Isolated Industrial Current Loop Using the IL300 Linear

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1 VISHAY SEMICONDUCTORS Optocouplers and Solid-State Relays Application Note Isolated Industrial Current Loop Using the IL Linear INTRODUCTION Programmable logic controllers (PLC) were once only found in large manufacturing firms but now are used in small to medium manufacturing firms. PLCs are being retrofitted into manufacturing environments where temperature, pressure, and level sensor control signals are exposed to harsh electrical noise. The connection between these sensors and the controller requires the use of high noise immunity communication technology. One solution to this communication problem is the analog current loop. A current loop is an interface technique that converts a process sensor s output to a DC current signal. When compared to voltage control techniques, a current loop receiver s low input resistance offers higher noise immunity. Current loops have the added advantage of better accuracy, because they eliminate sensor signal errors introduced by communication line resistance. Electrical noise can be reduced further by providing isolation between the current loop receiver or transmitter and the process controller. An isolated receiver and transmitter can be constructed using the IL linear optocoupler. This application note will describe how to design a line powered isolated current loop receiver and transmitter. It will discuss the design process and show circuit variations compatible with common current loop pseudo-standards. Sensor XMTR Line RCVR V out Process controller 111 Power supply Fig. 1 - Isolated Transmitter and Receiver Current Loop CURRENT LOOP ELEMENTS A current loop typically consists of a transmitter, a receiver, and a DC power supply. The highest insulation and noise immunity is achieved when an isolated transmitter and an isolated receiver are used as shown in figure 1. However, there are many situations where only one end of the loop can be isolated. Figures and illustrate combinations of isolated and non-isolated current loop elements. Isolated current loop transmitters and receivers commonly require separate isolated power supplies in addition to the standard loop voltage supply. The designs in this application note derive their power from the DC supply found in the loop. Commonly the loop power supply is an isolated voltage supply whose output voltage will range from 1 V to V. Thus only a single isolated power supply is needed to power the loop. CURRENT LOOP CONVENTIONS The ma to ma current loop is the most common pseudo-standard. This convention defines a ma loop current as the sensor s zero reference. The full scale of the sensor output corresponds to a ma loop current, representing a minimum to maximum current ratio of 1:. The sensor s signal output commonly has a zero reference of 1 V and a full scale of V which also corresponds to a 1: signal ratio and a V span. Figure shows the transmitter s output loop current as a function of input sensor voltage. Other conventions include sensor signal spans of V, where the sensor s zero reference is V, and full scale is V (figure ). Rev. 1., 1-Oct-11 1 Document Number: 1 ARE SUBJECT TO SPECIFIC DISCLAIMERS, SET FORTH AT

2 Application Note Isolated Industrial Current Loop Using the IL Linear Sensor XMTR Line RCVR V out Process Controller Power Supply 11 Fig. - Isolated Transmitter and non-isolated Receiver Current Loop Sensor XMTR Line RCVR V out Process Controller 11 Power Supply Fig. - Non-Isolated Transmitter and Isolated Receiver Figures and show the transmitter transfer function. The loop current (IL) is the product of the sensor voltage ( ) times the transmitter trans conductance, milli-siemens. The receiver in Figure has a trans resistance of Ω, while for Figure it is 1. Ω (ma) = ms x 1 Sensor Voltage - Fig. - 1 V to V, ma to ma Current Loop Transfer Fig. - V to V, ma to ma Current Loop Transfer Rev. 1., 1-Oct-11 Document Number: 1 ARE SUBJECT TO SPECIFIC DISCLAIMERS, SET FORTH AT (ma) = ma. ms x 1 11 Sensor Voltage -

3 Application Note Isolated Industrial Current Loop Using the IL Linear CURRENT LOOP TRANSMITTER Figure shows an isolated current loop transmitter with a 1 V to put and a ma to ma output. The sensor section consists of an optical feedback amplifier (U1, IL) that converts the sensor voltage ( ) to an output photocurrent (I P ). The output amplifier, U, operates as a current controlled current sink. The equation for the line current ( ) as a function of the output photocurrent (I P ) is given below: I P x R I O = (1) The equation for the output photocurrent, I P, as a function of the sensor voltage is given below: x K I P = () Combining equations 1 and results in the complete transmitter DC transfer relationship with K the IL s transfer gain. I O K x R = () x 1 V to V, ma to ma TRANSMITTER DESIGN The design of the 1 V to put, ma to ma output isolated current loop transmitter starts with analyzing the isolated current to current converter. This amplifier (U), a National Semiconductor LM1 operational amplifier, was chosen for its high output current and ability to operate from a single supply. The input sensor amplifier controls the output photocurrent (I P ). I P develops a voltage across R at the inverting input of U, forcing a loop current to flow through. Thus I o times is equal to the voltage developed across R times IP (Equation ). Equation shows that resistors R and set U s current gain. I P x R = I O x () I P x R Current Gain = () Vin U1 N9 OP9 1pF 1 IL 1Ω 1 V K1 K Sensor Input IP1 IP Isolated Line U LM1 R Output Io 11 Sensor Connection A current gain of is selected, with equal to Ω. From equation, R is kω. Equation 1 shows that a loop current of ma to ma requires an output photocurrent (I P ) of 1 μa to μa. The last design step is to determine the input resistor () by rearranging Equation. The trans conductance, I o / of Figure, is milli-siemens (ms). The remaining variable is the IL s transfer gain, K. The part to part variation of the Fig. - Isolated 1 V to V, ma to ma Transmitter transfer gain offers a range of. to 1.. With K = 1, is calculated to be 1 kω from equation. See figure for the spread of versus the guaranteed range of K. Thus a kω, 1 turn potentiometer will compensate for the full distribution of K. Rev. 1., 1-Oct-11 Document Number: 1 ARE SUBJECT TO SPECIFIC DISCLAIMERS, SET FORTH AT

4 Application Note Isolated Industrial Current Loop Using the IL Linear - Input Resistor - kω Fig. - vs. K for Isolated 1 V to V, ma to ma Transmitter V to V, ma to ma TRANSMITTER DESIGN A current loop transmitter conforming to the pseudostandard of V to put to ma to ma output can be designed using the general circuit topology in figure. With K - Transfer Gain - K/K1 = K x kω ms x kω () = 1 kω for K = 1.. the addition of a bias source (V ref ) ma of line current will flow when = V. The LM1 offers an integrated mv band gap reference source with voltage follower buffer amplifier. The LM1 s voltage reference and differential amplifier make it uniquely qualified as the output current amplifier. Figure shows the schematic of a current transmitter including a bias source, U. By inspection and using Equation, the transmitter current transfer function can be determined. The transfer function for figure is given in equation. x K x R I O V ref = x R () x R x This equation shows that the loop current is the sum of the sensor controlled signal ( ) and current provided by the bias source (V ref ). The bias source consists of a voltage follower (U) that buffers a mv band gap reference. This voltage reference is converted to a current source by the R resistor. The value of R can be calculated from equation, when = V, and I o = ma. I ref V ref = R V ref I O = x R when V R in = V V ref x R R = () I O x 11 Sensor Input U1 N9 OP9 IL 1-1 pf 1 Ω V CC K1 K Sensor Connection U IP1 IP LM1 - U - Fig. - Isolated V to V, ma to ma Transmitter Rev. 1., 1-Oct-11 Document Number: 1 ARE SUBJECT TO SPECIFIC DISCLAIMERS, SET FORTH AT R 1 V ref R Isolated Line Output I o

5 Application Note Isolated Industrial Current Loop Using the IL Linear Given the current gain, R/ =, = V, and I o = ma, R is calculated to be kω. The input resistor () sets the trans conductance (ΔI P /Δ ) of the input amplifier. The current transmitter s trans conductance equals the trans conductance of the input amplifier times the current gain of the output amplifier. The transmitter incremental trans conductance is calculated given a Δ of V, ( V to V), and ΔIo of 1 ma ( ma to ma). A transmitter trans conductance. milli-siemens results. 1 V to V, ma to ma TRANSMITTER PERFORMANCE The transmitter described in figure was constructed and evaluated for accuracy and linearity as a function of input sensor voltage and ambient temperature. The transmitter was calibrated by adjusting for 1 ma loop current with an input voltage of V at T A = C. Figure 1 shows the percent error deviation from the expected loop current. This circuit offers a typical accuracy of ±. % over a temperature range of C to C. Note that the temperature performance appears to follow a parabolic contour.. - Input Resistor - K 1 1 Percent Error - %..1. C C C C K - IL Transfer Gain Fig. 9 - vs. K for Isolated V to V, ma to ma Transmitter ΔI P Δ = K ΔI O K = x R Δ (9) V in x K x R = ΔI O (1) Assume an output amplifier current gain of (R = kω, = Ω), a typical K = 1, and a transmitter trans conductance of. ms. Substituting R,, and K into Equation 1, can be determined. = 1. x kω ms x Ω (11) = 1 kω Figure 11 shows the relationship of as a function K. See Table 1 for the component values for each design. Isolated transmitter resistor values, K = 1. V to V to ma to ma 1 V to V to ma to ma 1 kω 1 kω R kω INF R kω kω kω Ω - Sensor Input Voltage - V Fig. 1 - Percent Error vs. Input Sensor Voltage 1 V to V, ma to ma Transmitter Fig Linearity Error vs. Input Sensor Voltage 1 V to V, ma to ma Transmitter Many industrial controllers have calibration techniques that can compensate for temperature imposed accuracy errors. These techniques are only valid if the transmitter exhibits a high degree of linearity. Figure 11 shows the linearity error for the transmitter. The linearity error is expressed as a deviation in parts per million (ppm) from a best fit linear regression at each temperature. Figure 11 shows a typical linearity of ppm to - ppm over a C to C temperature range. Rev. 1., 1-Oct-11 Document Number: 1 ARE SUBJECT TO SPECIFIC DISCLAIMERS, SET FORTH AT Linearity Error - ppm Sensor Input Voltage - V.. C C C C....

6 Application Note Isolated Industrial Current Loop Using the IL Linear V to V, ma to ma TRANSMITTER PERFORMANCE The transmitter in figure was constructed and evaluated for accuracy and linearity as a function of input sensor voltage and ambient temperature. The transmitter was calibrated by adjusting R for ma loop current with an input voltage of zero volts (. V). The resistor is then adjusted for 1 ma loop current with an input voltage of. V at T A = C. Figure 1 shows the percent error deviation from the expected loop current. This circuit offers a typical accuracy of. % over a temperature range of C to C. Note that the temperature performance appears to follow a parabolic contour. Percent Error - % Fig. 1 - Percent Error vs. Input Sensor Voltage V to V, ma to ma Transmitter Figure 1 shows the linearity error for the transmitter. The linearity error is expressed as a deviation in parts per million (ppm) from a best fit linear regression at each temperature. Figure 1 shows a typical linearity of ppm to - 1 ppm over a C to C temperature range. Linearity Error - ppm C C C C.. - Sensor Input Voltage - V Sensor Input Voltage - V... C C C C. Fig. 1 - Percent Error vs. Input Sensor Voltage V to V, ma to ma Transmitter.. CURRENT LOOP RECEIVER The sensor controlled, current loop signal is converted to a voltage by the current loop receiver. The receiver s conversion gain and output voltage span is determined by the adopted current loop standard. A ma to ma loop current is commonly converted to a 1 V to V output signal. The receiver design in this section conforms to this standard. Signal conversion and isolation are provided by an IL, linear optocoupler. The circuit is loop current powered. The isolation feature and the receiver s low operating voltage drop permits multiple receivers within the loop. RECEIVER OPERATION The isolated current loop receiver consists of two sections. They include a loop current to photocurrent current amplifier, U1, and an output trans resistance amplifier, U. Figure 1 shows a simplified schematic. The receiver s linearity and stability are insured by using optical feedback within the loop current to photocurrent amplifier. Fig. 1 - Isolated Current Loop Receiver The optical feedback amplifier provides precise control of the LED s output flux. A bifurcated optical signal path within the IL provides an equally well controlled photocurrent for the output trans resistance amplifier. The loop current to photocurrent current amplifier consists of a single-supply micro-powered differential control amplifier, U1, and an LED current shunt regulator, Q1. Shunt control of the LED current was chosen to accommodate the receiver s need for a low supply voltage operation. The current loop receiver circuit functions as follows. The loop current ( ) flows into the junction of U1 s V cc ( and R). U1 s supply current (IU1) is substantially smaller than the loop current and will be omitted in the analysis. The loop current is divided at the juncture of and R. The sum of the currents flowing in each leg is equal to the loop current. The individual currents (I q and ) are determined by the required LED current to generate the needed photocurrent (I P1 ) connected to the control network at U1. Figure 1 shows the I q and relationships for the receiver. Rev. 1., 1-Oct-11 Document Number: 1 ARE SUBJECT TO SPECIFIC DISCLAIMERS, SET FORTH AT I UI Q1 U1 V gs V b P D1 V a I P1 I q K1 R R LED V1 I R K P D I P R U 1 V o

7 Application Note Isolated Industrial Current Loop Using the IL Linear I - Current - ma 1 1 (ma) = -. ma. x (ma) I q (ma) =. ma. x (ma) = I q I q The output current, I P, is converted to a voltage by the trans resistance amplifier U. The output voltage gain equation is shown below. = I P x R (1) V O Combining equations 1 and 1 results in the current loop transfer gain solution, V o / (equation 19). V O R = x R x K (19) 1 Fig. 1 - LED Current Shunt Control The total loop current flows into the junction, V1. This current, IR, develops a voltage across R. Under initial conditions, this positive voltage appearing at the inverting input of U1 will force U1 s output towards the negative rail. This V gs forces Q1 into cut-off. Under this condition the LED current ( ) equals the loop current ( ). This rise in LED current generates an optical flux which falls on the feedback photodiode (PD1) and generates a photocurrent (I P1 ). This photocurrent will rise to a value where voltage developed across equals the voltage across R. This satisfies the differential amplifier requirement of V a = V b. U1 s output provides the control signal for Q1 s gate, forcing it into conduction and shunting excess loop current away from the LED current path. The feedback control relationship is shown in equation 1. I P1 x = I R x R; I R I P1 x = x R (1) I P1 I P I P Where: I P1 = feedback photocurrent K1 = feedback gain P = output photocurrent K = output gain K = transfer gain (K/K1) With equations 1 and 1, solve for I P. R I P = x I L x K (1) The transfer gain can be written from equation 1. I P 1 = K1 x (1) = K x (1) = I P1 x K (1) R = x K (1) 1 LED CURRENT SHUNT OPERATION The differential amplifier, U1, provides the control signal to the LED current shunt regulator. U1 s output is connected to the gate of an n-channel FET, Q1. This transistor is the control element of the LED current shunt regulator. The regulator consists of a network made up of the series connection of the FET and, in parallel with the series connection of the IL s LED and R. The amplifier s output signal controls the FET s drain to source resistance, R q. As the gate voltage is increased, the FET resistance will decrease causing a larger percentage of the loop current to be diverted away from the LED signal path. Thus a rising control voltage, V gs, causes the LED current to decrease. A Siliconix TN1L enhancement low-voltage FET was selected as the control device for two reasons. First, with I q ma, the FET s gate voltage should be less than V. The TN1L control characteristics as a function of loop current are shown in figure 1. Second, the FET s dynamic resistance should be in the same order of magnitude as the IL s LED dynamic resistance. The dynamic resistance of both the LED and FET are shown in figure I 1 q. V gs Fig. 1 - TN1L Gate Voltage vs. Drain Current Rev. 1., 1-Oct-11 Document Number: 1 ARE SUBJECT TO SPECIFIC DISCLAIMERS, SET FORTH AT I ds - Drain Source Current - ma V gs - Gate to Source Voltage - V

8 Application Note Isolated Industrial Current Loop Using the IL Linear R - Resistance - Ω R LED ( Ω) = 1. (A)^-1.1 R LED (Ω) ~ 1 (A) R FET (Ω) =. I ds (A) ^ -.9 TN1L FET IL LED Fig. 1 - Dynamic Resistance vs. Current The shunt regulator includes a series resistor in each leg of the network. These resistors are included in the design for two reasons: first, to provide a measure of current overload protection for the LED and FET, and second to set the initial control conditions for the network. The design equations are given below: L = I q x () V n V n Where: = loop current I q = Q1 drain current = LED forward current R FET = Q1 dynamic resistance R LED = LED dynamic resistance V n = Voltage across the control network Combining equations, 1, and : I q x ( R FET ) = x ( R LED R) () Replacing I q in terms of and setting to zero gives equation. = R FET - R LED - R () The LED and FET dynamic resistance equations are substituted into EQ. = (. x ( - ) -.9 ) R () This transcendental equation is best solved by iterative techniques. CURRENT LOOP RECEIVER DESIGN The current loop receiver design is divided into two sections. The first is the shunt regulator; the second is the feedback control amplifier. The shunt regulator design relies on equation and intuitive selection of an LED operating 1 I - Current - ma = I q x ( R FET ) (1) = x ( R LED R) () 1 point. The LED forward current is bounded by the loop current range which is ma to ma. The selection of and R is determined by solving equation when the LED current, = 1 ma, for a loop current equal to ma. This point is selected to provide sufficient FET current control range given the initial value range of K1 and its temperature dependence. Under the and IL conditions selected, Equation will provide the resistance range for and R. R - = Ω () Equation shows that R is greater than, and the recommended difference is Ω. Given this guidance, a 1 Ω resistor is selected for R. A larger value than the recommended Ω is selected for. A Ω resistor is used providing for greater LED current limiting. Given = Ω and R = 1 Ω, the LED current is calculated equation at loop current extremes. At = ma, the LED current ( ) is equal to 1. ma, while for a loop current of ma, = 9. ma. The next part of the design is selecting the resistors, R and, surrounding the feedback control amplifier. Recall that R is the loop current sense resistor and should be valued less than 1 Ω. For this design example, R = Ω. equation shows the relationship of in terms of circuit variables. R x = () x K1 Figure 1 shows the nonlinear nature of the feedback gain, K1, for the IL. The worst case condition occurs when the loop current is at its minimum, = ma. Under this condition = 1. ma. Figure 1 can be used to determine K1 under these conditions. The figure shows that at = 1. ma, K1 equals.. 1. Fig. 1 - LED Current and Feedback Gain vs. Feedback Photocurrent Substituting these values into equation, can be determined. Ω x ma = () 1. ma x. = 9. kω, a 1 kω resistor is selected. Rev. 1., 1-Oct-11 Document Number: 1 ARE SUBJECT TO SPECIFIC DISCLAIMERS, SET FORTH AT - LED Current - ma K1 I P1 - Feedback Photocurrent - µa K1 - Feedback Gain - I P1 /

9 Application Note Isolated Industrial Current Loop Using the IL Linear The final section of the design centers on the selection of the trans resistance of the output amplifier shown in figure 19. The feedback resistor (R) combined with the operation of the output amplifier (U) converts the IL s output photocurrent (I P ) into the output voltage (V o ). The output voltage span (ΔV o ) will be 1 V to V, given a loop current span (Δ ) of 1 ma. This relationship substituted into equation 19 can be used to solve for R. ΔV O x R = (9) Δ x K x R ΔV O = V O max. - V O min. Δ = max. - I () L min. ( V O max. - V O min. ) x R = (1) ( max. - min. ) x K x R ( V - 1 V) x 1 kω R = () ( ma - ma) x 1. x Ω R = 1 kω The final circuit of the isolated current loop receiver is shown in figure 19. The circuit is completed by adding two diodes placed in series with the loop. The diode, D, is a protection device which will block current flow if the receiver s loop voltage source is improperly connected. The diode, D1, performs two functions: (1) a visual indicator of loop current flow, () functions as a V drop in the loop. This voltage drop is needed to provide supply head room for the control of the shunt regulator FET. RECEIVER PERFORMANCE ma to ma LOOP CURRENT, 1 V to V OUTPUT The receiver in Figure 19 was constructed and evaluated for accuracy and linearity as a function of input loop current and ambient temperature. The receiver was calibrated by adjusting R for. V output with a loop current of 1. ma at T A = C. Figure shows the percent error deviation from the expected output voltage. This circuit offers a typical accuracy of. % to -. % over a temperature range of C to C. Note that the temperature performance appears to follow a linear temperature characteristic. Figure 1 shows a typical temperature coefficient of 1 ppm/ C. Many industrial controllers have calibration techniques that can compensate for temperature imposed accuracy errors. These techniques are only valid if the receiver exhibits a high degree of linearity. Figure 1 shows the receiver s linearity error as a deviation in parts per million (ppm) from a best fit linear regression at each temperature. Figure 1 shows a typical linearity of ppm to -1 ppm over a C to C temperature range. CONCLUSION Isolated current loops offer the industrial control designer the peace of mind that electrical noise and grounding problems will not influence the sensor signal. This application note has shown the design technique and results to construct a line powered ma to ma current loop receiver. It also presented two isolated current loop transmitters, one conforming to the 1 V to put and a second to the V to put standard. The performance data in this application note shows that the receiver and transmitter easily conform to a bit operation over a C to C operating range. 19 Line D* 1N91 Line *C1 1 pf OP9 D1 LDH1111 Ω TN1L 1 kω Q1 R R 1 Ω 1 Fig Isolated Current Loop Receiver Rev. 1., 1-Oct-11 9 Document Number: 1 ARE SUBJECT TO SPECIFIC DISCLAIMERS, SET FORTH AT K1 IP1 * optional IL K IP R R kω 1 kω V CC OUTPUT OP9 V CC = ± 9 V - V CC *C 1 pf GAIN Q1 Siliconix OP9 Analog Devices

10 Application Note Isolated Industrial Current Loop Using the IL Linear.. C Percent Error - % C C C 1 1 Linearity Error - ppm C C C C 11 1 Fig. - Percent Error vs. Loop Current ma to ma Receiver Fig. 1 - Linearity Error vs. Loop Current ma to ma Receiver Rev. 1., 1-Oct-11 1 Document Number: 1 ARE SUBJECT TO SPECIFIC DISCLAIMERS, SET FORTH AT

Isolated Industrial Current Loop Using the IL300 Linear Optocoupler Appnote 54

Isolated Industrial Current Loop Using the IL Linear Optocoupler by Bob Krause Introduction Programmable Logic Controllers (PLC) were once only found in large manufacturing firms but now are used in small