Resource sharing and area optimization in Verilog - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare a 4-bit register named 'data_reg'.

Verilog

reg [[1]:0] data_reg;

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

A1

B4

C7

D3

Attempts:

3 left

💡 Hint

Common Mistakes

Using 4 instead of 3 as the upper index.

Confusing the bit width with the index range.

The register is 4 bits wide, so the index goes from 3 down to 0.

2fill in blank

medium

Complete the code to assign the sum of 'a' and 'b' to 'result' using a shared adder.

Verilog

assign result = a [1] b;

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A-

B+

C*

D&

Attempts:

3 left

💡 Hint

Common Mistakes

Using '-' which subtracts instead of adds.

Using '&' which is a bitwise AND, not addition.

The '+' operator adds the two inputs 'a' and 'b'.

3fill in blank

hard

Fix the error in the always block sensitivity list to optimize resource sharing.

Verilog

always @([1]) begin
  sum <= a + b;
end

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Aposedge clk

Ba or b

Cposedge clk or posedge reset

Dnegedge reset

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'a or b' causes combinational logic, not resource sharing.

Including reset unnecessarily in this context.

Using 'posedge clk' ensures the block triggers on clock edge, enabling synchronous resource sharing.

4fill in blank

hard

Fill both blanks to create a shared multiplier module instance with inputs 'x' and 'y'.

Verilog

multiplier mul_inst(.[1](x), .[2](y));

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Aa

Bb

Cinput1

Dinput2

Attempts:

3 left

💡 Hint

Common Mistakes

Using generic names like 'input1' and 'input2' that don't match the module.

Swapping the input names.

The multiplier module expects inputs named 'a' and 'b' to connect to 'x' and 'y'.

5fill in blank

hard

Fill all three blanks to implement a resource sharing logic that selects between addition and subtraction.

Verilog

assign result = sel ? a [1] b : a [2] b;

always @(posedge clk) begin
  stored_result <= result [3] 1;
end

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

A+

B-

C+=

D-=

Attempts:

3 left

💡 Hint

Common Mistakes

Mixing up '+' and '-' in the ternary expression.

Using '-=' instead of '+' for increment.

The code uses '+' when sel is true, '-' when false, and increments stored_result with '+'.

Practice

1. What is the main goal of resource sharing in Verilog designs?

easy

A. To make the code more complex and harder to read

B. To increase the clock speed of the design

C. To add more hardware components for better performance

D. To use the same hardware for multiple tasks to save space

Solution

1. Step 1: Understand resource sharing concept and main benefit

   Resource sharing means using one hardware block for different tasks instead of multiple blocks. This reduces the total hardware area, saving space and cost.

2. Final Answer:

   To use the same hardware for multiple tasks to save space -> Option D

3. Quick Check:

   Resource sharing = save space [OK]

Hint: Resource sharing means reusing hardware to save space [OK]

Common Mistakes:

• Thinking resource sharing always increases speed
• Confusing resource sharing with adding more hardware
• Believing resource sharing makes code more complex

2. Which Verilog construct is commonly used to implement resource sharing for arithmetic operations?

easy

A. A single always block with a shared adder and multiplexers

B. Instantiating multiple modules for each arithmetic unit

C. Using multiple assign statements for each operation

D. Multiple always blocks with separate adders

Solution

1. Step 1: Identify resource sharing method and Verilog construct

   Resource sharing uses one hardware unit shared by multiple operations, controlled by multiplexers. A single always block with a shared adder and multiplexers allows selecting inputs dynamically.

2. Final Answer:

   A single always block with a shared adder and multiplexers -> Option A

3. Quick Check:

   Shared adder + mux = resource sharing [OK]

Hint: Use multiplexers to select inputs for shared hardware [OK]

Common Mistakes:

• Using multiple adders instead of sharing one
• Not using multiplexers to control input selection
• Instantiating separate modules unnecessarily

3. Consider the following Verilog snippet for resource sharing:

reg [3:0] a, b, c, result;
reg sel;
always @(*) begin
  case(sel)
    1'b0: result = a + b;
    1'b1: result = b + c;
  endcase
end

What is the output in result when a=4, b=3, c=5, and sel=1?

medium

A. 8

B. 5

C. 7

D. Syntax error

Solution

1. Step 1: Understand case selection and calculate

   When sel=1, result = b + c. b=3 and c=5, so result = 3 + 5 = 8.

2. Final Answer:

   8 -> Option A

3. Quick Check:

   sel=1 means result = b+c = 8 [OK]

Hint: Check sel value to pick correct operation [OK]

Common Mistakes:

• Adding a + b instead of b + c when sel=1
• Confusing variable values
• Assuming syntax error due to case statement

4. Identify the error in this Verilog code snippet intended for resource sharing:

reg [7:0] x, y, z;
reg sel;
always @(posedge clk) begin
  if(sel == 0)
    z = x + y;
  else
    z = x - y;
end

medium

A. Missing sensitivity list for combinational logic

B. Using blocking assignment (=) inside a clocked always block

C. Incorrect use of if-else statement

D. No error, code is correct

Solution

1. Step 1: Check assignment type and identify error

   Inside a clocked always block, non-blocking assignment (<=) should be used for registers. The code uses blocking assignment (=), which can cause simulation mismatches and timing issues.

2. Final Answer:

   Using blocking assignment (=) inside a clocked always block -> Option B

3. Quick Check:

   Clocked block needs <=, not = [OK]

Hint: Use <= in clocked always blocks for registers [OK]

Common Mistakes:

• Using = instead of <= in sequential logic
• Confusing sensitivity list for clocked blocks
• Thinking if-else is incorrect syntax

5. You want to optimize area by sharing a multiplier between two operations in a Verilog design. Which approach best balances area saving and performance?

hard

A. Use combinational logic to duplicate multiplier outputs without sharing hardware

B. Instantiate two separate multipliers to run operations in parallel

C. Use one multiplier with a multiplexer to select inputs and schedule operations sequentially

D. Replace multiplier with addition to save area

Solution

1. Step 1: Understand trade-off and identify best approach

   Sharing one multiplier saves area but may reduce speed due to sequential operation. Using one multiplier with multiplexers and scheduling operations sequentially balances area saving and acceptable performance.

2. Final Answer:

   Use one multiplier with a multiplexer to select inputs and schedule operations sequentially -> Option C

3. Quick Check:

   Shared multiplier + mux + scheduling = area optimization [OK]

Hint: Share hardware with mux and schedule operations sequentially [OK]

Common Mistakes:

• Using multiple multipliers wastes area
• Duplicating outputs without sharing hardware
• Replacing multiplier with addition changes functionality