Datasheet

NCP1422
http://onsemi.com
11
TYPICAL APPLICATION CIRCUIT
LBI/EN
FB
LBO
REF BAT
GND
LX
OUT
NCP1422
R4
330 k
R2 200 k
Shutdown
Open Drain
Input
Low Battery
Open Drain
Output
C3
200 nF
R1
350 k
C2
33 F
+
V
OUT
= 3.3 V
800 mA
C1
22 F
V
IN
L
6.5 H
Figure 23. Typical Application Schematic for 2 Alkaline Cells Supply
R3
220 k
C4
10 p*
*Optional
GENERAL DESIGN PROCEDURES
Switching mode converter design is considered a
complicated process. Selecting the right inductor and
capacitor values can allow the converter to provide
optimum performance. The following is a simple method
based on the basic first−order equations to estimate the
inductor and capacitor values for NCP1422 to operate in
Continuous Conduction Mode (CCM). The set component
values can be used as a starting point to fine tune the
application circuit performance. Detailed bench testing is
still necessary to get the best performance out of the circuit.
Design Parameters:
V
IN
= 1.8 V to 3.0 V, Typical 2.4 V
V
OUT
= 3.3 V
I
OUT
= 500 mA
V
LB
= 2.0 V
V
OUT−RIPPLE
= 40 mV
p−p
at I
OUT
= 500 mA
Calculate the feedback network:
Select R2 = 200 k
R1 + R2
ǒ
V
OUT
V
REF
* 1
Ǔ
R1 + 200 k
ǒ
3.3 V
1.20 V
* 1
Ǔ
+ 350 k
Calculate the Low Battery Detect divider:
V
LB
= 2.0 V
Select R4 = 330 k
R3 + R4
ǒ
V
LB
V
REF
* 1
Ǔ
R3 + 300 k
ǒ
2.0 V
1.20 V
* 1
Ǔ
+ 220 k
Determine the Steady State Duty Ratio, D, for typical
V
IN
. The operation is optimized around this point:
V
OUT
V
IN
+
1
1 * D
D + 1 *
V
IN
V
OUT
+ 1 *
2.4 V
3.3 V
+ 0.273
Determine the average inductor current, I
LAVG,
at
maximum I
OUT
:
I
LAVG
+
I
OUT
1 * D
+
500 mA
1 * 0.273
+ 688 mA
Determine the peak inductor ripple current, I
RIPPLE−P,
and calculate the inductor value:
Assume I
RIPPLE−P
is 20% of I
LAVG
. The inductance of the
power inductor can be calculated as follows:
L +
V
IN
t
ON
2I
RIPPLE*P
+
2.4 V 0.75 S
2 (137.6 mA)
+ 6.5 H
A standard value of 6.5 H is selected for initial trial.
Determine the output voltage ripple, V
OUT−RIPPLE,
and
calculate the output capacitor value:
V
OUT−RIPPLE
= 40 mV
P−P
at I
OUT
= 500 mA
C
OUT
u
I
OUT
t
ON
V
OUT*RIPPLE
* I
OUT
ESR
COUT
where t
ON
= 0.75 S and ESR
COUT
= 0.05 ,
C
OUT
u
500 mA 0.75 S
45 mV * 500 mA 0.05 
+ 18.75 F