Datasheet

Data Sheet ADP2441
Rev. A | Page 23 of 32
DESIGN EXAMPLE
Consider an application with the following specifications:
 V
IN
=24 V ± 10%
 V
OUT
= 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,
k10
A60
6.0

STRING
REF
BOTTOM
I
V
R










REF
REF
OUT
BOTTOMTOP
V
VV
RR
k3.73
V6.0
V6.0V5
k10 










TOP
R
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
V
IN
Duty Cycle
24 V (Nominal) D
NOMINAL
= 20.8%
26 V (10% Above Nominal) D
MIN
= 19%
22 V (10% Less than Nominal) D
MAX
= 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,
SW
FREQ
f
R
500,92

R
FREQ
= 132 kΩ
Soft Start Capacitor
For a given soft start time, the soft start capacitor can be calculated
using Equation 5,
SS
SS
SS
REF
C
I
t
V

REF
SSSS
SS
V
t
I
C


nF10
V6.0
ms6A1



SS
C
Inductor Selection
Select the inductor by using Equation 9.
SW
IN
OUT
IN
OUT
IDEAL
fV
VVV
L





)(3.3
H3.18H66.18
kHz700V24
V)524(V53.3




IDEAL
L
In Equation 9, V
IN
= 24 V, V
OUT
= 5 V, I
LOAD(MAX)
= 1 A, and f
SW
=
700 kHz, which results in L = 18.66 µH. When L = 18 H (the
closest standard value) in Equation 8, I
L
= 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
SW
PP
OUT
MININ
fV
DDI
C



)1(
_
where:
V
PP
= 50 mV.
I
OUT
= 1 A.
D = 0.23.
f
SW
= 700 kHz.
Therefore,
F9.4
kHz700V05.0
)22.01(22.0A1
_





MININ
C
Choosing an input capacitor of 10 µF with a voltage rating of
50 V ensures sufficient capacitance over voltage and temperature.