DC CHOPPER DRIVES - Philadelphia University · Power Electronics and Drives 845 CHAPTER TWELVE DC...

Power Electronics and Drives

845

CHAPTER TWELVE

DC CHOPPER DRIVES

12.1 INTRODUCTION

The principle of operation of d.c. chopper with passive loads (R and R-L

loads) is discussed in much details in Chapter Four. Also the operation of

the chopper with load consisting back emf is discussed in the same chapter

briefly. In the present chapter, the operation of the chopper with the d.c.

motors shall be discussed in details. The analytical properties of the voltage

and current of the motor with d.c. chopper drive are presented.

The d.c. motor is considered a SISO (Single Input and Single Output)

system having torque-speed characteristics compatible with most mech-

anical loads. This makes a d.c. motor controllable over a wide range of

speeds by proper adjustment of the terminal voltage.

12.2 ONE-QUADRANT DC CHOPPER DRIVE

A simple chopper-fed d.c. motor drive is shown in Fig.12.1. The basic

principle behind d.c. motor speed control is that the output speed of a d.c.

motor can be varied by controlling armature voltage for speed below and up

to rated value keeping field voltage constant. The armature voltage can be

controlled by controlling the duty cycle of the converter (here the converter

used is a d.c. chopper).

In Fig. 12.1, the converter output gives the d.c. output voltage Va required

to drive the motor at the desired speed. In this diagram, the d.c. motor is

represented by its equivalent circuit consisting of inductor La and resistor

Ra in series with the ideal source back emf (Ea). The thyristor T1 is triggered

by a pulse width modulated (PWM) signal to control the average motor

voltage.


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Fig.12.1 Chopper-fed d.c. motor drive.

Theoretical waveforms illustrating the chopper operation are shown in

Fig.12.2. In this case the average armature voltage is a direct function of the

chopper duty cycle γ, i.e. Vav = γ Vd .

Fig.12.2 Load current and voltage waveforms with motor load:

(a) Continuous armature current, (b) Discontinuous armature

current.


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Like the operation of the d.c. chopper with the passive inductive load

discussed in Chapter Four. Two modes of operation may result when the

chopper operates with d.c. motor load, namely, continuous armature current

operation and discontinuous armature current operation modes as illustrated

in Fig.12.2. In both cases, the armature voltage and current waveforms are

different in shape and each has its own analytical properties as will be

discussed in the following subsections.

12.2.1 Armature Voltage Waveform Analysis for

Continuous Armature Current Operation

The armature voltage waveform applied to the d.c. motor for

continuous current operating mode shown in Fig.12.2(a) is given by

| |

In order to analyse all equations in terms of (ωt) rather than the time (t)

which gives more convenient and simple way for harmonic analysis, let the

repetition periodicity , Fig.12.2, be designated as radians,

hence

The ON period of the chopper is, ( ) ,

therefore, Eq.(12.1) can be re-written as

|

|

Hence, the average value of the armature voltage is

∫

∫

[ ]

The rms value of the armature voltage waveform is given by

√

∫

√

[ ]

√

The ripple factor, defining the ratio of the a.c. components to the average

value, is given by


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√

√

√

The ripple factor = 0 for full conduction when .

The harmonic analysis of the armature voltage waveform is the same as that

given in equations (4.52) – (4.56) in Chapter Four – Part I.

12.2.2 Armature Voltage Waveform Analysis for

Discontinuous Armature Current Operation

With some values of the duty cycle , the armature current may falls to

zero during part of the cycle, as illustrated in Fig.12.2(b), this will occur

when switch T1 and diode DFW are both off. In this case the armature is

disconnected from the supply and any emf generated in the armature

winding will then be appeared across the armature terminals. Therefore, for

the interval or ,where is the time at which the

current becomes zero and is the diode extinction angle in Fig.12.2(b), it is

seen that there is an armature voltage . If we assume that is

constant then

|

|

The average value of this voltage is

∫

(∫

∫

)

(

)

For full conduction and Eq.(12.7) reduced to Eq.(12.2). The rms

load voltage is

√

∫

∫


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√

(

)

The Fourier components of the armature voltage waveform with

discontinuous current are obtained by substituting Eq.(12.1) into Fourier

integrals:

∫

∫

to result :

From which, the fundament component, (n =1), of the armature voltage, is

given by

√ [ ]

[

]

12.3 ANALYTICAL PROPERTIES OF THE

ARMATURE CURRENT WAVEFORM

When the chopper is loaded with a d.c. motor, having the equivalent

circuit representation as in Fig.12.1, the armature current equation is,


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This differential equation represents the armature current behaviour in both

the continuous and discontinuous current modes as will be discussed here in

after.

(A) Continuous armature current operation

The analytical properties of the armature current of the d.c. motor is very

similar to that of the chopper operation with back emf given in chapter Four-

Part I. Hence, equations (4.29) to (4.34) are also applied to the motor with

some modification including designating T, the repetition periodicity

in Fig.12.2 , as radians and as .

With switch T1 - ON ( and DFW - OFF)

At , in Fig.12.2(a) , and , solution of the

first order linear differential equation (12.14) gives the result :

For the interval ,

( ⁄ )

⁄

where ⁄

When , (ton) in Fig.12.2 (a), , ,

substituting these values into Eq. (12.15) gives

( ⁄ )

⁄

Equation (12.16) is not time dependent and remains true after T1 switches

off.

With switch T1 - OFF ( and DFW -ON)

At , in Fig.12.2(a) , and , hence in the

interval the load current is given by

( ⁄ )

⁄

But at , , equation (12.17) may then be re-written as

( ⁄ )

⁄

The simultaneous solution of Eq.(12.16) and Eq. (12.18) yields


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( ⁄

⁄)

( ⁄

⁄ )

Equations (12.19) and (12.20) are used for the solution of equations (12.15)

and (12.17) to determine the value for the armature current at any

instant of the cycle during continuous conduction mode.

To find the average armature current, we can use the following basic

equation,

∫

For continuous current operation equations (12.15) and (12.18) are

substituted into Eq.(12.21) for the intervals and

respectively. Hence, the average armature current in the

steady-state is given by

where

Va = average armature voltage = γ Vd

Ia = average armature current

Ea = the internal generated voltage (back emf) is given by: Ea= Ke φ n

Now solving for the motor’s speed using the motor general equation,

Va = Ea +Ia Ra ,

or for the chopper drive in the steady-state ,

where n is the speed in rpm.

and the motor torque is Tm = KT Ia φ

KT = Torque constant = 9.55 Ke .


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At starting, n = 0. The starting torque Tst may be found as:

The speed-torque characteristics of a separately-excited d.c. motor is shown

in Fig.12.3 for different values of the duty cycle γ. It is clear that, as the

duty cycle is reduced, the no load speed and the starting torque are reduced

accordingly and the characteristic lines shift downward in parallel manner.

Fig.12.3 Speed-torque characteristics of a d.c. motor with d.c. chopper drive.

(B) Discontinuous armature current operation

Under different loading and duty cycle value conditions, the motor current

may fall to zero resulting in the discontinuous current shown in

Fig.12.2(b), this waveform can be divided into two parts that may be

analysed according to the conduction of switch T1, whether it is on or off

as in (A) above. In this case the maximum current is different from the value

obtained with continuous operation and it occurs for a different value

of . The minimum current for discontinuous operation is,

Speed

(n)

Torque

Tst

Tm 0

1

2

3

1 > 2 > 3

Starting

torque


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With switch T1 - ON, (and diode DFW - OFF)

In Fig.12.2 (b), the current of Eq.(12.14) can be found for the interval

by considering that , and at .

This gives

( ⁄ )

Since maximum value of occurs at ( ton in Fig.12.2(b))

then,

( ⁄ )

where ImaxD is the maximum value of the motor current for discontinuous

operation.

With switch T1 - OFF (and diode DFW - ON)

When T1 in Fig.12.1 is switched off, the motor voltage falls to zero

due to the conduction through diode DFW, the circuit differential equation

(12.14) is then modified to

This has the solution given in Eq.(12.17) except that the maximum current

is now given by Eq.(12.26) to result in,

( ⁄ )

( ⁄ ) ⁄

The current extinction angle x

In Fig.12.2 (b), in order to find the current extinction angle x, let it occurs at

at which ia (ωt) = 0. Now if we substitute these values into

Eq.(12.28) and solve for x, the following expression will be obtained:

[ ⁄

( ⁄ )]


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Hence, Eq.(12.27) defines the current in the region , where

x is found from Eq. (12.29), and this equation is indeterminate if

The critical value of the duty cycle

Let switch T1 in Fig.12.1 have a conduction period of particular value

radians that represent the boundary between continuous and

discontinuous operation. Then, from Eq.(4.5),

At the particular value of conduction the minimum current Imin = 0

defined by Eq.(12.20),

( ⁄

⁄ )

From this equation we can write,

⁄

⁄

The relationship between

and is discussed in chapter Four and

shown in Fig.4.12 with the factor ⁄ ⁄ as parameter. If a

circuit operate with a specified value of , defined by Eq.(12.30), then the

criteria for continuous or discontinuous operation of a d.c. motor are the

same as that given in Fig.4.12 in Chapter Four so that ,

, Continuous current

Discontinuous current

where is defined by Eq.(12.30 ).

It should be noted that the ratio ⁄ is the periodic time of the overall

cycle, the parameter ⁄ of the 'state of conduction' curves, Fig.4.12, is

The electrical time constant of armature circuit depends on the motor

armature inductance La which is considered as an important parameter in

determining current continuity. For small motors is of the order 10-50

mH and for larger motors the inductance is smaller, being typically 2-10

mH.


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12.4 POWER INPUT, SUPPLY CURRENT, AVERAGE ARMATURE

CURRENT AND TORQUE CALCULATIONS

The average power in a chopper circuit is transferred from the constant

supply Vd to the load by the combination of with average component of

the input current , hence

The instantaneous input current flows only while switch T1 is

conducting in Fig.12.2,

The average value of is defined by

∫

For continuous conduction, substituting Eq.(12.15) and Eq.(12.20) into

Eq.(12.36) gives

(

)

(

)

( )

( )

For discontinuous conduction, substituting Eq.(12.25) into Eq. (12.36) gives

[

]

Example 12.1

The speed of a separately-excited d.c. motor with Ra = 1.2 Ω and La = 30

mH, is to be controlled using class-A thyristor chopper as shown in

Fig.12.4. The d.c. supply Vd = 120 V. By ignoring the effect of the armature

inductance La , it is required to:

(a) Find the no load speed and starting torque of the motor when the

duty cycle =1.


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(b) Draw the speed-torque characteristics for the motor when the duty

cycle γ = 1. The motor design constant KeΦ has a value of 0.042

V/rpm.

(c) Find the speed of the motor n (rpm) when a torque of 8 Nm is

applied on the motor shaft and the duty cycle is set to = 0.5.

Fig. 55.4 Thyristor chopper drive.

Solution

The average armature voltage for is

The motor’s speed:

At no load Td = 0, hence

At starting, n = 0. The starting torque Tst may be found as:


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Fig. 12.5 Speed-torque characteristics.

(b) At γ = 0.5 ,

At γ = 0.5 , TL = 8 Nm

Note: KT = Torque constant = 9.55 Ke

Example 12.2

A d.c. motor is driven from a class-A d.c. chopper with source voltage of

220 V and at frequency of 1000 Hz. Determine the range of duty cycle to

obtain a speed variation from 0 to 2000 rpm while the motor delivered a

constant load of 70 Nm. The motor details as follows:

1kW, 200 V, 2000 rpm, 80% efficiency, Ra = 0.1 Ω, La = 0.02 H, and = 0.54 V/rad /s.

Torque (N.m)

1428.1

857.13

γ =0.5

40

no =2857 rpm Speed

(rpm)

0 8 20

γ = 1


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Solution

(

)

For m = 0

Hence the range of is .

Example 12.3

In the microcomputer-controlled class-A IGBT transistor d.c. chopper

shown in Fig.12.6, the input voltage Vd = 260 V, the load is a separately-

excited d.c. motor with Ra = 0.28 Ω and La = 30 mH. The motor is to be

speed controlled over a range 0 – 2500 rpm , provided that the load torque is

kept constant and requires an armature current of 30 A .

(a) Calculate the range of the duty cycle γ required if the motor design

constant KeΦ has a value of 0.10 V/rpm.


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(b) Find the speed of the motor n (rpm) when the chopper is switched

fully ON such that the duty cycle γ = 1.0.

Fig.12.6 IGBT d.c. chopper drive.

Solution

(a) With steady-state operation of the motor, the armature inductance La

behaves like a short circuit and therefore has no effect at all.

At stand still n = 0, and therefore , Ea = 0 , hence from Eq.(12.22),

At full speed n = 2500 rpm ,

For separately-excited d.c. motor,

Therefore the range of the duty cycle γ will be:

Similarly


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(b) When the chopper is switched fully on, i.e. γ =1, then

.

At this condition,

|

Example 12.4

A separately-excited d.c. motor has the following parameters:

Ra = 0.5 Ω , La = 5.0 mH , Ke Φ = 0.078 V/rpm.

The motor speed is controlled by a class-A d.c. chopper fed from an ideal

200 V d.c. source. The motor is driven at a speed of 2180 rpm by switching

on the thyristor for a period of 4 ms in each overall period of 6 ms .

(a) State whether the motor will operate in continuous or discontinuous

current mode,

(b) Calculate the extinction angle of the current if it exist,

(c) Sketch the armature voltage and current waveforms,

(d) Calculate the maximum and minimum values of the armature current,

(e) Calculate the average armature voltage and current.

Solution

(a) To find whether the motor operates in continuous or discontinuous

current modes ,we have to find the values of γ and γ’ :

At speed of 2180 rpm , The critical value of γ’ will be , (using Eq.( 12. 31 ))

’

’


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From which γ’ = 0.0.8829 , therefore , γ’ > γ , hence the motor is operating

in discontinuous current mode.

(b) The extinction angle x of the current is calculated from Eq.(12.29) as,

[ ⁄

( ⁄ )]

[ ⁄

( ⁄ )]

From which

(c) The armature voltage and current waveforms are shown in Fig.12.7.

(d) The maximum and minimum values of the armature currents are:

Imin = 0 , since it is discontinuous.

ImaxD is calculated from Eq.(12.26) as,

( ⁄ )

Fig.12.7 Armature voltage and current waveforms.

Example 12.5

A separately-excited d.c. motor with Ra = 0.1 Ω and La = 20 mH, is to be

controlled using class-A thyristor chopper. The d.c. supply is a battery with

Vd = 400 V. The motor voltage constant is 5 V.s/rad. In the steady-state


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operation the average armature current Ia = 100 A and it is assumed to be

continuous and ripple-free.

(a) For a duty cycle of 0.5, it is required to calculate (i) the input power

to the motor, (ii) the speed of the motor, (iii) the developed torque.

Mechanical, battery and semiconductor losses may be neglected.

(b) If the duty cycle of the chopper is varied between 20% and 80%, find

the difference in speed resulting from this variation.

Solution

(a) Input power to the motor, speed of the motor and the developed torque

are calculated as follows:

(i) For continuous current operation the input power is

(ii) Speed of the motor can be calculated as,

The voltage across the armature circuit

The induce voltage

To find the speed n in rpm

(iii) The torque produced by the motor,

(b) For duty cycle of 20%,

For duty cycle of 80%,


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⁄

Hence the difference in speed is

Example 12.6

A class-A d.c. chopper operating at a frequency of 500 Hz and feeding a

separately-excited d.c. motor from 200 V d.c. source. The load torque is 35

Nm and speed is 1000 rpm. Motor resistance and inductance are 0.15 Ω and

1.0 mH respectively. The emf and torque constant of motor are 1.6 V/rad/s

and 1.4 Nm /A respectively. Find (a) Maximum and minimum values of

motor armature current, and (b) Variation of armature current. Neglect

chopper losses.

Solution

(a) Let duty cycle =

Vd = 200 V

Average armature current Ia = T / Kφ = 35/1.4 = 25 A

Back emf Ea = K φ = 1.6 (950 2 /60) = 159.16 V

200 γ = 159.16 + 25 0.15 = 162.29 V

γ = 0.8145

T = 1/500 = 2 ms

ton = γT = 2 0.8145 = 1.629 ms

toff = 2 - 1.629 = 0.371 ms

From Eq.s (12.19) and (12.20) ,The maximum and minimum currents are

calculated as

Let:

⁄

Hence Eq.(12.19) and (12.20) can be re-written as


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( ⁄

⁄)

and

( ⁄

⁄ )

(

)

(

)

(

)

(

)

(b) Variation of armature current =

PROBLEMS

12.1 A permanent-magnet d.c. motor with Ra = 5 Ω and La = 300 mH, is to speed

controlled using class-A d.c. chopper. The d.c. power supply is 120 V,

armature resistance Ra = 0.5 Ω and La = 20 mH. The motor constant

Ke ϕ = 0.05 V/rpm. The motor is assumed to drive a constant load torque on

the motor which required an average armature current of 20 A. Assuming

continuous current drawn by the motor, determine:


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(a) The ranges of the chopper drive duty cycle.

(b) The range of speed control.

[Ans: (a) γ = (0.083 – 1) , (b) n = (0 – 2200) rpm]

12.2 A separately-excited d.c motor with Ra = 5 Ω and La = 300 mH, is to speed

controlled using class-A d.c. chopper. The d.c. supply is 120 V. (a) It is

required to draw the speed torque characteristics for the motor for duty

cycle, = 1. The motor constant Ke ϕ = 0.7 (V /rpm). (b) Find the speed of

the motor (ω) at a torque of 5 Nm applied on the motor shaft when the duty

cycle used is ( = 0.5).

[Ans: (b) 34.69 rpm]

12.3 A d.c. motor is driven by a single-quadrant d.c. chopper with source voltage

of 240 V.The chopper operates at a frequency of 400 Hz. The motor is

loaded such that it draws contiuous and ripple-free current of 80 A and runs

at speed of 500 rpm. The armature resistance is Ra = 0.25 Ω and the machine

constant Keφ = 0.1 V/rpm. It is required to:

(a) Draw waveforms of the armature voltage, armature current and the

current drawn from the source.

(b) Determine the duty cycle γ and the ON-time ton of the chopper.

(c) Determine the power developed by the motor, power absorbed by

the armature resistance and the power drawn from the source.

[Ans (b) γ = 0.291, ton = 0.729 ms, (c) 4000 W, 1600 W, 5600 W ]

12.4 A class-A, single-quadrant d.c. chopper is used to feed a d.c. shunt motor

motor which has the following parameters:

Ra = 0.25 Ω, La = 15 mH, and Ke ϕ = 0.209 V/rpm.

The chopper is supplied from an ideal battery source of voltage = 500 V. At

a certain operating condition, the motor runs at speed of 9000 rpm and the

mechanical output power is 9000 W. At the stated condition, the armature

current is continuous with peak-to-peak ripple of 8 A. It is required to

determine:

(a)The chopper duty cycle.

(b)The approximate switching frequency (neglect the effect of Ra).

(c)The power lost in Ra .

[Ans: (a) 0.4, (b) 1039 Hz, (c) 600 W]

12.5 A small electric train is driven by a 1250 rpm, 400 V d.c. series motor has an

armature resistance Ra = 0.05 Ω and a series field resistance Rs = 0.1 Ω. The

motor rated full load current is 150 A. The train is to be speed controlled

using single-quadrant d.c. chopper fed from 500 V ideal battery. The


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chopper frequency varies between 100 Hz and 1800 Hz but the ON time

remains constant at 0.5 ms. At stand still, the motor current is limited to

200 A. Calculate:

(a) The chopper switching frequency and the supply current drawn

when the motor is at standstill.

(b) The chopper switching frequency and the supply current drawn at

rated output.

[Ans: (a) 120 Hz, 12 A, (b) 1600 Hz, 120 A]

12.6 A permanent-magnet d.c. motor is fed by a class-A d.c. chopper. The

machine has negligible armature resistance and an armature inductance of

5 mH with machine constant Ke ϕ = 0.262 V/rpm. The chopper is fed from

a 400 V ideal battery source and operating with a switching frequency of

1000 Hz. It is required to:

(a) Determine the speed of the motor if the duty cycle is 50% and

the torque is 100 Nm.

(b) The torque is now reduced to 20 Nm and the duty cycle is

maintained at 50%. Show that the armature current is

discontinuous and calculate the new motor speed.

(c) For the condition in (b) determine the duty cycle required to

bring the motor speed back to the value calculated in (a).

[ Ans: (a) 764 rpm, (b) 849 rpm, (c) 44.7 % ]

12.7 A d.c. supply with Vd = 250 V supplies power to a separately excited d.c.

motor via a class-A thyristor chopper. The motor has an armature circuit

resistance of 0.25 Ω and inductance of 10 mH. The chopper is fully ON at

the rated motor speed of 1500 rpm when the armature current is 25 A. If the

speed is to be reduced to 1000 rpm, with the load torque constant, calculate

the necessary duty cycle.

[Ans : γ = 0.675]

12.8 A 100 V battery supplies power to a d.c. separately-excited motor, with

Rα = 0.2 Ω and Lα =1 mH, via a class-A d.c. chopper operating at 300 Hz.

With a duty cycle of 0.75 the motor back emf is 70 V, calculate the average

values of the load voltage and current, the average value of input current,

the output power and the approximate efficiency of the motor.

[ Ans: Vav= 75 V, Iav = 25 A, Isav = 20.89 A , Po = 1750 W , η = 84 % ]


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12.9 A separately-excited d.c. motor with Ra = 0.3 Ω and La = 15 mH is to be

speed controlled over a range 0 – 1500 rpm using class-A MOSFET d.c.

chopper fed from ideal d.c. source with Vd =300 V as shown in Fig.12.8(a).

The motor load is a conveyer belt which exerts a constant torque on the

motor, as shown in Fig.12.8 (b), and requires an average armature current

of 30A. The machine design constant ke φ = 0.17 V/rpm.

(a) Calculate the required range of the duty cycle of the chopper.

(b) Calculate the speed of the motor when the chopper is fully ON.

(a)

(b)

Fig.12.8.

[Ans : (a) γ Range : 0.03 – 0.91, (b) 1712 rpm ]

12.10 A 240 V, 150 kW, 500 rpm separately-excted d.c. motor is controlled by a

class-A d.c. chopper. The d.c. supply for the chopper is an ideal battery

of 310 V. The chopper is operates with a frequency of 1000 Hz. The motor

parameters are as follows:

Ra = 0.04 Ω , La = 1 mH , KT = 4.17 V/rad/s

If the motor is running at 300 rpm with 0.55 duty cycle, determine the

average current taken by the motor and the average load produced by the


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motor using steady-state approximation analysis and assume continuous

current operation. Neglect any switching losses.

[Ans : 991.8 A , 4138.1 Nm]

12.11 A 20 kHz d.c. chopper is used to drive a separately-excited d.c. motor from

240 V source. Resistance and inductance of armature circuit are 0.6 Ω and

4mH, respectively. At rated conditions, armature current is 10 A and back

emf is 174 V,

(a) Find the duty cycle of the chopper .

(d) Find the required duty cycle of chopping to reduce motor speed to

80% of rated speed, assuming that the armature current at this speed

remains at rated value.

[Ans: (a) 0.75 , (b) 0.605]

12.12 A class-A, single-quadrant d.c. chopper is used for dynamic braking of

a separately-excited d.c. motor having armature resstance Ra = 0.2 Ω, and

inductance La of 30 mH. The braking resistance used is 10 Ω. The voltage

constant is 1.3 V/A.rad /s, and the armature current is 100 A. If the field

current is 2 A and the duty cycle of chopper is 0.5; find : (a) Average

voltage across chopper, (b) Power dissipated in braking resistance, and

(c) Motor speed.

[Ans : (a) 500 V ,(b) 15 kW, (c) 1910 rpm]

12.13 A separately-excited d.c. motor is fed from an ideal d.c. source of 500 V

through a single-quadrant d.c. chopper. The armature resistance is 0.08 Ω

and armature current is 150 A. The voltage and torque constants are 1.3

V/A.rad /s and 1.4 Nm /A2 respectively. The field current is 1.5 A. The

duty cycle of chopper is set to 0.75. Determine: (a) The input power to the

motor,(b) The speed of the motor, (c) The output torque.

[Ans : (a) 33.75 kW, (b)782.3 rpm , (c) 390 Nm]

DC CHOPPER DRIVES - Philadelphia University · Power Electronics and Drives 845 CHAPTER TWELVE DC...

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