Datasheet

ManualsBrandsTEXAS INSTRUMENTS ManualsDev KitsPCB design board

= F7.4 PC

IN

fI

SWLED

xx

=

I

RMSIN-

tt

OFFON

x

mA486ns700s29.1kHz503A02.1

=

xPxx

=

I

RMSIN-

2CC

MINININ

=x=

-

PF

64.3

mV720

C

MININ

=

-

= =

PF

82.1

s29.1A02.1 Px

v

PPIN

'

-

tI

ONLED

x

t

ON

=

1

t

OFF

-

kHz503

1

=

-700 ns

=

s29.1 P

f

SW

2.2 PF

C

O

=

=

C

O

xC

MINO-

1.75

=

PF

2.2

1

PF

1.27

=

C

MINO

=

-

=

1

C

MINO-

2xS x f

SW

x Z

C

m250kHz5032 :xxSx

==

Z

C

-'

-

i

PPL

'i

PPLED-

'x

-

ir

PPLEDD

=

250 m:

x: mA502

- mA50mA450

:

=

2.0R

SNS

R5

SNS

x

V

ADJ

I

LED

=

-

2

=

2

-

A02.1

=

mA445

2.05 :x

V24.1

i

PPL

'

-

=

R

SNS

= =

:203.0

x A22.15

V24.1

V

ADJ

x

-

I5

MAXL

LM3409, LM3409HV, LM3409-Q1

www.ti.com

SNVS602J –MARCH 2009–REVISED MAY 2013

(83)

The closest 1% tolerance resistor is 0.2 Ω therefore I

LED

is:

(84)

The chosen component from step 3 is:

(85)

4. OUTPUT CAPACITANCE

Assume r

D

= 2 Ω and determine Z

C

:

(86)

Solve for C

O-MIN

and :

(87)

Choose C

O

:

(88)

The chosen component from step 5 is:

(89)

5. INPUT CAPACITANCE

Determine t

ON

:

(90)

Solve for C

IN-MIN

:

(91)

Choose C

IN

:

(92)

Determine I

IN-RMS

:

(93)

The chosen component from step 5 is:

(94)

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