ACT512

Rev 4, 13-Feb-14

- 8 -

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Copyright © 2014 Active-Semi, Inc.

Active-Semi Proprietary―For Authorized Recipients and Customers

ActiveQR

TYPICAL APPLICATION

Design Example

The design example below gives the procedure for

12V/2A flyback converter using ACT512. Refer to

application circuit Figure 4, the design for an

adapter application starts with the following

specification:

The operation for the circuit shown in Figure 4 is as

follows: the rectifier bridge D1−D4 and the capacitor

C1/C2 convert the AC line voltage to DC bus

voltage. This voltage supplies the primary winding

of the transformer T1 and the startup circuit of R7/

R8 and C4 to VDD pin of ACT512. The primary

power current path is formed by the transformer’s

primary winding, Q1, and the current sense resistor

R9. The resistors R3, R2, diode D5 and capacitor

C3 create a snubber clamping network that protects

Q1 from damage due to high voltage spike during

Q1’s turn off. The network consisting of capacitor

C4, diode D6 and resistor R4 provides a VDD

supply voltage for ACT512 from the auxiliary

winding of the transformer. The resistor R4 is

optional, which filters out spikes and noise to makes

VDD more stable. C4 is the decoupling capacitor of

the supply voltage and energy storage component

for startup. During power startup, the current

charges C4 through startup resistor R7/R8 from the

rectified bus

voltage. The diode D8 and the

capacitor C5/L2/C6 rectify filter the output voltage.

The resistor divider consists of R15 and R16

programs the output voltage.

Since a bridge

rectifier and bulk input capacitors are used, the

resulting minimum and maximum DC input voltages

can be calculated:

47µF electrolytic capacitors.

The maximum duty cycle is set to be 45% at low

line voltage 90VAC and the circuit efficiency is

estimated to be 86%. Then in CCM the primary to

EF25 core is selected for the transformer. The core

minimum Ae is 0.51cm^2. The minimum turn of the

primary winding is:

VDD voltage is set to 13V, base on the data we can

get primary, secondly and auxiliary turns:

We set DCM/CCM boundary is 185Vac,then the

duty at full load is:

The peak current of primary is:

The primary inductance is:

V

90

F

47

86

.

0

)

ms

5

.

3

47

2

1

(

24

2

90

2

C

)

t

f

2

1

(

P

2

V

2

V

2

IN

C

L

OUT

2

MIN

_

INAC

MIN

_

INDC

≈

-

-

-

μ

η

×

×

×

×

×

=

×

=

(2)

V

375

)

V

265

(

2

V

2

V

AC

AC

)

MAX

(

IN

DC

)

MAX

(

IN

=

×

=

×

=

(3)

136

.

6

12

)

45

.

0

1

(

90

45

.

0

V

)

D

1

(

V

D

N

N

o

max

_

CCM

min

_

IN

max

_

CCM

s

p

=

×

−

×

=

×

−

×

=

(4)

T

42

51

.

0

2500

75000

10

45

.

0

90

)

cm

gaus

(

Ae

B

f

10

D

V

N

8

2

min

max

max

8

max

_

CCM

min

_

in

p

=

×

×

×

×

=

×

×

×

×

×

≥

Δ

(5)

22

.

0

56

12

9

414

.

1

185

56

12

N

V

N

V

N

V

D

p

o

s

INDC

p

o

CCM

=

×

+

×

×

×

=

×

+

×

×

=

(8)

(9)

A

97

.

0

86

.

0

22

.

0

414

.

1

185

2

12

2

D

V

I

V

2

I

DCM

in

o

o

ppk

=

×

×

×

×

×

=

×

×

×

×

=

η

(10)

Input Voltage Range

90VAC - 265VAC, 50/60Hz

24W

12V

2A

2.3-2.6A

System Efficiency CV,

η

0.86

(7)

T

10

N

,

T

9

N

,

T

56

N

a

s

p

=

=

=

(6)

mH

8

.

0

75000

97

.

0

22

.

0

414

.

1

185

f

I

D

V

L

sw

ppk

uty

indc

m

=

×

×

×

=

×

×

=

1

.

1

45

.

0

12

7

.

0

13

V

V

V

V

N

N

sec

_

d

o

aux

_

d

dd

s

A

=

+

+

=

+

+

=