LT1375/ LT1376
U
W
U U
APPLICATIONS INFORMATION
Example: with L = 2µH, VOUT = 5V, and VIN(MAX) = 15V,
must withstand continuous fault conditions. If maxi-
mum load current is 0.5A, forinstance, a 0.5A inductor
may not survive a continuous 1.5A overload condition.
Dead shorts will actually be more gentle on the induc-
tor because the LT1376 has foldback current limiting.
2
1.5 500 •103 2 •10−6 15
(
)
( )
IOUT MAX
=
= 338mA
(
)
2 5 15 − 5
( )(
)
2. Calculate peak inductor current at full load current to
ensure that the inductor will not saturate. Peak current
can be significantly higher than output current, espe-
cially with smaller inductors and lighter loads, so don’t
omit this step. Powdered iron cores are forgiving
because they saturate softly, whereas ferrite cores
saturate abruptly. Other core materials fall in between
somewhere. The following formula assumes continu-
ous mode of operation, but it errs only slightly on the
high side for discontinuous mode, so it can be used for
all conditions.
The main reason for using such a tiny inductor is that it is
physically very small, but keep in mind that peak-to-peak
inductorcurrentwillbeveryhigh. This willincreaseoutput
ripplevoltage.Iftheoutputcapacitorhas tobemadelarger
to reduce ripple voltage, the overall circuit could actually
wind up larger.
CHOOSING THE INDUCTOR AND OUTPUT CAPACITOR
For most applications the output inductor will fall in the
range of 3µH to 20µH. Lower values are chosen to reduce
physical size of the inductor. Higher values allow more
output current because they reduce peak current seen by
the LT1376 switch, which has a 1.5A limit. Higher values
also reduce output ripple voltage, and reduce core loss.
Graphs intheTypicalPerformanceCharacteristics section
show maximum output load current versus inductor size
andinputvoltage. Asecondgraphshows coreloss versus
inductor size for various core materials.
V
V −V
OUT
(
)
OUT IN
IPEAK =IOUT
+
2 f L V
( )( )( )
IN
V = Maximum input voltage
IN
f = Switching frequency, 500kHz
3. Decide if the design can tolerate an “open” core geom-
etry like a rod or barrel, which have high magnetic field
radiation, orwhetheritneeds aclosedcorelikeatoroid
to prevent EMI problems. One would not want an open
core next to a magnetic storage media, for instance!
This is atoughdecisionbecausetherods orbarrels are
temptingly cheap and small and there are no helpful
guidelines to calculate when the magnetic field radia-
tion will be a problem.
When choosing an inductor you might have to consider
maximum load current, core and copper losses, allowable
component height, output voltage ripple, EMI, fault cur-
rent in the inductor, saturation, and of course, cost. The
following procedure is suggested as a way of handling
thesesomewhatcomplicatedandconflictingrequirements.
1. Choose a value in microhenries from the graphs of
maximumloadcurrentandcoreloss.Choosingasmall
inductor with lighter loads may result in discontinuous
mode of operation, but the LT1376 is designed to work
well in either mode. Keep in mind that lower core loss
means higher cost, at least for closed core geometries
like toroids. The core loss graphs show both absolute
loss andpercentloss fora5Woutput,soactualpercent
losses must be calculated for each situation.
4. Start shopping for an inductor (see representative
surface mount units in Table 2) which meets the re-
quirements of core shape, peak current (to avoid satu-
ration), average current (to limit heating), and fault
current(iftheinductorgets toohot, wireinsulationwill
melt and cause turn-to-turn shorts). Keep in mind that
allgoodthings likehighefficiency,lowprofile,andhigh
temperature operation will increase cost, sometimes
dramatically. Get a quote on the cheapest unit first to
calibrate yourself on price, then ask for what you really
want.
Assume that the average inductor current is equal to
load current and decide whether or not the inductor
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Linear [ Linear ]
