VOLTAGE/CURRENT CONVERTER

AM422

analog microelectronics

April 99

9/10

TYPICAL THREE–WIRE APPLICATION (0.5 – 4.5V)

Example 3: Output current range IOUT = 4...20mA, input voltage range VIN = 0.5...4.5V

The transfer function of the output current IOUT is (equation 2)

II

I

I

V

R

R

RR

OUT

SET

IN

SET

IN

=+

=+

+













2

0

2

12

For the 0.5 to 4.5V application (Figure 8) a change of the input voltage from

∆VIN = 4V should be

displayed over a change of the output current

∆IOUT = 16mA.

With the voltage change at pin 4 (

∆VPIN4 = 800mV, equation 7) the relationship of R1/R2, using

equation 8, becomes then

∆

∆

V

V

R

RR

PIN

IN

42

12

800

4

=

+

=

mV

V

⇒

R

R

VV

V

IN

PIN

PIN

1

2

4

4

4

=

−

=

∆∆

∆

With that relation an offset current ISET from

II

I

I

SET

OUTmin

INmin

OUTmin

=−

=

− 2mA = 2mA

has to be adjusted (equation 10). With R3 = R4 the set resistor RSET can be calculated (equation 4)

R

RR I

V

SET

SET

REF

≈

4

04

The values of the external devices than can be calculated as follows (V

REF = 5V )

R0 = 25

Ω

R3 = R4 = 33k

Ω

RSET

≈ 1.32kΩ

R5 = 40

Ω

R1/R2

≈ 4

R1

≈ 68kΩ

R2

≈ 18kΩ

RL = 0...500

Ω

C1 = 2.2

µF

C

1

R

4

R

3

V

REF

R

0

V

S

I

OUT

R

L

Ground

3

4

7

G = 1

Amplifier

2

1

VREF

2

8

V

SET

V

OFFSET

R

1

R

2

R

SET

5

V

I

6

5/10 V Reference

AM422

−1

R

5

D

1

T

1

V

IN

Figure 8