Chopper Drives
Transcript of Chopper Drives
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ELEC 482 Module # 5 Outline
TOPIC: DC Chopper Drives
Key educational goals:
Evaluate and identify the different types of dc choppers (dc-dc converter) configurations
to drive a separately excited dc motor.
Reading/Preparatory activities for class
i) Textbook:
Chapter 5: DC motor control using a dc chopper.
5.1. Basic equations of a dc motor
5.2 DC chopper drives
ii) Power-point file: DC_chopper_drives.
Questions to guide your reading and think about ahead of time.
1. What are the two main conduction modes in a class A chopper?
2. How does the critical duty ratio (cycle) determine the conduction mode of a class A
chopper?
3. What is a class B (two quadrant) chopper?
4. What is a four quadrant chopper?
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ELEC 482 Module # 5 Outline
TOPIC: DC Chopper Drives
The main concepts for today
1. Evaluate how minimum and maximum currents are determined in continuous anddiscontinuous conduction mode of a class A chopper.
2. Identify the parameters that control the critical duty ratio (cycle).
3. Analyze how to compute the efficiency of a class A chopper drive.
4. Contrast the working of the class A, class B and the four quadrant chopper drives.
Summary
The knowledge gained from this module helps in analyzing and designing choppers (dc-dc
converters) for separately excited dc machines.
For next time
We will discuss in the next module how to control the separately excited dc motor using
single phase and three phase line commutated converters supplied from the standard utility
AC supply. Then we will compare their performance with the dc chopper controlled
drives.
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ELEC 482 Module # 5 Outline
TOPIC: DC Chopper Drives
Sample test/exam questions/problems to help you study
1) A separately excited dc motor has armature resistance of 0.2 , and armatureinductance of 1 mH. The switching time period of the chopper is 3 ms with TON =
1 ms. If the back emf is 10 V and the supply voltage is 100 V, find the following:
(i) Average load voltage(ii) Average load current(iii) Maximum value of load current(iv) Minimum value of load current(v) The drive efficiency
2) A separately excited dc motor has armature resistance of 0.2 , and armatureinductance of 1 mH. The switching time period of the chopper is 3 ms with TON =2 ms. If the back emf is 85 V and the supply voltage is 100 V, find the following:
(vi) Average load voltage(vii) Average load current(viii) Maximum value of load current(ix)
Minimum value of load current
(x) The drive efficiency
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Continuous conduction mode (CCM) in a class A chopper
Boundary between the continuous conduction mode (CCM) anddiscontinuous conduction mode (DCM) in a
Class A chopper
Discontinuous conduction mode (DCM) in a class A chopper
Numerical examples on CCM and DCM in a class A chopper
Class B Chopper ( Multiquadrant or 2 quadrant chopper)
forward motoring, regeneration/braking.
Class B Chopper (Alternative configuration) forward motoring,regeneration/braking.
Four quadrant chopper.
Chopper Drives
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Chopper (DC-DC converter) driven
dc drives
Vs
S
DC M
isiL
iD
+
vL
-DC }=Ra
La
eb
Class A chopper drive
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Continuous conduction mode (CCM)
Imax
vL(t)
iL(t)
TONTOFF
Imin
t
tt = 0
VsT = TON+TOFF
Vs
S
DC M
isiL
iD
+
vL
-DC }=Ra
La
eb
CCM implies that iLdoes not go to zero
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Definitions
d= duty cycle;
= = + c = chopping frequency = 1/T
( ) = 1 0 = 1 0 + 0 =1
= (1)
( ) = 1 20 = 1 20 = = (2)Ripple factor:
.. = ( )2 ( )2
(
)
(3)
Substituting equations (1) and (2) in (3),
.. = 1 (4)Smaller the value ofdlarger is the ripple factor.
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Harmonics in load voltage
= ( ) + (=1 + ()) (5)
= 1 20 = = 20 (2 ) (6)
= 1 20 = = 20 1 (2 ) (7)
RMS value of the fundamental ( = 1)switching component = 1( ) = 12+122 (8)
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Example 1Question:
= 100
,
= 1ms,
= 2.5 ms. Calculate
(
),
(
),
1(
), ripple factor RF.
Solution:
= =1
2.5= 0.4.
(
) =
= 100
0.4 = 40V.
( ) = = 100 0.4 = 63.25 V .
R.F.= 1 = 10.40.4 = 1.225.
1 = (2 ) =
100
sin0.8 = 18.71V.1 = 1 (2 ) =
100
1 cos(0.8) = 57.58V.
Also
1( ) = 12+122 = 18.71
2+57.582
2= 42.81 V.
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Example 2Question:
A separately excited dc motor with
= 0.3
,
= 15 mH is controlled by a chopper over a
speed range of 0-2000 rpm. The dc supply voltage is 220V. = constantand requires an( ) = 25A. Calculate the range of if = 0.96 Vsrad .
Solution:
At = 0; = N
60 2 = 0V; ( ) = + =25 0.3 + 0 =7.5 V. = 7.5220 = 0.034.
At = 2000; = 200060 2 0.96 = 201.06V.
( ) = + = 7.5+201.06 = 208.56 V. = 208.56220 = 0.948.
0.034 0.948.
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Continuous Conduction Mode (CCM)
Imax
vL(t)
iL(t)
TONTOFF
Imin
t
tt = 0
Vs
Vs
S
DC M
is iL
iD
+
vL
-
DC }=Ra
La
eb
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CCM(When S is ON, D is OFF)
= +
+ ; [Ignoring speed ripple, = ]
Taking Laplace transformation
() = () + 0 +
or = () + +
or = ( )
+ +
Solving for current by taking inverse Laplace transform
= . 1 + where = =Armature time constant.
Maximum value of current can be obtained by substituting = = in the above equation.
=
. 1
+
(9)
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CCM(When S is OFF, D is ON)
In this case, = 0 since the freewheeling diode is conducting. Shifting the origin of time axis to = = the initial value of current is instead of .Therefore, the final expression for current can be written as
= . 1 +
Minimum value of current can be obtained by substituting
=
= (1
)
in the above
equation.
= . 1 (1)
+ (1) (10)Solving the two simultaneous equations (9) and (10) the expressions for and can beobtained.
= .11 (11) = .
11
(12)
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Boundary between CCM and DCM
TON
vL(t)
iL(t)
t = 0
TOFF
Vs
Imax
Imin = 0
t
t
Vs
S
DC M
is iL
iD
+
vL
-
DC }=Ra
La
eb
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Boundary between CCM and DCM(2)
At the boundary is zero. Equating to zero in (12) and replacing dwith d'(critical dutyratio),
. 1
1 = 0
=
1
1
= ln 1 + 1 (13)Using equation (13) the critical duty ratio can be evaluated. This is because
From (12)
= ln 1 + + 1 (14) > 0, for the converter to be in CCM > .Thus for a given chopper frequency or a given time period and a given d
>d'
implies CCM andd
d', the drive is in CCM.Average load voltage = = 100 0.333 = 33.3 VAverage load current =
=33.310
0.2= 116.5
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Example 2 (CCM) (2)
Maximum value of load current is given by
= .1
1
or = 1000.2
.10.333 0.003
0.005
1
0.003
0.005
100.2
= 500 0.181110.4512
50 = 150.70A.
Minimum value of load current is given by
= . 11
or = 1000.2
.0.333 0.003
0.005 10.0030.005 1
10
0.2= 500 0.22116
0.8221 50 = 84.51 A.
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Example 2 (CCM) (3)Drive efficiency = =
. .
[Neglecting mechanical losses]
The average value of switch current has to be computed. The switch current waveform isgiven below.
TON
is
T
Imin
Imax
The average value can be calculated using
= 1 ()
. 1 + 0
or =
+
1 1
or =10010
0.2
0.001
0.003+
100100.2
0.005
0.003
0.001
0.005 1 84.650.003
0.005 0.001
0.005 1 AOr = 150 136 + 25.57 = 39.57 A
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Example 2 (CCM) (4)
Plot of motor voltage and current under CCM
Substituting the values of supply voltage, back emf, average load current and average
switch current the drive efficiency is found to be equal to
= = . . =
10 116.5100 39.57 = 0.2944
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Example 3 (DCM)
A separately excited dc motor has armature resistance of 0.2 , and
armature inductance of 1 mH. The switching time period of the chopper
is 3 ms with TON = 2 ms. If the back emf is 85 V and the supply voltage
is 100 V, find the following:
(i) Average load voltage(ii) Average load current(iii) Maximum value of load current(iv) Minimum value of load current(v) The drive efficiency
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Example 3 (DCM) (2)Solution:
Checking for mode of operation:
The operating duty ratio of the chopper is = = 0.666The critical duty ratio dis obtained using the following relation
= ln 1 + 1 =
0.005
0.003ln 0.0030.005 1 85
100+ 1
d= 0.8832
Since d
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Example 3 (DCM) (3)
t(ms)
TON
tx
2.2802. 03.
T
L(t)v
iL(t)
VsEb
t( ms)
Average load voltage can be calculated from the above waveform.
= 1
0+
1
0. +
1
= 86.98 V.
= ( ) =86.9885
0.2= 9.9 A.
=
.
1
=
100850.2
.
1
0.6660.003
0.005
= 24.71 A.
= 0 A. = 1
. 1
0
or = +
1
or = 100850.2 0.0020.003 + 100850.2 0.0050.003 0.002
0.005 1 = 8.79 A
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Example 3 (DCM) (4)
=
=
.
. =
859.9
1008.79= 95.73 %.
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Multi-quadrant Chopper
DC
DC
Eb
Vs
LaRa
S2
S1D2
D1
S1D1
S2 D2
vL
iL
vL
iL
Class B chopper: [Two quadrant chopper]
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Class B chopper: Forward motoring
DC
DC
Eb
Vs
LaRa
S1
vL
iL
DC
DC
Eb
Vs
LaRa
D1vL
iL
In forward motoring, energy flows from Vs to Eb. The circuit functions like a Class A
( one quadrant) chopper.
S1 D1
vL
iL
S1 ON D1 ON
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Class B chopper: Regeneration/Braking
DC
DC
Eb
Vs
LaRa
S2
vL
iL
DC
DC
Eb
Vs
LaRa
D2
vL
iL S2 D2
vL
iL
S2 ON D2 ON
In this mode, energy flows from Eb to Vs. The energy stored in La when S2 is
on aids in transferring the energy from Eb to Vs once S2 is switched-off and
D2 is turned-on.
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Multi-quadrant chopper(2)
DCEb
Vs
LaRa
S2
S1
D2
D1
iL DC
vL
S1D1 S2D2
vL
iL
Another Class B chopper: [ Two quadrant chopper]
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Class B chopper: Forward motoring (2)
EbLa
Ra
S2
S1
iL DC
vL
S1 S2
vL
iL
In forward motoring, S1 and S2 are ON.
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Class B chopper: Braking/Plugging
D1D2
vL
iL
In this mode, D1 and D2 conduct reversing the potential across the motor terminals.
Eb and Vs get connected in series and quickly decreases the armature current.
D1
DCEb
LaRa
iL DC
vL
Vs
D2
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Class B chopper: Alternative switch
sequence
DC
Eb
Vs
LaRa
S2
S1
D2
D1
iL
DCvL
Another switch sequence for this converter can be :
S1S2; S1D1; D1D2; S2D2; D1D2
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Four quadrant chopper
DCEb
Vs
LaRa
S2
S1
D2
D1
iL DC
vL
D3 D4
vL
iL
S3D3
S4D4
S1 S2
S3 S4 D1 D2