AIC1550
∆IL
V
OUT is substituted by 1.8V in equation (2) as
= IOUT,MAX
+
ID,MAX(ON)
2
V
1.8
OUT
250mA
2
L1>
=
= 3.33µH
= 800mA +
= 925mA
0.54
0.54
Let L1 = 6.8µH, and choose CF = 12pF, R1 =
820kΩ.
ID,avg(ON) = (1−D)×IOUT
Co calculated by the following formula can
improve circuit stability.
1.8
= (1−
)× 800mA
4.2
= 457.14mA
1
1
≅
R1× CF
L1× C
According the datas above, the Schottky diode,
SS12, from GS is recommend.
O
Therefore,
For feedback resistors, choose R2=390kΩ and R1
can be calculated as follow:
2
2
(
R1× CF
)
(
820k × 12pF
)
= 12µF
C
=
=
O
L1
6.8µ.
1.8V
0.75
R1 =
− 1 × 390kΩ = 546kΩ ; use 560kΩ
Say, CO is 22µF. Then, R2 can be decided by
equation (8) as
Fig. 22 shows the application circuit of AIC1550,
and Fig. 23 to 26 show the layout diagrams of it.
V
R1
R2
1.8
OUT
=
− 1 =
− 1 = 1.4
V
0.75
ref
2. Ceramic capacitors application:
So, R2 = 560kΩ.
Of the same AIC1550 application above, except
for ceramic capacitor used, Co, R1, and R2 can
be calculated as following formulas. And the same
values of load current and output voltage at
800mA and 1.8V respectively are used.
Note: Schottky diode, SS12 from GS, is still
required in this application.
V
= 2.5V to 6.5V
IN
V
= 1.8V
OUT
L1
10 F
8
1
LX
VIN
BP
µ
7
6
2
3
**
GND
SYNC/
MODE
BP
+
D1
CF
SS12
SHDN
FB
C
IN
R1
560K
10P
5
4
Optional
10µF
RT
C
BP
0.1 F
+
*C
O2
µ
AIC1550
4.7 F
µ
*C
O1
33 F
µ
R2
390K
* Note: CO1 can be omitted if CO2 in 10 F Ceramic
µ
CIN: NIPPON 10µF/6V Tantalum capacitor
CO1: NIPPON 33µF/6V Tantalum capacitor
L: TDK SLF6025-100M1R0
** Note: Efficiency can boost 2% to 4% if D1 is connected.
D1: GS SS12
Fig. 23 AIC1550 Application Circuit (Tantalum capacitor application)
13
AIC [ ANALOG INTERGRATIONS CORPORATION ]
