ADP2443ACPZN-R7 datasheet(20/25 Pages) AD

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Data Sheet

ADP2443

Rev. 0 | Page 19 of 24

DESIGN EXAMPLE

GND

PGND

COMP

VREG

RT/SYNC

PGOOD

RAMP

PVIN

ADP2443

BST

CCP

3.3pF

20kΩ

280kΩ

RRAMP

1.5MΩ

RBOT

3kΩ

RTOP

22kΩ

COUT

47µF

16V

CIN

10µF

50V

VOUT = 5V

VIN = 24V

6.8µH

CBST

0.1µF

CVREG

1µF

2.7nF

CSS

22nF

Figure 39. Schematic for Design Example

This section describes the procedures for selecting the external

components based on the example specifications that are listed

in Table 7. See Figure 39 for the schematic of this design example.

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

Parameter

Symbol

Specification

Input Voltage

VIN

VIN = 24.0 V ± 10%

Output Voltage

VOUT

VOUT = 5 V

Output Current

IOUT

IOUT = 3 A

Output Voltage Ripple

∆VOUT_RIPPLE

∆VOUT_RIPPLE = 50 mV

Load Transient

ILOAD

±5%, 0.5 A to 2.5 A, 2 A/μs

Switching Frequency

fSW

fSW = 600 kHz

OUTPUT VOLTAGE SETTING

Choose a 22 kΩ resistor as the top feedback resistor (RTOP),

and calculate the bottom feedback resistor (RBOT) by using the

following equation:



















OUT

TOP

BOT

To set the output voltage to 5 V, the resistors values are as

follows: RTOP = 22 kΩ and RBOT = 3 kΩ.

FREQUENCY SETTING

To set the switching frequency to 600 kHz, connect a 280 kΩ

resistor from the RT/SYNC pin to GND.

INDUCTOR SELECTION

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

the maximum output current. Use the following equation to

estimate the inductor value:

OUT











)

(

where:

VIN = 24 V.

VOUT = 5 V.

D = 0.208.

ΔIL = 0.9 A.

fSW = 600 kHz.

This calculation results in L = 7.33 μH. Choose the standard

inductor value of 6.8 μH.

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

using the following equation:

OUT









)

(

This calculation results in ΔIL = 0.97 A.

Use the following equation to calculate the peak inductor current:

OUT

PEAK







This calculation results in IPEAK = 3.49 A.

Use the following equation to calculate the rms current flowing

through the inductor:

OUT

RMS







This calculation results in IRMS = 3.013 A.

Based on the calculated current value, select an inductor with

a minimum rms current rating of 3.013 A and a minimum

saturation current rating of 3.49 A.

However, to protect the inductor from reaching its saturation

point under the current-limit condition, the inductor must be

rated for at least a 5.1 A saturation current for reliable operation.

Based on the requirements described previously, select a 6.8 μH

inductor, such as the FDVE1040-6R8M from Toko, which has

a 20.2 mΩ DCR and an 7.1 A saturation current.

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