ZHCSAW6K March   2009  – April 2019 LM2840 , LM2841 , LM2842

PRODUCTION DATA.  

  1. 1 特性
  2. 2 应用
  3. 3 说明
    1.     Device Images
      1.      典型应用电路
  4. 4 修订历史记录
  5. 5 Pin Configuration and Functions
    1.     Pin Functions
  6. 6 Specifications
    1. 6.1 Absolute Maximum Ratings
    2. 6.2 ESD Ratings
    3. 6.3 Recommended Operating Conditions
    4. 6.4 Thermal Information
    5. 6.5 Electrical Characteristics
    6. 6.6 Typical Characteristics
  7. 7 Detailed Description
    1. 7.1 Overview
    2. 7.2 Functional Block Diagram
    3. 7.3 Feature Description
      1. 7.3.1 Protection
    4. 7.4 Device Functional Modes
      1. 7.4.1 Continuous Conduction Mode
  8. 8 Application and Implementation
    1. 8.1 Application Information
    2. 8.2 Typical Applications
      1. 8.2.1 Design Requirements
      2. 8.2.2 Detailed Design Procedure
        1. 8.2.2.1 Custom Design With WEBENCH® Tools
        2. 8.2.2.2 Setting the Output Voltage
        3. 8.2.2.3 Inductor Selection
        4. 8.2.2.4 Input Capacitor
        5. 8.2.2.5 Output Capacitor
        6. 8.2.2.6 Bootstrap Capacitor
        7. 8.2.2.7 Soft-Start Components
        8. 8.2.2.8 Shutdown Operation
        9. 8.2.2.9 Schottky Diode
      3. 8.2.3 Application Curves
      4. 8.2.4 Other Application Circuits
  9. 9 Power Supply Recommendations
  10. 10Layout
    1. 10.1 Layout Guidelines
    2. 10.2 Layout Example
  11. 11器件和文档支持
    1. 11.1 器件支持
      1. 11.1.1 第三方米6体育平台手机版_好二三四免责声明
      2. 11.1.2 开发支持
        1. 11.1.2.1 使用 WEBENCH® 工具创建定制设计
    2. 11.2 文档支持
      1. 11.2.1 相关文档
    3. 11.3 相关链接
    4. 11.4 接收文档更新通知
    5. 11.5 社区资源
    6. 11.6 商标
    7. 11.7 静电放电警告
    8. 11.8 Glossary
  12. 12机械、封装和可订购信息

封装选项

机械数据 (封装 | 引脚)
散热焊盘机械数据 (封装 | 引脚)
订购信息

8.2.2.2 Setting the Output Voltage

The output voltage is set using the feedback pin and a resistor divider connected to the output as shown in 典型应用电路. The feedback pin voltage 0.765 V, so the ratio of the feedback resistors sets the output voltage according to Equation 3:

Equation 3. VOUT = 0.765 V (1 + (R1 / R2))

Typically R2 is given as 100 Ω to 10 kΩ for a starting value. To solve for R1 given R2 and VOUT, use Equation 4:

Equation 4. R1 = R2 ((VOUT / 0.765 V) – 1)
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