Chopper Drives

7/28/2019 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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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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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

Chopper Drives

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