APPLICATION BULLETIN ® Mailing Address: PO Box 11400 • Tucson, AZ 85734 • Street Address: 6730 S. Tucson Blvd. • Tucson, AZ 85706 Tel: (602) 746-1111 • Twx: 910-952-111 • Telex: 066-6491 • FAX (602) 889-1510 • Immediate Product Info: (800) 548-6132 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. 1 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. 6 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 7 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 5 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. © 4 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. 14 15 FIGURE 2. Equal-Gain Summing Inverting Current-to-Current Converter. AB-031 Printed in U.S.A. April, 1991 16 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. 2 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. 4

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