LTC1875 LINER [Linear Technology], LTC1875 Datasheet - Page 16

• Manufacturer Part Number: LTC1875
• Description: 15mA Quiescent Current 1.5A Monolithic Synchronous Step-Down Regulator
• Manufacturer: LINER [Linear Technology]

APPLICATIO S I FOR ATIO
LTC1875
4. Place the small-signal components away from high
5. For optimum load regulation and true sensing, the top
Design Example
As a design example, assume the LTC1875 is used in a
single lithium-ion battery-powered cellular phone applica-
tion. The V
down to about 2.65V. The load current requirement is a
maximum of 1.5A but most of the time it will be on 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),
16
frequency switching nodes. In the layout shown in
Figure 8, all of the small-signal components have been
placed on one side of the IC and all of the power
components have been placed on the other.
of the output resistor divider should connect indepen-
dently to the top of the output capacitor (Kelvin connec-
tion), staying away from any high dV/dt traces. Place
the divider resistors near the LTC1875 in order to keep
the high impedance FB node short.
L
f
1
IN
I
L
will be operating from a maximum of 4.2V
V
OUT
U
1–
U
V
V
OUT
IN
W
U
(3)
Substituting V
f = 550kHz in equation (3) gives:
A 4.7 H inductor works well for this application. For good
efficiency choose a 2A inductor with less than 0.125
series resistance.
C
temperature and C
0.125 . In most applications, the requirements for these
capacitors are fairly similar.
For the feedback resistors, choose R2 = 412k. R1 can then
be calculated from equation (2) to be:
Figure 9 shows the complete circuit along with its effi-
ciency curve.
IN
R
L
will require an RMS current rating of at least 0.75A at
1
550
V
0 8
OUT
kHz
.
2 5
OUT
.
•
–
V
450
1
= 2.5V, V
OUT
•
mA
R
2 875 5
will require an ESR of less than
1
–
IN
2 5
4 2
= 4.2V, I
. ,
.
.
k use
V
V
4 09
887
.
L
= 450mA and
k
H
1875f