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LTC3448

3448f

APPLICATIO S I FOR ATIO

and high load currents is important. Output voltage is

1.8V. With this information we can calculate L using

Equation (1),

OUT

= () ∆()

−

⎛

⎝⎜

⎞

⎠⎟

(3)

Substituting VOUT = 1.8V, VIN = 4.2V, ∆IL = 240mA and

f = 1.5MHz in Equation (3) gives:

MHz

−

⎛

⎝⎜

⎞

⎠⎟

=µ

1 5

240

286

)

A 2.2

µH inductor works well for this application. For best

efficiency choose a 720mA or greater inductor with less

than 0.2

Ω series resistance.

CIN will require an RMS current rating of at least 0.3A ≅

ILOAD(MAX)/2 at temperature and COUT will require an ESR

of less than 0.25

Ω. In most cases, a ceramic capacitor will

satisfy this requirement.

For the feedback resistors, choose R1 = 316k. R2 can

then be calculated from Equation (2) to be:

OUT

1 632

−

⎛

⎝⎜

⎞

⎠⎟

Figure 9 shows the complete circuit along with its effi-

ciency curve.

Figure 9a

Figure 9b

VIN

632k

CIN: TAIYO YUDEN JMK212BJ475MG

COUT: TAIYO YUDEN JMK212BJ475MG

*MURATA LQH32CN2R2M11

316k

CIN

4.7

µF

CER

VIN

2.7V

TO 5.5V

LTC3448

FREQ

SYNC

2.2

µH*

3448 F09a

RUN

VOUT

MODE

VFB

GND

COUT

µF

CER

22pF

VOUT

1.8V

VOUT

100mV/DIV

AC COUPLED

ILOAD

100mA/DIV

500mA/DIV

VIN = 3.6V

VOUT = 1.8V

ILOAD = 100µA TO 200mA

µs/DIV

3448 F09c

VOUT

100mV/DIV

AC COUPLED

ILOAD

250mA/DIV

500mA/DIV

VIN = 3.6V

VOUT = 1.8V

ILOAD = 50mA TO 600mA

µs/DIV

3448 F09d

Figure 9c

Figure 9d

LOAD CURRENT (A)

100

0.0001

0.01

0.1

3448 F09b

0.001

VIN = 3.6V

VOUT = 1.8V

TA = 25°C

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