PID function block for process control in PLC Programming - Mini Project: Build & Apply

PID Function Block for Process Control

📖 Scenario: You are working in a factory where machines need to keep their temperature steady. To do this, you use a PID controller. A PID controller helps machines adjust their heating or cooling to keep the temperature just right.In this project, you will create a simple PID function block in a PLC program. This block will take the current temperature and the desired temperature, then calculate the control output to keep the machine stable.

🎯 Goal: Build a PID function block that calculates the control output based on the error between the desired and current temperature. You will set up the input variables, configure the PID parameters, write the calculation logic, and finally output the control signal.

📋 What You'll Learn

Create input variables for current temperature and setpoint

Add PID parameters: proportional gain (Kp), integral gain (Ki), and derivative gain (Kd)

Implement the PID calculation logic inside the function block

Output the control signal as the result of the PID calculation

💡 Why This Matters

🌍 Real World

PID controllers are widely used in factories to keep machines running smoothly by automatically adjusting controls like temperature, speed, or pressure.

💼 Career

Understanding PID function blocks is essential for automation engineers and technicians working with PLCs to maintain and optimize industrial processes.

1

DATA SETUP: Define input variables

Create two variables called current_temp and setpoint with initial values 20.0 and 25.0 respectively. These represent the current temperature and the desired temperature.

PLC Programming

# Define current temperature and setpoint variables
# Your code here

Hint

Use VAR block to declare variables with type REAL and assign the given values.

2

CONFIGURATION: Add PID parameters

Add three variables called Kp, Ki, and Kd with values 2.0, 0.5, and 1.0 respectively. These are the PID gains.

PLC Programming

VAR
    current_temp : REAL := 20.0;
    setpoint : REAL := 25.0;
    
    // Add PID gain variables Kp, Ki, and Kd here
    # Your code here
END_VAR

Hint

Declare the PID gain variables inside the VAR block with the given values.

3

CORE LOGIC: Implement PID calculation

Write the PID calculation logic using variables error, integral, and derivative. Calculate error as setpoint - current_temp. Update integral by adding error. Calculate derivative as the difference between error and previous_error. Then calculate output as Kp * error + Ki * integral + Kd * derivative. Store error in previous_error for next cycle.

PLC Programming

error := setpoint - current_temp;
integral := integral + error;
derivative := error - previous_error;
output := Kp * error + Ki * integral + Kd * derivative;
previous_error := error;
# Your code here

Hint

Follow the PID formula step by step using the variables given.

4

OUTPUT: Display the control output

Write a statement to output the value of output. This shows the control signal calculated by the PID.

PLC Programming

// Output the control signal
// Your code here

Hint

Return or output the variable output to see the control value.

Practice

1. What is the main purpose of a PID function block in process control?

easy

A. To keep a process variable close to a desired target value

B. To store data for later analysis

C. To display process values on a screen

D. To manually adjust valves and switches

Solution

1. Step 1: Understand the role of PID blocks

   PID blocks automatically adjust outputs to keep a process variable near a setpoint.

2. Step 2: Compare options to this role

   Only To keep a process variable close to a desired target value describes this automatic control function; others describe unrelated tasks.

3. Final Answer:

   To keep a process variable close to a desired target value -> Option A

4. Quick Check:

   PID controls process variable = keep near target [OK]

Hint: PID blocks control variables automatically, not manually [OK]

Common Mistakes:

• Thinking PID stores data instead of controlling
• Confusing PID with display or manual control
• Assuming PID only measures values

2. Which of the following is the correct way to declare a PID function block instance named pid1 in a PLC program?

easy

A. pid1 : PID;

B. pid1 = PID();

C. PID pid1;

D. pid1 -> PID();

Solution

1. Step 1: Recall PLC function block declaration syntax

   In PLC languages like Structured Text, function blocks are declared with name : Type;.

2. Step 2: Match options to this syntax

   pid1 : PID; matches the correct syntax; others use incorrect assignment or C-like syntax.

3. Final Answer:

   pid1 : PID; -> Option A

4. Quick Check:

   Function block declaration uses colon and semicolon [OK]

Hint: Use colon and semicolon to declare function blocks [OK]

Common Mistakes:

• Using equal sign instead of colon
• Using C or Java style declarations
• Using arrow operator incorrectly

3. Given the following PLC code snippet using a PID block:

pid1.Kp := 2.0;
pid1.Ki := 0.5;
pid1.Kd := 0.1;
pid1.SetPoint := 100;
pid1.ProcessVariable := 90;
pid1();
Output := pid1.ControlOutput;

What is the expected behavior of Output after this code runs once?

medium

A. Output will be negative to decrease the process variable

B. Output will be zero because PID needs multiple cycles

C. Output will be a positive value to increase the process variable

D. Output will equal the SetPoint value

Solution

1. Step 1: Analyze PID inputs and error

   The error is SetPoint (100) minus ProcessVariable (90) = 10, positive error.

2. Step 2: Understand PID output direction

   Positive error means output should increase to raise the process variable, so output is positive.

3. Final Answer:

   Output will be a positive value to increase the process variable -> Option C

4. Quick Check:

   Positive error -> positive output to correct [OK]

Hint: Positive error means output pushes variable up [OK]

Common Mistakes:

• Assuming output is zero on first run
• Confusing output direction with error sign
• Thinking output equals setpoint directly

4. Identify the error in this PID block usage snippet:

pid1.Kp := 1.5;
pid1.Ki := 0.3;
pid1.Kd := 0.05;
pid1.SetPoint := 50;
pid1.ProcessVariable := 55;
Output := pid1.ControlOutput;
pid1();

medium

A. Assigning ProcessVariable before SetPoint

B. Setting Kp, Ki, Kd values too low

C. Missing declaration of pid1

D. Calling the PID block function pid1() after reading output

Solution

1. Step 1: Check order of function call and output read

   The PID block must be called (executed) before reading ControlOutput to update output.

2. Step 2: Identify the mistake in code order

   Here, Output is assigned before pid1() is called, so output is stale or invalid.

3. Final Answer:

   Calling the PID block function pid1() after reading output -> Option D

4. Quick Check:

   Call PID block before reading output [OK]

Hint: Always call PID block before reading output [OK]

Common Mistakes:

• Reading output before calling PID block
• Confusing parameter order
• Ignoring function block declaration

5. You want to tune a PID block to control temperature in a tank. The temperature rises too slowly and overshoots the target. Which tuning change should you make to reduce overshoot and improve response?

hard

A. Decrease Kp and increase Ki

B. Decrease Kp and increase Kd

C. Increase Kp and decrease Kd

D. Increase Ki and Kd equally

Solution

1. Step 1: Understand overshoot and slow rise causes

   High Kp causes overshoot; low Kd means less damping; slow rise means Kp might be too low.

2. Step 2: Choose tuning to reduce overshoot and improve speed

   Decreasing Kp reduces overshoot; increasing Kd adds damping to reduce oscillations and overshoot.

3. Final Answer:

   Decrease Kp and increase Kd -> Option B

4. Quick Check:

   Lower Kp + higher Kd = less overshoot, better control [OK]

Hint: Lower Kp and raise Kd to reduce overshoot [OK]

Common Mistakes:

• Increasing Kp causes more overshoot
• Ignoring derivative gain effect
• Changing Ki without addressing overshoot