CURRENT CONVERTER IC
TYPICAL THREE–WIRE APPLICATION (0/4–20mA)
AM402
C
1
R
2
10
11
R
1
12
13
16
V
REF
V
S
R
0
1
2
3
AM 402
5/10 V reference
9
V
IA
I
14
7
T
1
D
1
R
5
I
OUT
R
L
Ground
V
IN
8
5
R
3
R
4
Figure 7
Used in a three–wire circuit, pin 2 (VCC) is connected to pin 1 (RS+) and ground pin 14 (GND) is
connected to
Ground
(Figure 7). The Gain
G
IA
is adjusted by external resistors
R
1
and
R
2
and can be
calculated by
G
IA
=
1
+
R
1
R
2
⇒
R
1
R
2
=
G
IA
−
1
Hence, the transfer–function of the output current
I
OUT
becomes
I
OUT
=
V
IN
G
IA
R
0
+
I
SET
with the current
I
SET
adjusted by external resistors
R
3
and
R
4
.
R
3
V
R
4
V
REF
I
SET
=
REF
⋅
⇒
=
−
1
2
R
0
R
3
+
R
4
R
4
2
R
0
I
SET
The supply voltage must be chosen with respect to the load resistor
R
L
described by the following
equation
V
S
≥
I
OUTmax
R
L
+6V
Example 1:
Output current range 4...20mA
The values of the external devices (
V
IN
=
0
K
50 mV ,
V
REF
=
5V ,
G
IA
=
8 ) are as follows
R
0
= 25Ω
R
1
= 33kΩ
R
2
= 4.7kΩ
R
3
= 100kΩ
R
4
= 0...5kΩ
R
5
= 40Ω
R
L
= 0...500Ω
C
1
= 2.2µF
Example 2:
Output current range 0...20mA
The values of the external devices (
V
IN
=
0
K
250 mV ,
V
REF
=
5V ,
G
IA
=
2 ) are as follows
R
0
= 25Ω
R
1
= 22kΩ
R
2
= 22kΩ
R
5
= 40Ω
R
3
,
R
4
not used (SET =
GND)
R
L
= 0...500Ω
C
1
= 2.2µF
analog microelectronics
April 99
7/8
AME [ ANALOG MICROELECTRONICS ]
