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ECE 442 Power Electronics 1
Series-Resonant Inverter
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)
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
+ + + =
=
= −
∫
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
−
=
−
=
 
= − ÷
 
+
= =
+
=
=
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
− −
−
−
=
 +
− + = ÷
 
=
=
=
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
ECE 442 Power Electronics 7
ECE 442 Power Electronics 8
Operation in Mode 2 – T1, T2 Both OFF
2 1
2 2 1
2
2
( ) 0
( )
( )m
C C
C C C
i t
v t V
v t V V
=
=
= =
ECE 442 Power Electronics 9
t2m
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
+ + + =
=
= − = −
∫
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
−
−
=
= −
− +
=
≤ ≤
∫
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
−
−
= = =
= = + +
=
−
=
−
+ =
ECE 442 Power Electronics 13
ECE 442 Power Electronics 14
Summary -- Series Resonant Inverter
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
ECE 442 Power Electronics 16
Series Resonant Inverter
Coupled Inductors
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.
ECE 442 Power Electronics 18
Half-Bridge Series Resonant Inverter
Note:
L1 = L2
C1 = C2
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.
ECE 442 Power Electronics 20
Full-Bridge Series-Resonant Inverter
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.
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
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
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
= −
= −
=
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−
=
 
+ ÷
 
=
 
× + ÷
 
=
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
= = =
= + = + =
= + = + =
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
−
−
−
− −
+
= = =
=
= =
+
= × ×
×
=
ECE 442 Power Electronics 28
Sketch the instantaneous load current,
capacitor voltage, and dc supply current
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
 
= + ÷
 
=
∫ ∫
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
ECE 442 Power Electronics 31
Determine the output power
( )
22
44.1 (2)
3,889
o o
o
P I R
P W
= =
=
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
=
= =
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
= =
=
= =
∫
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

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  • 1. ECE 442 Power Electronics 1 Series-Resonant Inverter
  • 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
  • 7. ECE 442 Power Electronics 7
  • 8. ECE 442 Power Electronics 8 Operation in Mode 2 – T1, T2 Both OFF 2 1 2 2 1 2 2 ( ) 0 ( ) ( )m C C C C C i t v t V v t V V = = = =
  • 9. ECE 442 Power Electronics 9 t2m
  • 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 − − = = = = = + + = − = − + =
  • 13. ECE 442 Power Electronics 13
  • 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