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NCP3125
when using electrolytic capacitors, a lower ripple current
will result in lower output ripple due to the higher ESR of
electrolytic capacitors. The ratio of ripple current to
I PK + I OUT @ 1 )
ra
2
3 4.6 A + 4 A @ 1 )
30%
2
(eq. 8)
ra + D I (eq. 5)
maximum output current is given in Equation 5.
Iout
D I = Ripple current
I OUT = Output current
ra = Ripple current ratio
Using the ripple current rule of thumb, the user can
establish acceptable values of inductance for a design using
Equation 6.
I OUT = Output current
I PK = Inductor peak current
ra = Ripple current ratio
A standard inductor should be found so the inductor will
be rounded to 5.6 m H. The inductor should also support an
RMS current of 4.01 A and a peak current of 4.6 A.
The final selection of an output inductor has both
mechanical and electrical considerations. From a
mechanical perspective, smaller inductor values generally
V OUT
I OUT @ ra @ F SW
5.7 m H +
SlewRate LOUT + 3 1.53 +
L OUT
L OUT +
D
F SW
I OUT
L OUT
ra
18
16
14
@ (1 * D) 3
3.3 V
4 A @ 30% @ 350 kHz
= Duty ratio
= Switching frequency
= Output current
= Output inductance
= Ripple current ratio
V IN = 12 V
(eq. 6)
@ (1 * 27.5%)
correspond to smaller physical size. Since the inductor is
often one of the largest components in the regulation system,
a minimum inductor value is particularly important in space
constrained applications. From an electrical perspective, the
maximum current slew rate through the output inductor for
a buck regulator is given by Equation 9.
V IN * V OUT A  12 V * 3.3 V
m s 5.6 m H
(eq. 9)
L OUT = Output inductance
V IN = Input voltage
V OUT = Maximum output voltage
Equation 9 implies that larger inductor values limit the
regulator ’s ability to slew current through the output
inductor in response to output load transients. Consequently,
6
V IN = 8 V
V IN = 4.2 V
V OUT (1 * D)
L OUT @ F SW
1.2 A +
12
10
8
Selected
4
2
10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40
CURRENT RIPPLE RATIO (%)
Figure 21. Inductance vs. Current Ripple Ratio
When selecting an inductor, the designer must not exceed
the current rating of the part. To keep within the bounds of
the part’s maximum rating, a calculation of the RMS and
peak inductor current is required.
output capacitors must supply the load current until the
inductor current reaches the output load current level.
Reduced inductance to increase slew rates results in larger
values of output capacitance to maintain tight output voltage
regulation. In contrast, smaller values of inductance increase
the regulator ’s maximum achievable slew rate and decrease
the necessary capacitance, at the expense of higher ripple
current. The peak ? to ? peak ripple current for NCP3125 is
given by the following equation:
Ipp + 3
(eq. 10)
3.3 V   (1 * 27.5%)
5.6 m H @ 350 kHz
D = Duty ratio
F SW = Switching frequency
Ipp = Peak ? to ? peak current of the inductor
1 ) ra 3
I RMS + I OUT @
2
12
4.01 A + 4 A * 1 )
30% 2
12
(eq. 7)
L OUT = Output inductance
V OUT = Output voltage
From Equation 10 it is clear that the ripple current increases
as L OUT decreases, emphasizing the trade ? off between
dynamic response and ripple current.
I OUT
I RMS
ra
= Output current
= Inductor RMS current
= Ripple current ratio
The power dissipation of an inductor falls into two
categories: copper and core losses. The copper losses can be
further categorized into DC losses and AC losses. A good
first order approximation of the inductor losses can be made
using the DC resistance as shown below:
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