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EET 424 POWER ELECTRONIC FOR ENERGY SYSTEM 
Tutorial Solution: inverter 
Some of the equations for the solutions are not given, please refer to the textbook, “Power Electronics, Daniel W. 
Hart” 
 
1) The square wave inverter of figure below has Vdc=125V, an output frequency of 60Hz, 
and a resistive load of 12.5Ω. Sketch the curents in the load, each switch, and the 
source, and determine the average and rms values of each. 
 
 
 
 
 
 
Solution 
Load: 0, 10 .
Switches: 5 ., 10 0.5 7.07 .
Source : 10 . (See Example 2-4)
avg rms
avg rms m
avg rms
I I A
I A I I D A
I I A
 
   
 
 
 
EET 424 POWER ELECTRONIC FOR ENERGY SYSTEM 
2) A square-wave inverter has a dc source of 96V and an output frequency of 60Hz.the 
load is a series RL load with R=5Ω and L=100mH. When the load is first energized, a 
transient precedes the steady-state waveform by equation, 
 
 
 
a) Determine the peak value of the steady state current. 
b) Assuming zero initial current, determine the maximum current thet occurs during 
the transient. 
 
Solution 
 
 
/2
max /2
max
/
/
1/2.4
1 0.1 1 96
; 0.02 .; ; 19.2 .
1 5 60 5
0.341
19.2 3.94 .
1.66)
) . 8 1: ( )
(0) 0
( ) 1
( / 2) 19.2 1 6.54 .
T
dc dc
T
tdc
o
dc
o
tdc
o
V Ve L
I s T A
R e R R
I A
V
b From Eq i t Ae
R
V
i A
R
V
i t e
R
i T e A










 
       
 
 
  
 
  
   
 
  
 
 
 
 
 
 
 
 
 
 
 
 
t /dc dc
min
o
t T/2 /dc dc
max
V V T
I e for 0 t
R R 2
i (t)
V V T
I e for t T.
R R 2
 
 
  
     
 

 

       
  
EET 424 POWER ELECTRONIC FOR ENERGY SYSTEM 
3) The dc source supplying an inverter with a bipolar PWM output is 96V. the load is an RL 
series combination with R=32Ω and L=24mH. The output has a fundamental frequency 
of 60Hz. 
a. Specify the amplitude modulation to provide a 54V rms fundamental frequency 
output. 
b. If the frequency modulation ratio is 17, determine the total harmonic distortion 
of the load current. 
 
Solution 
 
 
1 1,
1
0
2 54 2 76.8 .
76.8
0.8
96
32 2 60 .024 32 9.05
rms
a
dc
n
V V V
V
m
V
Z R jn L jn jn 
  
  
     
 
From Table 8-3, 
 n Vn/Vdc Vn Zn In=Vn/Zn 
 1 0.8 76.8 33.3 2.30 
mf 17 0.82 78.7 157 0.50 
mf - 2 15 0.22 21.1 139 0.151 
mf + 2 19 0.22 21.1 175 0.121 
 
 
 
2 2 2
0.50 0.151 0.121
2 2 2
0.23 23%
2.30
2
THD
     
      
     
   
 
 
ma=1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 
n = 1 1.00 0.90 0.80 0.70 0.60 0.50 0.40 0.30 0.20 0.10 
n = mf 0.6 0.71 0.82 0.92 1.01 1.08 1.15 1.20 1.24 1.27 
n = mf ± 2 0.32 0.27 0.22 0.17 0.13 0.09 0.06 0.03 0.02 0.00 
 
 
 
 
 
EET 424 POWER ELECTRONIC FOR ENERGY SYSTEM 
 
 
 
4) The dc source supplying an inverter with a bipolar PWM output is 250V. the load is an RL 
series combination with R=20Ω and L=50mH. The output has a fundamental frequency 
of 60Hz. 
a. Specify the amplitude modulation to provide a 160V rms fundamental frequency 
output. 
b. If the frequency modulation ratio is 31, determine the total harmonic distortion 
of the load current. 
 
Solution 
 
 
1 1,
1
0
2 160 2 226.3 .
226.3
0.9
250
20 2 60 .050 20 18.85
rms
a
dc
n
V V V
V
m
V
Z R jn L jn jn 
  
  
     
 
 
From Table 8-3, 
 n Vn/Vdc Vn Zn In=Vn/Zn 
 1 0.9 225 27.5 8.18 
mf 31 0.71 178 585 0.305 
mf-2 29 0.27 67 547 0.122 
mf+2 33 0.27 67 622 0.108 
 
 
2 2 2
0.305 0.122 0.108
2 2 2
0.044 4.4%
8.18
2
THD
     
      
     
   
 
 
 
 
 
 
 
EET 424 POWER ELECTRONIC FOR ENERGY SYSTEM 
 
 
5) Design an inverter that has a PWM output across an RL series load with R=10Ω and 
L=20mH. The fundamental frequency of the output voltage must be 120V rms at 60Hz, 
and the total harmonic distortion of the load current must be less than 8%. Specify the 
dc input voltage, the amplitude modulation ratio ma, and the switchng frequency. 
 
Solution 
 
 
 
 
1
1
1
1
0
0
0.9,
120 2
189 .
0.9
Using Table 8-3, at , 0.71 189 134 .
8%, 0.08
120 2 120 2
13.6 .
12.510 2 60 0.020
0.08 13.6 1.09 .
134
123
1.09
123 123
377 0.
a
m
dc
a
f mf
mf
mf
mf
mf f
mf
f
Let m
V
V V
m
n m V V
for THD I I
V
I A
Z j
I A
V
Z m L
I
m
L




  
  
 
   

 
    
 
 
16.4
020
Choose odd integer 19 or greater for m .f

 
 
 
 
 
 
 
 
 
 
 
 
 
EET 424 POWER ELECTRONIC FOR ENERGY SYSTEM 
 
6) A six step three phase inverter has a 250 V source nd an output frequency of 60 Hz. A 
balance Y-connected load consists of a series 25 Ω resistance and 20 mH inductance in 
each phase. Determine: 
a. The rms value of the 60 Hz component of load current 
b. The THD of the load current. 
Solution 
 
1,
1
1
1
1
1,
2 2 500
) 2 cos cos 3 159 .
3 3 3 3
159 159
6.09 .
26.125 377 0.020
4.31 .
2
dc
L N
rms
V
a V V
V
I A
Z j
I
I A
 
 

      
          
      
   

 
 
 
 
 
7) A six step three phase inverter has an adjustable dc input. The load is a balanced Y 
connection with a series RL combination in each phase, with R = 5Ω and L = 50mH. The 
output frequency frequency is to be varied between 30 and 60 Hz. 
a. Determine the range of the dc input voltage required to maintain the 
fundamental frequency component of current at 10A rms. 
 
Solution 
  
 
  
1 1 1 1
1,
1,
1 1 1 1
30 , 10.7 , 10 2 10.7 151 .
2 2
2 cos cos 0.637
3 3 3
151
237 .
0.637 0.637
60 , 19.5 , 10 2 19.5 276
276
433 .
0.637
dc
L N dc
L N
dc
dc
At f Hz Z V I Z V
V
V V
V
V V
At f Hz Z V I Z V
V V
 



     
    
       
    
  
     
 