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WS3412

Active PFC Non-isolated Buck LED Driver

Shenzhen Winsemi Microelectronics Co., Ltd 

WS3412 Product Description

Function Description

The WS3412 is designed for active PFC non-isolated buck

LED driver, which integrates 550V power MOSFET and

using SOP 8 package, producing up to 260mA output

current. It operates under valley switching mode,

automatically adapting to the variation of the inductance and

the output voltage.

It can achieve excellent constant current performance with

very few external components.

Start up

The start current is very low, Typ. 70uA(Max. 100uA). Under

the design system of 85VAC, the startup resistor is:

R = 85 * 2 = 1.2M

100

Chip Supply

strongest power supply):

D=80/260/1.414=0.218

The power consumption of R4:

P=(80-9)*(80-9)/58*(1-0.218)=68mW

Sence Resistor

The WS3412 is designed for active PFC non-isolated buck

LED driver. It operates under valley switching mode and

can achieve high accuracy constant current performance

with very few external components. The peak current of

inductor is continuous detected. CS terminal is connected

internal of the chip, and compared with the internal 200mV.

Internal Amplifier’s output COMP adjust the on-time, making

the average value os CS equal to 200mV after the system is

stable. In addition, a 1V cycle-by-cycle over current

protection is set up inside CS pin.

LED output current:

I LED

=

0.2V

R CS

After startup, the output voltage should supply the chip,

rectifier diodes D6 need to use fast recovery diodes.Current

FB Voltage Detection

limiting resistor R4 is calculated as:

R4

=

(1 − )D * VLED − 9

400uA

FB voltage determines the working status of the system,

when FB is greater then 1.6V(typ.), WS3412 will

automatically considered as output over voltage protection.

Where, D is duty cycle, 400uA is the normal operation

The system will enter extremely energy efficient hiccup

current of the chip, Vled is output load voltage. The

mode. Output over voltage protection voltage as follows:

consumption of R4 is:

PR 4

=

(VLED − 9)2

R4

* (1 −

D)

VOVP

= 1.6 *

R2 + R3

R3

R2,R3,please refer to the typical application diagram, in

For example:

which R3=1k(no more than 2k). use 1.3 instead of constant

Requirements:180~260V input voltage, 36~80V output,

1.6 in the above formula in the design of system. Assuming

240mA output current.

Vovp=90V, we got R2=56k from the above formula,

The R4 design of above program should be met:

choosing 60k for R2 (larger as far as possible).

1. Supply Problem when Min. input AC voltage 180V and

Because VFB2 is between 1.3 and 1.9, choose 1.9 to

Min. output voltage 36V(which is the weakest power supply):

calculate the withstand voltage of C4,

D=36/180/1.414=0.141,R4=(1-0.141)*(36-9)/400uA=58k

Vovp=1.9*(1+60)/1=116V

2.Power consumption problem of R4 when Max. input AC

The withstand voltage of C4 shoule larger than the above

voltage 260V and Max. output voltage 80V(which is the

value, 200V capacitor could be used.

WIN SEM I M ICROELECTRON ICS

WIN SEM I M ICROELECTRON ICS

WIN SEM I M ICROELECTRON ICS

www.winsemi.com Tel : 0755-82506288 Fax : 0755-82506299

WIN SEM I M ICROELECTRON ICS

WIN SEM I M ICROELECTRON ICS

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