LTC1877EMS8#TR Linear Technology, LTC1877EMS8#TR Datasheet - Page 13

• Manufacturer Part Number: LTC1877EMS8#TR
• Description: IC BUCK SYNC ADJ .6A 8MSOP
• Manufacturer: Linear Technology
• Type: Step-Down (Buck)r

Specifications of LTC1877EMS8#TR

• Internal Switch(s): Yes
• Synchronous Rectifier: Yes
• Number Of Outputs: 1
• Voltage - Output: 0.8 ~ 10 V
• Current - Output: 600mA
• Frequency - Switching: 550kHz
• Voltage - Input: 2.65 ~ 10 V
• Operating Temperature: -40°C ~ 85°C
• Mounting Type: Surface Mount
• Package / Case: 8-MSOP, Micro8™, 8-uMAX, 8-uSOP,
• Lead Free Status / RoHS Status: Contains lead / RoHS non-compliant
• Power - Output: -

APPLICATIONS INFORMATION
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. 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),
Substituting V
f = 550kHz in Equation (3) gives:
L
L
=
=
( )
550
f
( )
1
Δ
kHz
I
L
2 5
.
OUT
(
120
V
V
OUT
220pF
= 2.5V, V
mA
***SANYO POSCAP 6TPA47M
**TAIYO YUDEN CERAMIC LMK325BJ106MN
412k
⎛
⎜
⎝
*SUMIDA CD54-150
1
1
2
3
4
)
−
⎛
⎜
⎝
RUN
I
V
GND
TH
1
FB
V
V
−
Figure 8. Single Lithium-Ion to 2.5V/0.3A Regulator from Design Example
OUT
LTC1877
IN
SYNC/MODE
IN
2 5
4 2
OPTIONAL
.
.
⎞
⎟
⎠
PLL LPF
= 4.2V, ΔI
R
V
V
R1
C
⎞
⎟ =
⎠
887k
SW
20pF
V
IN
C
C
C2
C1
8
7
6
5
15 3
R2
15μH*
.
10μF**
L
μ
CER
=120mA and
Figure 7. LTC1877 Layout Diagram
H
1877 F08a
1
2
3
4
RUN
I
V
GND
TH
FB
C
+
IN
LTC1877
SYNC/MODE
(3)
PLL LPF
V
2.7V TO 4.2V
V
2.5V
IN
OUT
V
IN
SW
47μF***
8
7
6
5
A 15μH inductor works well for this application. For best
effi ciency choose a 1A inductor with less than 0.25Ω
series resistance.
C
at temperature and C
0.25Ω. 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 ef-
fi ciency curve.
IN
R
will require an RMS current rating of at least 0.15A
2
L1
=
⎛
⎜
⎝
95
90
85
80
75
70
65
60
55
50
BOLD LINES INDICATE
HIGH CURRENT PATHS
V
0.1
0 8
OUT
V
.
IN
V
+
IN
= 3.0V
= 4.2V
–
1877 F07
C
1.0
1 1 875 5
OUT
OUTPUT CURRENT (mA)
⎞
⎟
⎠
R
OUT
V
OUT
–
+
=
10
will require an ESR of less than
V
+
–
IN
. ;
k use
100
V
V
L = 15μH
IN
OUT
= 3.6V
= 2.5V
1877 F08b
887
1000
LTC1877
k
13
1877fa