R

V

1

D

V

V

IN

-

D

´

R

2

R + R

1

2

V = V

R

D

IN

2

=

´

®

= 4.167 kW

V

= V

G + V

=

G + V

OUT V

D

REF

-

-

REF

´

´

V

IN

´

R

2

R + R

1

2

V

= V

G + V

= (I

R )

G + V

OUT I

D

REF

IN

3

-

-

REF

´

´

´

RG = 10.4 NŸ

v

+

R3 =

20

Ÿ

R2 = 4.12 NŸ

R1 = 100 NŸ

INA826S

VOUT

15 V

15 V

RL = 10 NŸ

RO = 100 Ÿ

CO = 1.59 nF

±10 V

±20 mA

2.5 V ± 2.3 V

V+

V

VREF

REF3225

5 V

Copyright © 2017, Texas Instruments Incorporated

28

INA826S

SBOS770A – MAY 2017 – REVISED JUNE 2017

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Product Folder Links: INA826S

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Copyright © 2017, Texas Instruments Incorporated

8.2 Typical Application

Figure 74 shows a three-terminal, programmable-logic controller (PLC) design for the INA826S. This PLC

reference design accepts inputs of ±10 V or ±20 mA. The output is a single-ended voltage of 2.5 V ± 2.3 V (or

200 mV to 4.8 V). Many PLCs typically have these input and output ranges.

Figure 74. Three-Terminal Analog Input for PLCs

8.2.1 Design Requirements

This design has the following requirements:

•

Supply voltage: ±15 V, 5 V

•

Inputs: ±10 V, ±20 mA

•

Output: 2.5 V, ±2.3 V

8.2.2 Detailed Design Procedure

There are two modes of operation for the circuit shown in Figure 74: current input and voltage input. This design

requires R1 >> R2 >> R3. Given this relationship, Equation 2 calculates the current input mode transfer function.

where

•

G represents the gain of the instrumentation amplifier

(2)

Equation 3 shows the transfer function for the voltage input mode.

(3)

R1 sets the input impedance of the voltage input mode. The minimum typical input impedance is 100 kΩ. 100 kΩ

is selected for R1 because increasing the R1 value also increases noise. The value of R3 must be extremely

small compared to R1 and R2. 20 Ω for R3 is selected because that resistance value is much smaller than R1 and

yields an input voltage of ±400 mV when operated in current mode (±20 mA).

Equation 4 can be used to calculate R2 given VD = ±400 mV, VIN = ±10 V, and R1 = 100 kΩ.

(4)