L3380 Ver la hoja de datos (PDF) - Unisonic Technologies
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L3380 Datasheet PDF : 9 Pages
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L3380
APPLICATION CIRCUIT INFORMATION (Cont.)
CMOS IC
5. Charge stores in C3 during charging up is given by ∆Q = IC • TOFF
we can write
∆Q
=
(IL
−
IO
)
•
1−
f
d
6. Output ripple voltage is given by
VPP = ∆UC + ESR • (IL − IO )
(ESR: equivalent series resistance of the output capacitor)
VPP
=
∆Q
C
+
ESR • (IL
−
IO )
Then we give the following example about choosing external components by considering the design parameters.
Design parameters: U IN =1.5V Uo =2.1V IO =200mA VPP =100mV f=300KHZ ICR=0.2
Assume U D and U S are both 0.3V, the duty ratio is
d = UO + UD −UIN = 2.1+ 0.3 −1.5 = 0.429
UO +UD −US 2.1+ 0.3 − 0.3
In order to generate the desired output current and ICR, the value of inductor should meets the following formula
L≤
(1– d)2(UO+UD-UIN)
(1– 0.429)2(2.1V+0.3V-1.5V)
=
= 24.5uH
ICR • IO • f
0.2×0.2A×300000HZ
Calculate the average current and the peak current of inductor
IL
= IO
1− d
= 0.2A
1 − 0.429
=
0.35 A
I PK
=
IL (1+
1
2
ICR)
= 0.35A× (1+
1 × 0.2)
2
=
0.385 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
∆Q
=
(IL
−
I
O
)
•
1
−
f
d
=
(0.35A − 0.2A)× 1− 0.429
300000HZ
= 0.286uC
Assume the ESR of C3 is 0.15Ω, determine the value of C3
C≥
∆Q
=
0.286 ×10−6 C
= 3.69uF
VPP − ESR • (IL − IO ) 0.1− 0.15Ω × (0.35A − 0.2 A)
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.
UNISONIC TECHNOLOGIES CO., LTD
www.unisonic.com.tw
7 of 9
QW-R502-099,A
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