Datasheet
LTC3560
12
3560fb
standby mode, requiring only 2mA. Effi ciency at both
low and high load currents is important. Output voltage
is 2.5V. With this information we can calculate L using
equation (1),
L =
1
f
()
I
L
()
V
OUT
1
V
OUT
V
IN
(3)
Substituting V
OUT
= 2.5V, V
IN
= 4.2V, ∆I
L
= 320mA and
f = 2.25MHz in equation (3) gives:
L =
2.5V
2.25MHz(320mA)
1
2.5V
4.2V
1.4μH
A 1.5µH inductor works well for this application. For best
effi ciency choose a 960mA or greater inductor with less
than 0.2 series resistance.
APPLICATIONS INFORMATION
C
IN
will require an RMS current rating of at least 0.4A ≅
I
LOAD(MAX)
/2 at temperature and C
OUT
will require an ESR
of less than 0.1Ω. In most cases, a ceramic capacitor will
satisfy this requirement.
For the feedback resistors, choose R1 = 309k. R2 can
then be calculated from equation (2) to be:
R2 =
V
OUT
0.6
1
R1= 978.5k; use 976k
Figure 6 shows the complete circuit along with its ef-
fi ciency curve.
Figure 4. LTC3560 Layout Diagram
Figure 5. LTC3560 Suggested Layout
RUN
LTC3560
GND
SW
6
L1
R2
R1
C
FWD
BOLD LINES INDICATE HIGH CURRENT PATHS
V
IN
V
OUT
3560 F04
4
5
1
3
+
–
2
SYNC/MODE
V
FB
V
IN
C
IN
C
OUT
LTC3560
GND
3560 F05
PIN 1
V
OUT
V
IN
VIA TO V
OUT
SW
VIA TO V
IN
VIA TO GND
C
OUT
C
IN
L1
R2
C
FWD
R1