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LF347 Fiches technique(PDF) 9 Page - Texas Instruments |
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LF347 Fiches technique(HTML) 9 Page - Texas Instruments |
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9 / 22 page ![]() V RF A = RI - V 1.8 A = 3.6 0.5 = - - V VOUT A = VIN Vsup+ + VOUT RF VIN RI Vsup- 9 LF347, LF347B www.ti.com SLOS013C – MARCH 1987 – REVISED MARCH 2016 Product Folder Links: LF347 LF347B Submit Documentation Feedback Copyright © 1987–2016, 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 LF347 has four independent amplifiers that have very low input bias current which allow using higher resistance resistors in the feedback network. The upper input common mode range typically goes to the positive supply rail. The lower common mode range does not include the negative supply rail; it must be at least 4-V greater. Output resistance is 200 Ω to protect the device from accidental shorts. 9.2 Typical Application A typical application for an operational amplifier is 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 6. Inverting Amplifier 9.2.1 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. 9.2.2 Detailed Design Procedure Determine the gain required by the inverting amplifier: (1) (2) When the desired gain is determined, choose a value for RI or RF. Choosing a value in the k Ω range is desirable because the amplifier circuit uses currents in the milliamp range. This ensures the part does not draw too much current. For this example, choose 10 k Ω for RI which means 36 kΩ is used for R, as determined by Equation 3. (3) |
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