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APPLICATIONS INFORMATION
INTV
CC
60k
20k
INTV
CC
THERMAL
SENSOR
L1
0.33µH
(0.32m DCR)
R2
R1
931
4.64k
V
OUT
+
C
C
OUT3
OUT4
470µF
100µF
×2
×2
Figure 16. High Efficiency Dual Phase 400kHz, 1.5V/60A Step-Down Converter with Optional Thermal Balancing
The frequency is set by biasing the FREQ pin to 1V (see
Figure 12).
The inductance values are based on a 35% maximum
ripple current assumption (10.5A for each channel). The
highest value of ripple current occurs at the maximum
input voltage:
V
OUT
L =
1–
f • ∆I
V
L(MAX)
IN(MAX)
This design will require 0.33µH. The Würth 744301033,
0.32µH inductor is chosen. At the nominal input voltage
(12V), the ripple current will be:
V
OUT
∆I
=
1–
L(NOM)
f •L
4.7µF
V
D1
CMDSH-3
RUN1,2
M1
ILIM
ENTMPB
BSC050NE2LS
TG1
C
0.1µF
B1
BOOST1
SW1
M2
EXTV
BSC010NE2LSI
BG1
TAVG
R
10k
TRSET1
TRSET1
SNSA1
220nF
SNS1
220nF
SNSD1
TCOMP1
V
OSNS1
V
OSNS1
I
TH1
TK/SS1
R
TAVG
5k
V
OUT
V
OUT
V
IN(NOM)
For more information
INTV
IN
CC
PHASMD
D2
CLKOUT
CMDSH-3
PGOOD
IFAST
MODE/PLLIN
BSC050NE2LS
TG2
C
0.1µF
LTC3875
B2
BOOST2
SW2
CC
BSC010NE2LSI
BG2
PGND
R
10k
TRSET2
TRSET2
+
+
SNSA2
220nF
–
–
SNS2
220nF
+
+
SNSD2
TCOMP2
+
FREQ
–
+
V
OSNS2
–
V
OSNS2
I
TH2
TK/SS2
100k
0.1µF
It will have 10A (33%) ripple. The peak inductor current
will be the maximum DC value plus one-half the ripple
current, or 35A.
The minimum on-time occurs at the maximum V
should not be less than 90ns:
t
=
ON(MIN)
V
DCR sensing is used in this circuit. If C1 and C2 are chosen
to be 220nF, based on the chosen 0.33µH inductor with
0.32mΩ DCR, R1 and R2 can be calculated as:
L
R1=
DCR •C1
L
R2 =
DCR •C2 • 5
www.linear.com/LTC3875
LTC3875
V
IN
4.5V TO 20V
10µF
270µF
×4
50V
THERMAL
M3
SENSOR
L2
0.33µH
(0.32m DCR)
M4
R1
R2
4.64k
931
R
30.1k
B
C
1.5nF
OUT1
R
A
100µF
20k
×2
10k
V
1.5V
OUT
=
( )
(
f
20V 400kHz
IN(MAX)
= 4.69k
= 937Ω
V
OUT
1.5V
60A
+
C
OUT2
470µF
×2
3875 F16
, and
IN
= 187ns
)
3875fb
35
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