ACT512

Rev 4, 13-Feb-14

- 9 -

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Copyright © 2014 Active-Semi, Inc.

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TYPICAL APPLICATION CONT’D

Determining the value of the current sense resistor

(R7) uses the maximum current in the design. So

the input primary maximum current at maximum

load:

Since the ACT512 internal current limit is set to

0.96V, the design of the current sense resistor is

given by:

The voltage feedback resistors are selected

according to the design. Because the line UVLO is

75VDC, the upper feedback resistor is given by:

The lower feedback resistor is selected as:

When selecting the output capacitor, a low ESR

electrolytic capacitor is recommended to minimize

ripple from the current ripple. The approximate

equation for the output capacitance value is given

by:

Two 680µF electrolytic capacitors are used to

further reduce the output ripple.

PCB Layout Guideline

Good PCB layout is critical to have optimal

performance. Decoupling capacitor (C4) and

feedback resistor (R5/R6) should be placed close to

VDD and FB pin respectively. There are two main

power path loops. One is formed by C1/C2, primary

winding, mosfet transistor and current sense

resistor (R9). The other is secondary winding,

rectifier D8 and output capacitors (C5/C6). Keep

these loop areas as small as possible. Connecting

high current ground returns, the input capacitor

ground lead, and the ACT512 GND pin to a single

point (star ground configuration).

(11)

(12)

Ω

87

.

0

1

.

1

96

.

0

I

V

R

OCP

_

p

CS

CS

≈

=

=

(13)

Ω

k

9

.

54

mA

2

.

0

56

10

60

I

N

N

V

R

UVLO

_

FB

p

A

UVLO

_

INDC

UP

_

FB

≈

×

×

=

×

×

=

(14)

Ω

Ω

k

7

.

11

k

9

.

54

2

.

2

1

.

1

)

45

.

0

12

(

2

.

2

R

V

N

N

)

V

V

(

V

R

UP

_

FB

FB

S

A

D

OUT

LOW

_

FB

FB

≈

×

×

+

=

+

=

-

-

F

533

mV

50

k

75

2

V

f

I

C

RIPPLE

sw

OUT

OUT

μ

=

×

=

×

=

(15)

A

1

.

1

86

.

0

75

8

.

0

12

6

.

2

2

f

L

V

I

2

I

sw

P

OUT

OCP

_

OUT

OCP

_

p

=

×

×

×

×

=

×

×

×

×

=

η