Resource sharing and area optimization in Verilog - Step-by-Step Execution

Concept Flow - Resource sharing and area optimization

Start Design

↓

Identify Similar Operations

↓

Check Timing Constraints

↓

Can Share?

No→Use Separate Resources

Yes↓

Implement Shared Resource

↓

Optimize Area

↓

Verify Functionality & Timing

↓

End Design

This flow shows how to find similar operations in a design, decide if they can share hardware resources to save area, implement sharing, and verify the design.

Execution Sample

Verilog

module shared_adder(
  input wire clk,
  input wire [3:0] a, b, c, d,
  output reg [5:0] sum1, sum2
);
  reg [4:0] temp;
  always @(posedge clk) begin
    temp <= a + b;
    sum1 <= temp + c;
    sum2 <= temp + d;
  end
endmodule

This Verilog module shares one adder resource to compute sums in two steps, reducing hardware area.

Execution Table

Step	clk edge	temp (a+b)	sum1 (temp+c)	sum2 (temp+d)	Action
1	rising	a+b	old temp + c	old temp + d	Compute temp = a+b; sum1 and sum2 hold old values
2	rising	new a+b	temp + c	old temp + d	Update sum1 with temp+c; sum2 holds old value
3	rising	new a+b	temp + c	temp + d	Update sum2 with temp+d
4	rising	new a+b	temp + c	temp + d	Stable outputs; next cycle begins
Exit	-	-	-	-	Continuous operation on clock edges

💡 Design runs continuously on clock edges; resource sharing reduces area by reusing 'temp' adder.

Variable Tracker

Variable	Start	After clk 1	After clk 2	After clk 3	Final
temp	undefined	a+b	new a+b	new a+b	new a+b
sum1	undefined	old temp + c	temp + c	temp + c	temp + c
sum2	undefined	old temp + d	old temp + d	temp + d	temp + d

Key Moments - 3 Insights

Why does 'sum1' update before 'sum2' even though they use the same 'temp'?

Because 'sum1' uses the updated 'temp' value immediately after it's computed, while 'sum2' uses 'temp' in the next clock cycle, as shown in steps 2 and 3 of the execution_table.

How does sharing the 'temp' adder save hardware area?

Instead of having two separate adders for 'a+b' and 'temp+c'/'temp+d', the design reuses one adder in sequence, reducing the total hardware needed, as seen in the code and variable_tracker.

What happens if timing constraints are too tight for sharing?

If the clock period is too short, the shared resource cannot complete all operations in time, so separate resources must be used, as indicated in the concept_flow decision step.

Visual Quiz - 3 Questions

Look at the execution_table at step 2, what is the value of sum1?

A temp + c

B old temp + c

C a + b + c

D undefined

💡 Hint

Refer to execution_table row 2 under 'sum1 (temp+c)' column.

At which clock edge does sum2 first get updated with the new temp value?

A clk 2

B clk 3

C clk 1

D Never

💡 Hint

Check execution_table row 3 for sum2 update timing.

If we remove sharing and use two adders, how would the variable 'temp' change?

A It would hold two different values simultaneously

B It would remain the same

C It would be eliminated

D It would update slower

💡 Hint

Look at variable_tracker and consider that 'temp' stores the shared adder result.

Concept Snapshot

Resource sharing in Verilog means using one hardware block for multiple operations to save area.
Identify similar operations, check timing, then share resources if possible.
This reduces chip size but may add timing complexity.
Always verify functionality and timing after optimization.

Full Transcript

This visual execution trace shows how resource sharing and area optimization work in Verilog. The design uses one adder to compute multiple sums in sequence, saving hardware area. The execution table tracks how variables 'temp', 'sum1', and 'sum2' update on clock edges. Key moments clarify why sum1 updates before sum2 and how sharing reduces hardware. The quiz tests understanding of variable values at specific steps and the effect of removing sharing. The concept snapshot summarizes the main idea: share hardware to save area while ensuring timing and correctness.

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