LTC3410
W U U
APPLICATIO S I FOR ATIO
U
Substituting VOUT = 3V, VIN = 4.2V, ∆IL = 100mA of less than 0.5Ω. In most cases, a ceramic capacitor will
and f = 2.25MHz in Equation (3) gives:
satisfythisrequirement. FromTable2, CapacitanceSelec-
tion, COUT = 10µF and CIN = 4.7µF.
3V
3V
⎛
⎜
⎝
⎞
⎟
⎠
L=
1−
= 3.8µH
For the feedback resistors, choose R1 = 301k. R2 can
then be calculated from equation (2) to be:
2.25MHz(100mA)
4.2V
A 4.7µH inductor works well for this application. For best
efficiency choose a 350mA or greater inductor with less
than 0.3Ω series resistance.
V
0.8
⎛
⎜
⎝
⎞
⎠
OUT
R2=
−1 R1= 827.8k;use 825k
⎟
Figure 6 shows the complete circuit along with its
efficiency curve.
CIN will require an RMS current rating of at least 0.125A ≅
I
LOAD(MAX)/2 at temperature and COUT will require an ESR
4.7µH*
V
IN
4
1
3
6
V
OUT
3V
2.7V
V
SW
LTC3410
RUN
IN
††
10pF
C
TO 4.2V
IN
†
C
4.7µF
CER
OUT
10µF
CER
V
FB
825k
GND
2, 5
†TAIYO YUDEN JMK212BJ106
††TAIYO YUDEN JMK212BJ475
*MURATA LQH32CN4R7M23
301k
3410 F06a
Figure 6a
100
90
80
70
60
50
40
30
20
V
OUT
100mV/DIV
AC COUPLED
I
L
200mA/DIV
I
LOAD
200mA/DIV
V
V
= 3.6V
= 4.2V
IN
IN
10
0
0.1
1
10
(mA)
100
1000
20µs/DIV
3410 F06c
I
V
V
LOAD
= 3.6V
= 3V
LOAD
IN
OUT
3410 F06b
I
= 100mA TO 300mA
Figure 6b
Figure 6c
3410fb
13
Linear [ Linear ]
