Datasheet
LTC3735
24
3735fa
APPLICATIONS INFORMATION
Design Example
As a design example, assume V
IN
= 12V (nominal), V
IN
= 21V (max), V
OUT
= 1.5V, I
MAX
= 35A, and f = 350kHz
(each phase).
The inductance value is chosen first based on a 40% ripple
current assumption. The highest value of ripple current
occurs at the maximum input voltage. The minimum
inductance for 40% ripple current is:
L ≥
V
OUT
f • ∆I
• 1–
V
OUT
V
IN
=
1.5V
350kHz • 40%•17.5A
( )
•
1–
1.5V
21V
= 0.57µH
Using L = 0.6µH, a common “off-the-shelf” value results
in 38%ripple current. The peak inductor current will be
the maximum DC current plus one half of the ripple cur-
rent, or 21A.
Tie the FREQSET pin to 1.2V, resistively divided down from
PV
CC
to have 350kHz operation for each phase.
The minimum on-time also occurs at maximum input
voltage:
t
ON(MIN)
=
V
OUT
V
IN
• f
=
1.5V
21V • 350kHz
= 204ns
which is larger than 150ns, the typical minimum on time
of the LTC3735.
R
SENSE1
and R
SENSE2
can be calculated by using a con-
servative maximum sense voltage threshold of 40mV and
taking into account of the peak current:
R
SENSE
=
40mV
21A
= 0.002Ω
The power loss dissipated by the top MOSFET can be cal-
culated with equations 3 and 7. Using a Fairchild FDS7760
as an example: R
DS(ON)
= 8mΩ, Q
G
= 55nC at 5V V
GS
, C
RSS
= 307pF, V
TH(MIN)
= 1V. At maximum input voltage with
T
J
(estimated) = 85°C at an elevated ambient temperature:
P
TOP
=
1.5V
21V
•
35A
2
2
• 1+ 0.005 • 85°C – 25°C
( )
( )
•
0.008Ω +
21V
2
•17.5A
2
• 350kHz • 307pF •
2Ω •
1
5V – 1V
+
1
1V
= 1.26W
Equation 4 gives the worst-case power loss dissipated
by the bottom MOSFET (assuming FDS7760 and T
J
=
85°C again):
P
BOT
=
21V – 1.5V
21V
•
35A
2
2
•
1+ 0.005 • 85°C – 25°C
( )
( )
• 0.008Ω
= 2.95W
Therefore it is necessary to have two FDS7760s in parallel
to split the power loss.
A short-circuit to ground will result in a folded back cur-
rent of about:
I
SC
=
25mV
0.002Ω
+
1
2
•
200ns • 21V
0.6µH
= 16A
The worst-case power dissipation by the bottom MOSFET
under short-circuit conditions is:
P
BOT
=
1
350kHz
– 200ns
1
350kHz
• 16A
( )
2
•
1+ 0.005 • 85°C – 25°C
( )
( )
• 0.008Ω
=
2.48W
which is less than normal, full load conditions.
The nominal duty cycle of this application is equation 1:
DC =
1.5V
12V
= 12.5%