ece 442 power electronics1 series-resonant inverter

ECE 442 Power Electronics 1

Series-Resonant Inverter


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)


Operation in Mode 1 – Fire T1

11 1

1

1(0)

(0) 0

(0)

C S

C C

diL Ri i dt v Vdt C

i

v V


21 1

12 2

2

11

0

1

( ) sin

1

4

( ) sin

2

RtL

r

r

s c

t r

ts cr

r

i t A e t

R

LC L

V VdiA

dt L

V Vi t e t

L

R

L


To find the time when the current is maximum, set the first derivative = 0

1

1

1

0

sin cos 0

.....

tan

tan

1tan

2

t ts cr r r

r

rr m

r mr m

rm

r

di

dt

V Ve t e t

L

t

tt

t


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

tts c

C r Cr

tC s C r r r r s

mr

C m C s C s

v t i t dt VC

V Vv t e t dt V

C L

v t V V e t t V

t t

v t V V V e V

The current i1 becomes = 0 @ t=t1m


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


t2m


Operation in Mode 3 – Fire T2

3

3 2 1

33 3

3

1(0) 0

(0) 0

(0)

C

C C C

diL Ri i dt vdt C

i

v V V


1

3 1

1

3

3

3

0

3

( ) sin

1( )

( sin cos )( )

0 ( )m

C tr

r

t

C C

tC r r r

Cr

r

Vi t e t

L

v t i dt VC

V e t tv t

t t


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 Ve

eV V

eV V V


Summary -- Series Resonant Inverter


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 qr

qr

t t

f f

t

The maximum possible output frequency is


Series Resonant Inverter Coupled Inductors


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.


Half-Bridge Series Resonant Inverter

Note:

L1 = L2

C1 = C2


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.


Full-Bridge Series-Resonant Inverter


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.


Example 8.1 – Analysis of the Basic Resonant Inverter

• L1 = L2 = L = 50μH

• C = 6μF• R = 2Ω

• Vs = 220V

• fo = 7kHz

• tq = 10μs


1 12 12 2 122 2

2 2

6

1 10 2 1054,160 /

4 50 6 4 50

8619.82

1116

220,000

2 (2 50 10 )

r

rr

rr

Rrad s

LC L

f Hz

T sf

R

L

Determine the resonant frequency

The resonant frequency in Hz


Determine the turn-off time toff

0

43,982 54,160

13.42

offr

off

off

t

t

t s


Determine the maximum permissible frequency

max

max6

max

1

2

1

2 10 1054,160

7352

qr

f

t

f

f Hz


Determine the peak-to-peak capacitor voltage

20

54.16

1

1

220100.4

220 100.4 320.4

100.4 320.4 420.8

r

sC

C s C

pp C C

VV V

eeV V V V

V V V V


Determine the peak load current

1 1max

1

1

(0.02)(22.47) 61max

1max

( ) sin

1tan

1 54.16tan 22.47

54,160 20

220 100.4sin(54,160 22.47 10 )

0.05416 5070.82

mts Cm r m

r

rm

r

m

V Vi t t e t i

L

t

t s

i e

i A


Sketch the instantaneous load current, capacitor voltage, and dc supply current


Calculate the rms load current

2 21 3

0 0

1 1( ) ( )

44.1

T T

o

o

I i t dt i t dtT T

I A


2 7000

0

58 1006

t320.4exp 20000 t( ) sin 54160t( )

5416050 1006

2

d

1

2

44.091

158 22

0

2s

o o oI fi dt

Using MATHCAD,

Io = 44.1Amperes


Determine the output power

22 44.1 (2)

3,889o o

o

P I R

P W


Determine the average supply current

3,88917.68

220

os

s

PI

VW

I AV


Determine the average, peak, and rms thyristor currents

2

0

1( ) 17.68

70.82

44.131.18

2

T

A o

p

R

I i t dt AT

I A

AI A


7000

0

58 1006

t320.4exp 20000 t( ) sin 54160t( )

5416050 1006

2

d

1

2

31.177

rms Thyristor Current

44.131.18

2R

AI A

Using MATHCAD

ece 442 power electronics1 series-resonant inverter

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