se5230 ON Semiconductor, se5230 Datasheet - Page 10

• Manufacturer Part Number: se5230
• Description: Low Voltage Operational Amplifier
• Manufacturer: ON Semiconductor

REMOTE TRANSDUCER WITH CURRENT
TRANSMISSION
wires, but current transmission is the most beneficial when
the sensing of remote signals is the aim. It is further
enhanced in the form of 4.0 to 20 mA information which is
used in many controlïtype systems. This method of
transmission provides immunity from line voltage drops,
large load resistance variations, and voltage noise pickup.
The zero reference of 4mA not only can show if there is a
break in the line when no current is flowing, but also can
power the transducer at the remote location. Usually the
transducer itself is not equipped to provide for the current
transmission. The unique features of the NE5230 can
provide high output current capability coupled with low
power consumption. It can be remotely connected to the
transducer to create a current loop with minimal external
components. The circuits for this are shown in Figures 6
and 7. Here, the part is configured as a voltageïtoïcurrent,
or transconductance amplifier. This is a novel circuit that
takes advantage of the NE5230’s large openïloop gain. In
AC applications, the load current will decrease as the
openïloop gain rolls off in magnitude. The low offset
voltage and current sinking capabilities of the NE5230 must
also be considered in this application.
transistor configuration. The inverting input is equivalent to
the “base,” the point where V
meet is the “emitter,” and the connection after the output
diode meets the V
is essential to keep the output from saturating in this
configuration. From here it can be seen that the base and
emitter form a voltageïfollower and the voltage present at
R
input.
currentïfollower and the current flowing through R
equivalent to the current through R
sets up the current loop. Therefore, the following equation
can be formulated for the working current transmission line.
The load current is:
and proportional to the input voltage for a set R
current is constant no matter what load resistance is used
while within the operating bandwidth range of the op amp.
When the NE5230’s supply voltage falls past a certain point,
the current cannot remain constant. This is the “voltage
compliance” and is very good for this application because of
the near rail output voltage. The equation that determines the
voltage compliance as well as the largest possible load
resistor for the NE5230 is as follows:
C
R Lmax +
There are many ways to transmit information along two
The NE5230 circuit shown in Figure 6 is a pseudo
must equal the input voltage present at the inverting
Also,
[ V remote supply * V CC min * V IN max ]
the
CC
pin is the collector. The output diode
emitter
I L +
EE
V IN
R C
and the nonïinverting input
and
I L
L
and the amplifier. This
collector
NE5230, SA5230, SE5230
C
. Also, the
form
http://onsemi.com
(eq. 2)
(eq. 3)
C
is
a
10
(approximately 1.8 V) that will still keep the part
operational. As an example, when using a 15 V remote
power supply, a current sensing resistor of 1.0 W, and an
input voltage (V
20 mA. Furthermore, a load resistance of zero to
approximately 650 W can be inserted in the loop without any
change in current when the bias currentïcontrol pin is tied
to the negative supply pin. The voltage drop across the load
and line resistance will not affect the NE5230 because it will
operate down to 1.8 V. With a 15 V remote supply, the
voltage available at the amplifier is still enough to power it
with the maximum 20 mA output into the 650 W load.
NOTES:
Figure 7. The Same Type of Circuit as Figure 6, but
Where V
Transconductance Amp with 4.0 ï 20 mA Current
1. I
2. R
V
Figure 6. The NE5230 as a Remote Transducer
T
R
A
N
S
D
U
C
E
R
IN
+
ï
OUT
L MAX
V
IN
= V
CC min
for Sourcing Current to the Load
3
2
Transmission Output Capability
≈
IN/RC
V
+
NE5230
ï
3
2
REMOTE
IN
+
NE5230
ï
4
7
is the worstïcase power supply voltage
) of 20 mV, the output current (I
V
V
4
7
R
EE
CC
5
* 1.8V * V
C
V
V
R
I
EE
CC
OUT
5
C
6
6
+ I
INMAX
OUT
I
OUT
20mA
R
For R
4mA
I
OUT
L
V
+
ï
C
R
REMOTE
POWER
SUPPLY
L
= 1.0 W
+
ï
20mV
4mV
V
V
IN
CC
L
) is