NCP1410_05 ONSEMI [ON Semiconductor], NCP1410_05 Datasheet - Page 12

• Manufacturer Part Number: NCP1410_05
• Description: 250 mA Sync-Rect PFM Step-Up DC-DC Converter with Low-Battery Detector
• Manufacturer: ONSEMI [ON Semiconductor]

GENERAL DESIGN PROCEDURES
magic to most engineers, some complicate empirical
formulae are available for reference usage. Those formulae
are derived form the assumption that the key components,
i.e. power inductor and capacitors are available with no
tolerance. Practically, its not true, the result is not a matter
of how accurate the equations you are using to calculate the
component values, the outcome is still somehow away from
the optimum point. In below a simple method base on the
most basic first order equations to estimate the inductor and
capacitor values for NCP1410 operate in Continuous
Conduction Mode is introduced. The component value set
can be used as a starting point to fine tune the circuit
operation. By all means, detail bench testing is needed to get
the best performance out of the circuit.
Design Parameters:
Calculate the feedback network:
Calculate the Low Battery Detect divider:
Switching mode converter design is considered as black
V
V
I
V
V
Select R
V
Select R
OUT
IN
OUT
LB
OUT−RIPPLE
LB
= 1.8 V to 3.0 V, Typical 2.4 V
= 2.0 V
= 2.0 V
= 200 mA (250 mA max)
= 3.3 V
R FB1 + R FB2
R FB1 + 200 K
R LB1 + R LB2
R LB1 + 330 K
FB2
LB2
= 200 K
= 330 K
= 40 mV
P−P
V REF
V OUT
1.19 V
V REF
V LB
1.19 V
2.0 V
3.3 V
at I
OUT
* 1
* 1
* 1 + 225 K
* 1 + 355 K
= 250 mA
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NCP1410
12
operation will be optimized around this point:
maximum I
calculate the inductor value:
power inductor can be calculated as in below:
I
Standard value of 22 mH is selected for initial trial.
calculate the output capacitor value:
V
where t
order to achieve the specified ripple level at conditions
stated. Practically, a one level larger capacitor will be used
to accommodate factors not take into account in the
calculation, therefore a capacitor value of 33 mF is selected.
RIPPLE−P
OUT−RIPPLE
Determine the Steady State Duty Ratio, D for typical V
Determine the average inductor current, I
Determine the peak inductor ripple current, I
Assume I
Determine the output voltage ripple, V
From above calculation, you need at least 23.33 mF in
C OUT u
C OUT u
L +
ON
I LAVG +
2 I RIPPLE * P
D + 1 *
= 0.20 x 344 mA = 68.8 mA
V IN
OUT
RIPPLE−P
= 1.4 mS and ESR
40 mV * 250 mA
V OUT−RIPPLE * I OUT
= 40 mV
:
t ON
250 mA
1 * D
I OUT
V OUT
is 20% of I
V IN
V OUT
V IN
P−P
+
+
+ 1 * 2.4 V
at I
I OUT
2.4 V
+
1 * 0.273
1.4 mS
250 mA
2(68.8 mA)
OUT
COUT
1 * D
LAVG
1
3.3 V
0.1 W
= 250 mA
t ON
1.4 mS
= 0.1 W,
, the inductance of the
+ 344 mA
+ 0.273
+ 23.33 mF
ESR COUT
OUT−RIPPLE
+ 24.4 mH
RIPPLE−P
LAVG
and
and
IN
at
,