LT1374
U
W U U
APPLICATIONS INFORMATION
MAXIMUM OUTPUT LOAD CURRENT
Note that there is less load current available at the higher
input voltage because inductor ripple current increases.
This is not always the case. Certain combinations of
inductor value and input voltage range may yield lower
available load current at the lowest input voltage due to
reduced peak switch current at high duty cycles. If load
current is close to the maximum available, please check
maximum available current at both input voltage
extremes. To calculate actual peak switch current with a
given set of conditions, use:
Maximum load current for a buck converter is limited by
the maximum switch current rating (IP) of the LT1374.
This current rating is 4.5A up to 50% duty cycle (DC),
decreasing to 3.7A at 80% duty cycle. This is shown
graphically in Typical Performance Characteristics and as
shown in the formula below:
IP = 4.5A for DC ≤ 50%
IP = 3.21 + 5.95(DC) – 6.75(DC)2 for 50% < DC < 90%
DC = Duty cycle = VOUT/VIN
VOUT V − V
(
)
IN
OUT
Example: with VOUT = 5V, VIN = 8V; DC = 5/8 = 0.625, and;
ISW(MAX) = 3.21 + 5.95(0.625) – 6.75(0.625)2 = 4.3A
ISW PEAK =IOUT
+
(
)
2 L f V
( )( )(
)
IN
For lighter loads where discontinuous operation can be
used, maximum load current is equal to:
Current rating decreases with duty cycle because the
LT1374hasinternalslopecompensationtopreventcurrent
mode subharmonic switching. For more details, read Ap-
plicationNote19.TheLT1374isalittleunusualinthisregard
because it has nonlinear slope compensation which gives
better compensation with less reduction in current limit.
2
I
f L V
IN
( ) ( )( )(
)
P
IOUT(MAX)
=
Discontinuous mode 2 V
V − V
(
)(
)
OUT
IN
OUT
Example: with L = 1.2µH, VOUT = 5V, and VIN(MAX) = 15V,
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 peak-to-peak inductor current. The following
formula assumes continuous mode operation, implying
that the term on the right is less than one-half of IP.
2
4.5 500 •103 1.2 •10−6 15
(
)
( )
IOUT MAX
=
= 1.82A
(
)
2 5 15 − 5
( )(
)
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. Thiswillincreaseoutput
ripplevoltage.Iftheoutputcapacitorhastobemadelarger
to reduce ripple voltage, the overall circuit could actually
wind up larger.
V
V − V
IN OUT
(
OUT)(
)
IOUT(MAX)
=
IP −
Continuous Mode
2 L f V
( )( )(
)
IN
For the conditions above and L = 3.3µH,
5 8 − 5
( )(
)
CHOOSING THE INDUCTOR AND OUTPUT CAPACITOR
IOUT MAX) = 4.3 −
(
2 3.3 •10−6 500•103
8
( )
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 LT1374 switch, which has a 4.5A limit. Higher values
also reduce output ripple voltage, and reduce core loss.
GraphsintheTypicalPerformanceCharacteristicssection
show maximum output load current versus inductor size
andinputvoltage. Asecondgraphshowscorelossversus
=4.3 − 0.57= 3.73A
AtVIN =15V, dutycycleis33%, soIP isjustequaltoafixed
4.5A, and IOUT(MAX) is equal to:
5 15 − 5
( )(
)
4.5 −
2 3.3 •10−6 500•103 15
( )
= 4.5 −1= 3.5A
inductor size for various core materials.
1374fb
10
Linear [ Linear ]
