APPLICATION BULLETIN
®
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4-20mA TO 0-20mA CONVERTER AND
CURRENT SUMMING CURRENT-TO-CURRENT CONVERTERS
By R. Mark Stitt and David Kunst (602) 746-7445
Current loops have become the standard for signal transmission in the process control industry. Current loops are
insensitive to noise and are immune to errors from line
impedance. Burr-Brown offers a complete line of monolithic
4mA to 20mA current loop transmitters and receivers.
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R2
2
IIN
3
A1
IOUT
XTR101
General purpose two-wire 4-20mA current-loop transmitter.
This transmitter has an instrumentation amplifier input and
two 1mA current sources for transducer excitation and
offsetting.
R1
IOUT = –IIN • R2/R1
NOTE: The minus sign in the equation means that the IIN, IOUT currents
flow as shown.
FIGURE 1. Basic Inverting Current-to-Current Converter.
XTR103
Two-wire RTD 4-20mA current-loop transmitter. Similar to
XTR101, but with internal linearization circuitry for direct
interface to RTDs (Resistance Temperature Detectors). The
XTR103 along with an RTD forms a precision temperature
to 4-20mA current loop transmitter.
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For Figure 1:
IOUT = –IIN • R2/R1
Notice that, like its cousin the inverting voltage amplifier,
I2
Self-contained 4-20mA receiver. Conditions and offsets 420mA input signals to give a precision 0-5V output. Contains precision voltage reference, 75Ω precision sense resistor and ±40V common-mode input range difference amplifier.
8
9
10
I1
RCV420
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Application Bulletin Number 31
INVERTING CURRENT-TO-CURRENT
CONVERTERS
Figure 1 shows the basic inverting current-to-current converter. The input current all flows through R2 resulting in an
IIN • R2 voltage drop across R2. The op amp forces the same
voltage across R1 so that there is no voltage difference at the
op amp inputs. The IOUT current is therefore:
XTR110
Three-wire 4-20mA transmitter. The XTR110 converts a 05V or 0-10V high-level input into a 0-20mA or 4-20mA
current-soruce output.
11
R2
IN
12
A1
The 4mA to 20mA current loop is the most often used
standard. Since the minimum signal current is 4mA, the
transducer and transmitter can be powered by the same two
wires used for the current loop connection. This feature
eliminates the need for a remote power supply. Also, open
circuits are easy to detect since the signal goes to 0mA.
Some systems, however, use a 0 to 20mA current loop
standard instead. To interface to these systems, the 4-20mA
1991 Burr-Brown Corporation
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XTR output must be converted to 0-20mA. This bulletin
shows the suggested circuit and also discusses summing
current-to-current converters in general.
XTR104
Two-wire bridge 4-20mA current-loop transmitter. Similar
to XTR101, but with shunt regulator and linearization circuitry for direct interface to resistor transducer bridges.
©
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IOUT
13
R1
IOUT = (–I1 – I2 – ••• – IN) R2/R1
NOTE: The minus sign in the equation means that the IIN, IOUT currents
flow as shown.
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15
FIGURE 2. Equal-Gain Summing Inverting Current-to-Current Converter.
AB-031
Printed in U.S.A. April, 1991
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apply to the noninverting circuits.
I1
For Figure 4:
I2
IOUT = IIN(1 + R2/R1)
IN
RN
R3
Notice that, like its cousin the noninverting voltage amplifier, the gain is always greater than 1.0 so that the output
current must always be greater than the input current.
R2
A1
IOUT
As with the inverting summing current-to-current converter,
so long as the gain is the same, any number of currents can
be summed at the current-to-current converter input as
shown in Figure 5.
R1
IOUT = –I1 • R2/R1 – I2(R2 + R3)/R1 – ••• – IN(R2 + R3 + ••• + RN)/R1
NOTE: The minus sign in the equation means that the IIN, IOUT currents
flow as shown.
For Figure 5:
FIGURE 3. Arbitrary-Gain Summing Inverting Current-toCurrent Converter.
IN
A1
I2
the output current can be greater than or less than the input
current. Of course, the output can also be equal to the input
so the circuit can be used as a unity-gain current inverter
with R1 = R2.
I1
R2
So long as the gain is the same, any number of currents can
be summed at the current-to-current converter input as
shown in Figure 2.
R1
IOUT
IOUT = (I1 + I2 + ••• + IN)(1 + R2/R1)
For Figure 2:
IOUT = (–I1 – I2 – ••• – IN)R2/R1
FIGURE 5. Equal-Gain Summing Noninverting Currentto-Current Converter.
Any number of currents can be summed with arbitrary gain
by using separate gain-setting resistors as shown in Figure 3.
IOUT = (I1 + I2 + ••• + IN)(1 + R2/R1)
For Figure 3:
Likewise, any number of currents can be summed with
arbitrary gain by using separate gain-setting resistors as
shown in Figure 6.
IOUT = –I1 • R2/R1 – I2(R2 + R3)/R1 – ••• –
IN(R2 + R3 + ••• + RN)/R1
For Figure 6:
IOUT = I1(1 + R2/R1) + I2(R1 + R2 + R3)/R1 + ••• +
IN(R1 + R2 + R3 + ••• + RN)/R1
NONINVERTING CURRENT-TO-CURRENT
CONVERTERS
Figure 4 shows the circuit for the basic noninverting currentto-current converter. As before, all the input current flows
through R2 resulting in a IIN • R2 voltage drop across R2. In
this circuit, the op amp is connected as a unity-gain buffer
forcing the same voltage drop across R1. The output current
is the sum of the input current flowing through R2 and the
current flowing through R1. The following relationships
A1
IN
RN
I2
R3
IOUT = IIN(1 + R2/R1)
IIN
I1
A1
R2
R2
R1
R1
IOUT
IOUT
IOUT = I1(1 + R2/R1) + I2(R1 + R2 + R3)/R1 + ••• +
IN(R1 + R2 + R3 + ••• + RN)/R1
FIGURE 4. Basic Noninverting Current-to-Current
Converter.
FIGURE 6. Arbitrary-Gain Summing Noninverting Currentto-Current Converter.
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INVERTING AND NONINVERTING
CURRENT-TO-CURRENT CONVERTERS
The inverting and noninverting circuits can be used simultaneously either with equal gains or arbitrary gains as shown
in Figures 7 and 8:
For Figure 7:
IOUT = {(–I1A – I2A – ••• – INA)R2A/R1B} +
{(I1B + I2B + ••• + INB)(1 + R2B/R1B)}
I1A
4-20mA to 0-20mA CONVERTER USING BOTH
INVERTING AND NONINVERTING CURRENT-TOCURRENT CONVERTER CIRCUITS
One implementation of the 4-20mA to 0-20mA converter is
shown in Figure 9. From the Figure 4 and Figure 1 equations:
I2A
R2A
INA
INB
A1
I2B
4-20mA to 0-20mA CONVERTER
There are two ways to make a 4-20mA to 0-20mA converter
using current summing circuits and a REF200 100µA current source for offsetting. Since the polarity of the 0-20mA
output current must be the same as the 4-20mA input signal,
the noninverting circuit is used in the main signal path. The
REF200 100µA current source can be connected in either
polarity so either inverting or noninverting summing can be
used to offset the signal for 0mA out with 4mA in.
For Figure 9:
I1B
R2B
R1B
IOUT = I1(1 + R2/R1) – I2 • R3/R1
IOUT
IOUT = {(–I1A – I2A – ••• – INA)R2A/R1B} + {(I1B + I2B + ••• + INB)(1 + R2B/R1B)}
FIGURE 7. Equal-Gain Summing Inverting and
Noninverting Current-to-Current Converter.
For Figure 8:
IOUT = {–I1A • R2A/R1B – I2A(R2A + R3A)/R1B – ••• –
INA(R2A + R3A + ••• + RNA)/R1B} +
{I1B(1 + R2B/R1B) + I2B(R1B + R2B + R3B)/R1B + ••• +
INB(R1B + R2B + R3B + ••• + RNB)/R1B}
Where:
I1 = 4-20mA input current
I2 = 100µA REF200 offsetting current
Two equations can be written: one for IOUT = 0mA with IIN
= 4mA and one for IOUT = 20mA with IIN = 20mA. Since
there are three unknowns (R1, R2, and R3) and only two
equations, one resistor value must be selected first. A value
of 100Ω was selected as an arbitrary but adequate value for
R2. With R2 = 100Ω, the maximum voltage drop across it is
2V at 20mA. A smaller value for R2 will reduce the voltage
burden and power dissipation in the circuit, but since the
signal-to-noise ratio is reduced, it will be more sensitive to
op amp errors.
To solve for R1:
I1A
Notice that the current gain is:
I2A
(20mA – 0mA)/(20mA – 4mA) = (20mA)/(16mA)
= 1.25mA/mA
INA
RNA
R3A
R2A
Rewriting the Figure 4 equation in terms of gain:
GAIN = IOUT/IIN
GAIN = 1 + R2/R1
A1
INB
Solving the gain equation for R1:
RNB
I2B
R1 = R2/(GAIN – 1)
R3B
Substituting R2 = 100Ω:
R1B
I1B
R1 = 100Ω/(1.25 – 1)
R1 = 400Ω
R2B
IOUT
IOUT = {–I1A • R2A/R1B – I2A(R2A + R3A)/R1B – ••• –
INA(R2A + R3A + ••• + RNA)/R1B} +
{I1B(1 + R2B/R1B) + I2B(R1B + R2B + R3B)/R1B + ••• +
INB(R1B + R2B + R3B + ••• + RNB)/R1B}
FIGURE 8. Arbitrary-Gain Summing Inverting and
Noninverting Current-to-Current Converter.
Then, solving the Figure 9 equation for R3:
R3 = I1(R1 + R2)/I2
Substituting I1 = 4mA, I2 = 100µA:
R3 = 40(R1 + R2)
Substituting R1 = 100Ω, R2 = 400Ω:
R3 = 20.0kΩ
3
0.01µF
I2
REF200
100µA
I2
REF200
100µA
R3
20kΩ
+VS
–VS
+VS
I1
4mA to 20mA In
VIN
R2
100Ω
OPA177
VIN
I1
4mA to 20mA In
R1
400Ω
Op Amp
VOUT
OPA177
–VS
R3
19.5kΩ
Op Amp
VOUT
–VS
+VS
R1
400Ω
0.01µF
R2
100Ω
0mA to 20mA Out
0mA to 20mA Out
I Source
VOUT
I Source
VOUT
NOTE: This circuit has poor compliance with positive power-supply rail.
NOTE: This circuit has excellent compliance with positive power-supply
rail.
FIGURE 9. 4-20mA to 0-20mA Current-to-Current
Converter Using Inverting and Noninverting
Summing Circuits.
FIGURE 10. 4-20mA to 0-20mA Current-to-Current
Converter Using Only Noninverting Summing
Circuits.
4-20mA TO 0-20mA CONVERTER USING ONLY
NONINVERTING CURRENT-TO-CURRENT
CONVERTER CIRCUIT
The other possible circuit is shown in Figure 10. From the
Figure 6 equations:
FIGURE 9 CIRCUIT
FIGURE 10 CIRCUIT
OUTPUT
CURRENT
OP AMP VIN
OP AMP VOUT
OP AMP VIN
OP AMP VOUT
0mA
0.4
–1.6
–1.56
–1.56
20mA
2.0
0.0
0.04
0.04
NOTE: With the 0-20mA output connection node held at 0V.
For Figure 10:
TABLE I. Op Amp Input/Output Voltages for Figures 9
and 10 Circuits.
IOUT = I1(1 + R2/R1) + I2(R1 + R2 + R3)/R1
Where, as before:
I1 = 4-20mA input current
I2 = 100µA REF200 offsetting current
Both circuits have an op amp VOUT of approximately –1.6V
at one extreme. This limits the compliance to the negative
power-supply rail to SL + 1.6V where SL is the op amp
output negative swing limit. For example, if an op amp can
swing to –12V (VS = ±15V), its swing limit is SL = 3V. The
closest the circuit can swing to the negative rail is 3V + 1.6V
= 4.6V or VOUT = –10.4V (VS = ±15V).
The relationships for R1 and R2 are the same as for the
Figure 9 circuit: with R2 = 100Ω, R1 = 400Ω.
Solving the Figure 10 equation for R3:
R3 = (I1 – I2)(R1 + R2)/I2
Substituting I1 = 4mA, I2 = 100µA:
R3 = 39(R1 + R2)
The Figure 10 circuit has an advantage for compliance to the
positive power-supply rail. The Figure 9 circuit has a worstcase VIN = 2V. Also, the REF200 (with a min compliance of
2.5V) is connected between the input and +VS. This limits
the compliance to the positive power-supply rail to 2V +
2.5V = 4.5V.
Substituting R1 = 100Ω, R2 = 400Ω:
R3 = 19.5kΩ
WHICH 4-20mA TO 0-20mA
CONVERTER IS BETTER?
The only significant functional difference between the Figure 9 and Figure 10 converter circuits is the output voltage
compliance range. Output range of the current-to-current
converter circuit is limited by the op amp input and output
ranges and by the minimum compliance range of the REF200
current source. Voltages for the two circuits at the 0mA and
20mA output current extremes are shown in Table I.
The Figure 10 circuit has a worst-case VIN = 0.04V. Also,
the REF200 is connected between the input and –VS. This
allows the Figure 10 circuit to swing to the positive powersupply rail within the limits of the op amp input compliance.
Op amps are available with input compliance to +VS. Compliance of the Figure 10 circuit (using the OPA177) is:
+13V (+14V typ), –10.9V (–11.4V typ) with VS = ±15V.
The information provided herein is believed to be reliable; however, BURR-BROWN assumes no responsibility for inaccuracies or omissions. BURR-BROWN assumes
no responsibility for the use of this information, and all use of such information shall be entirely at the user’s own risk. Prices and specifications are subject to change
without notice. No patent rights or licenses to any of the circuits described herein are implied or granted to any third party. BURR-BROWN does not authorize or warrant
any BURR-BROWN product for use in life support devices and/or systems.
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