L3380 Datasheet, PDF (7/9 Pages) Unisonic Technologies – PWM STEP UP DC-DC CONTROLLER

L3380
APPLICATION CIRCUIT INFORMATION
(Cont.)
CMOS IC
5. Charge stores in C3 during charging up is given by ∆Q = I_C • T_OFF
we can write
∆Q
=
(I_L
−
I_O
)
•
1−
f
d
6. Output ripple voltage is given by
V_PP = ∆U_C + ESR • (I_L − I_O )
(ESR: equivalent series resistance of the output capacitor)
V_PP
=
∆Q
C
+
ESR • (I_L
−
I_O )
Then we give the following example about choosing external components by considering the design parameters.
Design parameters: U _IN =1.5V Uo =2.1V I_O =200mA V_PP =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
U_O +U_D −U_S 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(U_O+U_D-U_IN)
(1– 0.429)²(2.1V+0.3V-1.5V)
=
= ^24.5uH
ICR • I_O • f
0.2×0.2A×300000HZ
Calculate the average current and the peak current of inductor
I_L
= ^IO
1− d
= ^0.2A
1 − 0.429
=
0.35 A
I _PK
=
I_L (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
=
(I_L
−
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
V_PP − ESR • (I_L − I_O ) 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.




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