LTC3547 LINER [Linear Technology], LTC3547 Datasheet - Page 14

• Manufacturer Part Number: LTC3547
• Description: Dual Monolithic 300mA Synchronous Step-Down Regulator
• Manufacturer: LINER [Linear Technology]

APPLICATIO S I FOR ATIO
LTC3547
Design Example
As a design example, consider using the LTC3547 in a
portable application with a Li-Ion battery. The battery
provides a V
each channel requires a maximum of 300mA in active
mode and 2mA in standby mode. The output voltages are
V
Start with channel 1. First, calculate the inductor value
for about 40% ripple current (120mA in this example) at
maximum V
For the inductor, use the closest standard value of 4.7µH.
A 4.7µF capacitor should be more than suffi cient for this
output capacitor. As for the input capacitor, a typical value
of C
a Li-Ion battery is very low.
14
OUT1
L
IN
1
=
= 4.7µF should suffi ce, as the source impedance of
= 2.5V and V
2 25
.
IN
IN
MHz
. Using a derivation of Equation 1:
ranging from 2.8V to 4.2V. The load on
2 5
.
• (
1.8V AT 300mA
U
OUT2
100
V
90
80
70
60
50
40
30
20
10
120
0
0.1
V
OUT
= 1.8V.
V
mA
OUT2
U
= 1.8V
2.5V TO 5.5V
)
1
OUTPUT CURRENT (mA)
•
⎛
⎝ ⎜
C1, C2, C3: TAIYO YUDEN JMK316BJ475ML
1
V
C
4.7µF
−
IN
OUT2
W
10
2 5
4 2
.
.
C
F2
562k
V
V
R4
, 10pF
Figure 4b. Effi ciency vs Output Current
⎞
⎠ ⎟
100
C1
4.7µF
Figure 4a. Design Example Circuit
V
V
V
=
4.7µH
IN
IN
IN
L2
= 2.7V
= 3.6V
= 4.2V
3 7
U
R3
280k
. 5 5 µH
1000
RUN2
SW2
V
FB2
LTC3547
GND
V
IN
RUN1
SW1
V
FB1
The feedback resistors program the output voltage. To
maintain high effi ciency at light loads, the current in these
resistors should be kept small. Choosing 2µA with the
0.6V feedback voltage makes R1~300k. A close standard
1% resistor is 280k. Using Equation 4:
An optional 10pF feedback capacity (C
improve transient response.
Using the same analysis for channel 2 (V
the results are:
Figure 4 shows the complete schematic for this example,
along with the effi ciency curve and transient response.
L2 = 3.81µH
R3 = 280k
R4 = 560k
L1, L2: MURATA LQH32CN4R7M33
R
4.7µH
2
100
L1
90
80
70
60
50
40
30
20
10
0
=
R1
280k
0.1
V
⎛
⎝ ⎜
OUT
C
F1
V
0 6
887k
, 10pF
= 2.5V
OUT
R2
.
1
OUTPUT CURRENT (mA)
−
1
⎞
⎠ ⎟
10
•
R
C
4.7µF
V
2.5V AT 300mA
OUT1
1 887
OUT1
=
100
V
V
V
3547 F04a
IN
IN
IN
= 2.7V
= 3.6V
= 4.2V
k
3547 F04b
1000
F1
) may be used to
OUT2
= 1.8V),
3547fa