TPS40200-Q1
SLUS739D –SEPTEMBER 2006–REVISED JULY 2011
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Inductor Selection Criteria
The TPS40200 P-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 LSRs (internal resistance) for their various-sized ferrites.
In this application, the part must deliver a maximum current of 2.5 A. This requires that the output inductor’s
saturation current be above 2.5 A plus one-half 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-sized 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 one-half the peak current that must develop during a minimum on time. The conditions for
minimum on time are high line and low load.
LMAX is calculated using:
V
IN − VOUT
LMAX
=
× tON = 32 µH
Where:
VIN = 16 V
IPEAK
VOUT = 3.3 V
IPEAK = 0.25 A
tON = 0.686 μs
3.3 V
1
´
300 kHz 16 V
tON is given by
The inductor used in the circuit is the closest standard value of 33 μH. This is the maximum inductance that can
be used in the converter to deliver the minimum current while maintaining continuous conduction.
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