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20192119
FIGURE 5. Schematic for Design Example 1
R
ON
and t
ON
To select R
the expression relating t
to input voltage from
the Controlled On-time Overview section can be re-written as:
Minimum on-time occurs at the maximum V
, which is 24V x
110% = 26.4V. R
ON
is therefore calculated as:
R
ON
= (300 x 10
-9
x 26.4) / 1.34 x 10
-10
= 59105
The closest 1% tolerance resistor is 59.0 k
. The switching
frequency of the circuit can then be found using the equation
relating R
ON
to f
SW
:
f
SW
= 3.7 / (59000 x 1.34 x 10
-10
) = 468 kHz
USING AN OUTPUT CAPACITOR
The inductor will be the largest component used in this design.
Because the application does not require any PWM dimming,
an output capacitor can be used to greatly reduce the induc-
tance needed without worry of slowing the potential PWM
dimming frequency. The total solution size will be reduced by
using an output capacitor and small inductor as opposed to
one large inductor.
OUTPUT INDUCTOR
Knowing that an output capacitor will be used, the inductor
can be selected for a larger current ripple. The desired max-
imum value for
Δ
i
is ±30%, or 0.6 x 350 mA = 210 mA
.
Minimum inductance is selected at the maximum input volt-
age. Re-arranging the equation for current ripple selection
yields the following:
L
MIN
= [(26.4 – 3.7) x 300 x 10
-9
] / (0.6 x 0.35) = 32.4 μH
The closest standard inductor value is 33 μH. Off-the-shelf
inductors rated at 33 μH are available from many magnetics
manufacturers.
Inductor datasheets should contain three specifications which
are used to select the inductor. The first of these is the aver-
age current rating, which for a buck regulator is equal to the
average load current, or I
. The average current rating is given
by a specified temperature rise in the inductor, normally 40°
C. For this example, the average current rating should be
greater than 350 mA to ensure that heat from the inductor
does not reduce the lifetime of the LED or cause the LM3402
to enter thermal shutdown.
The second specification is the tolerance of the inductance
itself, typically ±10% to ±30% of the rated inductance. In this
example an inductor with a tolerance of ±20% will be used.
With this tolerance the typical, minimum, and maximum in-
ductor current ripples can be calculated:
Δ
i
L(TYP)
= [(26.4 – 3.7) x 300 x 10
-9
] / 33 x 10
-6
= 206 mA
P-P
Δ
i
L(MIN)
= [(26.4 – 3.7) x 300 x 10
-9
] / 39.6 x 10
-6
= 172 mA
P-P
Δ
i
L(MAX)
= [(26.4 – 3.7) x 300 x 10
-9
] / 26.4 x 10
-6
= 258 mA
P-P
The third specification for an inductor is the peak current rat-
ing, normally given as the point at which the inductance drops
off by a given percentage due to saturation of the core. The
worst-case peak current occurs at maximum input voltage
and at minimum inductance, and can be determined with the
equation from the Design Considerations section:
I
L(PEAK)
= 0.35 + 0.258 / 2 = 479 mA
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