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SP7663ERL/TR 数据表(PDF) 10 Page - Sipex Corporation

部件名 SP7663ERL/TR
功能描述  Wide Input Voltage Range 6A, 600kHz, Buck Regulator
PDF  19 Pages
Scroll/Zoom Zoom In 100%  Zoom Out
制造商  SIPEX [Sipex Corporation]
网页  http://www.sipex.com
标志 SIPEX - Sipex Corporation

SP7663ERL/TR 数据表(HTML) 10 Page - Sipex Corporation

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Feb2-07 Rev O
SP7663 Wide Input Voltage Range 6A, 600kHz Buck Regulator
© 2007 Sipex Corporation
APPLICATIONS INFORMATION
and provide low core loss at the high
switching frequency. Low cost powdered-
iron cores have a gradual saturation char-
acteristic but can introduce considerable
AC core loss, especially when the induc-
tor value is relatively low and the ripple
current is high. Ferrite materials, although
more expensive, have an abrupt satura-
tion characteristic with the inductance
dropping sharply when the peak design
current is exceeded. Nevertheless, they
are preferred at high switching frequen-
cies because they present very low core
loss while the designer is only required
to prevent saturation. In general, ferrite
or molypermalloy materials are a better
choice for all but the most cost sensitive
applications.
Optimizing Efficiency
The power dissipated in the inductor is equal
to the sum of the core and copper losses.
To minimize copper losses, the winding
resistance needs to be minimized, but this
usually comes at the expense of a larger
inductor.Corelosseshaveamoresignificant
contribution at low output current where the
copper losses are at a minimum, and can
typically be neglected at higher output cur-
rents where the copper losses dominate.
Core loss information is usually available
from the magnetics vendor. Proper induc-
tor selection can affect the resulting power
supplyefficiencybymorethan15%!
The copper loss in the inductor can be cal-
culatedusingthefollowingequation:
PL(Cu) = I
2
L(rMS)
• rwiNDiNg
where IL(rMS)istheRMSinductorcurrent
thatcanbecalculatedasfollows:
IL(rMS) = IOut(MAx)
+ (. IPP )2
3
IOut(MAx)
Inductor Selection
There are many factors to consider in se-
lecting the inductor including core material,
inductance vs. frequency, current handling
capability,efficiency,sizeandEMI.Inatypi-
cal SP7663 circuit, the inductor is chosen
primarilyforvalue,saturationcurrentandDC
resistance. Increasing the inductor value will
decrease output voltage ripple, but degrade
transientresponse.Lowinductorvaluespro-
vide the smallest size, but cause large ripple
currents,poorefficiencyandrequiremore
output capacitance to smooth out the larger
ripple current. The inductor must be able
to handle the peak current at the switching
frequencywithoutsaturating,andthecopper
resistance in the winding should be kept as
low as possible to minimize resistive power
loss.Agood compromise between size, loss
and cost is to set the inductor ripple current
to be within 20% to 40% of the maximum
output current.
The switching frequency and the inductor
operating point determine the inductor value
asfollows:
L
=
.
VOut • (ViN(MAx) - VOut)
ViN(MAx) ƒs • Kr • IOut(MAx)
where:
ƒs = switching frequency
Kr = ratio of the AC inductor ripple current
to the maximum output current
Thepeak-to-peakinductorripplecurrentis:
IPP
=
.
VOut • (ViN(MAx) - VOut)
ViN(MAx) ƒs •L
Once the required inductor value is selected,
theproperselectionofcorematerialisbased
onpeakinductorcurrentandefficiencyre-
quirements. The core must be large enough
not to saturate at the peak inductor current
IPeAk = IOut(MAx) +
IPP
2



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