Texas Instruments LM3402HV Evaluation Board LM3402HVEVAL/NOPB LM3402HVEVAL/NOPB 데이터 시트
제품 코드
LM3402HVEVAL/NOPB
'
i
F
x r
D
'
i
L
-
'
i
F
Z
C
=
R
SNS
=
2
V
IN
- V
O
0.2 x L
I
F
x L + V
O
x t
SNS
-
x t
ON
I
L(PEAK)
= I
F
+
'
i
L(MAX)
2
SNVS450E – SEPTEMBER 2006 – REVISED MAY 2013
Inductor datasheets should contain three specifications which are used to select the inductor. The first of these is
the average current rating, which for a buck regulator is equal to the average load current, or I
the average current rating, which for a buck regulator is equal to the average load current, or I
F
. 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.
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 inductor current ripples can be calculated:
In this example an inductor with a tolerance of ±20% will be used. With this tolerance the typical, minimum, and
maximum inductor 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
(23)
Δ
i
L(MIN)
= [(26.4 – 3.7) x 300 x 10
-9
] / 39.6 x 10
-6
= 172 mA
P-P
(24)
Δ
i
L(MAX)
= [(26.4 – 3.7) x 300 x 10
-9
] / 26.4 x 10
-6
= 258 mA
P-P
(25)
The third specification for an inductor is the peak current rating, 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
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
section:
(26)
I
L(PEAK)
= 0.35 + 0.258 / 2 = 479 mA
(27)
For this example the peak current rating of the inductor should be greater than 479 mA. In the case of a short
circuit across the LED array, the LM3402 will continue to deliver rated current through the short but will reduce
the output voltage to equal the CS pin voltage of 200 mV. Worst-case peak current in this condition is equal to:
circuit across the LED array, the LM3402 will continue to deliver rated current through the short but will reduce
the output voltage to equal the CS pin voltage of 200 mV. Worst-case peak current in this condition is equal to:
Δ
i
L(LED-SHORT)
= [(26.4 – 0.2) x 300 x 10
-9
] / 26.4 x 10
-6
= 298 mA
P-P
I
L(PEAK)
= 0.35 + 0.149 = 499 mA
(28)
In the case of a short at the switch node, the output, or from the CS pin to ground the short circuit current limit
will engage at a typical peak current of 735 mA. In order to prevent inductor saturation during these short circuits
the inductor’s peak current rating must be above 735 mA. The device selected is an off-the-shelf inductor rated
33 µH ±20% with a DCR of 96 m
will engage at a typical peak current of 735 mA. In order to prevent inductor saturation during these short circuits
the inductor’s peak current rating must be above 735 mA. The device selected is an off-the-shelf inductor rated
33 µH ±20% with a DCR of 96 m
Ω
and a peak current rating of 0.82A. The physical dimensions of this inductor
are 7.0 x 7.0 x 4.5 mm.
R
SNS
The current sensing resistor value can be determined by re-arranging the expression for average LED current
from the LED Current Accuracy section:
from the LED Current Accuracy section:
(29)
R
SNS
= 0.74
Ω
, t
SNS
= 220 ns
(30)
Sub-1
Ω
resistors are available in both 1% and 5% tolerance. A 1%, 0.75
Ω
resistor will give the best accuracy of
the average LED current. To determine the resistor size the power dissipation can be calculated as:
P
SNS
= (I
F
)
2
x R
SNS
P
SNS
= 0.35
2
x 0.75 = 92 mW
(31)
Standard 0805 size resistors are rated to 125 mW and will be suitable for this application.
To select the proper output capacitor the equation from Buck Regulators with Output Capacitors is re-arranged to
yield the following:
yield the following:
(32)
The target tolerance for LED ripple current is ±5% or 10%
P-P
= 35 mA
P-P
, and the LED datasheet gives a typical
value for r
D
of 1.0
Ω
at 350 mA. The required capacitor impedance to reduce the worst-case inductor ripple
current of 258 mA
P-P
is therefore:
Z
C
= [0.035 / (0.258 - 0.035] x 1.0 = 0.157
Ω
(33)
A ceramic capacitor will be used and the required capacitance is selected based on the impedance at 468 kHz:
24
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