什么是PT100传感器芯片? 3-线式PT100温度探头

PT100传感器芯片概述:
当PT100处于 0 摄氏度, 它的电阻是 100 欧姆, 这就是它被命名为 PT100 的原因. Its resistance will increase at an approximately uniform rate as the temperature rises. But the relationship between them is not a simple proportional relationship, but should be closer to a parabola. Since the isolation of PT100 resistance per degree Celsius is very small, within 1Ω, it is destined to have a more complicated circuit, because in actual use, the wire will be longer, there will be line resistance, and there will be interference, so it is more troublesome to read the resistance. PT100 usually has two-wire, three-wire and four-wire measurement methods, each with its own advantages and disadvantages. The more wires, the more complex the measurement circuit and the higher the cost, but the corresponding accuracy is better. There are usually several test schemes, using a dedicated IC for reading, or a constant current source, or an op amp to build. Dedicated ICs are naturally expensive, so this article uses an op amp to build and collect PT100 resistance values. The following figure is a partial picture of the PT100 scale:

2. Overview of the principle of three-wire PT100
The figure above is a three-wire PT100 preamplifier circuit. The PT100 sensor leads to three wires of exactly the same material, wire diameter and length, and the connection method is shown in the figure. A 2V voltage is applied to the bridge circuit composed of R14, R20, R15, Z1, PT100 and its wire resistance. Z1, Z2, Z3, D11, D12, D83 and each capacitor play a filtering and protection role in the circuit. They can be ignored during static analysis. Z1, Z2, Z3 can be regarded as short circuit, and D11, D12, D83 and each capacitor can be regarded as open circuit. From the resistor voltage divider, V3=2*R20/(R14+20)=200/1100=2/11 ……一个. From the virtual short, the voltage of pins 6 和 7 of U8B is equal to the voltage of pin 5 V4=V3 ……乙. From the virtual short circuit, we know that no current flows through the second pin of U8A, so the current flowing through R18 and R19 is equal. (V2-V4)/R19=(V5-V2)/R18 ……c. From the virtual short circuit, we know that no current flows through the third pin of U8A, V1=V7 ……d. In the bridge circuit, R15 is connected in series with Z1, PT100 and line resistance, and the voltage obtained by connecting PT100 and line resistance in series is added to the third pin of U8A through resistor R17, V7=2*(Rx+2R0)/(R15+Rx+2R0) ……e. From the virtual short circuit, we know that the voltage of the third pin and the second pin of U8A are equal, V1=V2 ……f. From abcdef, we get (V5-V7)/100=(V7-V3)/2.2. Simplified, we get V5=(102.2*V7-100V3)/2.2, 那是, V5=(204.4(Rx+2R0)/(1000+Rx+2R0) – 200/11)/2.2 ……克. The output voltage V5 in the above formula is a function of Rx. Let’s look at the influence of line resistance. Note that there are two V5s in the circuit diagram. In the context, we refer to the one on U8A. There is no relationship between the two. The voltage drop generated on the line resistance at the bottom of PT100 passes through the middle line resistance, Z2, and R22, and is added to the 10th pin of U8C. From the virtual disconnection, we know that V5=V8=V9=2*R0/(R15+Rx+2R0) ……一个. (V6-V10)/R25=V10/R26……乙. From the imaginary short circuit, we know that V10=V5……c. From the formula abc, we get V6=(102.2/2.2)V5=204.4R0/[2.2(1000+Rx+2R0)]……小时. From the equation group composed of formula gh, we know that if the values ​​of V5 and V6 are measured, Rx and R0 can be calculated. Knowing Rx, we can know the temperature by looking up the PT100 scale. 所以, we get two formulas, namely V6=204.4R0/[2.2(1000+Rx+2R0)] and V5=(204.4(Rx+2R0)/(1000+Rx+2R0) – 200/11)/2.2. V5 and V6 are the voltages we want to collect, which are known conditions. To get the final formula, we have to solve these two formulas. By the way, Z1, Z2 and Z3 are three three-terminal filter through-hole capacitors. The actual objects are shown in the figure below, with plug-in and surface mount versions.