Vsup+

+

VOUT

RF

VIN

RI

Vsup-

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16

TLV341, TLV341A, TLV342, TLV342S

SLVS568D – JANUARY 2005 – REVISED APRIL 2016

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Copyright © 2005–2016, 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 TLV34xx devices have rail-to-rail output and input range from ground to VCC – 0.6 V. CMOS inputs provide

very low input current. Shutdown capability is an option in dual amplifier version. Operation from 1.5 V to 5.5 V is

possible.

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 37. 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 voltage range.

For instance, this application scales a signal of ±0.5 V to ±1.8 V. Setting the supply at ± 2 V is sufficient to

accommodate this application. The supplies can power up in any order; however, neither supply can be of

opposite polarity relative to ground at any time; otherwise, a large current can flow though the input ESD diodes.

TI highliy recommends adding a series resistor to the grounded input to limit current in such an occurrence.

Vsup+ must be more positive than Vsup– at all times; otherwise, a large reverse supply current may flow.

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 k

Ω range is desirable

because the amplifier circuit uses currents in the mA 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 RF. This was determined by

Equation 3.

(3)