The solution to the problem of Siemens 200PID constant temperature control

1. Pay attention to distinguish whether the input terminal is connected to a voltage signal or a current signal; the output terminal is a current signal or

Is the voltage signal. The wiring method under different signals on the analog module.

2. Understand the relevant information of signal input components:

If you use a temperature transmitter, you must understand the measurement range of the temperature transmitter, such as 0~100℃; the output current range is 4~20mA; what is the graduation number, such as PT100; wiring diagrams, etc. Relevant input components; the wiring method of the output components on the analog module. Others, such as the accuracy required by the project, etc.

3. Confirm the PID setting value (VD204):

Assume that we will control the temperature at 23.5℃; taking unipolarity as an example, we should first determine whether the input signal is a 0-10V voltage signal or a 4-20mA current signal? , This is very important in PID setting.

If the voltage input signal of 0~10V corresponds to 0~32000, and the temperature range is 0~100℃, the setting value can be directly calculated: VD204=23.5/(100-0)=0.235;

If the current is 4-20mA, the corresponding value should be 6400-32000, and the temperature range should be 0-100°C, and the set value should be 0.388.

Reason: 0~32000 corresponds to 0~20mA in the analog module; 6400~32000 corresponds to 4~20mA corresponds to 0~100℃; this must be related to the calculation, 23.5℃ current calculation method:

(20-4): (100-0)=(X-4): 23.5;

Solve the equation: X=7.76(mA).

Setting value: VD204=7.76/20=0.388.

4. Confirmation of PID output value (VD208):

Take unipolarity as an example. Should the output signal be 0~10V voltage signal or 4~20mA current signal corresponding to 0~32000?

If the output signal AQW0 corresponds to a voltage signal, such as 0~10V, then

AQW0=(The real number VD208*32000 is converted into an integer);

If the output signal AQW0 corresponds to a current signal, such as 4-20 mA, then

AQW0=(The real number VD208*32000 is converted into an integer +6400).

5. Regarding the actual PID constant temperature control:

Through the computer experiment, it can be known that the PID constant temperature control is adjusted around the set value. If the set temperature is 23.5°C; when the temperature is lower than the set value, the heating steam control valve is always fully open; when the temperature reaches 23.5°C, the heating steam control valve starts to close gradually, and in the closing process , The temperature may still rise gradually, the greater the temperature deviation, the faster the closing speed; until all closed; when the temperature is lower than the set value again, the heating steam control valve will gradually open, the opening speed depends on the temperature deviation The larger the deviation, the faster the opening speed; until the temperature reaches the set value again. If the temperature does not reach the set value for a long time, the regulating function will open the regulating valve. This is the PID constant temperature control situation that I have observed. Therefore, we can carry out the necessary programming according to the actual control situation, and effectively use the PID control period when the set value is lower than the set value; cut off the PID control of the part higher than the set value, after the temperature is higher than the set value, you can produce according to It is required to simply close the heating valve partially or completely. In case the temperature rises too high. To obtain a superior temperature control effect.

The solution to the problem of Siemens 200PID constant temperature control

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