ADP2441ACPZ-R2 datasheet(23/32 Pages) AD

23 / 32 page

Data Sheet

ADP2441

Rev. A | Page 23 of 32

DESIGN EXAMPLE

Consider an application with the following specifications:



VIN =24 V ± 10%



VOUT = 5 V ± 1%



Switching frequency = 700 kHz



Load = 800 mA typical



Maximum load current = 1 A



Soft start time = 6 ms



Overshoot ≤ 2% under all load transient conditions

CONFIGURATION AND COMPONENTS SELECTION

Resistor Divider

The first step in selecting the external components is to

calculate the resistance of the resistor divider that sets the

output voltage.

Using Equation 2 and Equation 3,

kΩ

μA



STRING

REF

BOTTOM



















REF

OUT

BOTTOM

TOP

kΩ





















TOP

Switching Frequency

Choosing the switching frequency involves considering the

trade-off between efficiency and component size. Low

frequency improves the efficiency by reducing the gate losses

but requires a large inductor. The choice of high frequency is

limited by the minimum and maximum duty cycle.

Table 11. Duty Cycle

VIN

Duty Cycle

24 V (Nominal)

DNOMINAL = 20.8%

26 V (10% Above Nominal)

DMIN = 19%

22 V (10% Less than Nominal)

DMAX = 23%

Based on the estimated duty cycle range, choose the switching

frequency according to the minimum and maximum duty cycle

limitations, as shown in Figure 55. For example, a 700 kHz,

frequency is well within the maximum and minimum duty

cycle limitations.

Using Equation 4,

FREQ

500



RFREQ = 132 kΩ

Soft Start Capacitor

For a given soft start time, the soft start capacitor can be calculated

using Equation 5,

REF



REF





μA







Inductor Selection

Select the inductor by using Equation 9.

OUT

IDEAL









)

(

μH

kHz

700

)

(













IDEAL

In Equation 9, VIN = 24 V, VOUT = 5 V, ILOAD(MAX) = 1 A, and fSW =

700 kHz, which results in L = 18.66 μH. When L = 18 μH (the

closest standard value) in Equation 8, ΔIL = 0.314 A. Although

the maximum output current required is 1 A, the maximum

peak current is 1.6 A. Therefore, the inductor should be rated

for higher than 1.6 A current.

Input Capacitor Selection

The input filter consists of a small 0.1 μF ceramic capacitor

placed as close as possible to the IC.

The minimum input capacitance required for a particular load is

OUT

MIN









)

(

where:

VPP = 50 mV.

IOUT = 1 A.

D = 0.23.

fSW = 700 kHz.

Therefore,

μF

kHz

700

)

(











MIN

Choosing an input capacitor of 10 μF with a voltage rating of

50 V ensures sufficient capacitance over voltage and temperature.

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