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| 型号： | LT8582IDKD#TRPBF |
| 厂商： | Linear Technology |
| 文件页数： | 15/36页 |
| 文件大小： | 0K |
| 描述： | IC REG MULTI CONFIG ADJ 3A 24DFN |
| 标准包装： | 2,500 |
| 类型： | 升压（升压），反相，回扫，Sepic |
| 输出类型： | 可调式 |
| 输出数： | 2 |
| 输出电压： | 最高 42V |
| 输入电压： | 2.5 V ~ 22 V |
| PWM 型： | 电流模式 |
| 频率 - 开关： | 200kHz ~ 2.5MHz |
| 电流 - 输出： | 3A |
| 同步整流器： | 无 |
| 工作温度： | -40°C ~ 125°C |
| 安装类型： | 表面贴装 |
| 封装/外壳： | 24-WFDFN 裸露焊盘 |
| 包装： | 带卷 (TR) |
| 供应商设备封装： | 24-DFN（7x4） |
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LT8582
APPLICATIONS INFORMATION
Dual Inductor Inverting Converter Component
Selection – Coupled or Uncoupled Inductors
Table 3. Dual Inductor Inverting Design Equations
PARAMETERS/EQUATIONS
| V OUT | + 0.5V
V IN + | V OUT | + 0.5V – 0.3V
V IN
5V
L1
4.7μH
C1
2.2μF
L2
4.7μH
D1
30V, 2A
V OUT
–12V
550mA
Step 1: Inputs
Step 2: DC
Choose V IN , V OUT and f OSC to calculate equations
below.
DC ?
100k
SWA SWB
V IN LT8582 FBX
CHx
SHDN GATE
PG CLKOUT
RT V C
R FBX
143k
C OUT2
10μF
Step 3: L
L TYP =
L MIN =
(V IN – 0.3V) ? DC
f OSC ? 1A
(V IN – 0.3V) ? (2 ? DC – 1)
1.7A ? f OSC ? (1 – DC)
(1)
(2)
C IN
4.7μF
R T
53.6K
SYNC GND SS
0.1μF
47pF
14.7k
2.2nF
L MAX =
(V IN – 0.3V) ? DC
f OSC ? 0.18A
(3)
8582 F07
Figure 7. Dual Inductor Inverting Converter – The Component
Values Given Are Typical Values for a 1.5MHz, 5V to –12V
Inverting Topology Using Coupled Inductors
Due to its unique FBX pin, each channel of the LT8582 can
work in a dual inductor inverting configuration as shown in
?
?
?
?
?
Solve equations 1, 2 and 3 for a range of L
values
The minimum of the L value range is the
higher of L TYP and L MIN
The maximum of the L value range is L MAX
L = L1 = L2 for coupled inductors.
L = L1||L2 for uncoupled inductors.
I RIPPLE =
I OUT = ? 3A –
C OUT ≥
Figure7.ChangingtheconnectionsofL2andtheSchottky
diode in the SEPIC topology results in generating negative
output voltages. This configuration results in very low
output voltage ripple due to inductor L2 in series with
the output. Output disconnect is inherently built into this
topology because of capacitor C1.
Table 3 is a step-by-step set of equations to calculate
component values for the LT8582 when operating as a dual
inductor inverting converter. Input parameters are input
and output voltage and switching frequency (V IN , V OUT
Step 4: I RIPPLE
Step 5: I OUT
Step 6: D1
Step 7: C1
Step 8: C OUT
Step 9: C IN
(V IN – 0.3V) ? DC
f OSC ? L
? I RIPPLE ?
?
? 2 ? ? (1 – DC)
V R > V IN + |V OUT |; I AVG > I OUT
C1 ≥ 1μF; V RATING ≥ V IN + |V OUT |
I RIPPLE
8 f OSC 0.005 | V OUT |
C IN ≥ C VIN + C PWR ≥
and f OSC respectively). Refer to the Appendix for further
information on the design equations presented in Table 3.
3A ? DC
50 ? f O sc ? 0.005 ? V IN
+
I RIPPLE
8 ? f O sc ? 0.005 ? V IN
Variable Definitions:
V IN = Input Voltage
Step 10: R FBX
R FBX =
| V OUT | + 7mV
83.3μA
V OUT = Output Voltage
DC = Power Switch Duty Cycle
Step 11: R T
R T =
81.6
f OSC
–1; f OSC in MHz, R T in k Ω
f OSC
I OUT
= Switching Frequency
= Maximum Output Current
Note 1: Above equations use numbers good for many applications but
for more exact results use the equations from the appendix with numbers
I RIPPLE = Inductor Ripple Current
from the Electrical Characteristics.
Note 2: The final values for C OUT , and C IN may deviate from the above
equations in order to obtain desired load transient performance.
8582f
15
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