Trace width for current capacity in PCB Design - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the formula to calculate the trace width based on current capacity.

PCB Design

trace_width = current [1] k

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

A*

B+

C-

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Using addition or division instead of multiplication.

The trace width is proportional to the current multiplied by a constant factor k.

2fill in blank

medium

Complete the DAX measure to calculate total current capacity from a table named 'Traces'.

PCB Design

TotalCurrent = SUM(Traces[[1]])

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

AResistance

BWidth

CCurrent

DLength

Attempts:

3 left

💡 Hint

Common Mistakes

Summing width or resistance instead of current.

The measure sums the 'Current' column to get total current capacity.

3fill in blank

hard

Fix the error in this SQL query to select trace widths greater than 10 mils.

PCB Design

SELECT * FROM Traces WHERE Width [1] 10

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

A>

B<

C=

D<=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '=' or '<' which do not filter correctly.

The query filters traces with width greater than 10 mils using the '>' operator.

4fill in blank

hard

Fill both blanks to create a dictionary comprehension that maps trace names to widths for traces with current {{BLANK_1}} 5 and width {{BLANK_2}} 10.

PCB Design

{trace['name']: trace['width'] for trace in traces if trace['current'] [1] 5 and trace['width'] [2] 10}

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

A>

B<

C>=

D<=

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong comparison operators that reverse the filter logic.

The comprehension filters traces with current greater than 5 and width less than 10.

5fill in blank

hard

Fill all three blanks to create a DAX measure that calculates average trace width for traces with current {{BLANK_1}} 3, width {{BLANK_2}} 20, and length {{BLANK_3}} 50.

PCB Design

AvgWidth = AVERAGEX(FILTER(Traces, Traces[Current] [1] 3 && Traces[Width] [2] 20 && Traces[Length] [3] 50), Traces[Width])

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

A>

B<

C<=

D>=

Attempts:

3 left

💡 Hint

Common Mistakes

Mixing up comparison operators or using wrong logical conditions.

The measure filters traces with current greater or equal to 3, width less than 20, and length greater than 50, then averages their widths.

Practice

1. What is the main reason to choose the correct trace width in PCB design?

easy

A. To increase the signal speed on the trace

B. To make the PCB look visually appealing

C. To reduce the cost of copper used in the PCB

D. To ensure the trace can safely carry the required current without overheating

Solution

1. Step 1: Understand trace width purpose

   Trace width determines how much current a PCB trace can carry safely without overheating.

2. Step 2: Identify the main safety concern

   Overheating can damage the PCB and cause failure, so correct width prevents this.

3. Final Answer:

   To ensure the trace can safely carry the required current without overheating -> Option D

4. Quick Check:

   Trace width controls current safely [OK]

Hint: Trace width prevents overheating by carrying current safely [OK]

Common Mistakes:

• Thinking trace width is for aesthetics
• Assuming trace width reduces cost only
• Believing trace width affects signal speed directly

2. Which of the following is the correct way to express trace width in a PCB design specification?

easy

A. TraceWidth = 10mm + Current * 0.5

B. TraceWidth = 10 mils

C. traceWidth = '10mil'

D. Trace_Width = 10mil

Solution

1. Step 1: Recognize standard units for trace width

   Trace width is commonly specified in mils (thousandths of an inch) or mm, with a space before the unit.

2. Step 2: Identify correct syntax for specification

   TraceWidth = 10 mils uses a numeric value followed by a space and unit 'mils', which is clear and standard.

3. Final Answer:

   TraceWidth = 10 mils -> Option B

4. Quick Check:

   Standard unit with space = correct [OK]

Hint: Use numeric value plus unit with space (e.g., 10 mils) [OK]

Common Mistakes:

• Using quotes around units incorrectly
• Combining units without space
• Using non-standard unit names

3. Given a PCB trace carrying 3A current with a temperature rise limit of 10°C, which trace width is closest to the recommended value using IPC-2152 guidelines?

Options are based on typical IPC-2152 charts for external layers.

medium

A. 30 mils

B. 15 mils

C. 50 mils

D. 70 mils

Solution

1. Step 1: Refer to IPC-2152 chart for 3A current and 10°C rise

   According to IPC-2152, a 3A current with 10°C rise typically requires about 25-35 mils trace width on external layers.

2. Step 2: Match closest option

   Among given options, 30 mils fits best within the recommended range.

3. Final Answer:

   30 mils -> Option A

4. Quick Check:

   3A current needs ~30 mils width [OK]

Hint: Use IPC-2152 chart ranges for current and temp rise [OK]

Common Mistakes:

• Choosing too narrow trace causing overheating
• Picking overly wide trace wasting PCB space
• Ignoring temperature rise in calculation

4. A designer used the formula TraceWidth = Current * 10 to calculate trace width in mils. The trace carries 2A current. The calculated width is 20 mils. However, the trace overheats during testing. What is the likely error?

medium

A. The formula does not consider temperature rise and copper thickness

B. The current value should be squared in the formula

C. Trace width should be in mm, not mils

D. The formula should multiply current by 5, not 10

Solution

1. Step 1: Analyze the formula used

   The formula is a simple linear multiplication ignoring key factors like temperature rise and copper thickness.

2. Step 2: Understand why overheating occurs

   Without considering temperature rise and copper thickness, the trace width may be underestimated, causing overheating.

3. Final Answer:

   The formula does not consider temperature rise and copper thickness -> Option A

4. Quick Check:

   Missing temp rise and copper thickness causes errors [OK]

Hint: Always include temperature rise and copper thickness in calculations [OK]

Common Mistakes:

• Using oversimplified formulas
• Ignoring units consistency
• Assuming linear scaling is enough

5. You need to design a PCB trace for a 5A current on an internal layer with a maximum temperature rise of 20°C. Using IPC-2152 guidelines, which approach is best to determine the correct trace width?

hard

A. Use the external layer chart and double the width value

B. Use a fixed 50 mil width for all internal traces carrying above 4A

C. Use the internal layer chart and adjust for copper thickness and temperature rise

D. Calculate width by dividing current by temperature rise

Solution

1. Step 1: Identify correct IPC-2152 chart for internal layers

   Internal layers have different heat dissipation, so use the internal layer chart for accurate width.

2. Step 2: Adjust for copper thickness and temperature rise

   Consider copper thickness and max temperature rise to find precise trace width for 5A current.

3. Final Answer:

   Use the internal layer chart and adjust for copper thickness and temperature rise -> Option C

4. Quick Check:

   Internal layer needs specific chart and adjustments [OK]

Hint: Use internal layer IPC-2152 chart with copper and temp adjustments [OK]

Common Mistakes:

• Using external layer data for internal traces
• Applying fixed widths without calculation
• Ignoring copper thickness effect