ADP2165ACPZ-2.5-R7 datasheet(17/23 Pages) AD

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Data Sheet

ADP2165/ADP2166

Rev. B | Page 17 of 23

DESIGN EXAMPLE

ADP2166

BST

COMP

PGOOD

GND

VREG

TRK

SYNC

0.47µH

PGND

PVIN

CIN

47µF

16V

VPVIN = 5V

CBST

0.1µF

COUT1

100µF

6.3V

COUT2

100µF

6.3V

VOUT = 1.2V

RTOP

10kΩ

RBOT

10kΩ

680pF

36kΩ

CSS

22nF

AVIN

CVREG

1µF

CCP

4.7pF

Figure 31. Schematic for Design Example

This section describes the procedures for selecting the external

components based on the example specifications listed in Table 7.

See Figure 31 for the schematic of this design example.

Table 7. Step-Down DC-to-DC Regulator Requirements

Parameter

Specification

Input Voltage

VPVIN = 5.0 V ± 10%

Output Voltage

VOUT = 1.2 V

Output Current

IOUT = 6 A

Output Voltage Ripple

∆VOUT_RIPPLE = 12 mV

Load Transient

±5%, 1 A to 5 A, 2 A/µs

Switching Frequency

fSW = 1.2 MHz

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 1.2 V, the resistor values are as

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

FREQUENCY SETTING

To use the fixed 1.2 MHz switching frequency, connect the RT

pin to the VREG pin.

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

PVIN

∆

−

)

(

where:

VPVIN = 5 V.

VOUT = 1.2 V.

D = 0.24.

∆IL = 1.8 A.

fSW = 1.2 MHz.

This calculation results in L = 0.422 µH. Choose the standard

inductor value of 0.47 µH.

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

using the following equation:

ΔIL =

OUT

−

)

(

This calculation results in ∆IL = 1.617 A.

Use the following equation to calculate the peak inductor

current:

IPEAK = IOUT +

∆

This calculation results in IPEAK = 6.809 A.

Use the following equation to calculate the rms current flowing

through the inductor:

IRMS =

OUT

∆

This calculation results in IRMS = 6.018 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 6.9 A.

However, to protect the inductor from reaching its saturation

point under the current-limit condition, use an inductor that is

rated for at least a 9 A saturation current for reliable operation.

Based on the requirements described previously, select a 0.47 µH

inductor, such as the 744314047 from Würth, which has a 1.35 mΩ

DCR and a 20 A saturation current.

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