EENG441 SOLVED PROBLEMS

(INVERTERS, AC-DC CONVERTERS)

1. The single-phase full-bridge inverter shown below is operated in the quasi-square-wave mode at the

frequency f = 100 Hz with a phase-shift of β between the half-bridge outputs vao and vbo.

(a) With a purely resistive load R = 10 Ω, find β so that the average power supplied to the load is

Po,av = 2 kW.

(b) With a purely inductive load L = 20 mH and β = 2π/3,

i. Find the peak-to-peak value (Ipp) of the load current io.

ii. Find the amplitude of the fundamental component (Io1) of io.

Solution

(a) With a purely resistive load oo

vi

R ,

Instantaneous power: 2

( ) ( ) ( ) oo o o

vp t v t i t

R

2

,2

, ,

0 0

2

1 1 1( ) ( )

202

00 W

T T

o rms

o av o o o rms s

s

VP p t dt v t dt V V

T R T R

V

R

(b)

(i)

0 / ( )

/

2 2002

3 200 0.033.33 A

2

o ss o p

so p p p

spp p

di VL V t i t I t

dt L

Vt i I I I

L

VI I

L

(ii) 11 1 1 2

4 800sin / 2 sin( / 3)

200 0.0217.55 Ao s

o o o

V VI V I

L

+va0

+

Vs= 200 V

-vb0

vo

load

β

π

vo

ioIp

-Ip

ωt

2. The single-phase full-bridge inverter shown below is operated in the quasi-square-wave mode at

the frequency f = 50 Hz with a phase-shift of β = 2π/3 between the half-bridge outputs vao and vbo.

(a) Sketch the load voltage vo and find its total harmonic distortion (THD).

(b) With a purely inductive load L= 50 mH, sketch the load current io and find its peak-to-

peak value Ipp.

Solution

(a)

2

,

,

1,

1,

21 0.8165

3

2 2sin( / 2) 0.7797 31.084 %

o rms

o rms s s s

o rms

so rms s

VTHD V V V V

V

VV V THD

(b)

0 / ( )

/

2 2002 26.66 A

3 100 0.05

o ss o p

so p p p

spp p

di VL V t i t I t

dt L

Vt i I I I

L

VI I

L

+va0

+

Vs= 200 V

-vb0

vo

load

Vs

-Vs

β

π

ωt

vo

β

π

ωt

vo

ioIp

-Ip

3. The single-phase full-bridge inverter shown below is operated in the quasi-square-wave (QSW ) mode (phase

displacement control ) at the frequency f = 50 Hz , with phase shift between half-bridge output voltages

vao and vbo. The load is an R-L load with R = 5 and L = 10 mH.

(a) For = / 2 sketch vo and find its rms value.

(b) For = 2 / 3 find the rms values of load current harmonic components with orders n = 1, 3, 5, 7.

(i.e. find Ion,rms , n=1,3,5,7)

(c) Find approximately the total harmonic distortion (THD) of io using the results in part (b).

Solution

(a)

(b) ,

, is the impedance of the RL load at the frequency of the harmonic.on rms

on rms n

n

VI Z R jn L

Z

2 2 2

2 2

1, 1

3,

5,

4sin( / 3) ( ) 100 10 3.1416

2 21 sin( / 3) 155.94 V, 5 (3.1416) 5.905

3 0

2 2 5 sin(5 / 3) 31.19 V,

5

son n

so rms

o rms

so rms

VV n Z R n L L

n

Vn V Z

n V

Vn V

2 2

5

2 2

7, 7

1, 3, 5, 7,

5 (5 3.1416) 16.48

2 2 7 sin(7 / 3) 22.28 V, 5 (7 3.1416) 22.55

7

26.408 A, 0 A, 1.893 A,

so rms

o rms o rms o rms o rm

Z

Vn V Z

I I I I

0.988 As

(c)

11

2 2 222

,

3,5,71,

1 (1.893 0.988 )8.086 %

26.408on rms

no rms

THD II

+va0

+

Vs= 200 V

-vb0

vo

load

io

Vs

-Vs

β

π

ωt

vo

,

10.7071 141.42 V

2o rms s s sV V V V

4. The single-phase full-bridge inverter shown below is operated in the multiple pulse-width modulation mode

with two pulses (p=2) per half-cycle with pulse-width = π/3. The frequency of operation is fs.

(a) Sketch the output voltage (vo) waveform and find its rms value.

(b) With a purely inductive load L, sketch the load current (io) waveform and find its peak-to- peak value.

Solution

/

0

1

22

3

sp s

s s spp p

s s

VI V dt

L L

V V VI I

L f L f L

+va0

+

Vs= 200 V

-vb0

vo

load

io

Vs

-Vs

π

ωt

vo

2π

δ

Ip

I p

ωt

,

2200 163.3 V

3o rms s

pV V

5. The single-phase full-bridge inverter shown below is operated in the quasi-square-wave (QSW) mode (phase

displacement control ) at the frequency f = 100 Hz , with phase shift between half-bridge output voltages

va and vb . The load is an R-L load with R=10 and L = 20 mH.

(a) Find so that the fundamental amplitude of the load voltage vo is 200 V.

(b) For = /2 sketch vo and io (approximately).

(c) For = /2 find the total average power supplied to the load (do not find the fundamental average

power Po1).

Solution

(a) 1

1

4sin( / 2) 200 V 2sin ( / 4) 103.52s

o

VV

(b) Waveforms of io and vo:

(c) Total average power:

/

,

0 0

1 2( ) ( ) ( ) 2 100 200 rad./sec

T

o av o o s oP v t i t dt V i t dtT T

Note that the instantaneous power is zero in the interval t , since the voltage is zero. Also, the

instantaneous power is the same in the interval t . In order to be able to evaluate the above

integral, the load current must be solved as a function of time:

In the interval 0 t :

/

1 1(0) ( ) 20 A ( ) 20 20 2 mstso o o

V Li I i i t I e

R R

At 2.5/2

2 2 1 12.5 ms 20 20 14.27 0.28652

ot i I I I e I

(1)

In the interval 0 where t t t :

/

2 2(0) ( ) 0 A ( ) t

o o oi I i i t I e

At 2.5/2

1 1 2 22.5 ms 0.28652

ot i I I I e I

(2)

From (1) and (2) 1 23.78 A, 13.19 AI I

/ ( ) 20 23.78 0 /t

oi t e t

/

3

,

0

( ) 16.07 10 A.sec. 642.8 Wo o avi t dt P

+va0

+

Vs= 200 V

-vb0

vo

load

io

ωtβ

vo

Vs

-Vs

π

io

I1

-I2

I2

-I1

π+β

6. The three-phase half-bridge inverter shown below feeds a balanced Y-connected purely inductive load

having inductance L per phase, and is operated in the square-wave mode (pole voltages are square waves) at

the frequency f s.

(a) Sketch phase-to-neutral voltage van (show all important voltage and time values) and find its rms

value Van,rms.

(b) Find the peak-to-peak value (Ipp) of the line current ia in terms of L and fs.

Solution

(a) 0 0 0

1(2 )

3an a b cv v v v

1/22

2

,

0

1/22 2 2

1

2

1 1 2 1 2

3 3 3 3 3 3 3

an rms an

s s s s

V v d

V V V V

(b)

2

1

2 1

1( ) ( ) ( )

t

a a an

t

i t i t v t dtL

or in terms of angle 2

1

2 1

1( ) ( ) ( )a a ani i v d

L

/3

1 2 2 2

0

1 1 10, ,

3 3 3 9

san s p s p

Vv V I I V d I I

L L

/3

1 2 2 2 2

0

42 2 1 2, , 2

3 3

2

3 3 99 9(2 )

s san s s pp p

s

s

s

V

f L

V Vv V I I V d I I I

L L f L

a b c

+

van

+

ia

n

Vs

ωt

π 2π

-Vs / 3

2Vs / 3

-2Vs / 3

Vs / 3

van

π/3 2π/3

ωt

ππ/3 2π/3 2π

van

ia

-Ip

Ip

I2

-I2

7. The three-phase half-bridge inverter shown below feeds a balanced -connected resistive load, and is

operated in the square-wave mode (pole voltages are square waves) at frequency f s.

(a) Sketch line current ia (show all important current and time values) and find its rms value.

(b) Find the total average power supplied to the load (take R= 10).

Solution

(a) 0 0 00 0 0 0

2A C a b ca A C A a b C c a a

v v v v vi i i v v v v v v i

R R

1 2

2s sV VI I

R R

1/2

2 2

, 1 2

1 22 28.28 A

3 3

sa rms

VI I I

R

(b) Instantaneous power in phase A: 2

( ) AA

vp t

R Average power in phase A:

2

,2

0

1 1T

A rms

A A

VP v dt

R T R

0 0A a bv v v

2 2 2

,

2 / 3 2

3A rms s sV V V

Total three-phase average power:

223 8 kWs

av A

VP P

R

+

-

+

-

R

ia

R

R

a b c

vA

Vs = 200 V

iA iC

π 2ππ/3 2π/3

ωt

Vs

Vs

Vs

va0

vb0

vc0

I2

I1

I 1

I 2

ia

π/3 2π/3 π

Vs

Vs

2π/3

π 2π

ωt

vA

8. The three-phase half-bridge inverter shown below feeds a balanced Y-connected resistive-inductive load with

R = 10 Ω and L = 10 mH, and is operated in the square-wave mode (pole voltages are square waves) at

frequency f s = 200 Hz .

(a) Sketch phase a voltage van (show all important voltage and time values) and find its rms value.

(b) Find the amplitude of the fundamental component of the line current ia.

Solution

(a) As in Q.6

(b) ,1

,1

1

an

a

VI

Z where, Van,1 is the amplitude of the fundamental component of van and Z1 is the

impedance of the load at the fundamental frequency.

Van,1 can be found from the equation of van in terms of the pole voltages:

0 0 0 ,1 0,1 0,1 0,1

0,1 0,1 0,1

,1 ,1

1 2 1 1(2 )

3 3 3 3

2 2 2sin( ) sin( 120 ) sin( 240 )

2 2 22 1 1sin( ) sin( 120 ) sin( 240 ) sin( ) 127.32 V

3 3 3

an a b c an a b c

s s sa b c

s s san an

v v v v v v v v

V V Vv t v t v t

V V Vv t t t t V

Z

2 2 2

1 1

,1

10 (400 10 ) 16.06

7.93 Aa

R j L Z

I

a b c

+

van

+

ia

n

Vs=200 V

9. In the single-phase half-wave rectifier shown below, assume that the load current io= Ia is constant (load

inductance L is very large).

(a) Sketch the load voltage waveform v0 for = 90 and find its average value Vdc.

(b) For R = 10 Ω find the average and rms values of the diode current iD for = 45.

vs

(240 V rms)

R

io= Ia

+

_L (large)

_

vo

+iD

Solution

(a)

Average load voltage: 1 240 2

sin( ) 1 cos 541 cos90 .02 V2 2 2

mdc m

VV V d

(b) The diode current:

2

,

0

1 12 2

2

,

5.763 A

7.29

240 2(1 cos45 ) 92.21 V 9.221 A

2

1. 0.625 9.221

2 2

10.79065 9.22

2A1

2

dcdc a

D av D a

D rms D a

VV I

R

I i d I

I i d I

Ia

π

iD

α 2π 2π+α ωt

10. In the single-phase half-wave rectifier shown below, assume that the load current Ia is constant

(load inductance L is very large).

(a) Sketch waveforms of the load voltage v0 and the thyristor current iT for = 60.

(b) Find the average power supplied to the load Pav as a function of the firing angle . (c) The thyristor has a constant voltage of VT=1.2 V across it when it conducts. Find the

average power dissipated by the thyristor for = 90.

+

_

vs

(240 V rms)

L

+ _vT

io= Ia

R=20 L (large)

iT

R

+

vo

_

Solution

(a)

(b) 2 2

2max

2(1 cos )

4

dc dcav dc a a av

V V VP V I I P

R R R

(c) ,.Th T T avP V I where ,T avI is the average thyristor current.

,

,

240 2(1 cos90 ) 2.7 A

2 2 20

0.25 0.675 A 1.2 0.675 0.81 W

dcT av a a

T av a Th

VI I I

R

I I P

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04-400

-300

-200

-100

0

100

200

300

400

time (s)

volt

age (

V)

vo

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04-400

-300

-200

-100

0

100

200

300

400

time (s)

20*iT

11. A single-phase fully-controlled bridge converter supplies power to a highly-inductive load with resistance

R = 10 from a 240 V rms 50 Hz AC source, as shown below.

(a) Find the firing angle so that the average power supplied to the load is Pav = 1200 W.

(b) For = π/6 find the input power factor of the rectifier.

io

+

-

vs

(240 V rms)

T1

T2

T3

T4

+ vo

R=10 Ω

L (very large)

Solution

(a) The average power supplied to the load

2

max

22

2 2 240 2. cos cos 216.08cos

216.081200 cos cos 0.507 59.54

10

dcav dc dc dc

V VP V I V

R

(b) For = π/6 the source current is a square-wave as shown

0 0.005 0.01 0.015 0.02-400

-300

-200

-100

0

100

200

300

400

is

The input power factor is given by

PF =Disp.F.×D.F

11

1

. . cos . . : rms value of fundamental current

: rms value of total current

2 2 2 2. . cos 0.7797

ss

s

s

s a s a

IDisp F D F I

I

I

I I I I P F

12. In the single-phase half-controlled bridge rectifier shown below, the source voltage vs is 240 V rms. The

load resistance is R = 10 Ω.

(a) If the load inductance L is very large (load current constant), find the firing angle α so that the load

current is io=15 A.

(b) Sketch the voltage (vL) across the inductor for α = 60 (L is very large).

(c) For L = 200 mH, assuming that the voltage across L is as in part (b), find approximately the peak-to-

peak ripple in the load current io.

(Hint: Integrate equation for the inductor: oL

diL v

dt ; i0 is not to be taken as a constant now)

io

+

vs

T1T3

-

D2D4

+vo

-

R

L

Solution

(a)

15 A 1 cos 150 V

150cos 1 0.3884

240 267.14

dc ma dc

V VI V

R

(b) For α = 60

240 21 cos60 162.06 V dcV

The voltage across the load inductor:

( ) sinoL m dc

div t L V t V t

dt

1

( ) ( ) ( ) (1)

t

o o Li t i v dL

.max .min

In the interval , increases as 0. In , decreases as 0.

Therefore, ( ) and ( ).

o L o L

o o o o

t i v t i v

i i i i

162.06To find , ( ) 0 sin sin 0.4775 151.48

240 2

1 1In let ( ) ( ) sin (cos cos ) ( )

1 151.48240 2 cos(60 ) cos(151.48 162.06

62.

(1

83 180 3

)

dcL m dc

m

o o m dc m dc

o

Vv V V

V

t i i V V d V VL L

i

3.33 A

0 0.005 0.01 0.015 0.02-200

-150

-100

-50

0

50

100

150

200

13. In the single-phase half-controlled rectifier shown below, the source voltage is 240 V rms. The load: R=

10Ω, L is very large. Assume that the load current i0=Ia is constant.

(a) Sketch the load voltage waveform (v0) (use = 90) and show that its average is given by

(1 cos )mdc

VV

(b) Sketch the source current is for = 90 and find

(i) the fundamental amplitude of is.

(ii) the rms value of is.

io

+

vs

T1T3

R

L-

D2D4

is+

-

vo

Solution

a.

Average load voltage:

1

sin( ). 1 cosmdc m

VV V d

b.

(i) Fundamental amplitude of is: 1

4sin 0.9

4

aa

II I

1

240 21 cos (1 cos90 ) 108.04 V 10.8 A 9.72 Am dc

dc a

V VV I I

R

(ii) Rms of is :

1 1 12 2 2 2

2 2

,

0

1 1. . 7.64 A

2s rms s a aI i d I d I

π/2 π

Ia

2π

is

-Ia

ωt