ELEC 431
Resonant Converters/Inverters
P.K. Jain
Assignment #5
1. In the series resonant inverter of Figure 1, vi = 500V, Ts = 10µs. The load
parameters are: R = 1.2Ω, ωL = 10 Ω, 1/ωC = 10Ω. If the variable frequency
control is used to regulate the output, perform the following:
a) Sketch to scale the waveforms vo, io, iQ1 (current flowing through Q1), iD1
(current flowing through D1), and vAK,1. Higher harmonics other than
fundamental component may be neglected.
b) Calculate the RMS value of the switch current (S1).
c) Calculate the average value of diode current (D1).
d) Determine each switch if it is seeing zero voltage switching (ZVS) or zero
current switching (ZCS).
e) Determine RMS load voltage.
is
ar stu
ed d
vi y re
aC s
o
ou urc
rs e
eH w
er as
o.
co
m
2. If the operating frequency is varied from 25kHz to 300kHz, plot the output
voltage vo (RMS) as a function of frequency for (i) vi = 500V and (ii) vi = 1000V.
All other circuit parameters are same as in Q#1.
3. If MOSFET switches are used in Q#2, determine the range of frequency control if
vi varies from 500V to 1000V and the output voltage is kept constant to the same
value as obtained in Q#1.
4. Repeat Q#3 if MOSFETs were replaced by thyristors.
Q1
+
vi
_
Q2
D1
ωL
D2
1/ωC
io
+
Q3
D4
sh
Th
Q4
vs
_
D3
Figure 1
This study source was downloaded by 100000826074309 from CourseHero.com on 05-16-2021 02:59:20 GMT -05:00
https://www.coursehero.com/file/9392190/ELEC-431-Assignment-5-Solutions/
R
+
vo
_
ELEC 431
Resonant Converters/Inverters
P.K. Jain
Solutions for Q1:
Since L 
→
1
 10
C
→
L 
1
0
C
Hence, the operating frequency is equal to the resonant frequency, for series
resonant circuit, vo = vs and the voltage vs and the load current io are in phase.
4
4
vs  vi sin s t   500 sin  s t   636.6 sin  s t 

io 

vs
4
4

vi sin  s t  
500 sin  s t   530.5 sin  s t 
R 1.2
1.2
is
ar stu
ed d
vi y re
aC s
o
ou urc
rs e
eH w
er as
o.
co
m
a) See next page
b) iQ1, RMS 

1
Ts
Ts / 2
1
2
2 Ts
2
 iQ1dt 
0
Ts / 2
 i dt 
2
o
0
1
Ts
Ts / 2
 i dt
2
o
0
1
1 530.5
io, RMS 
 265.25 A
2
2
2
c) In this case, D1 does not conduct at all, hence iD1 = 0
=0
→
the average value of iD1
d) When the switches are MOSFETs, the switches are seeing ZCS. It is because during off
time, the voltage across the switch is 500V. At turn-on, the voltage is not zero, but the
current is zero at both turn-on and turn-off, so the switching is ZCS.
e) vo = vs = 636.6 sin s t 
Vo,max
2

636.6
 450.2 V
2
sh
Th
Hence, vo,RMS 
This study source was downloaded by 100000826074309 from CourseHero.com on 05-16-2021 02:59:20 GMT -05:00
https://www.coursehero.com/file/9392190/ELEC-431-Assignment-5-Solutions/
ELEC 431
Resonant Converters/Inverters
P.K. Jain
636.6V
vo
ωst
io
At resonance, vo =
vs
530.5A
is
ar stu
ed d
vi y re
aC s
o
ou urc
rs e
eH w
er as
o.
co
m
ωst
530.5A
iQ
ωst
iD1
ωst
vAK1
500V
ωst
π
sh
Th
2π
This study source was downloaded by 100000826074309 from CourseHero.com on 05-16-2021 02:59:20 GMT -05:00
https://www.coursehero.com/file/9392190/ELEC-431-Assignment-5-Solutions/
ELEC 431
Resonant Converters/Inverters
P.K. Jain
Solutions for Q2:
The resonant frequency is: f r 
Quality factor: Q 
vo

vs
r L
R

1
 100 kHz
Ts
10
 8.33
1.2
R

1 

R    s L 
 s C 

2
R


 1 

R   s  r L  r
 s  r C 
 r
2
vo

vs
→
2
2
1
f
f 
1  Q  s  r 
 fr fs 
2
2
4
2 2

vi 
vi
→
vi 
vs , RMS

2
2 2
vo
vo
2 2 vo



vi
 vs , RMS
vs , RMS
2 2
vo
0.9
0.9


→
2
vi
f
f
f 
f
1  Q 2  s  r 
1  8.332  s  r
 fr fs 
 fr fs



2
When fs = 25kHz,
→
fs
 0.25
fr
→
When fs = 300kHz,
→
fs
3
fr
→
When fs = fr,
→
K  f   0.9
For vi = 500V,
→
vo(25kHz) = 0.029x500 = 14.5V
vo(100kHz) = 0.9x500 = 450V
vo(300kHz) = 0.041x500 = 20.5V
→
vo(25kHz) = 0.029x1000 = 29V
vo(100kHz) = 0.9x1000 = 900V
vo(300kHz) = 0.041x1000 = 41.5V
Th
Hence,
is
ar stu
ed d
vi y re
aC s
o
ou urc
rs e
eH w
er as
o.
co
m
Since vs , RMS 
sh
For vi = 1000V,
0.9
1  69.410.25  4
2
0.9
 1
1  69.41 3  
3

This study source was downloaded by 100000826074309 from CourseHero.com on 05-16-2021 02:59:20 GMT -05:00
https://www.coursehero.com/file/9392190/ELEC-431-Assignment-5-Solutions/
K f 

2
=0.029
=0.041
ELEC 431
Resonant Converters/Inverters
P.K. Jain
vo
900V
450V
1000V
is
ar stu
ed d
vi y re
aC s
o
ou urc
rs e
eH w
er as
o.
co
m
500V
100kHz
300kHz
sh
Th
25kHz
This study source was downloaded by 100000826074309 from CourseHero.com on 05-16-2021 02:59:20 GMT -05:00
https://www.coursehero.com/file/9392190/ELEC-431-Assignment-5-Solutions/
f
ELEC 431
Resonant Converters/Inverters
P.K. Jain
Solutions for Q3 and Q4:
K fs  
From Q#2,
0.9
and
2
vo  K  f s vi
f
f 
1  69.41 s  r 
 fr fs 
When vi increases, in order to control the output voltage so as to keep it constant,
the switching frequency (fs) can either increase or decrease. Recall from Q1 that
at vi = 500V, the desired output voltage occurs at the resonant frequency.
vo  K  f r 500
At vi = 500V →
vo  K  f s 1000
At vi = 1000V →
vo  0.9500 -----
(11)
--------------------------------
(12)
→
Hence,
is
ar stu
ed d
vi y re
aC s
o
ou urc
rs e
eH w
er as
o.
co
m
When the output voltage is kept constant, we can combine (11) and (12):
1
0.9500  K  f s 1000
→
K  f s   x0.9
2
0.9
2

1
x0.9
2
f
f 
1  69.41 s  r 
 fr fs 
2
 f s f r  
1 
1  1  69.41  
4
 f r f s  

fs fr
  0.208
fr fs
vo
450V
1000V
2
 fs 
f 
   1  0.208 s 
 fr 
 fr 
500V
2
 fs 
f 
   0.208 s   1  0
 fr 
 fr 
Th
f s  0.208  0.208 2  4

fr
2
→
fs
 1.109
fr
sh
fr  fs
For MOSFETs:
→
→
f  1.109 f r  f r
fr  fs
For SCRs:
→
→
f  f r  0.901 f r
flow
or
fs
 0.901
fr
10.9kHz
10kHz
This study source was downloaded by 100000826074309 from CourseHero.com on 05-16-2021 02:59:20 GMT -05:00
https://www.coursehero.com/file/9392190/ELEC-431-Assignment-5-Solutions/
Powered by TCPDF (www.tcpdf.org)
fr
fhigh
f