Analog input configuration (4-20mA, 0-10V) in PLC Programming - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to read the analog input value from channel 1.

PLC Programming

AnalogValue := AI_Channel[1];

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

A0

B2

C1

D3

Attempts:

3 left

💡 Hint

Common Mistakes

Using 0 as the channel number which may not exist.

Confusing channel numbers with sensor numbers.

Channel numbering usually starts at 1 for analog inputs, so AI_Channel1 reads the first channel.

2fill in blank

medium

Complete the code to scale a 4-20mA input to a 0-100% value.

PLC Programming

ScaledValue := (RawInput - [1]) * 100 / ([2] - [1]);

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

A10

B0

C20

D4

Attempts:

3 left

💡 Hint

Common Mistakes

Using 0 instead of 4 as the lower limit.

Not subtracting the lower limit before scaling.

BLANK_1 is 4mA (lower limit) and BLANK_2 is 20mA (upper limit) for the 4-20mA signal. Subtracting 4 normalizes the input before scaling by the span (16mA) to 0-100%.

3fill in blank

hard

Fix the error in the voltage scaling formula for 0-10V input.

PLC Programming

ScaledVoltage := RawVoltage [1] 10;

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

A/

B*

C+

D-

Attempts:

3 left

💡 Hint

Common Mistakes

Multiplying instead of dividing.

Adding or subtracting instead of dividing.

To scale a 0-10V input to a 0-1 range, divide the raw voltage by 10.

4fill in blank

hard

Fill both blanks to convert a 4-20mA input to a percentage.

PLC Programming

Percentage := (RawCurrent [1] [2]) * 100 / 16;

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

A-

B4

C+

D20

Attempts:

3 left

💡 Hint

Common Mistakes

Adding instead of subtracting the lower limit.

Using the wrong range value.

Subtract 4mA from the raw current to normalize, then divide by 16mA (20-4) to get the fraction.

5fill in blank

hard

Fill all three blanks to create a dictionary mapping channel names to scaled values for 0-10V inputs.

PLC Programming

ScaledInputs := { [1]: (RawInputs[[2]] [3] 10) for [1] in Channels };

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

Ach

B'ch1'

C/

D*

Attempts:

3 left

💡 Hint

Common Mistakes

Using multiplication instead of division for scaling.

Using the loop variable incorrectly as a string.

Use 'ch' as the loop variable (BLANK_1 and BLANK_2 for key and access), and '/' (BLANK_3) to divide RawInputs[ch] by 10 to scale the voltage.

Practice

1. What is the typical current range used for analog input signals in industrial PLCs?

easy

A. 0-5V

B. 10-30mA

C. 4-20mA

D. 0-24V

Solution

1. Step 1: Understand common analog signal standards

   Industrial sensors commonly use 4-20mA current loops for analog signals because they are less affected by electrical noise.

2. Step 2: Compare options with standard ranges

   Among the options, only 4-20mA is the standard current range for analog inputs in PLCs.

3. Final Answer:

   4-20mA -> Option C

4. Quick Check:

   Standard analog current signal = 4-20mA [OK]

Hint: Remember: 4-20mA is the classic industrial current signal [OK]

Common Mistakes:

• Confusing voltage and current ranges
• Choosing 0-5V as standard current signal
• Thinking 10-30mA is common

2. Which of the following is the correct way to configure a 0-10V analog input in a PLC program?

easy

A. Set input range to 0-24V and scale raw value accordingly

B. Set input range to 4-20mA and scale raw value accordingly

C. Set input range to 0-5V and scale raw value accordingly

D. Set input range to 0-10V and scale raw value accordingly

Solution

1. Step 1: Identify the signal type

   For a 0-10V sensor, the PLC input must be configured to accept voltage signals from 0 to 10 volts.

2. Step 2: Apply correct scaling

   The raw input value must be scaled based on the 0-10V range to convert it into meaningful units.

3. Final Answer:

   Set input range to 0-10V and scale raw value accordingly -> Option D

4. Quick Check:

   0-10V input needs 0-10V range setting [OK]

Hint: Match sensor type with input range setting [OK]

Common Mistakes:

• Using 4-20mA range for voltage input
• Choosing 0-5V range for 0-10V sensor
• Ignoring scaling after setting range

3. Given a 4-20mA input scaled to 0-100 units, what is the scaled value if the raw input is 12mA?

medium

A. 50

B. 60

C. 40

D. 70

Solution

1. Step 1: Understand scaling formula for 4-20mA

   The formula to scale current (I) to units is: scaled = ((I - 4) / (20 - 4)) * 100

2. Step 2: Calculate scaled value for 12mA

   scaled = ((12 - 4) / 16) * 100 = (8 / 16) * 100 = 0.5 * 100 = 50

3. Step 3: Re-check calculation

   Wait, 12mA is halfway between 4 and 20, so scaled value should be 50.

4. Final Answer:

   50 -> Option A

5. Quick Check:

   Halfway current = 50 units [OK]

Hint: Use linear scaling: subtract 4mA, divide by 16mA range [OK]

Common Mistakes:

• Using 20mA as denominator instead of (20-4)
• Not subtracting 4mA offset
• Mixing up units and raw values

4. A PLC analog input configured for 0-10V is reading 5V, but the scaled value shows 100 units instead of 50. What is the most likely cause?

medium

A. Input range is incorrectly set to 0-5V

B. Sensor is faulty and outputs wrong voltage

C. Scaling formula uses 4-20mA range instead of 0-10V

D. PLC input channel is damaged

Solution

1. Step 1: Analyze the symptom

   5V input should scale to 50 units if range is 0-10V, but shows 100 units, meaning scaling is doubled.

2. Step 2: Identify configuration error

   If input range is set to 0-5V, 5V is max input, so scaled value is max (100 units), causing the error.

3. Final Answer:

   Input range is incorrectly set to 0-5V -> Option A

4. Quick Check:

   Wrong input range causes wrong scaling [OK]

Hint: Check input range matches sensor output voltage [OK]

Common Mistakes:

• Assuming sensor fault without checking config
• Confusing 4-20mA scaling with voltage scaling
• Ignoring input channel hardware issues

5. You have a sensor outputting 4-20mA representing 0-200°C. How do you configure the PLC to correctly scale the input to temperature?

hard

A. Set input range to 0-10V and scale using formula: Temp = (V / 10) * 200

B. Set input range to 4-20mA and scale using formula: Temp = ((I - 4) / 16) * 200

C. Set input range to 4-20mA and scale using formula: Temp = (I / 20) * 200

D. Set input range to 0-20mA and scale using formula: Temp = (I / 20) * 200

Solution

1. Step 1: Identify sensor output and temperature range

   The sensor outputs 4-20mA representing 0-200°C, so 4mA = 0°C and 20mA = 200°C.

2. Step 2: Apply correct scaling formula

   Subtract 4mA offset, divide by 16mA span, then multiply by 200°C: Temp = ((I - 4) / 16) * 200.

3. Step 3: Verify other choices

   The other choices use wrong input ranges or omit offset, causing incorrect scaling.

4. Final Answer:

   Set input range to 4-20mA and scale using formula: Temp = ((I - 4) / 16) * 200 -> Option B

5. Quick Check:

   Correct scaling subtracts 4mA offset and scales to 200°C [OK]

Hint: Always subtract 4mA offset before scaling [OK]

Common Mistakes:

• Not subtracting 4mA offset in formula
• Using voltage scaling for current input
• Assuming 0-20mA range instead of 4-20mA