I
=
L(RMS)
(I
L(avg)
2 + 1
12
(I
)
RIPPLE
2
) =
(I
)
OUT
2 + 1
12 (I
)
RIPPLE
2
(5)
OUTPUT CAPACITOR SELECTION
C
=
OUT(MIN)
I
L
TRAN(MAX)
2
(V
-V
) V
IN(MIN)
OUT
TRAN
when
<2
V
IN(MIN)
OUT
(6)
C
=
OUT(MIN)
I
L
TRAN(MAX)
2
V
OUT
TRAN
when
>2
V
IN(MIN)
OUT
(7)
ESR
=
MAX
V
-V
RIPPLE(total)
RIPPLE(cap)
I
RIPPLE
=
V
-
RIPPLE(total)
I
RIPPLE
(
I
RIPPLE
C
f
OUT
SW
)
(8)
PEAK CURRENT RATING OF THE INDUCTOR
I
=
CHARGE
V
C
OUT
T
SS
(9)
I
=I
L(PEAK)
OUT(MAX)
+ 1
2
I
RIPPLE
CHARGE
+
(10)
INPUT CAPACITOR SELECTION
SLVSA10A – SEPTEMBER 2009 – REVISED SEPTEMBER 2009.................................................................................................................................... www.ti.com
Using
Equation 5, the maximum RMS current in the inductor is about 6.01 A.
The selection of the output capacitor is typically driven by the output load transient response requirement. The
output capacitance (base) proposed is 2 x 22
F (or 4 X 10 F or 1 x 47 F) ceramic, providing a good balance
between ripple, cost and performance. Extra caps added should be electrolyte or far from base cap to have
considerable amount of ESR and not to affect compensation.
Equation 6 and
Equation 7 estimate the output capacitance required for a given output voltage transient
deviation.
For this example,
Equation 7 is used in calculating the minimum output capacitance.
Based on a 4-A load transient with a maximum 125-mV deviation (2.5% of set voltage), a minimum of 120-
F
output capacitance is required. We choose two 22-
F ceramic capacitors and two electrolytic 47 F in parallel for
a total capacitance of 138
F.
The output ripple is divided into two components. The first is the ripple generated by the inductor ripple current
flowing through the output capacitor’s capacitance, and the second is the voltage generated by the ripple current
flowing in the output capacitor’s ESR. The maximum allowable ESR is then determined by the maximum ripple
Based only on the 138-
F of capacitance, 1.25-A ripple current, 500-kHz switching frequency and a design goal
of 50-mV ripple voltage (1% of set voltage), we calculate a capacitive ripple component of 18 mV and a
maximum ESR of 25 m
. The X5R ceramic capacitors selected provide significantly less than 25-m of ESR.
With output capacitance known, it is now possible to calculate the charging current during start-up and determine
the minimum saturation current rating of the inductor. The start-up charging current is approximated by
Using the TPS65230 and TPS65231’s recommended 1.3-ms soft-start time, COUT = 188 F and VOUT = 5 V,
ICHARGE is found to be 720 mA. The peak current rating of the inductor is now found by Equation 10. For this example an inductor with a peak current rating of 7.3 A is required. Note however that the inductor will
need to withstand the current limit figure without a major reduction of its rated inductance.
The input voltage ripple is divided between capacitance and ESR. For this design, VRIPPLE(cap) = 60 mV (0.5% of
supply) and VRIPPLE(ESR) = 30 mV (0.25% of supply). The minimum capacitance and maximum ESR are
18
Copyright 2009, Texas Instruments Incorporated