AS1324 AMSCO [austriamicrosystems AG], AS1324 Datasheet - Page 11

• Manufacturer Part Number: AS1324
• Description: 1.5MHz, 600mA, DC/DC Step-Down Regulator
• Manufacturer: AMSCO [austriamicrosystems AG]

AS1324
Data Sheet - A p p l i c a t i o n I n f o r m a t i o n
Figure 25. Single Li-Ion 1.8V/600mA Regulator for Low Output Ripple
External Component Selection
Inductor Selection
For most applications the value of the external inductor should be in the range of 2.2 to 6.8µH as the inductor value
has a direct effect on the ripple current. The selected inductor must be rated for its DC resistance and saturation cur-
rent. The inductor ripple current ( Δ I
In
tion current of the inductor should be rated higher than the maximum inductor current as calculated with
3). This is recommended because the inductor current will rise above the calculated value during heavy load tran-
sients.
Where:
f = Switching Frequency (1.5 MHz typical)
L = Inductor Value
I
Δ
The recommended starting point for setting ripple current is Δ
The DC current rating of the inductor should be at least equal to the maximum load current plus half the ripple current
to prevent core saturation. Thus, a 720mA rated inductor should be sufficient for most applications (600mA + 120mA).
A easy and fast approach is to select the inductor current rating fitting to the maximum switch current limit of the con-
verter.
Note: For highest efficiency, a low DC-resistance inductor is recommended.
Accepting larger values of ripple current allows the use of low inductance values, but results in higher output voltage
ripple, greater core losses, and lower output current capability.
www.austriamicrosystems.com
Lmax
I
L =
Equation (EQ 2)
Peak to Peak inductor ripple current
= Maximum Inductor current
2.7 to 4.2V
V
IN
the maximum inductor current in PWM mode under static load conditions is calculated. The satura-
10µF
C
IN
L
) decreases with higher inductance and increases with higher V
V
EN
IN
4
1
Δ I
I
LMAX
L
AS1324-
=
Revision 1.03
V
=
2
18
OUT
I
OUTMAX
GND
×
I
1
----------------------- -
L
–
= 240mA (40% of 600mA).
L
V
--------------
+
3
SW
5
V
×
V
OUT
OUT
f
Δ I
------- -
IN
2
L
4.7µH
C
22µF
OUT
IN
600mA
V
1.8V
or V
OUT
Equation (EQ
OUT
.
11 - 20
(EQ 2)
(EQ 3)