LM25005 datasheet(14/28 Pages) TI

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L1 =

5V x (42V ± 5V)

0.5A x 300 kHz x 42V

= 29 PH

L1 =

VOUT x (VIN(max) ± VOUT)

IRIPPLE x FS x VIN(max)

IPK+

0 mA

IPK-

IRIPPLE

1/Fs

RT =

[(1 / 300 x 10

3) ± 580 x 10-9]

135 x 10

-12

LM25005

SNVS411C – JANUARY 2006 – REVISED MARCH 2013

www.ti.com

APPLICATION INFORMATION

EXTERNAL COMPONENTS

The procedure for calculating the external components is illustrated with the following design example. The Bill of

Materials for this design is listed in 5V, 2.5A Demo Board Bill of Materials. The circuit shown in Typical

Application Circuit and Block Diagram is configured for the following specifications:

•

VOUT = 5V

•

VIN = 7V to 42V

•

Fs = 300 KHz

•

Minimum load current (for CCM) = 250 mA

•

Maximum load current = 2.5A

R3 (RT)

RT sets the oscillator switching frequency. Generally, higher frequency applications are smaller but have higher

losses. Operation at 300KHz was selected for this example as a reasonable compromise for both small size and

high efficiency. The value of RT for 300KHz switching frequency can be calculated as follows:

(3)

The nearest standard value of 21 k

Ω was chosen for RT.

The inductor value is determined based on the operating frequency, load current, ripple current, and the

minimum and maximum input voltage (VIN(min), VIN(max)).

Figure 15. Inductor Current Waveform

To keep the circuit in continuous conduction mode (CCM), the maximum ripple current IRIPPLE should be less

than twice the minimum load current, or 0.5 Ap-p. Using this value of ripple current, the value of inductor (L1) is

calculated using the following:

(4)

(5)

This procedure provides a guide to select the value of L1. The nearest standard value (33 µH) will be used. L1

must be rated for the peak current (IPK+) to prevent saturation. During normal loading conditions, the peak current

occurs at maximum load current plus maximum ripple. During an overload condition the peak current is limited to

3.5A nominal (4.25A maximum). The selected inductor (see 5V, 2.5A Demo Board Bill of Materials) has a

conservative 6.2 Amp saturation current rating. For this manufacturer, the saturation rating is defined as the

current necessary for the inductance to reduce by 30%, at 20°C.

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