Datasheet

+
+
-
-
sw
i
i
)t(
1L
1L
R
v
L1
( )
t
IN
V
2L
R
i
)t(
2L
i
)t(
1D
+
-
v
C1
( )
t
v
D1
( )
t
v
L2
( )
t
on
R
i
)t(
2C
+
-
v
C2
( )
t
+
-
v
O
( )
t
i
)t(
1C
V
O
V
IN
L
1
D
1
C
1
C
2
R
2
R
1
C
3
R
3
C
5
C
4
L
2
C
6
1
2
3
6
5
4
LM2735
D
D
'
=
( )
V
o
V
IN
D
D
V
'
C1
=
( )
V
o
LM2735
SNVS485F –JUNE 2007–REVISED APRIL 2013 www.ti.com
Applying Charge balance on C1:
(53)
Since there are no DC voltages across either inductor, and capacitor C6 is connected to Vin through L1 at one
end, or to ground through L2 on the other end, we can say that
V
C1
= V
IN
(54)
Therefore:
(55)
This verifies the original conversion ratio equation.
It is important to remember that the internal switch current is equal to I
L1
and I
L2
. During the D interval. Design
the converter so that the minimum specified peak switch current limit (2.1A) is not exceeded.
Figure 33. SEPIC CONVERTER Schematic
Steady State Analysis with Loss Elements
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