Chapter 13old

8/7/2019 Chapter 13old

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DC choppers 352

Figure 13.3. First-quadrant dc chopper and two basic modes of chopper output

current operation: (a) basic circuit and current paths; (b) continuous load current;

and (c) discontinuous load current after t = tx.

LOAD

Vs

D2

T1I

vo

io

on

off

(a)

t

tt

t

tx

T

I

T

vo

i

tT tT

i

Vs

vo

I

I

I

I

I

Vs

E E

conducting devices

T Df T Df T Df T Df T Df

(b) (c)

ioio

on

(a)

vo

Vs

R L + E

T1

io

Vs

(b)

vo

R L + E

D2

io

off

(b) (c)

oV o

V

oI

oI

Power Electronics353

13.2.1 Continuous load current

Load waveforms for continuous load current conduction are shown in figure 13.3b.

The output voltage vo, or load voltage is defined by

( )for 0

0 for

s T

o

T

V t tv t

t t T

=

(13.1)

The mean load voltage is

( )

0 0

1 1T Tt to o s

T

s s

V v t dt V dt T T

tV V

T

= =

= =

(13.2)

where the switch on-state duty cycle = tT/Tis defined in figure 13.3b.

The rms load voltage is

( )

2 2

0 0

1 1T Tt trms o s

T

s s

V v t dt V dt T T

tV V

T

= =

= =

(13.3)

The output ac ripple voltage is

( ) ( ) ( )

2 2

2 2

1

r rms o

s s s

V V V

V V V

=

= = (13.4)

The maximum rms ripple voltage in the output occurs when = giving an rms

ripple voltage of Vs.

The output voltage ripple factor is2

2

1

11 1

r rms

o o

s

s

V VRF

V V

V

V

= =

= =

(13.5)

Thus as the duty cycle 1 , the ripple factor tends to zero, consistent with theoutput being dc, that is Vr= 0.

Steady-state time domain analysis of first-quadrant chopper

- with load back emf and continuous output current

The time domain load current can be derived in a number of ways.

First, from the Fourier coefficients of the output voltage, the current canbe found by dividing by the load impedance at each harmonic frequency.

Alternatively, the various circuit currents can be found from the timedomain load current equations.

i. Fourier coefficients: The Fourier coefficients of the load voltage are

independent of the circuit and load parameters and are given by


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DC choppers 354

( )

sin2

1 cos2

s

n

s

n

Va n

n

Vb n

n

=

=

(13.6)

Thus the peak magnitude and phase of the nth

harmonic are given by2 2

1tan

n n n

nn

n

c a b

ab

= +

=

Substituting expressions from equation (13.6) yields

1

2sin

sin2tan

1 cos2

s

n

n

Vc n

n

nn

n

=

= =

(13.7)

where

( )sinn n nv c n t = + (13.8)

such that

( ) ( )1

sino o n n

n

v t V c n t

=

= + + (13.9)

The load current is given by

( )( )

0 1 1

sin

n n

o o n nn

o n

n n no o

c n tV v Vi t i

R Z R Z

= = =

= = + = + (13.10)

where the load impedance at each frequency is given by

( )22

no Z R n L= +

ii. Time domain differential equations: By solving the appropriate time domain

differential equations, the continuous load current shown in figure 13.3b is defined

by

During the switch on-period, when vo(t)=Vs

oo s

di L Ri E V dt + + =

which yields

( ) 1 for 0t t

s

o T

V Ei t e I e t t

R

= +

(13.11)

During the switch off-period, when vo(t)= 0

0oo

di L Ri E

dt+ + =

which, after shifting the zero time reference to tT, in figure 13.3a, gives

( ) 1 for 0t t

o T

Ei t e I e t T t

R

= +

(13.12)


1

Ipp

Vs/R

0

0 1

on-state duty cycle

T/

25

5

2

1

1where (A)

1

1and (A)

1

T

T

t

s

T

t

s

T

V e EI

R Re

V e EI

R Re

=

=

(13.13)

The output ripple current, for continuous conduction, is independent of the backemfEand is given by

(1 ) (1 )

1

T Tt T t

s

T p p o

V e e I i I I

Re

+

= = =

(13.14)

which in terms of the on-state duty cycle, =tT/T, becomes( )1

(1 ) (1 )

1

TT

s

Tp p

V e eI

Re

=

(13.15)

Figure 13.4. Ripple current as a function of duty cycle =tT/T and ratio of cycle

period T (switching frequency, fs=1/T) to load time constant=L/R.Valid only for continuous load current conduction.

The peak-to-peak ripple current can be extracted from figure 13.4, which is a

family of curves for equation (13.15), normalised with respect to Vs/R. For a given

load time constant =L/R, switching frequency fs=1/T, and switch on-state duty

cycle , the ripple current can be extracted. This figure shows a number of

important features of the ripple current.

The ripple current reduces to zero as 0 and 1. Differentiation of equation (13.15) reveals that the maximum ripple

current p pI

occurs at = .


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DC choppers 358

Figure 13.5. Bounds of discontinuous load current with E>0.

The rms output voltage with discontinuous load current conduction is given by

( )

2 2 2

0

2 2

10

(V)

T x

T x

t t T

rms st t

x

s

V V dt d E dt T

T tV E

T

= + +

= +

(13.27)

The ac ripple current and ripple factor can be found by substituting equations

(13.26) and (13.27) into2 2

r rms oV V V= (13.28)

and2

1r rms

o o

V VRF

V V

= =

(13.29)

Steady-state time domain analysis of first-quadrant chopper

- with load back emf and discontinuous output current

i. Fourier coefficients: The load current can be derived indirectly by using the

output voltage Fourier series. The Fourier coefficients of the load voltage are

1

E/ Vs

00 1

switch on-state duty

T/ 0 1 2 5 10 ?

not

possib

le

continuo

usdis

continuo

us

E/Vs


( )

sin 2 sin 2

1 cos2 1 cos2

s xn

s xn

V E ta n n

Tn n

V E tb n n

Tn n

=

=

(13.30)

which using2 2

1tan

n n n

nn

n

c a b

ab

= +

=

give

( ) ( )1

sinoo n nn

v t V c n t

=

= + + (13.31)

The appropriate division by ( )22

n Z R n L= + yields the output current.

ii. Time domain differential equations: For discontinuous load current, 0.I

= Substituting this condition into the time domain equations (13.11) to (13.14) yields

equations for discontinuous load current, specifically:

During the switch on-period, when vo(t)=Vs,

( ) 1 for 0t

s

o T

V Ei t e t t

R

=

(13.32)

During the switch off-period, when vo(t)=0, after shifting the zero time reference

to tT,

( ) 1 for 0t t

o x T

Ei t e I e t t t

R

= +

(13.33)

where from equation (13.32), with t=tT,

1 (A)Tt

sV E

I eR

=

(13.34)

Aftertx, vo(t)=Eand the load current is zero, that is

( )0 for

o x

i t t t T = (13.35)The output ripple current, for discontinuous conduction, is dependent of the back

emfEand is given by equation (13.34), that is

1Tt

s

p p

V E I I e

R

= =

(13.36)

Since 0I

= , the mean output current for discontinuous conduction, is

( )

( )

-

0 0 0

1 11 1

x T x T t t tt t t t

so

o

o

V E EI i t dt e dt e I e dt R RT T

V E

R

= = + +

=


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DC choppers 360

1

(A)

x xs s

o

t tV E V E

T E TIRR R

+ = = (13.37)

The input and output powers are related such that2

oiin s out in out o rms P V I P I P P R E I = = =+ (13.38)

from which the average input current can be evaluated.

Alternatively the average input current, which is the switch average current, isgiven by

( )0

0

1

11

1

T

T

T

t

i switch o

tts

t

s s

I I i t dt T

V Ee dt

T R

V E V E e I

R T R T

= =

=

= =

(13.39)

The average diode current diodeI is the difference between the average output

current oI and the average input current, iI , that is

diode o i

x

I I I

tET

IT R

=

=

(13.40)

Alternatively, the average diode current can be found by integrating the diode

current given in equation (13.33), as follows

0

11

x Tt tt t

diode

x

EI e I e dt

T R

tE

TI

T R

= +

=

(13.41)

IfErepresents motor back emf, then electromagnetic energy conversion efficiency

is given byoo

in is

E I E I

P V I = = (13.42)

The chopper effective input impedance is given by

s

in

i

VZ

I= (13.43)

Example 13.1: DC chopper with load back emf (first quadrant)

A first-quadrant dc-to-dc chopper feeds an inductive load of 10 ohms resistance,

50mH inductance, and back emf of 55V dc, from a 340V dc source. If the chopper

is operated at 200Hz with a 25% on-state duty cycle, determine, with and without

(rotor standstill) the back emf:


i. the load average and rms voltages;ii. the rms ripple voltage, hence ripple factor;

iii. the maximum and minimum output current, hence the peak-to-peakoutput ripple in the current;

iv. the current in the time domain;v. the average load output current, average switch current, and average

diode current;

vi. the input power, hence output power and rms output current;vii. effective input impedance, (and electromagnetic efficiency forE> 0);viii. sketch the output current and voltage waveforms.

SolutionThe main circuit and operating parameters are

on-state duty cycle = period T= 1/fs = 1/200Hz = 5ms on-period of the switch tT= 1.25ms load time constant = L/R = 0.05H/10 = 5ms

Figure Example 13.1.Circuit diagram.

i. From equations (13.2) and (13.3) the average and rms output voltages are both

independent of the back emf, namely

= 340V = 85V

To s s

tV V V

T= =

240V = 120V rms

T

r s s

tV V V

T= =

=

ii. The rms ripple voltage hence ripple factor are given by equations (13.4) and

(13.5), that is

( )

( )

2 21

= 340V 1 - 147.2V ac

r rms o sV V V V = =

=

and

R L E

T1

10 50mH

+

D2 55V

340V

Vs

=T=5ms


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DC choppers 362

11

1- 1 3 1.732

r

o

VRF

V = =

= = =

No back emf,E = 0iii. From equation (13.13), withE=0, the maximum and minimum currents are

-1.25ms

5ms

-5ms

5ms

1 340V 111.90A

101 1

Tt

s

T

V e eI

Re e

= = =

14

1

1 340V 15.62A

10 11

Tt

s

T

V e eI

R ee

= = =

The peak-to-peak ripple in the output current is therefore

=11.90A - 5.62A = 6.28A

p p I I I

=

Alternatively the ripple can be extracted from figure 13.4 using T/=1 and = .

iv. From equations (13.11) and (13.12), withE= 0, the time domain load currentequations are

( ) 5 5ms

5ms

1

34 1 5.62

34 28.38 (A) for 0 1.25ms

t t

s

o

t t

ms

o

t

Vi e I e

R

i t e e

e t

= +

= +

=

( )

5ms11.90 (A) for 0 3.75ms

t

o

t

o

i I e

i t e t

=

=

v. The average load current from equation (13.17), withE= 0, is

85V= = 8.5A10

oo

VIR

=

The average switch current, which is the average supply current, is

( )

( )( )

340V - 0 5ms- 11.90A - 5.62A = 2.22A

10 5ms

s

i switch

V EI I I I

R T

= =

=

The average diode current is the difference between the average load current and

the average input current, that is


= 8.50A - 2.22A = 6.28A

diode o i I I I =

vi. The input power is the dc supply multiplied by the average input current, that is

=340V2.22A = 754.8W

754.8W

iin s

out in

P V I

P P

=

= =

From equation (13.18) the rms load current is given by

754.8W= = 8.7A rms

10

rms

out

o

PI

R=

vii. The chopper effective input impedance is

340V= = 153.2

2.22A

s

in

i

VZ

I=

Load back emf, E= 55Vi. and ii. The average output voltage, rms output voltage, ac ripple voltage, and

ripple factor are independent of back emf, provided the load current is continuous.

The earlier answers forE= 0 are applicable.

iii. From equation (13.13), the maximum and minimum load currents are-1.25ms

5ms

-5ms

5ms

1 340V 1 55V- = 6.40A

10 101 1

Tt

s

T

V e E eI

R Re e

= =

1

1 340V 1 55V0.12A

10 1 101

Tt

s

T

V e E eI

R R ee

= = =

14

The peak-to-peak ripple in the output current is therefore

= 6.4A - 0.12A = 6.28A

pp I I I

=

The ripple value is the same as the E= 0 case, which is as expected since ripplecurrent is independent of back emf with continuous output current.Alternatively the ripple can be extracted from figure 13.4 using T/=1 and =.

iv. The time domain load current is defined by


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DC choppers 364

( ) 5 5ms

5ms

1

28.5 1 0.12

28.5 28.38 (A) for 0 1.25ms

t t

s

o

t t

ms

o

t

V Ei e I e

R

i t e e

e t

= +

= +

=

( ) 5ms 5

5ms

1

5.5 1 6.4

5.5 11.9 (A) for 0 3.75ms

t t

o

t t

ms

o

t

Ei e I eR

i t e e

e t

= +

= +

= +

v. The average load current from equation (13.37) is

85V-55V= = 3A10

oo

V EIR

=

The average switch current is the average supply current,

( )

( )( )

340V - 55V 5ms- 6.40A - 0.12A = 0.845A

10 5ms

s

i switch

V EI I I I

R T

= =

=

The average diode current is the difference between the average load current and

the average input current, that is

= 3A - 0.845A = 2.155A

diode o i I I I =

vi. The input power is the dc supply multiplied by the average input current, that is

=340V0.845A = 287.3W

287.3Wiin s

out in

P V I

P P

=

= =

From equation (13.18) the rms load current is given by

287.3W - 55V3A= = 3.5A rms

10

rms

oout

o

P E II

R

=

vii. The chopper effective input impedance is


E= 0 E= 55V

I

Conducting device

T1 D2 T1 D2 T1 D2 T1 D2 T1 D2 T1 D2

I

tT

t

io

Vs 340V

o 1ms 5ms

t

vo

tT

t

io

Vs 340V

o 1ms 5ms

t

vo

I

I

E E=55V

TtT

oI

oI

11.9A

8.5A

5.62A6.4A

0.12A3A

oV oV85V 85V

io=6.28A

io=6.28A

340V= = 402.4

0.845A

s

in

i

VZ

I=

The electromagnetic efficiency is given by equation (13.22), that is

55V3A= 57.4%

287.3W

o

in

E I

P

=

=

viii. The output voltage and current waveforms for the first-quadrant chopper, with

and without back emf, are shown in the figure to follow.

Figure Example 13.1. Circuit waveforms.

Example 13.2: DC chopper with load back emf

- verge of discontinuous conduction

A first-quadrant dc-to-dc chopper feeds an inductive load of 10 ohms resistance,

50mH inductance, and back emf of 55V dc, from a 340V dc source. If the chopper

is operated at 200Hz with a 25% on-state duty cycle, determine:

i. the maximum back emf before discontinuous load current conductioncommences with =;

ii. with 55V back emf, what is the minimum duty cycle before discontinuousload current conduction; and

iii. minimum switching frequency at E=55V and tT = 1.25ms before dis-continuous conduction.


10/32


11/32

DC choppers 368

iv. the current in the time domain;v. the load average current, average switch current and average diode

current;

vi. the input power, hence output power and rms output current;vii. effective input impedance, and electromagnetic efficiency; and

viii. Sketch the circuit, load, and output voltage and current waveforms.

SolutionThe main circuit and operating parameters are

on-state duty cycle = period T= 1/fs = 1/200Hz = 5ms on-period of the switch tT= 1.25ms load time constant = L/R = 0.05H/10 = 5ms

Figure Example 13.3.

Circuit diagram.

Confirmation of discontinuous load current can be obtained by evaluating the

minimum current given by equation (13.13), that is

1

1

Tt

s

T

V e EI

R Re

=

1.25ms

5ms

5ms

5ms

340V e -1 100V= - = 5.62A - 10A = - 4.38A

10 10e -1

I

The minimum practical current is zero, so clearly discontinuous current periods

exist in the load current. The equations applicable to discontinuous load current

need to be employed.

The current extinction time is given by equation (13.24), that is

-1.25ms

5ms

1 1

340V - 100V= 1.25ms + 5ms 1 + 1 - e

100V

= 1.25ms + 2.13ms = 3.38ms

Tt

s

x T

V Et t n e

E

n

= + +

i. From equations (13.26) and (13.27) the load average and rms voltages are

R L E

T1

10 50mH

+

D2 100V

340V

Vs

=T=5ms


5ms - 3.38ms= 340V + 100V = 117.4V

5ms

x

o s

T tV V E

T

= +

2 2

2 25ms - 3.38ms= 340 + 100 = 179.3V rms

5ms

x

rms s

T tV V E

T

= +

ii. From equations (13.28) and (13.29) the rms ripple voltage, hence ripple factor,

are2 2

2 2= 179.3 - 117.4 = 135.5V ac

r rms oV V V=

135.5V= = 1.15

117.4V

r

o

VRF

V=

iii. From equation (13.36), the maximum and minimum output current, hence the

peak-to-peak output ripple in the current, are

-1.25ms

5ms

1

340V-100V= 1 - e = 5.31A

10

Tt

sV E

I eR

=

The minimum current is zero so the peak-to-peak ripple current isoi =5.31A.

iv. From equations (13.32) and (13.33), the current in the time domain is

( )

5ms

5ms

1

340V - 100V

110

24 1 (A) for 0 1.25ms

t

s

o

t

t

V Ei t e

R

e

e t

=

=

=

( )

5ms 5ms

5ms

1

100V1 5.31

10

15.31 10 (A) for 0 2.13ms

t t

o

t t

t

Ei t e I e

R

e e

e t

= +

= +

=

( ) 0 for 3.38ms 5msoi t t=


12/32

DC choppers 370

Figure Example 13.3. Circuit waveforms.

v. From equations (13.37) to (13.40), the load average current, average switch

current, and average diode current are

117.4V - 100V= = 1.74A10

oo

V EIR

=

3.38ms100V - 0.25

5ms 5ms5.31A - = 1.05A

5ms 10

x

diode

tE

TI I

T R

=

=

Conducting deviceT D T D T

vo

io

iD

iT

vT

0 1.25 3.37 5 6.25 8.37 10 11.25

time (ms) t

5.31A

Vs=340V

E=100V

240V

Vs=340V

5.31A

5.31A

117.4V

E=100VoV

oI

1.05ADI

0.69ATI

1.74A


=1.74A - 1.05A = 0.69A

i o diode I I I =

vi. From equation (13.38), the input power, hence output power and rms output

current are

2

340V0.69A = 234.6Wiin s

oin out o rms

P V I

P P I R E I

= =

= = +

Rearranging gives

/

= 234.6W - 100V0.69A / 10 = 1.29A

rmsoo in

I P RE I=

vii. From equations (13.42) and (13.43), the effective input impedance, and

electromagnetic efficiency forE> 0

340V= 493

0.69A

s

in

i

VZ

I

= =

100V1.74A= = 74.2%

340V0.69A

oo

in is

E I E I

P V I = =

viii. The circuit, load, and output voltage and current waveforms are plotted in

figure example 13.3.

13.3 Second-Quadrant dc chopper

The second-quadrant dc-to-dc chopper shown in figure 13.2b transfers energy from

the load, back to the dc energy source, called regeneration. Its operating principlesare the same as those for the boost switch mode power supply analysed in chapter

15.4. The two energy transfer modes are shown in figure 13.6. Energy is

transferred from the back emfEto the supply Vs, by varying the switch T2 on-stateduty cycle. Two modes of transfer can occur, as with the first-quadrant chopper

already considered. The current in the load inductor can be either continuous or

discontinuous, depending on the specific circuit parameters and operating

conditions.

In this analysis it is assumed that:

No source impedance; Constant switch duty cycle; Steady-state conditions have been reached; Ideal semiconductors; and No load impedance temperature effects.


13/32

DC choppers 372

13.3.1 Continuous inductor current

Load waveforms for continuous load current conduction are shown in figure 13.7a.

The output voltage vo, load voltage, or switch voltage, is defined by

( )0 for 0

for

T

o

s T

t tv t

V t t T

=

(13.44)

The mean load voltage is

( )

( )

0

1 1

1

T

T T

o o st

T

s s

V v t dt V dt T T

T tV V

T

= =

= =

(13.45)

where the switch on-state duty cycle = tT/Tis defined in figure 13.7a.

Figure 13.6. Stages of operation for the second-quadrant chopper:

(a) switch-on, boosting current and (b) switch-off, energy into Vs.

Alternatively the voltage across the dc source Vs is

1

1os

V V

=

(13.46)

Since 0 1, the step-up voltage ratio, to regenerate into Vs, is continuouslyadjustable from unity to infinity.

The average output current is

( )1o so

E VE VI

R R

= = (13.47)

The average output current can also be found by integration of the time domain

output current io. By solving the appropriate time domain differential equations, thecontinuous load current io shown in figure 13.7a is defined byDuring the switch on-period, when vo=0

o

o

di L Ri E

dt+ =

which yields

(a) (b)

R L

R L

Vs

T2

D1 ioff

ion

+ +

E Evo

io io

Vo

Io

II


( ) 1 for 0t t

o T

Ei t e I e t t

R

= +

(13.48)

During the switch off-period, when vo=Vs

o

o s

di L Ri V E

dt+ + =

which, after shifting the zero time reference to tT, gives

( ) 1 for 0t t

so TE Vi t e I e t T t

R

= +

(13.49)

where (A)

1

1and (A)

1

T

T

t T

s

T

T t

s

T

E V e eI

R Re

E V eI

R Re

+

=

=

(13.50)

Figure 13.7. Second-quadrant chopper output modes of current operation:

(a) continuous inductor current and (b) discontinuous inductor current.

The output ripple current, for continuous conduction, is independent of the back

emfEand is given by

(1 ) ( )

1

T TT t T t

s

Tp p

V e e e I I I

Re

+

+ += =

(13.51)

which in terms of the on-state duty cycle, =tT/T, becomes

t

t

TtT

io

Vs

vo

E E

I

t

t

T

I

vo

io

tT

I

I

Vs

I

I

I

tx

(a) (b)

Conducting devices

T2 D1 T2 D1 T2 D1 T2 D1 T2 D1

oVoV

oI

oI


14/32


15/32


16/32


17/32


18/32

DC choppers 382

on

T1

vo

Vs

R L + E

D2

off

D1

vo

Vs

R L + E

T2

on

off

I

Vo

Io

Vo

Io

II

T1 D1

vo

Vs

R L + E

QI io

QII io

T2D2

(a)

is

Vo

Io

III

(b) (c)

Vs

I

t

vo

io

is

t

t

Vs

E

I

I

I

I

oV

oV

oI

sIsI

E

I

I

t

t

t

o

o

o

o

o

o

Conducting devicesD1 T1 D2 T2 D1 T1 D1 T2 D 1 T 2

oI

I

tT T

txTtxD

tT T

Figure 13.8. Two-quadrant (I and II) dc chopper circuit where vo>0:

(a) basic two-quadrant dc chopper; (b) operation and waveforms for quadrant I;

and (c) operation and waveforms for quadrant II, regeneration into Vs.


Thus

0

1 TtT

o s s s

tV V dt V V

T T= = = (13.79)

The rms output voltage is also determined solely by the duty cycle,

2

0

1 Ttrms s

s

V V dt T

V

=

=

(13.80)

The output ac ripple voltage, hence ripple factor are given by equations (13.3) and

(13.5), and are independent of the load:

( )2 2 1r rms o sV V V V = = (13.81)

and

11r

o

VRF

V = = (13.82)

The Fourier series for the load voltage can be used to determine the load current at

each harmonic frequency as described by equations (13.6) to (13.10).

The time domain differential equations from section 13.2.1 are also valid, where

there is no zero restriction on the minimum load current value.

In a positive voltage loop, when vo(t)=Vs and Vs is impressed across the load, theload circuit condition is described by

( ) 1 for 0t t

s

o T

V Ei t e I e t t

R

= +

(13.83)

During the switch off-period, when vo=0, forming a zero voltage loop

( ) 1 for 0t t

o T

Ei t e I e t T t

R

= +

(13.84)

where

1where (A)

1

1and (A)1

T

T

t

s

T

t

sT

V e EI

R Re

V e EIR R

e

=

=

(13.85)

The peak-to-peak ripple current is independent ofE,( )1

(1 ) (1 )

1

TT

s

Tp p

V e eI

Re

=

(13.86)

The average output current,oI , may be positive or negative and is given by

( )( )

( )

0

1

(A)

To

o o

s

V Ei t dt I RT

V ER

= =

=

(13.87)

The direction of the net power flow between E and Vs determines the chopper


19/32


20/32

DC choppers 386

( )

( )

2 2

2 2

1

= 170 - 85 = 340V 1 - = 147.2V

r rms o sV V V V = =

1 11= - 1 = 1.732

r

o

VRF

V = =

iii. From equations (13.85) and (13.86), the maximum and minimum outputcurrent, hence the peak-to-peak output ripple in the current are given by

-1.25ms

5ms

-5ms

5ms

1.25ms

5ms

5ms

5ms

1 340V 1 - e 100V= - = 1.90A

10 101 1 - e

1 340V e - 1 100V= - = - 4.38A

10 101 e - 1

T

T

t

s

T

t

s

T

V e EI

R Re

V e EI

R Re

=

=

The peak-to-peak ripple current is thereforeo

i = 1.90A - - 4.38A = 6.28A p-p.

iv. The current in the time domain is given by equations (13.83) and (13.84)

( )

- -

5ms 5ms

- -

5 5ms

-

5ms

1

340V-100V= 1- - 4.38

10

= 24 1- - 4.38

= 24 - 28.38 for 0 1.25ms

t t

s

o

t t

t t

ms

t

V Ei t e I e

R

e e

e e

e t

= +

( )

5ms 5ms

5ms 5ms

5ms

1

100

1 1.9010

10 1 1.90

10 11.90 for 0 3.75ms

t t

o

t t

t t

t

Ei t e I e

R

V

e e

e e

e t

= +

= +

= +

= +

v. Since the maximum current is greater than zero (1.9A) and the minimum is less

that zero (- 4.38A), the current crosses zero during the switch on-time and off-time.

The time domain equations for the load current are solved for zero to give the cross

over times txTand txD, as given by equation (13.89), or solved from the time domainoutput current equations as follows.

During the switch on-time


( )-

5ms24 - 28.38 0 where 0 1.25ms

28.38= 5ms = 0.838ms

24

t

o xT

xT

i t e t t

t n

= = =

During the switch off-time

( ) 5ms10 11.90 0 where 0 3.75ms

11.90=5ms = 0.870ms10

(1.250ms + 0.870ms = 2.12ms with respect to switch turn-on)

t

o xD

xD

i t e t t

t n

= + = =

vi. The load average current, average switch current, and average diode current for

all devices;

( ) ( )

( )85V - 100V= -1.5A

10

os

o

V E V EI R R

= =

When the output current crosses zero current, the conducting device changes.

Table 13.1 gives the necessary current equations and integration bounds for the

condition 0, 0I I

> < . Table 13.1 shows that all four semiconductors are involvedin the output current cycle.

1

-1.25ms

5ms

0.838ms

11

124 - 28.38 0.081A

5ms

T

xT

t tts

Tt

t

V E I e I e dt

T R

e dt

= +

= =

10

-0.84ms

5ms

0

11

124 - 28.38 0.357A

5ms

xTt tt

sD

t

V E I e I e dt

T R

e dt

= +

= =

-

2

3.75ms5ms

0870ms

11

110 11.90 1.382A

5ms

T

xD

t tT t

Tt

t

E I e I e dt

T R

e dt

= +

= + =

20

0.8705ms

0

11

110 11.90 0.160A

5ms

xDt tt

D

tms

E I e I e dt

T R

e dt

= +

= + =

Check 1 1 2 2 - 1.5A + 0.080A - 0.357A - 1.382A + 0.160A = 0o T D T D I I I I I + + + + =


21/32


22/32


23/32


24/32


25/32


26/32


27/32


28/32


29/32


30/32


31/32

DC choppers 410 Power Electronics411


32/32

1 2

= 340V 1 - 2 = 240V rms

rms sV V=

From equation (13.119), the output ac ripple voltage is

( )

( )

2 1 2

= 2 340V 1 - 2 = 170V ac

r sV V =

iii. The average output current is given by equation (13.117)

( )

( )

2 1

340V 2 - 1 - 55V= = -22.5A

10

so

o

V EV EI

R R

= =

Since both the average output current and voltage are negative (-170V and -22.5A)

the chopper with a modulation depth of= , is operating in the third quadrant.

iv. The electromagnetic power developed by the back emfEis given by

( )55V -22.5A = -1237.5WoE P EI = =

v. The average output current is given by

( ) ( )( )2 1o so

V E V EI

R R

= =

when the mean current is -11.25A, = 0.415, as derived in part vi.

vi. Then, if the average current is halved to -11.25A

=55V - 11.25A10 = -57.5V

ooV E RI= +

The average output voltage rearranged in terms of the modulation depth gives

1

-57.5V= 1 + = 0.415

340V

o

s

V

V

= +

The switch on-time when < is given by

( )2 = 20.415 5ms = 2.07msTt T=

From figure 13.14b both T1 and T4 are turned on for 2.07ms, although, from table

13.3B, for negative load current, oI = -11.25A, the parallel connected freewheeldiodes D1 and D4 conduct alternately, rather than the switches (assuming 0oI

< ).The switches T1 and T4 are turned on for 1.25ms, while T2 and T3 are subsequently

turned on for 2.93ms.

vii. The electromagnetic power developed by the back emfEis halved and is given

by

( )55V -11.25A = -618.75WoE P E I = =

Reading list

Dewan, S. B. and Straughen, A.,Power Semiconductor Circuits,

John Wiley and Sons, New York, 1975.

Dubey, G.K., Power Semiconductor Controlled Drives,Prentice-Hall International, 1989.

Mohan, N., Undeland, T. M., and Robbins, W.P., Power Electronics: Converters,

Applications and Design,

John Wiley and Sons, New York, 2003.

Problems

13.1. The dc GTO thyristor chopper shown in figure 13.lc operates at 1 kHz and

supplies a series 5 and 10 mH load from an 84 V dc battery source. Derive

general expressions for the mean load voltage and current, and the load rms voltage

at an on-time duty cycle of. Evaluate these parameters for = 0.25.

[21 V, 4.2 A; 42 V]

13.2. The dc chopper in figure 13.lc controls a load ofR = 10 ,L = 10 mH and

40 V battery. The supply is 340 V dc and the chopping frequency is 5 kHz.

Calculate (a) the peak-to-peak load ripple current, (b) the average load current, (c)

the rms load current, (d) the effective input resistance, and (e) the rms switch

current.

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