Datasheet

ON
PEAK
OUTIN
MIN
t
I
VV
L ´
-
= =32 Hm
V16
V3.3
kHz300
1
´
TPS40200
www.ti.com SLUS659F –FEBRUARY 2006–REVISED MARCH 2012
Inductor Selection Criteria
The P-channel FET driver facilitates switching the power FET at a high frequency. This, in turn, enables the use
of smaller, less-expensive inductors as illustrated in this 300-kHz application. Ferrite, with its good high
frequency properties, is the material of choice. Several manufacturers provide catalogs with inductor saturation
currents, inductance values, and LSR’s (internal resistance) for their various-sized ferrites.
In this application, the device must deliver a maximum current of 2.5 A. This requires that the output inductor’s
saturation current be above 2.5 A plus ½ the ripple current caused during inductor switching. The value of the
inductor determines this ripple current. A low value of inductance has a higher ripple current that contributes to
ripple voltage across the resistance of the output capacitors. The advantages of a low inductance are a higher
transient response, lower DCR, a higher saturation current, and a smaller, less expensive part. Too low an
inductor however, leads to higher peak currents which ultimately are bounded by the overcurrent limit set to
protect the output FET or by output ripple voltage. Fortunately, with low ESR Ceramic capacitors on the output,
the resulting ripple voltage for relatively high ripple currents can be small.
For example, a single 1-μf, 1206 size, 6.3-V, ceramic capacitor has an internal resistance of 2 Ω at 1 MHz. For
this 2.5 A application, a 10% ripple current of 0.25 A produces a 50-mV ripple voltage. This ripple voltage may be
further reduced by additional parallel capacitors.
The other bound on inductance is the minimum current at which the controller enters discontinuous conduction.
At this point, Inductor current is zero. The minimum output current for this application is specified at 0.125 A. This
average current is ½ the peak current that must develop during a minimum on time. The conditions for minimum
on time are high line and low load.
Using:
where
• V
IN
= 16 V
• V
OUT
= 3.3 V
• I
PEAK
= 0.25 A
• t
ON
= 0.686 μs
• t
ON
is given by
(14)
The inductor used in the circuit is the closest standard value of 33 μH. This is the minimum inductance that can
be used in the converter to deliver the minimum current while maintaining continuous conduction.
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