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L3380L-18-AF5-R 数据表(PDF) 7 Page - Unisonic Technologies

部件名 L3380L-18-AF5-R
功能描述  PWM STEP UP DC-DC CONTROLLER
PDF  9 Pages
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制造商  UTC [Unisonic Technologies]
网页  http://www.utc-ic.com
标志 UTC - Unisonic Technologies

L3380L-18-AF5-R 数据表(HTML) 7 Page - Unisonic Technologies

  L3380L-18-AF5-R Datasheet HTML 1Page - Unisonic Technologies L3380L-18-AF5-R Datasheet HTML 2Page - Unisonic Technologies L3380L-18-AF5-R Datasheet HTML 3Page - Unisonic Technologies L3380L-18-AF5-R Datasheet HTML 4Page - Unisonic Technologies L3380L-18-AF5-R Datasheet HTML 5Page - Unisonic Technologies L3380L-18-AF5-R Datasheet HTML 6Page - Unisonic Technologies L3380L-18-AF5-R Datasheet HTML 7Page - Unisonic Technologies L3380L-18-AF5-R Datasheet HTML 8Page - Unisonic Technologies L3380L-18-AF5-R Datasheet HTML 9Page - Unisonic Technologies  
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L3380
CMOS IC
UNISONICTECHNOLOGIESCO.,LTD
7 of 9
www.unisonic.com.tw
QW-R502-099,A
APPLICATION CIRCUIT INFORMATION (Cont.)
5. Charge stores in C3 during charging up is given by
COFF
QI
T
∆ =•
we can write
1
()
LO
d
QI
I
f
∆=
6. Output ripple voltage is given by
()
PPC
L
O
VU
ESR
I
I
=∆
+
(ESR: equivalent series resistance of the output capacitor)
()
PPL
O
Q
VESR
I
I
C
=+
Then we give the following example about choosing external components by considering the design parameters.
Design parameters:
IN
U =1.5V Uo =2.1V
O
I =200mA
PP
V =100mV f=300KHZ ICR=0.2
Assume
D
U and
S
U are both 0.3V, the duty ratio is
2.1 0.3 1.5
0.429
2.1 0.3 0.3
OD
IN
OD
S
UU
U
d
UU
U
+−
+−
==
=
+−
+
In order to generate the desired output current and ICR, the value of inductor should meets the following formula
L ≤
(1– d)
2(U
O+UD-UIN)
ICR
IO f
=
(1– 0.429)
2(2.1V+0.3V-1.5V)
0.2×0.2A×300000HZ
= 24.5uH
Calculate the average current and the peak current of inductor
0.2
0.35
1
1 0.429
O
L
I
A
I
A
d
==
=
−−
11
(1
)
0.35
(1
0.2)
0.385
22
PK
L
I
IICR
A
A
=+
=
× + ×
=
So, we make a trial of choosing a 22uH inductor that allowable maximum current is lager than 0.385A.
Determine the delta charge stores in C3 during charging up
1
1 0.429
(
)
(0.35
0.2 )
0.286
300000
LO
d
QI
I
A
A
uC
fHZ
−−
∆=
=
×
=
Assume the ESR of C3 is 0.15Ω, determine the value of C3
6
0.286 10
3.69
(
)
0.1 0.15
(0.35
0.2 )
PP
L
O
QC
CuF
VESR
I
I
A
A
∆×
≥=
=
−•
Ω×
Therefore, a Tantalum capacitor with value of 10uF and ESR of 0.15Ω can be used as output capacitor. However,
the optimized value should be obtained by experiment.



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