LT1507
U
W U U
APPLICATIONS INFORMATION
pin (RDIV = R1/R2 ≤ 4k). The net result is that reductions
in frequency and current limit are affected by output
voltage divider impedance. Although divider impedance is
not critical, caution should be used if resistors are
increased beyond the suggested values and short-circuit
conditions will occur with high input voltage. High
frequency pickup will also increase and the protection
accordedbyfrequencyandcurrentfoldbackwilldecrease.
may not survive a continuous 1.5A overload condition.
Deadshorts(VOUT ≤1V)willactuallybemoregentleon
the inductor because the LT1507 has foldback current
limiting (see graph in Typical Performance Character-
istics).
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 a con-
tinuous mode of operation, but it errs only slightly on
the high side for discontinuous mode, so it can be used
for all conditions.
CHOOSING THE INDUCTOR AND OUTPUT CAPACITOR
For most applications the value of the inductor will fall in
the range of 2µH to 10µH. 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 LT1507 switch, which has a 1.5A limit. Higher
values also reduce output ripple voltage and reduce core
loss. Graphs in the Typical Performance Characteristics
section show maximum output load current versus induc-
torsizeandinputvoltage. Asecondgraphshowscoreloss
versus inductor size for various core materials.
V
(V – V
)
OUT IN
OUT
I
= I
+
OUT
PEAK
2(f)(L)(V )
IN
VIN = Maximum input voltage
f = Switching frequency = 500kHz
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.
3. Decide if the design can tolerate an “open” core geom-
etry like ferrite rods or barrels, which have high mag-
netic field radiation or whether it needs a closed core
like a toroid to prevent EMI problems. One would not
wantanopencorenexttoamagneticstoragemediafor
instance! This is a tough decision because the rods or
barrelsaretemptinglycheapandsmallandthereareno
helpful guidelines to calculate when the magnetic field
radiationwillbeaproblem. Thefollowingisanexample
of just how subtle the “B” field problems can be with
open geometry cores.
1. Choose a value in microhenries from the graphs of
Maximum Load Current and Inductor Core Loss for
3.3V Output. If you want to double check that the
chosen inductor value will allow sufficient load current,
go to the next section, Maximum Output Load Current.
Choosing a small inductor with lighter loads may result
in discontinuous mode of operation, but the LT1507 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. Type 52 powdered iron,
Kool Mµ and Molypermalloy are old standbys for tor-
oids in ascending order of price. A newcomer, Metglas,
gives very low core loss with high saturation current.
We had selected an open drum shaped ferrite core for
the LTC1376 demonstration board because the induc-
tor was extremely small and inexpensive. It met all the
requirements for current and the ferrite core gave low
core loss. When the boards came back from assembly,
many of them had somewhat higher than expected
output ripple voltage. We removed the inductors and
output capacitors and found them to be no different
than the good boards. After much head scratching and
hours of delicate low level ripple measurements on the
good and bad boards, I realized that the problem must
Assume that the average inductor current is equal to
load current and decide whether or not the inductor
must withstand continuous fault conditions. If maxi-
mum load current is 0.5A, for instance, a 0.5A inductor
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Linear [ Linear ]
