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.
ii.
Find the peak-to-peak value (Ipp) of the load current io.
Find the amplitude of the fundamental component (Io1) of io.
+
vo
+
Vs= 200 V
va0
-
load
vb0
Solution
(a) With a purely resistive load
vo
,
R
vo2
R
T
T
 V2
1
11
  po (t ) dt    vo2 (t ) dt   o ,rms
T 0
RT 0
R

po (t )  vo (t )io (t ) 
Instantaneous power:
Po ,av
io 
Vo ,rms  Vs


Vs2 

 2000 W   
R 
2
(b)
(i)
vo
io
Ip
L
π
β
ωt
dio
 Vs
dt
(ii) I o1 
Vo1
L
Vo1 
4Vs

sin   / 2   I o1 
Vs
t
L
Vs 
L
Vs 2
200
 2I p  

 33.33 A
 L 3 200  0.02
t   /  io  I p
 I pp
-Ip
0  t   /   io (t )   I p 
 I p  I p 
800
sin( / 3)  17.55 A
200 2  0.02
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-topeak value Ipp.
+
vo
+
Vs= 200 V
va0
vb0
load
Solution
(a)
vo
2
V

THD   o,rms   1
V

 o1,rms 
Vs
π
β
ωt
Vo1,rms 
-Vs
2 2Vs

Vo,rms  Vs
sin(  / 2)  0.7797Vs

2
 Vs
 0.8165Vs

3
 THD  31.084 %
(b)
L
vo
io
Ip
-Ip
0  t   /   io (t )   I p 
ωt
 I pp
Vs
t
L
Vs 
L
Vs 2
200
 2I p  

 26.66 A
 L 3 100  0.05
t   /  io  I p
π
β
dio
 Vs
dt
 I p  I p 
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).
+
+
Vs= 200 V
io
va0
vo
vb0
load
Solution
(a)
vo
Vs
Vo,rms  Vs
π
β

1
 Vs
 0.7071Vs  141.42 V

2
ωt
-Vs
(b) I on ,rms 
Von 
Von ,rms
Z n  R  jn L
Zn
4Vs
sin( n / 3)
n
 n 1
n3
Vo1,rms 
is the impedance of the RL load at the frequency of the harmonic.
Z n  R 2  ( n L) 2
2 2Vs

 L  100  102  3.1416 
sin( / 3)  155.94 V,
Z1  52  (3.1416) 2  5.905 
Vo 3,rms  0
2 2Vs
sin(5 / 3)  31.19 V,
Z 5  52  (5  3.1416) 2  16.48 
5
2 2Vs
n7
Vo 7,rms 
sin(7 / 3)  22.28 V,
Z 7  52  (7  3.1416) 2  22.55 
7
 I o1,rms  26.408 A,
I o 3,rms  0 A,
I o5,rms  1.893 A,
I o 7,rms  0.988 A
n5
Vo 5,rms 
1
(c) THD 
1
I o1,rms
1

 2 (1.8932  0.9882 ) 2
2
I
 8.086 %
  on ,rms  
26.408
 n 3,5,7

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.
+
+
Vs= 200 V
va0
io
vo
vb0
load
Solution
vo
Vs
π
δ
p
Vo,rms  Vs
2π

ωt
 200
2
 163.3 V
3
-Vs
Ip
Ip 
ωt
Ip
1
L
 /
V
s
dt 
0
 I pp  2 I p 
Vs
L

2Vs
V
V
 s  s
L
 fs L
3 fs L
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).
+
+
Vs= 200 V
io
vo
va0
vb0
load
Solution
(a)
Vo1 
(b)
4Vs

sin(  / 2)  200 V

Waveforms of io and vo:
  2sin1 ( / 4)  103.52
vo
io
Vs
I2
I1
β
-I1
π+β
π
ωt
-I2
-Vs
(c)
Total average power:
T
Po,av 
1
2
vo (t ) io (t )dt  Vs
T 0
T
 /

  2 100  200 rad./sec
io (t )dt
0
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   :
io (0)   I1
t
At
io () 
 

 2.5 ms
 2
In the interval 0  t     
Vs
 20 A
R
io  I 2

io (t )  20    I1  20 e  t /
 I 2  20   I1  20 e2.5/2  14.27  0.2865 I1
where t   t   :
io (0)  I 2
io ()  0 A
 io (t )  I 2et/
  

 2.5 ms io  I1  I1  I 2e 2.5/2  0.2865 I 2
At t  

2
From (1) and (2) I1  3.78 A, I 2  13.19 A
io (t )  20  23.78et /

0  t   /
 /

0
io (t )dt  16.07  103 A.sec.


Po,av  642.8 W
(2)
L
 2 ms
R
(1)
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.
+
a
Vs
b
c
+
ia
van
n
Solution
1
van  (2va 0  vb0  vc 0 )
3
(a)
van
2Vs / 3
 1
Van ,rms  
 2
Vs / 3
π/3
2π/3
π
2π
ωt

2
0 van d 
1/2
 1  1 2   2  2   1  2  
   Vs 
  Vs 
  Vs  
   3  3  3  3  3  3 
-Vs / 3
-2Vs / 3
(b)
1/2
2

2
Vs
3
van
ia
Ip
I2
π/3
2π/3
π
2π
ωt
-I2
-Ip

t
ia (t2 )  ia (t1 ) 
1  0, 2 
1 

3
, 2 
1 2
van (t ) dt
L t1

3
,
1
van  Vs
3
2
2
, van  Vs
3
3
ia (2 )  ia (1 ) 
or in terms of angle
  I2  I p 
 I2   I2 
1
L
 /3

0
1
L
2
Vs d
3
 /3
1
 3V
s
d
0
 I2 
1 2
v ( ) d
 L 1 an
 I2  I p 
Vs
9 L
Vs
4Vs
2Vs
 I pp  2 I p 

9 L
9(2 f s ) L 9 f s L
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).
+
Vs = 200 V
a
b
c
-
-
ia
vA
+
R
iA
R
iC R
Solution
ia  iA  iC 
(a)
v A  vC
R
v A  va 0  vb0
vC  vc 0  va 0
 ia 
2va 0  vb0  vc 0
R
va0
Vs
vb0
π/3 2π/3
π
Vs
vc0
Vs
ia
I2
I1 
I1
π/3
2π/3
π
2π
ωt
I1
Vs
R
I2 
2Vs
R
1/2
 
 1  2
I a ,rms    I12
 I 22 
3
3 
 
 2
Vs
 28.28 A
R
I2
(b) Instantaneous power in phase A:
vA  va 0  vb0

p A (t ) 
v A2
R
Average power in phase A: PA 
2
T
1  1 2  VA,rms
v
dt



A
R  T 0
R

vA
Vs
π
2π/3
2π
ωt
Vs
 2 / 3  2 2
VA2,rms  Vs2 
  Vs
   3
Total three-phase average power:
Pav  3PA 
2Vs2
 8 kW
R
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.
+
a
Vs=200 V
b
c
+
ia
van
n
Solution
(a)
As in Q.6
(b)
I a ,1 
Van ,1
Z1
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:
1
2
1
1
van  (2va 0  vb0  vc 0 )  van ,1  va 0,1  vb 0,1  vc 0,1
3
3
3
3
2Vs
2Vs
2V
va 0,1 
sin( t )
vb 0,1 
sin( t  120 )
vc 0,1  s sin( t  240 )



2V  2
1
1
 van ,1  s  sin( t )  sin( t  120 )  sin( t  240
 3
3
3
Z1  R  j L 

I a ,1  7.93 A
Z1  102  (400  102 ) 2  16.06 
 2V
)  s sin( t )
 
 Van ,1 
2Vs

 127.32 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.
io= Ia
+
iD
+
R
vs
vo
(240 V rms)
L (large)
_
_
Solution
(a)

Average load voltage:
Vdc 
V
1
240 2
Vm sin( ) d  m 1  cos   
1  cos 90   54.02 V

2 
2
2
(b) The diode current:
iD
Ia
α
Vdc 
I D ,av
π
ωt
V
240 2
(1  cos 45 )  92.21 V  I a  dc  9.221 A
2
R
2
1






iD .d  
 I a  0.625  9.221  5.763 A
2 0
 2 
1
I D ,rms
2π 2π+α
1
 1 
2
    2
   iD2 d   I a 
  0.79065  9.221  7.29 A
 2 
 2 

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.
iT
+
io= Ia
+
vT _
+
R=20 
L (large)
R
vs
vo
(240 V rms)
L
_
_
Solution
(a)
400
400
voltage (V)
300
300
vo
200
200
100
100
0
0
-100
-100
-200
-200
-300
-300
-400
0
0.005
0.01
0.015
0.02
0.025
0.03
0.035
0.04
-400
20*iT
0
0.005
time (s)
Vdc
R
Pav  Vdc I a
(c)
PTh  VT .I T ,av where I T ,av is the average thyristor current.
 
Ia
2
 I T ,av
 Pav 
0.015
0.02
time (s)
Vdc2
V2
 max
(1  cos  )2
R 4 2 R
(b)
I T ,av 
Ia 
0.01
Vdc 240 2

(1  cos90 )  2.7 A
R 2  20
 0.25  I a  0.675 A
 PTh  1.2  0.675  0.81 W
Ia 
0.025
0.03
0.035
0.04
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
+
T3
T1
vo
R=10 Ω
+
vs
L (very large)
(240 V rms)
-
T2
T4
Solution
(a) The average power supplied to the load
Vdc2
2V
2  240  2
Vdc  max cos  
cos   216.08cos
R


216.082
 1200 
cos2 

cos  0.507
   59.54
10
Pav  Vdc . I dc 
(b) For  = π/6 the source current is a square-wave as shown
400
is
300
200
100
0
-100
-200
-300
-400
0
0.005
0.01
0.015
0.02
The input power factor is given by
PF =Disp.F.×D.F
Disp.F .  cos 
D .F . 
I s1
Is
I s1 : rms value of fundamental current
I s : rms value of total current
I s1 
2 2

Ia
Is  Ia

P.F . 
2 2

cos   0.7797
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-topeak ripple in the load current io.
di
(Hint: Integrate equation for the inductor: L o  vL ; i0 is not to be taken as a constant now)
dt
io
+
T3
T1
+
R
-
L
vo
vs
D2
D4
-
Solution
(a)
Vdc
V
 15 A  Vdc  m 1  cos    150 V
R

150
 cos  
 1  0.3884    67.14
240 2
Ia 
(b)
For α = 60
Vdc 
240 2

1  cos60   162.06 V
The voltage across the load inductor:
200
vL (t )  L
150
100
dio
 Vm sin t  Vdc
dt

50
io ( t )  io ( ) 
0
-50

  t  
t
1
v ( ) d
 L  L
(1)
-100
-150
-200
0
0.005
0.01
0.015
0.02
In the interval   t   , io increases as vL  0. In   t     , io decreases as vL  0.
Therefore,
io. max  io (  ) and io. min  io ( ).
Vdc 162.06

 0.4775    151.48
Vm 240 2
To find  , vL (  )  0  Vm sin   Vdc
 sin  
In (1) let t  
1
1
Vm sin   Vdc  d  Vm (cos   cos  )  Vdc (    ) 

L 
L

 io (  )  io ( ) 
1 
 151.48     
240 2  cos(60 )  cos(151.48   162.06 
 
62.83 
3 
 180
 3.33 A
 io 
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
V
Vdc  m (1  cos  )

(b) Sketch the source current is for  = 90 and find
(i) the fundamental amplitude of is.
(ii) the rms value of is.
io
+
T3
T1
is
+
R
-
L
vs
-
D2
D4
vo
Solution
a.
Average load voltage:
Vdc 
b.
1


V

m
sin( ).d 
Vm

1  cos  
is
Ia
2π
π/2
ωt
π
-Ia
I1 
(i) Fundamental amplitude of is:
Vdc 
Vm

1  cos   
(ii) Rms of is : I s ,rms
 1

 2
240 2

(1  cos90 )  108.04 V
1
2
 
sin    0.9 I a

4
4Ia
1
 Ia 
1
Vdc
 10.8 A  I1  9.72 A
R
2  1  2
2
    2
0 i .d      I a .d   I a     7.64 A
2
s