PID function block for process control in PLC Programming - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare a PID function block instance named 'pid1'.

PLC Programming

VAR
  pid1 : [1];
END_VAR

Drag options to blanks, or click blank then click option'

ABOOL

BPID

CINT

DREAL

Attempts:

3 left

💡 Hint

Common Mistakes

Using basic data types like INT or BOOL instead of the PID block type.

The PID function block is declared by using the type 'PID'.

2fill in blank

medium

Complete the code to set the proportional gain (Kp) of the PID block to 2.5.

PLC Programming

pid1.[1] := 2.5;

Drag options to blanks, or click blank then click option'

AKp

BKd

CKi

DSP

Attempts:

3 left

💡 Hint

Common Mistakes

Confusing Kp with Ki or Kd parameters.

Kp is the proportional gain parameter of the PID block.

3fill in blank

hard

Fix the error in the code to assign the setpoint (SP) value 50 to the PID block.

PLC Programming

pid1.SP [1] 50;

Drag options to blanks, or click blank then click option'

ASP

BOP

CPV

D:=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '=' instead of ':=' for assignment.

In PLC programming, ':=' is used for assignment, not '='.

4fill in blank

hard

Fill both blanks to write the process variable (PV) and read the output (OP) of the PID block.

PLC Programming

pid1.[1] := currentPV;
controlSignal := pid1.[2];

Drag options to blanks, or click blank then click option'

APV

BSP

COP

DKp

Attempts:

3 left

💡 Hint

Common Mistakes

Mixing SP (setpoint) with PV or OP.

PV is the process variable input, OP is the output signal of the PID block.

5fill in blank

hard

Fill all three blanks to configure the PID block with proportional gain 3.0, integral gain 0.5, and derivative gain 0.1.

PLC Programming

pid1.[1] := 3.0;
pid1.[2] := 0.5;
pid1.[3] := 0.1;

Drag options to blanks, or click blank then click option'

AKp

BKi

CKd

DSP

Attempts:

3 left

💡 Hint

Common Mistakes

Confusing the order or names of gain parameters.

Kp, Ki, and Kd are the proportional, integral, and derivative gains respectively.

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