON)
Apply KVL for circuit of mode 1,
Vs = Ria +L
dia
dt
+E ...(1)
Solution of above equation can be written as:
ia =
Vs −E
R
+Ke−t/τ
...(2)
At t = 0, ia = I1 ⇒ K = I1 −
Vs −E
R
...(3)
Thus, From (2) & (3)
ia = I1e−t/τ
+
Vs −E
R
1−e−t/τ
...(4)
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8. Analysis during off-duty interval (TON < t < T)
At the end of duty interval,
t = δT, ia = I2
To make the analysis simple redefine time origin at: t = T −TON
Hence, KVL eq. (1) can be written as:
Ria +L
dia
dt
+E = 0
which gives the solution of the form same as eq. (4)
ia = I2e−t/τ
−
E
R
1−e−t/τ
...(5)
The max. current I2 can be found from the following equation:
I2 = I1e−δT/τ
−
Vs −E
R
1−e−δT/τ
...(6)
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9. Analysis...contd
Similarly, I1 can be determined using the following equation:
I1 = I2e−(1−δ)T/τ
−
E
R
1−e−(1−δ)T/τ
...(7)
Combined solution of equations (4) to (7) gives explicit solutions for I1
& I2 as:
I1 =
Vs
R
eδT/τ −1
eT/τ −1
−
E
R
...(8)
I2 =
Vs
R
1−e−δT/τ
1−e−T/τ
−
E
R
...(9)
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10. Current Ripple 1
The peak-to-peak ripple current is
∆I = I2 −I1 =
Vs
R
1−e−δT/τ +e−T/τ −e−(1−δ)T/τ
1−e−t/τ
Condition for maximum ripple is:
d∆I
dδ
= 0 ⇒ δ = 0.5
Thus, max. ripple (p-p) is given by:
∆Imax =
Vs
R
tanh
T
4τ
=
Vs
R
tanh
R
4fL
Vs
4fL
(when 4fL R)
Max. ripple decreases as f ↑ and L ↑
1
Refer for further details: Power Electronics by S. K. Mandal, McGraw Hill.
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11. Regenerative Braking with Chopper

12. Regenerative Braking...contd
1. Energy Storage Interval (0 < t < TON)
Transistor switch is ON during the interval.
Diode D is reverse biased.
The energy released by the motor is partly absorbed by the inductor,
partly dissipated as heat in the resistor and switch.
The current increases from ia1 to ia2, and va = 0
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13. Regenerative Braking...contd
2. Energy Transfer Interval (TON < t < T)
Transistor switch is opened at t = TON
Armature current starts decreasing and flows through diode D and source
V, and reduces from ia2 to ia1. Armature voltage va = V
The polarity of the inductor is reversed, and D becomes forward biased.
Average armature voltage is:
Va =
1
T
T
TON
V dt = V
T −TON
T
= δV
Hence, here δ = T−TON
T = TOFF
T
Average armature current, Ia = E−δV
R
δ is varied so that Ia is close to the max. permissible current for high
braking torque.
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14. Speed-Torque Characteristics

15. Chopper for Motoring & Braking

16. Chopper for Motoring & Braking...Contd
The previous two circuits can be combined to give a two quadrant
chopper which can provide motoring and braking operations in the
forward direction.
For forward motoring operation, Tr1 is operated with desired duty ratio.
During ON period, current flows into the motor through Tr1.
During OFF period, current freewheels through D1.
For braking operation, Tr2 is operated. During ON period current flows
through Tr2 and energy is stored in La
During OFF period, energy flows to the source through D2.
For servo drives, where fast transition between motoring and braking is
required, both transistor switches are controlled simultaneously.
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17. Dynamic Braking Using Chopper
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18. Dynamic Braking...contd
The armature circuit is connected with a shunt braking resistor through
the transistor switch.
The transistor switch is operated at high frequency with duty ratio δ = TON
T
When switch is ON, armature is shorted and the resistance between
armature terminals is zero
When switch is OFF, armature terminal resistance is equal to RB
Thus, effectively the resistance can be expressed as:
R = RB
T −TON
T
= RB(1−δ)
Above equation shows that the effective value of braking resistor can be
changed steplessly from 0 to RB as δ is controlled from 0 to 1.
R =
RB , δ = 0
0 , δ = 1
As the speed decreases, δ can be increased steplessly to brake the
motor at a constant maximum torque.
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19. Ex. 5.19
A 230 V, 960 rpm and 200 A separately excited dc motor has an armature
resistance of 0.02 Ω. The motor is fed from a chopper which provides both
motoring and braking operations. The source has a voltage of 230 V.
Assuming continuous conduction.
(i) Calculate duty ratio of chopper for motoring operation at rated torque and
350 rpm.
(ii) Calculate duty ratio of chopper for braking operation at rated torque and
350 rpm.
(iii) If maximum duty ratio of chopper is limited to 0.95 and maximum
permissible motor current is twice the rated, calculate the maximum
possible motor speed obtainable without field weakening and power fed to
the source.
(iv) If motor field is also controlled in (iii), calculate field current as a fraction of
its rated value for a speed of 1200 rpm.
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20. Solution
At rated operation, E =
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21. Solution
At rated operation, E = 230−200×0.02 = 226 V
(i) E at 350 rpm
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22. Solution
At rated operation, E = 230−200×0.02 = 226 V
(i) E at 350 rpm = 350
960 ×226 = 82.4 V
Motor terminal voltage Va = E +IaRa
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23. Solution
At rated operation, E = 230−200×0.02 = 226 V
(i) E at 350 rpm = 350
960 ×226 = 82.4 V
Motor terminal voltage Va = E +IaRa = 86.4 V
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24. Solution
At rated operation, E = 230−200×0.02 = 226 V
(i) E at 350 rpm = 350
960 ×226 = 82.4 V
Motor terminal voltage Va = E +IaRa = 86.4 V
Duty ratio
δ =
86.4
230
= 0.376
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25. Solution
At rated operation, E = 230−200×0.02 = 226 V
(i) E at 350 rpm = 350
960 ×226 = 82.4 V
Motor terminal voltage Va = E +IaRa = 86.4 V
Duty ratio
δ =
86.4
230
= 0.376
(ii)
Va = E −IaRa =
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26. Solution
At rated operation, E = 230−200×0.02 = 226 V
(i) E at 350 rpm = 350
960 ×226 = 82.4 V
Motor terminal voltage Va = E +IaRa = 86.4 V
Duty ratio
δ =
86.4
230
= 0.376
(ii)
Va = E −IaRa = 78.4 V
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27. Solution
At rated operation, E = 230−200×0.02 = 226 V
(i) E at 350 rpm = 350
960 ×226 = 82.4 V
Motor terminal voltage Va = E +IaRa = 86.4 V
Duty ratio
δ =
86.4
230
= 0.376
(ii)
Va = E −IaRa = 78.4 V
Duty ratio
δ =
78.4
230
= 0.34
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28. Solution...Contd
(iii) Maximum available armature voltage
Va = 0.95×230 = 218.5 V
Back EMF E = Va +2IaRa =
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29. Solution...Contd
(iii) Maximum available armature voltage
Va = 0.95×230 = 218.5 V
Back EMF E = Va +2IaRa = 218.5+200×2×0.02 = 226.5 V
Maximum permissible motor speed = 226.5
226 ×960 =
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30. Solution...Contd
(iii) Maximum available armature voltage
Va = 0.95×230 = 218.5 V
Back EMF E = Va +2IaRa = 218.5+200×2×0.02 = 226.5 V
Maximum permissible motor speed = 226.5
226 ×960 = 962 rpm
Assuming lossless chopper, power fed into the source
VaIa = 218.5×400 =
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31. Solution...Contd
(iii) Maximum available armature voltage
Va = 0.95×230 = 218.5 V
Back EMF E = Va +2IaRa = 218.5+200×2×0.02 = 226.5 V
Maximum permissible motor speed = 226.5
226 ×960 = 962 rpm
Assuming lossless chopper, power fed into the source
VaIa = 218.5×400 = 87.4 kW
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32. Solution...Contd
(iii) Maximum available armature voltage
Va = 0.95×230 = 218.5 V
Back EMF E = Va +2IaRa = 218.5+200×2×0.02 = 226.5 V
Maximum permissible motor speed = 226.5
226 ×960 = 962 rpm
Assuming lossless chopper, power fed into the source
VaIa = 218.5×400 = 87.4 kW
(iv) As in (iii) E = 226.5 V for which at rated field current speed = 960 rpm.
Assuming linear magnetic circuit, E will be inversely proportional to field
current. Field current as a ratio of its rated value = 960/1200 = 0.8
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33. See Yourself
Ex. 5.20
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34. Home Assignment
Problems 5.51 to 5.54
Q. Describe Microprocessor controlled electric drives in detail.
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35. 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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36. Choice between Rectifier and Chopper Control
If supply is ac, rectifier is preferred as it requires single stage conversion.
However it suffers from the following drawbacks:
(i) Current drawn by the motor contains considerable ripples which increases
heating and causes de-rating of the motor.
(ii) At low voltages (higher α), the power factor becomes poor.
(iii) Operation is discontinuous at low torque values, resulting in poor speed
regulation.
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37. Choice...contd
If supply is dc, chopper is the obvious choice. It has the following advantages:
(i) Chopper operates at high frequencies, therefore armature current ripples
are less which reduces heating and de-rating of the motor.
(ii) Usually, the operation is continuous resulting in better speed regulation.
Because of these advantages, sometimes the available ac supply is converted
into dc using a diode bridge and then a chopper is used to vary the voltage.
These schemes offer an added advantage that the input side power factor
remains good even at low speed (low voltages).
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38. Control of Electrical Drives
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39. Control of Electrical Drives
1. Open loop control
2. Closed loop (feedback) control
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40. Open Loop Control of a DC drive
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