2. ECE 442 Power Electronics 2
Operation
T1 fired, resonant pulse of current
flows through the load. The current
falls to zero at t = t1m and T1 is “self –
commutated”.
T2 fired, reverse resonant current
flows through the load and T2 is also
“self-commutated”.
The series resonant circuit must be
underdamped,
R2
< (4L/C)
3. ECE 442 Power Electronics 3
Operation in Mode 1 – Fire T1
1
1 1
1
1
(0)
(0) 0
(0)
C S
C C
di
L Ri i dt v V
dt C
i
v V
+ + + =
=
= −
∫
4. ECE 442 Power Electronics 4
2
1 1
1
2 2
2
1
1
0
1
( ) sin
1
4
( ) sin
2
R
t
L
r
r
s c
t r
ts c
r
r
i t Ae t
R
LC L
V Vdi
A
dt L
V V
i t e t
L
R
L
α
ω
ω
ω
ω
ω
α
−
=
−
=
= − ÷
+
= =
+
=
=
5. ECE 442 Power Electronics 5
To find the time when the current is
maximum, set the first derivative = 0
( )
1
1
1
0
sin cos 0
.....
tan
tan
1
tan
2
t ts c
r r r
r
r
r m
r m
r m
r
m
r
di
dt
V V
e t e t
L
t
t
t
t
α α
α ω ω ω
ω
ω
ω
α
ω
ω
α
ω
ω
− −
−
−
=
+
− + = ÷
=
=
=
6. ECE 442 Power Electronics 6
To find the capacitor voltage, integrate the
current
( )
( )
1
1
1
1
1
0
0
1
1 1
1
( ) ( )
1
( ) sin
...
( ) ( ) ( sin cos )/
0 ( )
( ) r
t
C c
t
ts c
C r C
r
t
C s C r r r r s
m
r
C m C s C s
v t i t dt V
C
V V
v t e t dt V
C L
v t V V e t t V
t t
v t V V V e V
α
α
απ
ω
ω
ω
α ω ω ω ω
π
ω
−
−
−
= −
+
= − ÷
= − + + +
≤ ≤
= = + +
∫
∫
The current i1 becomes = 0 @ t=t1m
10. ECE 442 Power Electronics 10
Operation in Mode 3 – Fire T2
3
3 2 1
3
3 3
3
1
(0) 0
(0) 0
(0)
C
C C C
di
L Ri i dt v
dt C
i
v V V
+ + + =
=
= − = −
∫
11. ECE 442 Power Electronics 11
1
3 1
1
3
3
3
0
3
( ) sin
1
( )
( sin cos )
( )
0 ( )m
C t
r
r
t
C C
t
C r r r
C
r
r
V
i t e t
L
v t i dt V
C
V e t t
v t
t t
α
α
ω
ω
α ω ω ω
ω
π
ω
−
−
=
= −
− +
=
≤ ≤
∫
12. ECE 442 Power Electronics 12
3 3 1
1 1
1
1
3
1
( )
( ) ( )
.
.
1
1
1
r
m
r
m
C C C C
C C S C S
C S z
z
C S z
C S C
v t V V V e
v t V V V e V
V V
e
e
V V
e
V V V
π
α
ω
π
α
ω
−
−
= = =
= = + +
=
−
=
−
+ =
14. ECE 442 Power Electronics 14
Summary -- Series Resonant Inverter
15. ECE 442 Power Electronics 15
To avoid a short-circuit across the main dc supply,
T1 must be turned OFF before T2 is turned ON,
resulting in a “dead zone”.
This “off-time” must be
longer than the turn-off
time of the thyristors, tq.
0
0 max
1
2
off q
r
q
r
t t
f f
t
π π
ω ω
π
π
ω
− = >
≤ =
+ ÷
The maximum possible
output frequency is
16. ECE 442 Power Electronics 16
Series Resonant Inverter
Coupled Inductors
17. ECE 442 Power Electronics 17
Improvement in performance
• When T1 turned ON,
voltage @ L1 is as
shown, voltage @ L2
in same direction,
adding to the voltage
@ C
• This turns T2 OFF
before the load
current falls to 0.
18. ECE 442 Power Electronics 18
Half-Bridge Series Resonant Inverter
Note:
L1 = L2
C1 = C2
19. ECE 442 Power Electronics 19
This configuration reduces the high-pulsed
current from the dc supply
• Power drawn from the
source during both
half-cycles of the
output.
• Half of the current is
supplied from the
associated capacitor,
half of the current is
supplied from the
source.
20. ECE 442 Power Electronics 20
Full-Bridge Series-Resonant Inverter
21. ECE 442 Power Electronics 21
Characteristics of the
full-bridge inverter
• This configuration
provides higher
output power.
• Either T1-T2 or T3-T4
are fired.
• Supply current is
continuous but
pulsating.
22. ECE 442 Power Electronics 22
Example 8.1 – Analysis of the
Basic Resonant Inverter
• L1 = L2 = L = 50μH
• C = 6μF
• R = 2Ω
• Vs = 220V
• fo = 7kHz
• tq = 10μs
23. ECE 442 Power Electronics 23
1 1
2 12 2 122 2
2 2
6
1 10 2 10
54,160 /
4 50 6 4 50
8619.8
2
1
116
2
20,000
2 (2 50 10 )
r
r
r
r
r
R
rad s
LC L
f Hz
T s
f
R
L
ω
ω
π
µ
α −
×
= − = − = ÷ ÷
× ×
= =
= =
= = =
× ×
Determine the resonant frequency
The resonant frequency in Hz
24. ECE 442 Power Electronics 24
Determine the turn-off time toff
0
43,982 54,160
13.42
off
r
off
off
t
t
t s
π π
ω ω
π π
µ
= −
= −
=
25. ECE 442 Power Electronics 25
Determine the maximum permissible
frequency
max
max
6
max
1
2
1
2 10 10
54,160
7352
q
r
f
t
f
f Hz
π
ω
π−
=
+ ÷
=
× + ÷
=
26. ECE 442 Power Electronics 26
Determine the peak-to-peak capacitor
voltage
20
54.16
1
1
220
100.4
220 100.4 320.4
100.4 320.4 420.8
r
s
C
C s C
pp C C
V
V V
ee
V V V V
V V V V
απ π
ω
= = =
= + = + =
= + = + =
27. ECE 442 Power Electronics 27
Determine the peak load current
1 1max
1
1
(0.02)(22.47) 6
1max
1max
( ) sin
1
tan
1 54.16
tan 22.47
54,160 20
220 100.4
sin(54,160 22.47 10 )
0.05416 50
70.82
mts C
m r m
r
r
m
r
m
V V
i t t e t i
L
t
t s
i e
i A
α
ω
ω
ω
ω α
µ
−
−
−
− −
+
= = =
=
= =
+
= × ×
×
=
28. ECE 442 Power Electronics 28
Sketch the instantaneous load current,
capacitor voltage, and dc supply current
29. ECE 442 Power Electronics 29
Calculate the rms load current
2 2
1 3
0 0
1 1
( ) ( )
44.1
T T
o
o
I i t dt i t dt
T T
I A
= + ÷
=
∫ ∫
30. ECE 442 Power Electronics 30
2 7000⋅
0
58 10
06−
⋅
t
320.4 exp 20000− t⋅( )⋅ sin 54160 t⋅( )⋅
54160 50⋅ 10
06−
⋅
2
⌠
⌡
d
⋅
1
2
44.091=
1
58
22
0
2
s
o o o
I f i dt
µ
∫ =
Using MATHCAD,
Io = 44.1Amperes
31. ECE 442 Power Electronics 31
Determine the output power
( )
22
44.1 (2)
3,889
o o
o
P I R
P W
= =
=
32. ECE 442 Power Electronics 32
Determine the average supply current
3,889
17.68
220
o
s
s
P
I
V
W
I A
V
=
= =
33. ECE 442 Power Electronics 33
Determine the average, peak, and rms
thyristor currents
2
0
1
( ) 17.68
70.82
44.1
31.18
2
T
A o
p
R
I i t dt A
T
I A
A
I A
= =
=
= =
∫
34. ECE 442 Power Electronics 34
7000
0
58 10
06−
⋅
t
320.4 exp 20000− t⋅( )⋅ sin 54160 t⋅( )⋅
54160 50⋅ 10
06−
⋅
2
⌠
⌡
d
⋅
1
2
31.177=
rms Thyristor Current
44.1
31.18
2
R
A
I A= =
Using MATHCAD