VOLTAGE/CURRENT CONVERTER
TYPICAL THREE–WIRE APPLICATION (0.5 – 4.5V)
C
1
R
3
R
SET
5
6
AM422
V
REF
8
AM422−1
V
SET
V
REF
2
R
4
Sensor
0.5−4.5V
G
=1
V
OFFSET
V
S
5/10 V Reference
1
R
0
V
V
IN
4
2
Amplifier
R
1
R
2
7
I
3
T
1
D
1
R
5
I
OUT
R
L
Ground
Figure 8
Example 3:
Output current range
I
OUT
= 4...20mA, input voltage range
V
IN
= 0.5...4.5V
The transfer function of the output current
I
OUT
is (equation 2)
V
R
2
I
OUT
=
I
SET
+
I
IN
=
I
SET
+
IN
2
R
0
R
1
+
R
2
For the 0.5 to 4.5V application (Figure 8) a change of the input voltage from
∆V
IN
= 4V should be
displayed over a change of the output current
∆I
OUT
= 16mA.
With the voltage change at pin 4 (∆V
PIN4
= 800mV, equation 7) the relationship of
R
1
/R
2
, using
equation 8, becomes then
∆
V
PIN
4
R
2
800mV
=
=
∆
V
IN
R
1
+
R
2
4V
With that relation an offset current
I
SET
from
I
SET
=
I
OUTmin
−
I
INmin
=
I
OUTmin
−
2mA = 2mA
has to be adjusted (equation 10). With
R
3
=
R
4
the set resistor
R
SET
can be calculated (equation 4)
R
SET
≈
4
R
0
R
4
I
SET
V
REF
R
SET
≈
1.32kΩ
R
2
≈
18kΩ
⇒
R
1
∆
V
IN
− ∆
V
PIN
4
=
=
4
∆
V
PIN
4
R
2
The values of the external devices than can be calculated as follows (V
REF
=
5V )
R
0
= 25Ω
R
1
/R
2
≈
4
C
1
= 2.2µF
R
3
=
R
4
= 33kΩ
R
1
≈
68kΩ
R
5
= 40Ω
R
L
= 0...500Ω
analog microelectronics
April 99
9/10
AME [ ANALOG MICROELECTRONICS ]
