Why synthesis rules matter in Verilog

Introduction

Synthesis rules help turn your Verilog code into real hardware correctly. They make sure your design works as expected when built on a chip.

When writing Verilog code that will be turned into hardware circuits.

When you want your design to be efficient and use less chip area.

When you need your hardware to run at the right speed without errors.

When debugging why your hardware behaves differently than your code.

When learning how to write hardware-friendly Verilog code.

Syntax

Verilog

// Synthesis rules are guidelines, not code syntax.
// Example: Avoid using delays or unsupported constructs in Verilog.
// Use always blocks with proper sensitivity lists for combinational logic.

always @(*) begin
  // combinational logic here
end

Synthesis rules are not written as code but are guidelines to write synthesizable Verilog.

Following these rules ensures your code can be converted into hardware by synthesis tools.

Examples

This is a simple combinational logic block that synthesis tools understand well.

Verilog

always @(*) begin
  out = a & b;
end

This is a clocked process for flip-flops, which synthesis tools map to registers.

Verilog

always @(posedge clk) begin
  q <= d;
end

Delays (#5) are ignored by synthesis and only work in simulation, so avoid them.

Verilog

// Bad example: Using delays which synthesis tools ignore
always @(posedge clk) begin
  q <= #5 d;
end

Sample Program

This module shows a simple AND gate written in a way synthesis tools can convert to hardware.

Verilog

module simple_and(
  input wire a,
  input wire b,
  output reg out
);

  // Combinational logic following synthesis rules
  always @(*) begin
    out = a & b;
  end

endmodule

Important Notes

Always write Verilog code that synthesis tools can understand to avoid unexpected hardware.

Simulation-only features like delays or initial blocks do not translate to hardware.

Use proper sensitivity lists and avoid latches by assigning all outputs in combinational blocks.

Summary

Synthesis rules guide writing Verilog that becomes real hardware correctly.

Following these rules avoids bugs and inefficient hardware.

Good synthesis code uses proper always blocks and avoids simulation-only constructs.

Practice

1. Why is it important to follow synthesis rules when writing Verilog code?

easy

A. Because synthesis rules ensure the code can be converted into real hardware

B. Because synthesis rules make the code run faster in simulation

C. Because synthesis rules help the code look cleaner

D. Because synthesis rules allow the code to run on any software

Solution

1. Step 1: Understand synthesis purpose

   Synthesis converts Verilog code into hardware circuits.

2. Step 2: Connect rules to hardware creation

   Following synthesis rules ensures the code can be correctly transformed into hardware.

3. Final Answer:

   Because synthesis rules ensure the code can be converted into real hardware -> Option A

4. Quick Check:

   Synthesis rules = hardware conversion [OK]

Hint: Synthesis rules = code to hardware conversion [OK]

Common Mistakes:

• Confusing synthesis rules with simulation speed
• Thinking synthesis rules only affect code style
• Believing synthesis rules affect software compatibility

2. Which of the following Verilog code snippets follows synthesis rules for combinational logic?

easy

A. always @(posedge clk) begin out = in1 & in2; end

B. always @(negedge clk) begin out <= in1 | in2; end

C. initial begin out = 1'b0; end

D. always @(*) begin out = in1 & in2; end

Solution

1. Step 1: Identify combinational logic sensitivity list

   Combinational logic uses always @(*) to react to input changes.

2. Step 2: Check each option

   always @(*) begin out = in1 & in2; end uses always @(*) with blocking assignment, correct for combinational logic.

3. Final Answer:

   always @(*) begin out = in1 & in2; end -> Option D

4. Quick Check:

   Combinational logic uses always @(*) [OK]

Hint: Combinational logic uses always @(*) blocks [OK]

Common Mistakes:

• Using clock edges for combinational logic
• Using non-blocking assignments in combinational always blocks
• Using initial blocks for synthesis

3. What will be the output of this Verilog code after synthesis if clk is a clock and reset is active high?

module test(input clk, reset, input d, output reg q);
  always @(posedge clk or posedge reset) begin
    if (reset) q <= 0;
    else q <= d;
  end
endmodule

medium

A. q follows d asynchronously ignoring reset

B. q resets to 0 when reset is high, else q follows d on clock edge

C. q never resets and holds previous value

D. q resets to 1 when reset is high

Solution

1. Step 1: Analyze sensitivity list

   The always block triggers on rising edge of clk or reset.

2. Step 2: Understand reset behavior

   If reset is high, q is set to 0 immediately (asynchronous reset). Otherwise, q follows d on clock edge.

3. Final Answer:

   q resets to 0 when reset is high, else q follows d on clock edge -> Option B

4. Quick Check:

   Asynchronous reset sets q=0, else q=d [OK]

Hint: Check sensitivity list for async reset behavior [OK]

Common Mistakes:

• Assuming reset is synchronous
• Ignoring reset in output behavior
• Confusing reset value as 1 instead of 0

4. Identify the synthesis error in this Verilog code:

always @(posedge clk) begin
  if (enable)
    count = count + 1;
end

medium

A. Using blocking assignment inside sequential always block

B. Missing sensitivity to reset signal

C. Using non-blocking assignment instead of blocking

D. No error, code is synthesizable

Solution

1. Step 1: Check assignment type in sequential block

   Sequential blocks triggered by clock edges should use non-blocking assignments (<=).

2. Step 2: Identify error in code

   The code uses blocking assignment (=) inside a clocked always block, which can cause incorrect synthesis.

3. Final Answer:

   Using blocking assignment inside sequential always block -> Option A

4. Quick Check:

   Sequential always blocks need non-blocking assignments [OK]

Hint: Use <= in clocked always blocks, not = [OK]

Common Mistakes:

• Confusing blocking and non-blocking assignments
• Ignoring reset signals in counters
• Assuming blocking assignments are always okay

5. You want to design a 4-bit counter that resets asynchronously and counts up on each clock edge only when enable is high. Which of the following code snippets follows synthesis rules correctly?

hard

A. always @(posedge clk or posedge reset) begin
if (reset) count = 0;
else if (enable) count <= count + 1;
end

B. always @(posedge clk) begin
if (reset) count = 0;
else if (enable) count = count + 1;
end

C. always @(posedge clk or posedge reset) begin
if (reset) count <= 0;
else if (enable) count <= count + 1;
end

D. always @(posedge clk) begin
if (reset) count <= 0;
else if (enable) count <= count + 1;
end

Solution

1. Step 1: Check asynchronous reset sensitivity

   Asynchronous reset requires reset in sensitivity list (posedge reset).

2. Step 2: Verify assignment types and conditions

   always @(posedge clk or posedge reset) begin if (reset) count <= 0; else if (enable) count <= count + 1; end uses non-blocking assignments (<=) and includes reset in sensitivity list, correctly implementing async reset and enable control.

3. Final Answer:

   always @(posedge clk or posedge reset) begin if (reset) count <= 0; else if (enable) count <= count + 1; end -> Option C

4. Quick Check:

   Async reset in sensitivity + non-blocking assignments [OK]

Hint: Async reset needs posedge reset in sensitivity list [OK]

Common Mistakes:

• Using blocking assignments in sequential logic
• Omitting reset from sensitivity list for async reset
• Mixing blocking and non-blocking assignments