Drives lec 21_22_Chopper Controlled DC Drives

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. 21 & 22
Mohd. Umar Rehman
[email protected]
April 3, 2019
EEE-413N L-21 & 22 April 3, 2019 1 / 27

2. Unit-3
DC Drives-II
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3. Chopper Controlled DC DrivesF
a
Fig. Class-A/Step-down/Buck Chopper
S: Self-commutated power electronic switch
DF: freewheeling diode

4. Chopper Modesa a
F
Mode 1: S is ON, Mode 2: S is OFF
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5. Chopper Waveformsa

6. Motoring Control
Motoring Control is achieved using a step-down chopper as shown in the
above g.
The fully controlled switch S (MOSFET/BJT/IGBT) is operated at high
frequencies (~kHZ)
IfTis the time period of the chopper &TONis the ON period of the
switch, then duty ratio is dened as:d=TON=T
When S is closed, armature currentiaincreases and when S is open,ia
decreases. In steady state,iauctuates between two extremesI1&I2
Average output voltage is given by:V0=dVs
Motor attains a constant speedwmwith back EMFE=Kwm
Speed-torque equation:
wm=
dV
K

R
K
2
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=
VsE
R
+Ke
t=t
:::(2)
Att=0;ia=I1)K=I1
VsE
R
:::(3)
Thus, From (2) & (3)
ia=I1e
t=t
+
VsE
R

1e
t=t

:::(4)
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8. Analysis during off-duty interval(TON<t<T)
At the end of duty interval,
t=dT;ia=I2
To make the analysis simple redene time origin at:t
0
=TTON
Hence, KVL eq. (1) can be written as:
Ria+L
dia
dt
0
+E=0
which gives the solution of the form same as eq. (4)
ia=I2e
t=t

E
R

1e
t=t

:::(5)
The max. currentI2can be found from the following equation:
I2=I1e
dT=t

VsE
R

1e
dT=t

:::(6)
EEE-413N L-21 & 22 April 3, 2019 8 / 27

9. Analysis...contd
Similarly,I1can be determined using the following equation:
I1=I2e
(1d)T=t

E
R

1e
(1d)T=t

:::(7)
Combined solution of equations (4) to (7) gives explicit solutions forI1
&I2as:
I1=
Vs
R
"
e
dT=t
1
e
T=t
1
#

E
R
:::(8)
I2=
Vs
R
"
1e
dT=t
1e
T=t
#

E
R
:::(9)
EEE-413N L-21 & 22 April 3, 2019 9 / 27

10. Current Ripple
1
The peak-to-peak ripple current is
DI=I2I1=
Vs
R
1e
dT=t
+e
T=t
e
(1d)T=t
1e
t=t
Condition for maximum ripple is:
dDI
dd
=0)d=0:5
Thus, max. ripple (p-p) is given by:
DImax=
Vs
R
tanh
T
4t
=
Vs
R
tanh
R
4fL
'
Vs
4fL
(when4fLR)
Max. ripple decreases asf"andL"
1
Refer for further details:Power Electronicsby S. K. Mandal, McGraw Hill.EEE-413N L-21 & 22 April 3, 2019 10 / 27

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 fromia1toia2, andva=0
EEE-413N L-21 & 22 April 3, 2019 12 / 27

13. Regenerative Braking...contd
2. Energy Transfer Interval(TON<t<T)
Transistor switch is opened att=TON
Armature current starts decreasing and ows through diode D and source
V, and reduces fromia2toia1. Armature voltageva=V
The polarity of the inductor is reversed, and D becomes forward biased.
Average armature voltage is:
Va=
1
T
Z
T
TON
V dt=V
TTON
T
=dV
Hence, hered=
TTON
T
=
TOFF
T
Average armature current,Ia=
EdV
R
dis varied so thatIais close to the max. permissible current for high
braking torque.
EEE-413N L-21 & 22 April 3, 2019 13 / 27

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,Tr1is operated with desired duty ratio.
During ON period, current ows into the motor throughTr1.
During OFF period, current freewheels throughD1.
For braking operation,Tr2is operated. During ON period current ows
throughTr2and energy is stored inLa
During OFF period, energy ows to the source throughD2.
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 ratiod=
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 toRB
Thus, effectively the resistance can be expressed as:
R=RB
TTON
T
=RB(1d)
Above equation shows that the effective value of braking resistor can be
changed steplessly from 0 toRBasdis controlled from 0 to 1.
R=
(
RB;d=0
0;d=1
As the speed decreases,dcan be increased steplessly to brake the
motor at a constant maximum torque.
EEE-413N L-21 & 22 April 3, 2019 18 / 27

19. Ex. 5.19
A 230 V, 960 rpm and 200 A separately excited dc motor has an armature
resistance of 0.02W. 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 eld weakening and power fed to
the source.
(iv)If motor eld is also controlled in (iii), calculate eld current as a fraction of
its rated value for a speed of 1200 rpm.
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20. Solution
At rated operation, E=
2302000:02=226 V
(i) Eat 350 rpm
=
350
960
226=82:4 V
Motor terminal voltageVa=E+IaRa
=86:4 V
Duty ratio
d=
86:4
230
=0:376
(ii)
Va=EIaRa=
78:4 VDuty ratio
d=
78:4
230
=0:34
EEE-413N L-21 & 22 April 3, 2019 20 / 27

21. Solution
At rated operation, E=
2302000:02=226 V
(i) Eat 350 rpm
=
350
960
226=82:4 V
Motor terminal voltageVa=E+IaRa
=86:4 V
Duty ratio
d=
86:4
230
=0:376
(ii)
Va=EIaRa=
78:4 VDuty ratio
d=
78:4
230
=0:34
EEE-413N L-21 & 22 April 3, 2019 20 / 27

22. Solution
At rated operation, E=
2302000:02=226 V
(i) Eat 350 rpm
=
350
960
226=82:4 V
Motor terminal voltageVa=E+IaRa
=86:4 V
Duty ratio
d=
86:4
230
=0:376
(ii)
Va=EIaRa=
78:4 VDuty ratio
d=
78:4
230
=0:34
EEE-413N L-21 & 22 April 3, 2019 20 / 27

23. Solution
At rated operation, E=
2302000:02=226 V
(i) Eat 350 rpm
=
350
960
226=82:4 V
Motor terminal voltageVa=E+IaRa
=86:4 V
Duty ratio
d=
86:4
230
=0:376
(ii)
Va=EIaRa=
78:4 VDuty ratio
d=
78:4
230
=0:34
EEE-413N L-21 & 22 April 3, 2019 20 / 27

24. Solution
At rated operation, E=
2302000:02=226 V
(i) Eat 350 rpm
=
350
960
226=82:4 V
Motor terminal voltageVa=E+IaRa
=86:4 V
Duty ratio
d=
86:4
230
=0:376
(ii)
Va=EIaRa=
78:4 VDuty ratio
d=
78:4
230
=0:34
EEE-413N L-21 & 22 April 3, 2019 20 / 27

25. Solution
At rated operation, E=
2302000:02=226 V
(i) Eat 350 rpm
=
350
960
226=82:4 V
Motor terminal voltageVa=E+IaRa
=86:4 V
Duty ratio
d=
86:4
230
=0:376
(ii)
Va=EIaRa=
78:4 VDuty ratio
d=
78:4
230
=0:34
EEE-413N L-21 & 22 April 3, 2019 20 / 27

26. Solution
At rated operation, E=
2302000:02=226 V
(i) Eat 350 rpm
=
350
960
226=82:4 V
Motor terminal voltageVa=E+IaRa
=86:4 V
Duty ratio
d=
86:4
230
=0:376
(ii)
Va=EIaRa=
78:4 VDuty ratio
d=
78:4
230
=0:34
EEE-413N L-21 & 22 April 3, 2019 20 / 27

27. Solution
At rated operation, E=
2302000:02=226 V
(i) Eat 350 rpm
=
350
960
226=82:4 V
Motor terminal voltageVa=E+IaRa
=86:4 V
Duty ratio
d=
86:4
230
=0:376
(ii)
Va=EIaRa=
78:4 VDuty ratio
d=
78:4
230
=0:34
EEE-413N L-21 & 22 April 3, 2019 20 / 27

28. Solution...Contd
(iii) Maximum available armature voltage
Va=0:95230=218:5 V
Back EMF E=Va+2IaRa=
218:5+20020:02=226:5 V
Maximum permissible motor speed=
226:5
226
960=
962 rpm
Assuming lossless chopper, power fed into the source
VaIa=218:5400=
87:4 kW(iv) As in (iii)E=226:5 Vfor which at rated eld current speed=960 rpm.
Assuming linear magnetic circuit,Ewill be inversely proportional to eld
current. Field current as a ratio of its rated value=960=1200=0:8
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29. Solution...Contd
(iii) Maximum available armature voltage
Va=0:95230=218:5 V
Back EMF E=Va+2IaRa=
218:5+20020:02=226:5 V
Maximum permissible motor speed=
226:5
226
960=
962 rpm
Assuming lossless chopper, power fed into the source
VaIa=218:5400=
87:4 kW(iv) As in (iii)E=226:5 Vfor which at rated eld current speed=960 rpm.
Assuming linear magnetic circuit,Ewill be inversely proportional to eld
current. Field current as a ratio of its rated value=960=1200=0:8
EEE-413N L-21 & 22 April 3, 2019 21 / 27

30. Solution...Contd
(iii) Maximum available armature voltage
Va=0:95230=218:5 V
Back EMF E=Va+2IaRa=
218:5+20020:02=226:5 V
Maximum permissible motor speed=
226:5
226
960=
962 rpm
Assuming lossless chopper, power fed into the source
VaIa=218:5400=
87:4 kW(iv) As in (iii)E=226:5 Vfor which at rated eld current speed=960 rpm.
Assuming linear magnetic circuit,Ewill be inversely proportional to eld
current. Field current as a ratio of its rated value=960=1200=0:8
EEE-413N L-21 & 22 April 3, 2019 21 / 27

31. Solution...Contd
(iii) Maximum available armature voltage
Va=0:95230=218:5 V
Back EMF E=Va+2IaRa=
218:5+20020:02=226:5 V
Maximum permissible motor speed=
226:5
226
960=
962 rpm
Assuming lossless chopper, power fed into the source
VaIa=218:5400=
87:4 kW(iv) As in (iii)E=226:5 Vfor which at rated eld current speed=960 rpm.
Assuming linear magnetic circuit,Ewill be inversely proportional to eld
current. Field current as a ratio of its rated value=960=1200=0:8
EEE-413N L-21 & 22 April 3, 2019 21 / 27

32. Solution...Contd
(iii) Maximum available armature voltage
Va=0:95230=218:5 V
Back EMF E=Va+2IaRa=
218:5+20020:02=226:5 V
Maximum permissible motor speed=
226:5
226
960=
962 rpm
Assuming lossless chopper, power fed into the source
VaIa=218:5400=
87:4 kW(iv) As in (iii)E=226:5 Vfor which at rated eld current speed=960 rpm.
Assuming linear magnetic circuit,Ewill be inversely proportional to eld
current. Field current as a ratio of its rated value=960=1200=0:8
EEE-413N L-21 & 22 April 3, 2019 21 / 27

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.
Submission Date: 24
th
April 2019
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35. Home Assignment
Problems 5.51 to 5.54
Q. Describe Microprocessor controlled electric drives in detail.
Submission Date: 24
th
April 2019
EEE-413N L-21 & 22 April 3, 2019 23 / 27

36. Choice between Rectier and Chopper Control
If supply is ac, rectier 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 (highera), 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
1.Open loop control
2.Closed loop (feedback) control
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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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