Resource sharing and area optimization in Verilog

Introduction

Resource sharing helps use fewer parts in a design. Area optimization makes the design smaller and simpler.

When you want to save chip space by using one part for many tasks.

When your design has similar operations that can share hardware.

When you want to reduce power by using fewer components.

When you need to lower the cost of your hardware.

When you want to make your design easier to fit on a small chip.

Syntax

Verilog

module example(input clk, input [3:0] a, b, output reg [4:0] result);
  reg [4:0] temp;
  always @(posedge clk) begin
    temp <= a + b; // shared adder used here
    result <= temp + 1;
  end
endmodule

Resource sharing means using the same hardware block for multiple operations at different times.

Area optimization often involves reusing logic inside always blocks or combining operations.

Examples

This example shows one adder used for two operations controlled by sel.

Verilog

module shared_adder(input clk, input [3:0] x, y, input sel, output reg [4:0] out);
  reg [4:0] sum;
  always @(posedge clk) begin
    if (sel)
      sum <= x + y;  // first use of adder
    else
      sum <= y + x;  // second use of same adder
    out <= sum;
  end
endmodule

Here, the adder is used first for a + b and then the result is added to c to save area.

Verilog

module area_opt(input clk, input [3:0] a, b, c, output reg [4:0] result);
  reg [4:0] temp;
  always @(posedge clk) begin
    temp <= a + b;  // reuse adder
    result <= temp + c;
  end
endmodule

Sample Program

This program shows a simple design where one adder is shared to add a + b and then add c. The testbench runs the design and prints the result.

Verilog

module resource_sharing_example(input clk, input [3:0] a, b, c, output reg [5:0] result);
  reg [5:0] temp;
  always @(posedge clk) begin
    temp <= a + b;  // use adder once
    result <= temp + c;  // reuse adder result
  end
endmodule

// Testbench
module test;
  reg clk = 0;
  reg [3:0] a = 3, b = 4, c = 5;
  wire [5:0] result;
  resource_sharing_example uut(clk, a, b, c, result);
  always #5 clk = ~clk;
  initial begin
    #10;
    $display("Result: %d", result);
    #10 $finish;
  end
endmodule

Important Notes

Resource sharing can slow down your design if you reuse hardware too much because operations happen one after another.

Area optimization is a trade-off between size and speed.

Always check timing to make sure your design still works well after sharing resources.

Summary

Resource sharing uses the same hardware for multiple tasks to save space.

Area optimization makes your design smaller and cheaper.

Sharing resources can affect speed, so balance is important.

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