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LT1013DMD

LT1013DMD首页预览图
型号: LT1013DMD
PDF文件:
  • LT1013DMD PDF文件
  • LT1013DMD PDF在线浏览
功能描述: Dual Precision Operational Amplifier
PDF文件大小: 1524.94 Kbytes
PDF页数: 共36页
制造商: TI1[Texas Instruments]
制造商LOGO: TI1[Texas Instruments] LOGO
制造商网址: http://www.ti.com
捡单宝LT1013DMD
PDF页面索引
120%
Vsup+
+
VOUT
RF
VIN
RI
Vsup-
Copyright © 2016, Texas Instruments Incorporated
20
LT1013
,
LT1013D
,
LT1013M
,
LT1013AM
SLOS018I MAY 1988REVISED JULY 2016
www.ti.com
Product Folder Links: LT1013 LT1013D LT1013M LT1013AM
Submit Documentation Feedback Copyright © 1988–2016, Texas Instruments Incorporated
8 Application and Implementation
NOTE
Information in the following applications sections is not part of the TI component
specification, and TI does not warrant its accuracy or completeness. TI’s customers are
responsible for determining suitability of components for their purposes. Customers should
validate and test their design implementation to confirm system functionality.
8.1 Application Information
The LT1013x operational amplifiers are useful in a wide range of signal conditioning applications where high DC
accuracy is needed.
8.2 Typical Application
A typical application for an operational amplifier in an inverting amplifier. This amplifier takes a positive voltage
on the input and makes it a negative voltage of the same magnitude. In the same manner, it also makes negative
voltages positive.
Figure 27. Application Schematic
8.2.1 Design Requirements
The supply voltage must be chosen such that it is larger than the input voltage range and output range. For
instance, this application scales a signal of ±0.5 V to ±1.8 V. Setting the supply at ±12 V is sufficient to
accommodate this application.
8.2.2 Detailed Design Procedure
Determine the gain required by the inverting amplifier using Equation 1 and Equation 2:
(1)
(2)
Once the desired gain is determined, choose a value for RI or RF. Choosing a value in the kΩ range is desirable
because the amplifier circuit will use currents in the milliamp range. This ensures the part does not draw too
much current. This example chooses 10 kΩ for RI, which means 36 kΩ is used for RF. This was determined by
Equation 3.
(3)
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