ADP2442-EVALZ Analog Devices, ADP2442-EVALZ Datasheet - Page 24

• Manufacturer Part Number: ADP2442-EVALZ
• Description: Power Management IC Development Tools Eval board
• Manufacturer: Analog Devices
• Type: DC/DC Converters, Regulators & Controllersr
• Series: ADP2442r

Specifications of ADP2442-EVALZ

• Rohs: yes
• Tool Is For Evaluation Of: ADP2442
• Factory Pack Quantity: 1

DESIGN EXAMPLE
Consider an application with the following specifications:
•
•
•
•
•
•
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 1 and Equation 2,
Switching Frequency
Choosing the switching frequency involves consideration of 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
24 V (Nominal)
26 V (10% Above Nominal)
22 V (10% Less than Nominal)
Based on the estimated duty cycle range, choose the switching
frequency according to the minimum and maximum duty cycle
limitations, as shown in Figure 58. For example, a 700 kHz, fre-
quency is well within the maximum and minimum duty cycle
limitations.
Using Equation 3,
ADP2442
IN
V
V
Switching frequency: 700 kHz
Load: 800 mA typical
Maximum load current: 1 A
Overshoot ≤ 2% under all load transient conditions
R
R
R
R
R
FREQ
IN
OUT
BOTTOM
TOP
TOP
FREQ
: 24 V ± 10%
: 5 V ± 1%
= 132 kΩ
=
=
=
10
R
92
BOTTOM
=
kΩ
f
,
SW
500
I
STRING
V
×
REF




×
5




V
V
0
=
−
6 .
OUT
0
60
V
V
6 .
0
REF
6 .
μA
−
V
V




REF
=
=
10
73




kΩ
3 .
Duty Cycle
D
D
D
kΩ
NOMINAL
MIN
MAX
= 19%
= 23%
= 20.8%
Rev. 0 | Page 24 of 36
Inductor Selection
Select the inductor by using Equation 7.
In Equation 7, V
700 kHz, which results in L = 18.66 µH. When L = 18 μH (the
closest standard value) in Equation 6, ΔI
the maximum output current that is 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
where:
V
I
D = 0.23.
f
Therefore,
Choosing an input capacitor of 10 µF with a voltage rating of
50 V ensures sufficient capacitance over voltage and temperature.
Output Capacitor Selection
Select the output capacitor by using Equation 12 and Equation 13
Equation 12 is based on the output voltage ripple (ΔV
which is 1% of the output voltage.
Equation 13 calculates the capacitor selection based on the
transient load performance requirement of 2%. Perform these
calculations, then use the equation that yields the larger
capacitor size to select a capacitor.
SW
OUT
PP
= 700 kHz.
= 50 mV.
= 1 A.
C
C
C
C
L
L
IDEAL
IDEAL
OUT
OUT
IN
IN
_
_
MIN
MIN
(
(
MIN
MIN
=
=
=
3
=
)
)
3
3 .
3 .
≅
≅
I
1
24
IN
OUT
×
×
∆
8
A
. 0
= 24 V, V
V
5
×
I
V
V
05
×
OUT
V
OUT
×
f
×
PP
V
. 0
SW
×
D
V
700
IN
(
22
×
(
STEP
×
24
×
×
×
×
f
×
(
1 (
(
SW
700
kHz
V
−
∆
)
1 (
f
OUT




−
IN
SW
V
) 5
−
f
D
RIPPLE
kHz
SW
V
∆
−
= 5 V, I
. 0
)
I
V
22
=
L
×
OUT
18
∆
)
−
3
V
≈
.
∆
66
)
LOAD (MAX)
DROOP
I
4
L
L
μH
9 .
= 0.314 A. Although
×
μF
ESR
≈




18
Data Sheet
)
= 1 A, and f
3 .
μH
RIPPLE
),
SW
=