L3380
CMOS IC
APPLICATION CIRCUIT INFORMATION
The following equations show the relation of the basic design parameters.
1. Refer to the application circuit, the increasing inductor current when the switch is turn on is given by the following
equation
∆iL+
=
1
L U LTON
=
1
L
(U
IN
−US )
d
f
(U IN :input voltage ; U S :transistor saturation voltage)
The decreasing inductor current when the switch is turn off can derive by the equation below
∆iL−
=
1
L U TL OFF
=
−
1
L
(U
O
+UD
−
U
IN
)
1
−
f
d
(U D :diode forward voltage)
according to ∆iL+ + ∆iL− = 0 ,the duty ratio is given by
d = UO + U D −U IN
UO +UD −US
2. The average current flowing through the inductor is
IL
=
IO
1− d
3. We note that IO = (1− d )IL
then
we
can
write: IO
=
(1− d )
IL
∆ iL
• ∆iL
substituting
∆iL
=
1
L U TL OFF
∆ iL
for equation above, output current is given by
IO
=
(1 −
d)•
1
ICR
•
1
L U TL OFF
( ICR = ∆iL )
IL
IO
=
(1 −
d)•
1
ICR
•
1
L
(Uo + U D
− U IN
)1−
f
d
IO=(1-d
)2•
UO+UD
ICR •
– UIN
L• f
derive that
(1– d)2(UO+UD-UIN)
L=
ICR • IO • f
4. The peak current of the inductor is given by
I PK
=
IL
+
1
2
∆ iL
I PK
=
IL
+1
2
∆iL
IL
• IL
according to ICR = ∆iL derive that
IL
I PK
=
IL
+
1
2
ICR • IL
Then derive the following equation for peak current of inductor
I PK
=
IL (1+
1
2
ICR)
UNISONIC TECHNOLOGIES CO., LTD
www.unisonic.com.tw
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QW-R502-099,A