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Data Sheet

ADP2387

Rev. C | Page 19 of 25

DESIGN EXAMPLE

ADP2387

BST

COMP

PGOOD

GND

VREG

ILIM

PGND

PVIN

VIN = 12V

CSS

22nF

CIN

10µF

25V

COUT1

100µF

6.3V

COUT2

47µF

6.3V

RILIM

44.2kΩ

2.2µH

CBST

0.1µF

VOUT = 3.3V

RTOP

10kΩ

RBOT

2.21kΩ

CCP

4.7pF

1.2nF

44.2kΩ

100kΩ

CVREG

1µF

Figure 34. Schematic for Design Example

This section describes the procedures for selecting the external

components, based on the example specifications that are listed

in Table 8. See Figure 34 for the schematic of this design example.

Table 8. Step-Down DC-to-DC Regulator Requirements

Parameter

Specification

Input Voltage (VIN)

12.0 V ± 10%

Output Voltage (VOUT)

3.3 V

Output Current (IOUT)

6 A

Output Voltage Ripple (∆VOUT_RIPPLE)

33 mV

Load Transient

±5%, 1 A to 5 A, 2 A/µs

Switching Frequency (fSW)

600 kHz

OUTPUT VOLTAGE SETTING

Choose a 10 kΩ resistor as the top feedback resistor (RTOP),

and calculate the bottom feedback resistor (RBOT) by

RBOT = RTOP ×













− 6

OUT

To set the output voltage to 3.3 V, the resistors values are as

follows: RTOP = 10 kΩ, and RBOT = 2.21 kΩ.

FREQUENCY SETTING

To set the switching frequency to 600 kHz, connect a 100 kΩ

resistor from the RT pin to GND.

CURRENT-LIMIT THRESHOLD SETTING

Connect a 44.2 kΩ resistor between ILIM pin and GND to set

the current-limit threshold at 9 A.

INDUCTOR SELECTION

The peak-to-peak inductor ripple current, ∆IL, is set to 30% of

the maximum output current. To estimate the inductor value,

use the following equation:

L =

OUT

∆

−

)

(

where:

VIN = 12 V.

VOUT = 3.3 V.

D = 0.275.

∆IL = 1.8 A.

fSW = 600 kHz.

This calculation results in L = 2.215 µH. Choose the standard

inductor value of 2.2 µH.

Calculate the peak-to-peak inductor ripple current by using the

following equation:

ΔIL =

OUT

−

)

(

This calculation results in ∆IL = 1.81 A.

To calculate the peak inductor current, use the following equation:

IPEAK = IOUT +

∆

This calculation results in IPEAK = 6.905 A.

To calculate the rms current flowing through the inductor, use

the following equation:

IRMS =

OUT

∆

This calculation results in IRMS = 6.023 A.

Based on the calculated current value, select an inductor with

a minimum rms current rating of 6.03 A and a minimum

saturation current rating of 6.91 A.

However, to protect the inductor from reaching its saturation

point under the current-limit condition, rate the inductor for at

least a 9.2 A saturation current for reliable operation.

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