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ADP2386

Data Sheet

Rev. A | Page 18 of 24

DESIGN EXAMPLE

ADP2386

BST

COMP

PGOOD

GND

SYNC

VREG

PGND

PVIN

VIN = 12V

CSS

22nF

COUT1

100µF

6.3V

COUT2

47µF

6.3V

CVREG

1µF

CBST

0.1µF

100kΩ

2.2µF

VOUT = 3.3V

1.2nF

CCP

4.7pF

44.2kΩ

RTOP

2.21kΩ

RTOP

10kΩ

CIN

10µF

25V

Figure 33. 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 33 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 using the

following equation:

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.

INDUCTOR SELECTION

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

the maximum output current. Use the following equation to

estimate the inductor value:

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.

The peak-to-peak inductor ripple current can be calculated by

using the following equation:

ΔIL =

OUT

−

)

(

This calculation results in ∆IL = 1.81 A.

Use the following equation to calculate the peak inductor current:

IPEAK = IOUT +

∆

This calculation results in IPEAK = 6.905 A.

Use the following equation to calculate the rms current flowing

through the inductor:

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, the inductor should be

rated for at least a 9.6 A saturation current for reliable operation.

Based on the requirements described previously, select a 2.2 μH

inductor, such as the FDVE1040-2R2M from Toko, which has

a 6.8 mΩ DCR and a 11.4 A saturation current.

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