Impedance of HART Transmitters Nesebar, Inc.

Transcription

1 Impedance of HART Transmitters Nesebar, Inc. A 2Wire 420 ma Process Transmitter is essentially a current regulator. The compliance impedance of the regulator is often tens of megohms near DC but drops rapidly with increasing frequency. If the transmitter also implements HART, there is a minimum impedance requirement out to several khz, to avoid loading at HART signal frequencies. This paper analyzes the impedance of HART transmitters and examines two common current regulator topologies in light of HART requirements. The impedance magnitude versus frequency is calculated for several cases. Note: The original of this document is in MathCAD Version 3. You may request a copy of the original MathCAD file for your own analyses. Contact Nesebar, Inc. Introduction Process transmitters evolved in response to the need to signal a process variable over a long distance without loss of accuracy due to wiring resistance. Such transmitters are essentially DC current regulators. Modern transmitters often add HART digital signaling to the analog signal, which extends the basic current regulator idea to higher frequencies. HART documents still refer to these devices as "transmitters" and further classify them as "high impedance." Since the accuracy of the 420 ma signal is determined, in part, by the compliance impedance of the transmitter current regulator, the impedance magnitude near DC is usually highoften in the tens of megohms. If the transmitter is a HART transmitter, the impedance must also be reasonably high at HART signaling frequencies. This paper discusses HART impedance requirements, presents two generic transmitter circuits, and analyzes the impedance for both of these circuits. It is found that a Series Current Regulator circuit may be preferred, since it allows impedance requirements to be satisfied more easily. HART Requirements HART documents model the transmitter as a parallel R and C and require that R>00 kohm and C<5000 pf. Although this is a gross oversimplification of the active transmitter circuit, the usual interpretation is: If the smallsignal impedance magnitude of the transmitter exceeds that of a 00 kohm resistor in parallel with a 5000 pf capacitor at HART frequencies, then the requirement is met. To provide some margin, this should probably hold true out to 0 khz. The impedance requirement is illustrated in Figure.

2 Page Acceptable Region Z (ohm) K 5000 pf Frequency (Hz) > Figure Impedance Magnitude of RC and Acceptable Region for HART The impedance magnitude is plotted (red curve) for the RC circuit shown. The HART transmitter impedance magnitude should be above this curve in the area labeled "Acceptable Region." Need For Voltage Regulator In 2wire transmitters a small portionusually about 3.5 maof the regulated current provides operating power to a block of transmitter circuitry. This block typically includes lowpower integrated circuits and implements sensing, signal conditioning, communication, etc. While the voltage at the field terminals might range from 0 to 40 volt, the lowpower block usually runs on a regulated supply voltage of 3 to 5 volt. Therefore, in addition to the current regulator, transmitters commonly need a voltage regulator as well. Tale of Two Topologies Although there are many ways of implementing the combined current and voltage regulators, most of them can probably be reduced to one of two basic topologies, shown in simplified form in Figure 2. For either current regulator, the compliance impedance of interest is the impedance seen across the field terminals. For the Series Current Regulator (upper circuit), the Current Control Element is in series with the load (lowpower block of circuitry). For the Shunt Current Regulator, the Current Control Element is in parallel with the load. For either current regulator, the heavy gray line shows the approximate path of the 3.5 ma of load current. The heavy green line shows the path of the additional (0.5 ma to 6.5 ma) of current that is just dumped by the current regulator. For convenience the two current regulator topologies will be called SE for Series and SH for Shunt.

3 Page 3 Current Control Element Field Terminals R 2 HART Signal Shunt Voltage Regulator R Opamp Load Rs To D/A Field Terminals HART Signal Series Voltage Regulator R 2 R Opamp Current Control Element Load Rs To D/A Figure 2 Series (Upper Circuit ) and Shunt (Lower Circuit) Current Regulators An NChannel MOSFET is shown as the Current Control Element in Figure 2. However, an NPN transistor with an emitter resistor can also be used and will be discussed later. Regardless of the type of transistor, it is subject to as much as 40 volt and a power dissipation of as much as 0.6 wattrequirements that virtually dictate a separately packaged power transistor as the Current Control Element.

4 Page 4 Keep in mind that the circuits of Figure 2 may not be completely functional as shown. Levelshifting, for example, may be needed between the opamp output and the MOSFET to produce a working circuit. Or the MOSFET may have to be a depletionmode device. Such techniques are common and are outside of the scope of the present discussion, which is concerned primarily with AC behavior. The MOSFET of Figure 2 should be chosen to operate in saturation. Most of the time this probably happens by default, since the device is operated at a drain current well below its capability. (This results in a small gate drive, which results in saturated operation.) An example is the BSP89, a surfacemount MOSFET in an SOT223 package. This has a continuous drain current rating of 375 ma and power dissipation of.5 watt. The gate drive above threshold needed to conduct 20 ma is roughly 0.5 volt, which will be well below the drainsource voltage. The two current regulators of Figure 2 are similar. Current is sensed across Rs, which is typically a few tens of ohms. The DC current is set by the voltage at the node "To D/A." The HART signal is generated by ACcoupling a current into the RR2 node. The impedance at the field terminals is a function of the opamp openloop gain. The openloop gain must remain high enough at higher frequencies to meet the HART impedance requirement. Notice that the SE load current flows through the Current Control Element, while the SH load current does not. Consequently, the drain current of the SE MOSFET ranges from 4 ma to 20 ma, while the drain current of the SH MOSFET ranges from 0.5 ma to 6.5 ma. The SH (Lower Circuit) of Figure 2 also models some 3 and 4wire transmitters. In these circuits the series voltage regulator and load are removed, since power and other transmitter functions are being provided elsewhere. The model remains the same, since the series voltage regulator appears approximately as an open circuit. Now, however, the SH MOSFET drain current will range from 4 ma to 20 mathe same as for the SE MOSFET. The SH also models circuits in which the series voltage regulator is a switching regulator.

5 Page 5 Current Regulator Analysis The AC model of the current regulator is given in Figure 3. This applies to either current regulator of Figure 2. This reduction to a single circuit is possible because, to a good approximation, the input to the shunt voltage regulator is an AC shortcircuit, and the input to the series voltage regulator is an AC opencircuit. To visualize the opencircuit nature of the series regulator, notice that the input current stays almost constant (and equal to the load current) regardless of the applied input voltage. Impedance Z r, consisting of a parallel R and C, has been added to the AC model of Figure 3 and was not explicitly shown in Figure 2. These are general elements that model actual effects in either current regulator. They can be retained or removed, as needed, by setting their values appropriately. The capacitor, for example, models a bypass capacitor typically used in the SH. It is placed across the input to the series voltage regulator and may be large enough to affect the Field Terminals R 2 gm*v R Rs R Opamp V g0 C Current Control Element Z r Figure 3 Series and Shunt Current Regulators AC Equivalent Circuit desired impedance. The SE usually does not have this capacitor. (Keep in mind that this is an AC model and that R is not necessarily an actual resistor connected from drain to source of the MOSFET.) The analysis of the impedance at the field terminals is straightforward: Zin = R R 2 g m A 0 Rs R ( ) g o Z r ( s τ ) = R R 2 g m A 0 Rs R ( ) g o sc R ( s τ )

6 Page 6 where A 0 is the opamp DC gain, τ is the opamp time constant. (The opamp is modeled as a single pole.) The usual practice of discarding any small terms that do no include the opamp gain has been applied. The transconductance is proportional to the square root of current, so that this expression becomes Zin = R R 2 K I D A 0 Rs R ( ) g o sc R ( s τ ) where K is a constant. Notice that the impedance depends on g 0, the MOSFET output conductance. Output conductance is generally not available in device data sheets. In the saturation region it is proportional to drain current, which means that, if we operate a highcurrent power MOSFET at only a few ma, g 0 will be quite lowprobably on the order of 00 micromho or less for most devices. For purposes of this analysis it is assumed that R = kohm and that R effectively swamps the effect of g 0. This may not be entirely true for the SE. If not, then the calculated field terminal impedance will be smaller than the actual impedance. Dropping the output conductance, the expression for impedance becomes Zin = ( ) R R 2 K I D A 0 Rs R R sc ( s τ ) Current Regulator Calculations has The BSP89 described above will be used in the example calculations. This K = 0.67 mho / root amp Additional representative component values to be used in the analyses are R = 00K R 2 = 0K R S = 25 A 0 = 0 6 τ = Since impedance is lowest at the lowest drain current, only the lowest drain current is used for each regulator. This is 4 ma for the SE and 0.5 ma for the SH.

7 Page 7 The analysis results are plotted in Figure 4 below. The HART requirement is included for comparison. Capacitor C = 00 pf for SE and C = 0. microfarad for SH. I D = 4 ma for SE and I D = 0.5 ma for SH. The large SH capacitance causes the impedance magnitude to bend down above khz and fail the HART impedance requirement. At lower frequencies the higher SE impedance due to the higher quiescent drain current is evident. If the SH topology must be used, we are tempted to correct Z by using an opamp with a higher bandwidth (smaller τ). However, this must be done carefully to avoid instability SE. 0 7 SH. 0 6 HART Z (ohm) Frequency (Hz) > Figure 4 Analysis Results Compared to HART Requirement Bipolar Transistor With Emitter Resistor A bipolar transistor with an emitter resistor can also be used as the current control element. The comparison with the MOSFET is shown in Figure 5. The previous analysis applies to the bipolar transistor if g m is replaced with and R E we ignore the base current. At first it appears that we can simply reduce R E to a low value to get a very high g m and a correspondingly high compliance impedance. However, the bipolar transistor has an internal emitter resistance that must be added to R E. The internal emitter resistance depends on the emitter current and ( 26mV) is given by R EE =. (We have also ignored a small temperature I E

8 Page 8 Controlled Current Controlled Current Control Voltage R E Control Voltage Figure 5 Bipolar and MOSFET Transistors Used as Current Control Element K 2 I E dependence here.) Thus g m = =. For very small R R E R EE K 2 R E I E E the expressions for g m and Zin become similar to those for the MOSFET, except for the square root dependence on the controlled current. For I E values of 4 ma and 0.5 ma, the corresponding R EE values are 6.5 ohm and 52 ohm. The key observation is that we have almost the same situation as with the MOSFET: As the controlled current is reduced toward zero, the compliance impedance also approaches zero. The SE has a higher impedance than the SH, due to the higher quiescent collector current. The complete expression for the bipolar impedance is Zinb = K 2 R E I E K 2 I E A 0 Rs R ( ) ( R R 2) R sc ( s τ ) A Darlington transistor is sometimes used to reduce the amount of base current that must be supplied by the opamp. This does not change the expression for R EE of the main control transistor and, therefore, does not remove the I E dependence in the expression for Zin. Representative component values for the bipolar analyses are R = 00K R 2 = 0K R S = 25 A 0 = 0 6 τ = R E = 50

9 Page 9 The analysis results, which are similar to those for the MOSFET, are presented in Figure 6. The collector current is 4 ma for SE and 0.5 ma for SH. The shunt capacitance, C, is 00 pf for SE and 0. ufd for SH. Again, the SE meets the HART requirement while the SH does not. At low frequencies the SE impedance is greater than the SH impedance, due to the higher collector current. And at higher frequencies the shunt capacitance causes the SH curve to bend downward. It appears that, even if the shunt capacitance were to be set equal to 00 pf for SH, the impedance magnitude would still have very little margin at high frequencies Z (ohm) SE SH HART Frequency (Hz) > Figure 6 Bipolar Analysis Results Compared to HART Requirement SPICE Confirmation The BSP89 MOSFET and FZT05A bipolar transistor were used. The SPICE models for these devices are those supplied by their respective vendors. The impedance magnitude was found for the SE and SH cases analyzed above. For the bipolar transistor an external 50 ohm resistor was used for RE. A behavioral opamp model was used. The results, presented in Figures 7 and 8, agree with the equations and analyses above.

10 Page 0 MOSFET Impedance Magnitude Z (ohm) Freq (Hz) Figure 7 MOSFET Impedance Magnitude Found With SPICE Bipolar Impedance Magnitude 0 7 Z (ohm) Freq (Hz) Figure 8 Bipolar Impedance Magnitude Found With SPICE