LT1956/LT1956-5
W U U
APPLICATIO S I FOR ATIO
U
current without affecting the frequency compensation it
provides.
iron cores are forgiving because they saturate softly,
whereas ferrite cores saturate abruptly. Other core mate-
rials fall somewhere in between. The following formula
assumes continuous mode of operation, but errs only
slightly on the high side for discontinuous mode, so it can
be used for all conditions.
Maximum load current would be equal to maximum
switch current for an infinitely large inductor, but with
finite inductor size, maximum load current is reduced by
one half of peak-to-peak inductor current (ILP-P). The
following formula assumes continuous mode operation,
implying that the term on the right is less than one half
of IP.
VOUT V – V
ILP-P
2
(
)
IN
OUT
IPEAK = IOUT
EMI
+
= IOUT +
2•V • f •L
IN
IOUT(MAX) Continuous Mode
Decide if the design can tolerate an “open” core geometry
like a rod or barrel, which have high magnetic field
radiation, or whether it needs a closed core like a toroid to
prevent EMI problems. This is a tough decision because
the rods or barrels are temptingly cheap and small and
there are no helpful guidelines to calculate when the
magnetic field radiation will be a problem.
V
+ V V – V
2 V f L
( )( IN)( )( )
– V
ILP-P
2
(
F)(
)
OUT
IN
OUT F
= IP –
= IP –
For VOUT = 5V, VIN(MAX) = 8V, VF(DI) = 0.63V, f = 500kHz
and L = 10µH:
5 + 0.63 8 – 5 – 0.63
(
)(
)
IOUT(MAX) = 1.5 –
2 8 500•103 10•10–6
( )( )
Additional Considerations
(
)(
)
After making an initial choice, consider additional factors
such as core losses and second sourcing, etc. Use the
experts in Linear Technology’s Applications department if
you feel uncertain about the final choice. They have
experience with a wide range of inductor types and can tell
you about the latest developments in low profile, surface
mounting, etc.
= 1.5 – 0.17 = 1.33A
Note that there is less load current available at the higher
inputvoltagebecauseinductorripplecurrentincreases.At
VIN = 15V and using the same set of conditions:
5 + 0.63 15 – 5 – 0.63
(
)(
)
IOUT(MAX) = 1.5 –
2 15 500•103 10•10–6
( )( )
(
)(
)
MAXIMUM OUTPUT LOAD CURRENT
= 1.5 – 0.35 = 1.15A
Maximum load current for a buck converter is limited by
themaximumswitchcurrentrating(IP).Thecurrentrating
fortheLT1956is1.5A. Unlikemostcurrentmodeconvert-
ers, the LT1956 maximum switch current limit does not
fall off at high duty cycles. Most current mode converters
suffer a drop off of peak switch current for duty cycles
above 50%. This is due to the effects of slope compensa-
tion required to prevent subharmonic oscillations in cur-
rent mode converters. (For detailed analysis, see Applica-
tion Note 19.)
To calculate peak switch current with a given set of
conditions, use:
ILP-P
2
ISW(PEAK) = IOUT
= IOUT
+
+
V
OUT
+ V V – V
2 V f L
( )( IN)( )( )
– V
OUT F
(
F)(
)
IN
Reduced Inductor Value and Discontinuous Mode
The LT1956 is able to maintain peak switch current limit
over the full duty cycle range by using patented circuitry to
cancel the effects of slope compensation on peak switch
If the smallest inductor value is of the most importance to
a converter design, in order to reduce inductor size/cost,
discontinuous mode may yield the smallest inductor
1956f
11
Linear [ Linear ]
