ltc1877ems8-trpbf Linear Technology Corporation, ltc1877ems8-trpbf Datasheet - Page 13

• Manufacturer Part Number: ltc1877ems8-trpbf
• Description: High Efficiency Monolithic Synchronous Step-down Regulator
• Manufacturer: Linear Technology Corporation

APPLICATIONS
2. Does the V
3. Does the (+) plate of C
4. Keep the switching node SW away from sensitive small
Design Example
As a design example, assume the LTC1877 is used in a
single lithium-ion battery-powered cellular phone applica-
tion. The input voltage will be operating from a maximum
of 4.2V down to about 2.7V. The load current requirement
is a maximum of 0.3A but most of the time it will be in
standby mode, requiring only 2mA. Efficiency at both low
and high load currents is important. Output voltage is
2.5V. With this information we can calculate L using
equation (1),
LTC1877. The power ground traces should be kept
short, direct and wide. The signal ground and power
ground should converge to a common node in a star-
ground configuration.
resistors? The resistive divider R1/R2 must be con-
nected between the (+) plate of C
possible? This capacitor provides the AC current to the
internal power MOSFETs.
signal nodes.
FB
pin connect directly to the feedback
U
INFORMATION
U
IN
connect to V
OPTIONAL
R
R1
C
OUT
W
C
C
C2
C1
and signal ground.
R2
IN
Figure 7. LTC1877 Layout Diagram
as closely as
1
2
3
4
U
RUN
I
V
GND
TH
FB
C
IN
LTC1877
SYNC/MODE
PLL LPF
V
IN
SW
8
7
6
5
Substituting V
f = 550kHz in equation (3) gives:
A 15 H inductor works well for this application. For best
efficiency choose a 1A inductor with less than 0.25
series resistance.
C
temperature and C
0.25 . In most applications, the requirements for these
capacitors are fairly similar.
For the feedback resistors, choose R1 = 412k. R2 can
then be calculated from equation (2) to be:
Figure 8 shows the complete circuit along with its effi-
ciency curve.
IN
L
R
L
will require an RMS current rating of at least 0.15A at
2
L1
550
f
BOLD LINES INDICATE
HIGH CURRENT PATHS
V
1
0 8
kHz
OUT
.
I
L
2 5
+
.
OUT
(
V
120
–
V
OUT
C
1 1 875 5
OUT
= 2.5V, V
OUT
mA
R
V
OUT
1
+
–
)
will require an ESR of less than
1877 F07
1
V
V
+
–
V
IN
OUT
IN
IN
2 5
4 2
. ;
.
.
k use
= 4.2V, I
V
V
15 3
887
.
LTC1877
L
k
=120mA and
H
13
(3)