PT100温度测量电路

电阻温度检测器（RTD）除了用于测量温度的热电偶，仪器仪表工程师经常使用电阻温度检测器或RTD。

这些设备的直流电阻变化（几乎）作为线性温度的函数。

或许其中最常见的是PT100，铂为基础的传感器，其电阻在0℃，正是100欧姆，（见表1）。

由于传感器的温度升高其电阻也是如此，在一个合理的线性方式。

表1显示了一个PT100传感器的电阻随温度的变化。

而温度系数略有不同在一个很宽的温度范围内，（通常为0.0036至0.0042欧姆/ º C），它可以被认为是合理恒定在50或100 º C范围内。

普遍接受的平均温度系数为0.00385欧姆每ºC。

据此，PT100往往可以在不超过这个范围线性化使用提供相应的系数进行评估。

这个装置也能承受的温度范围很广，从-200到800 º C的能力，以及一些应用中的温度系数的变化可以容忍的。

此外，PT100提供了稳定和可重复的温度特性。

对于给定的基极电阻R O，一个RTD电阻在T º C为：或ααooRTRTTTTRTR-=--+=)())(1()(... (1)其中R O是基极电阻对应到T O，（在0 º C 100欧姆）和是温度系数（每º C0.00385Ohms）。

因此，R（100℃）= 138.5欧姆。

这种近似提供了相当良好的温度估计高达约300℃，如图1所示，在此之后，非线性就不言而喻了。

图1。

RTD线性模型与实际特性方程（1）假设，在RTD的非线性特性可以忽略不计，即该设备完全是线性的，而许多应用这种近似是可以接受的，这里需要一个更精确的非线性模型，必须使用，如公式概述（ 2）。

))100(1()(32TTCBTATRTRo-+++=（2）其中：A = 3.908E - 3，B = - 5.775E - 7和C = - 4.183E - 12 T <0，C =为T 0> 0。

温度信息可以从一个RTD通过测量其电阻，或者通过应用已知的电流并测量产生的电压，反之亦然。

护理时所采取的缪斯穿过一个RTD为内部I 2 R加热电流也会影响设备的性能。

在何种程度上发生这种情况取决于有问题的RTD的物理尺寸，因此它可以多少热量消散之前，其显着高于环境温度上升。

对于小型设备感电流必须保持相当低，通常比3毫安少。

小（厚片）PT100设备显示在图2。

图2。

一个PT100温度传感器厚膜建设图3。

样品PT100探头热电阻一般有一个小的热质，因此可以表现出快速反应，在温度急剧变化。

这可以在过程控制应用。

信息编码技术。

仪器仪表应用经常使用可编程逻辑控制器（PLC）来存储和处理数据，因此在检测设备模拟输出信号必须为AD转换器缩放的PLC输入卡适当关注。

这通常是由传感器来完成驱动电路。

有几个标准电压由制造商使用的范围，这些包括0至1，0至5和0至10伏，每到所需要的RTD温度检测范围对应。

除了电压源的信号，这也是通常使用一个电流源进行编码的模拟信息。

这种方法显着的运营商提供过电压噪声因为这两种常见模式和正常模式感应电压的免疫力，可以在不显着破坏的电流流过的耐受性。

四到二十毫安的电流回路中经常使用的传输距离超过中等，例如从一个工厂到另一个侧面，传达模拟信息。

循环变送器普遍建立使所需温度范围的下限对应4mA和上端至20mA。

因此，应循环会断裂，在目前的总损失，故障可以很容易地检测出来。

有效的模拟信号编码为016毫安，从产地转移的4mA电流。

该温度对应这些电流（通常为跨度已知）的范围是由用户，谁必须相应方案的发射机。

有些环变送器供电的4mA电流分量，而有些则需要外部电源。

RTD的驱动电路。

在图4所示的原理图设计，接口PT100到PLC模拟输入卡。

它提供两个输出信号，一个0-5伏的电压信号和4 - 20mA 电流信号。

该电路采用惠斯登电桥的安排，以得出一个正电压，正比于超出了基极电阻R O，它对应于所需温度范围的下限，（在这种情况下，0℃）传感器电阻增加。

图4。

A为PT100温度测量电路。

苏氨酸RTD是包含在一个惠斯登电桥的安排（有时是4桥配置），它从一个分裂电源供电。

但是，在这种电路的电压供应不太平等的。

负轨固定在0.265伏，而积极的轨道设置，以便对RTD的顶部侧电压为零，即让桥梁清空。

需要空桥的电压会有所不同，取决于温度RTD的。

因此温度信息被编码在正电源的潜力。

桥的左侧由两个相同的电阻，这在他们的工会产生的共模电压只含有相关信息的RTD 温度。

这种技术的一个特别好的特点是，输出是真正与电阻ΔR 线性，此外，输出电压是接地的 。

这意味着，没有任何共模电压的情况下，必须由差分放大器拒绝。

电路分析 。

图5显示了简化的惠斯通电桥和调零放大器。

在RTD 是由（R 0 +ΔR ）为代表的地方ΔR 表示电阻随温度变化;桥的上臂设置为 R O ，（基阻力，对应到 T O ）。

该归零放大器的目的是推动在反相端的电压由零 V +适当调整。

特别是，当RTD 温度为 T O ，然后ΔR= 0和 V - = V +。

图5。

简化的惠斯通电桥和归零放大器 。

通过应用叠加原理，我们得到：02)(2=++++-+rR r R V r R R V o o o o δδδ 使 V +变为：)1(o R r V V δ+-=-+ 在上图中，V O 是 V +和 V 平均- ，因此，我们发现：这个结果是令人满意的，尤其是线性的，因为在V O 有没有共同的模式组件。

o o o R r V R r V V V 22)/1(δδ----=⎥⎦⎤⎢⎣⎡+-=如果265.0-=-V 伏*，然后 oo R r V 2265.0δ=，因为我们知道，一个PT100电阻为100欧姆138.5℃，然后 o R r /δ = 0.385在100℃，因此我们发现，CmV V o o 10051= ，或者在V O 增加了0.51mV /℃率 。

这个值很小，并且为了实现更方便的输出电平放大器是必需的，如图4所示。

对于在100 ° C 的5伏输出潜力的增益要求是：015.98051.05==G*（0.265伏的选择，以限制在RTD 的自热效应作为桥梁当前的结果。

）非反相放大器如图4所示提供此增益，可修剪使用量程调整电位器。

除了 跨度， 偏移调整还提供了电路，（见图4）。

这是为了使用户能够匹配的桥右侧的电阻（即R O ，并 ） 当 T = T O 。

这将确保桥梁在温度T o 和由此V O （T O ）= 0平衡 。

这两个调整，使用精确的校准10圈电位器。

电路校准 。

该电路可以方便地校准使用固定校准电阻 如下：1. 更换一个100欧姆的电阻RTD 校准（R O ）。

调整偏移电位器，直到输出电压变为零。

2. 适合在1中插入的部分地方138.5欧姆的电阻校准电阻 。

调整量程电位器，直到输出电压等于5伏。

3. 重复步骤1和2，直到每个电位器不需要进一步的调整。

（因为这些调整的每一个影响外，校准过程是一个反复的一个。

）4 - 20mA 电流输出。

最后，关于4 - 20mA 电流信号词，这是驱动电路从0-5伏直流增益由放大器产生的输出。

4 - 20mA 电流源使用一个运算放大器和双极型晶体管（BJT ），发射器连接，使潜在的被反馈到反相输入端（见图4）。

集电极电流是通过提供一个很好的近似到非反相输入端的发射极电阻，R E 除以电压给定的。

当测得的温度为0℃，因而增益放大器输出为零伏特，前者电压变为（1.25）/ 2伏（2部是由于10K ：10K 分压器）。

由此产生的回路电流，因此必须： 。

因此 欧姆。

另一方面，当测量温度为100 摄氏度 ，从增益放大器输出变为5伏，电流源控制电压变为（1.25 +5）/ 2伏。

输出电流，因此成为 毫安 ，（如预期）。

因此，综上所述，回路电流之间的线性变化4mA 和20MA ，随着温度的变化介于0和100℃。

表1。

PT100电阻作为温度函数英文原文：Operational Amplifier ApplicationsThe Resistive Temperature Detector (RTD)In addition to thermocouples for measuring temperature, instrumentation engineers frequently use Resistive Temperature Detectors or RTDs. These are devices whose DC resistance varies (almost) linearly as a function of temperature. Perhaps the most common of these is the PT100, a platinum based sensor whose resistance at 0ºC is exa ctly 100 Ohms, (see Table 1). As the sensor’s temperature increases so does its resistance, in a reasonably linear manner. Table 1 shows the variation in resistance of a PT100 sensor with temperature. While the temperature coefficient varies slightly over a wide range of temperatures, (typically 0.0036 to 0.0042 Ohms/ºC), it can be considered reasonably constant over a 50 or 100 ºC range. The commonly accepted average temperature coefficient is 0.00385 Ohms per ºC. Accordingly the PT100 can often be used without linearization over such a range provided the appropriate coefficient is evaluated. This device is also capable of withstanding a wide range of temperatures, from -200 to 800ºC, and for some applications the variations in temperature coefficient can be tolerated. Further, the PT100 provides stable and reproducible temperature characteristics.For a given base resistance R o , the resistance of an RTD at T ºC is given by:Or ααo o R T R T T T T R T R -=--+=)())(1()(00 (1)Where R o is the base resistance corresponding to T o , (100Ohms at 0 ºC) and αis the temperature coefficient, (0.00385Ohms per ºC). Thus R(100 ºC ) = 138.5 Ohms . This approximation provides quite a good estimate of temperature up to about 300 ºC, as shown in Figure 1, thereafter the nonlinearity becomes evident.Figure 1. Linear RTD model vs. the actual characteristicEquation (1) assumes that the nonlinearities in the RTD characteristic are negligible, ie that the device is entirely linear, and while for many applications this approximation is acceptable, where more precision is required a nonlinear model must be used, as outlined in Equation (2).))100(1()(32T T C BT AT R T R o -+++= (2)Where: A = 3.908E-3, B = -5.775E-7 and C = -4.183E-12 for T<0 and C = 0 for T>0.Temperature information can be obtained from an RTD by measuring its resistance; either by applying a known current and measuring the resulting voltage or vice versa. Care muse be taken when passing a current through an RTD as internal I 2R heating will also affect the device’s resistance. The degree to which this occurs depends on the physical size of the RTD in question, and therefore how much heat it can dissipate before its temperature rises significantly above ambient. For small devices sense currents must be kept quite low, typically less than 3mA. A small (thick film) PT100 device appears in figure 2.Figure 2. A Thick Film PT100 Temperature Sensor ConstructionFigure 3. Sample PT100 probesRTDs generally have a small thermal mass and therefore can exhibit a fast response to rapid changes in temperature. This can be useful in process control applications.Information Coding Techniques.Instrumentation applications frequently use Programmable Logic Controllers (PLCs) to store and process data, and therefore the analogue output signals of sensing equipment must be scaled appropriately for the A-D converter input card of the PLC concerned. This is generally accomplished by the sensor driving circuitry. There are several standard voltage ranges used by manufacturers; these include 0 to 1, 0 to 5 and 0 to 10 volts, each corresponding to the desired range of temperatures detected by the RTD.In addition to the voltage source based signals, it is also common to use a current source to carry encoded analogue information. This method offers significant noise immunity over voltage carriers, since both common mode and normal mode induced voltages can be tolerated without significantly corrupting the current flowing. Four to twenty mA current loops are frequently used over moderate transmission distances, for example from one side of a factory to the other, to convey analogue information.The loop transmitter is generally set up so that the lower end of the required temperature range corresponds to 4mA and the upper end to 20mA. Thus should the loop become broken, resulting in a total loss of current, the fault can be readily detected. Effectively the analogue signal is encoded as a 0-16mA, current shifted from the origin by 4mA. The range of temperatures that correspond to these currents (usually known as the span) is determined by the user, who must program the transmitter accordingly. Some loop transmitters are powered by the 4mA current component, while others require an external power supply.An RTD Drive Circuit.The schematic shown in Figure 4 is designed to interface a PT100 to a PLC analogue input card. It offers two output signals; a 0-5 volt voltage signal and a 4-20mA current signal. The circuit uses a Wheatstone bridge arrangement to derive a positive voltage, proportional to the increase in sensor resistance beyond the base resistance R o, which corresponds to the lower end of the desired temperature range, (in this case 0 ºC).Figure 4. A Temperature Measuring Circuit for the PT100.Thr RTD is included in a Wheatstone bridge arrangement (sometimes known as a quarter bridge configuration), which operates from a split power supply. However in this circuit the voltage supplies are not quite equal. The negative rail is fixed at 0.265 volts while the positive rail is set so that the voltage on the top side of the RTD is zero,i.e. so that the bridge is nulled.The voltagerequired to null the bridge will vary, depending on the temperature of the RTD. Therefore temperature information is encoded in the positive supply potential .The left hand side of the bridge consists of two identical resistors, which at their union generate a common mode voltage containing information relating only to the temperature of the RTD. A particularly good feature of this technique is the fact that the output is truly linear with the resistance δr, and in addition, the output voltage is ground referenced . This means that there is no common mode voltage present that must be rejected by the differential amplifier.Circuit Analysis.Figure 5 shows the simplified Wheatstone bridge and nulling amplifier. The RTD is represented by (R 0+δr), where δr represents the resistance variation with temperature; the upper arm of the bridge is set to R o , (the base resistance, corresponding to T o ). The purpose of the nulling amplifier is to drive the voltage at the inverting terminal to zero , by adjusting V + appropriately. In particular, when the RTD temperature is T o, then δr = 0 and V - =V +.Figure 5. Simplified Wheatstone Bridge and Nulling Amplifier .By applying the principle of superposition we obtain:02)(2=++++-+rR r R V r R R V o o o o δδδ So that V + becomes:)1(oR r V V δ+-=-+ In the figure above, V o is the average of V + and V -, thus we find:This result is particularly satisfying since V o is linear in r δ and there is no common mode component.If 265.0-=-V volts*, then oo R r V 2265.0δ=, and since we know that the resistance of a PT100 is 138.5 Ohms at 100o C, then o R r /δ= 0.385 at 100o C, and thus we find that CmV V o o 10051=, or alternatively V o increases at a rate of 0.51mV/o C . This value is quite small, and in order to achieve more convenient o o o R r V R r V V V 22)/1(δδ----=⎥⎦⎤⎢⎣⎡+-=output level an amplifier is required, as shown in Figure 4. For an output potential of 5 volts at 100o C the gain required will be:015.98051.05==G* (0.265 volts was chosen so as to limit the self heating effect in the RTD, as a result of the bridge current.)The non inverting amplifier shown in Figure 4 provides this gain, which can be trimmed using the Span adjustment potentiometer.In addition to the span, an offset adjustment is also provided in the circuit, (see Fig 4). This is intended to enable the user to match the resistors on the right hand side of the bridge, (i.e. R o and r R o δ+) when T = T o . This will ensure that the bridge is balanced at temperature T o and thus V o (T o ) = 0.Both these adjustments use 10 turn potentiometers for precise calibration.Circuit Calibration.The circuit can be easily calibrated using fixed calibration resistances as follows:1. Replace the RTD with a 100 Ohm calibration resistance (R o ). Adjust the Offset potentiometeruntil the output voltage becomes zero.2. Fit a 138.5 Ohm calibration resistance in place of the resistance inserted in part 1. Adjust theSpan potentiometer until the output voltage equals 5 volts.3. Repeat steps 1 and 2 until each potentiometer requires no further adjustment. (Because eachof these adjustments affects the other, the calibration process is an iterative one.)4-20mA Current Output.Finally, a word about the 4-20mA current signal; this circuit is driven from the 0-5 volt DC output generated by the gain amplifier. The 4-20mA current source uses an operational amplifier and a Bipolar Junction Transistor (BJT), connected so that the emitter potential is fed back to the inverting input terminal, (see Fig 4). The collector current is to a good approximation given by the voltage supplied to the non-inverting input terminal divided by the emitter resistance, R E .When the measured temperature is 0o C, and thus the gain amplifier output is zero volts, the former voltage becomes (1.25)/2 volts (Division by 2 is due to the 10k:10k potential divider). The resulting loop current therefore must be: mA R v I Eo 0.4225.1==. Thus 25.156=E R Ohms. On the other hand when the measured temperature is 100o C, the output from the gain amplifier becomes 5 volts, and the current source controlling voltage becomes (1.25 +5)/2 volts. The output current therefore becomes 202)525.1(=+=Eo R v I mA , (as expected). So in summary, the loop current varies linearly between 4mA and 20mA, as the temperature varies between 0 and 100o C.KNE222 University of TasmaniaG . Vertigan Page 112009 Table 1. PT100 Resistance as a function of Temperature。