參數(shù)資料
型號(hào): LT3433IFE
廠商: LINEAR TECHNOLOGY CORP
元件分類: 穩(wěn)壓器
英文描述: High Voltage Step-Up/Step-Down DC/DC Converter
中文描述: 0.9 A SWITCHING REGULATOR, 230 kHz SWITCHING FREQ-MAX, PDSO16
封裝: 4.40 MM, PLASTIC, TSSOP-16
文件頁(yè)數(shù): 10/12頁(yè)
文件大?。?/td> 183K
代理商: LT3433IFE
10
LT3433
3433ia
Ripple current:
=
+
(
)
O
(
)
I
V
V
DC
L f
L P P
(
OUT
F
)
2
1
Inductor Selection
The primary criterion for inductor value selection in LT3433
applications is the ripple current created in that inductor.
Design considerations for ripple current are the amount of
output ripple and the ability of the internal slope compen-
sation waveform to prevent current mode instability.
The LT3433 maximizes available dynamic range using a
slope compensation generator that generates a continu-
ously increasing slope as duty cycle increases. The slope
compensation waveform is calibrated at 80% duty cycle to
compensate for ripple currents up to 12.5% of I
MAX
, or
~60mA.
Ripple current can be calculated as:
=
+
(
)
O
(
)
I
V
V
DC
L f
L P P
(
OUT
F
)
2
1
This relation can be used to determine minimum induc-
tance sizes for various values of V
OUT
using the DC = 80%
calibration:
L
MIN
= (V
OUT
+ 1.5V) (1 – 0.8) 60mA 200kHz)
V
OUT
4V
5V
9V
12V
L
MIN
92
μ
H
108
μ
H
175
μ
H
225
μ
H
APPLICATIOU
W
U
U
Discontinuous operation occurs when the ripple current in
the inductor is greater than twice the load current (I
LOAD
)
in buck mode, or greater than I
LOAD
/(1 – DC) during
bridged mode. Current mode instability is not a concern
during discontinuous operation so inductor values smaller
than L
MIN
can be used. If such a small inductor is used,
however, it must be assured that the converter never
enters continuous operation at duty cycles greater than
50% to prevent current mode instability.
Design Example
V
IN(MIN)
= 4V, V
OUT
= 5V, L = 150
μ
H
Using V
F
= 0.75V yields:
DC = (V
OUT
+ 2V
F
)/(V
OUT
+ V
IN
+ 2V
F
– V
SWH
– V
SWL
)
= (5V + 1.5V)/(4V + 5V + 1.5V – 0.6V – 0.5V)
= 0.69
I
L
= (V
OUT
+ 2V
F
) (1 – DC) (L f
0
)
–1
= (5V + 1.5V) (1 –0.69) (150
μ
H 200kHz)
–1
= 67mA
I
LOAD(MAX)
= I
SW(MAX)
(1 – 1.1 DC)
= [0.5A – (1/2 0.07)](1 – 1.1 0.69) = 0.112A
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