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LP358 bảng dữ liệu(PDF) 9 Page - Texas Instruments

tên linh kiện LP358
Giải thích chi tiết về linh kiện  Ultra-Low Power Dual Operational Amplifiers
PDF  22 Pages
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Trang chủ  https://www.ti.com
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LP358 bảng dữ liệu(HTML) 9 Page - Texas Instruments

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V
RF
A
RI
V
1.8
A
3.6
0.5
V
VOUT
A
VIN
Vsup+
+
VOUT
RF
VIN
RI
Vsup-
Copyright © 2016, Texas Instruments Incorporated
9
LP358, LP2904
www.ti.com
SLOS475A – AUGUST 2005 – REVISED MAY 2017
Product Folder Links: LP358 LP2904
Submit Documentation Feedback
Copyright © 2005–2017, Texas Instruments Incorporated
9 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.
9.1 Application Information
The LP358 and LP2904 operational amplifiers are useful in a wide range of signal conditioning applications due
to the wide VCC range. Inputs can be powered before VCC for flexibility in multiple supply circuits.
9.2 Typical Application
A typical application for an operational amplifier in an inverting amplifier. This amplifier takes a positive voltage
on the input, and makes the voltage a negative voltage. In the same manner, the amplifier also makes negative
voltages positive.
Figure 4. Application Schematic
9.2.1 Design Requirements
The supply voltage must be selected 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
9.2.2 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, select a value for RI or RF. Selecting a value in the kΩ range is desirable
because the amplifier circuit uses currents in the milliamp range. This ensures the part does not draw too much
current. This example uses 10 kΩ for RI, so 36 kΩ is used for RF; this is determined by Equation 3.
(3)



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