Synthesizable vs non-synthesizable constructs in VHDL - When to Use Which

The Big Idea

Discover why your perfect simulation might never become a real chip--and how to fix that!

The Scenario

Imagine you want to design a digital circuit using VHDL. You write code that looks perfect and simulates well on your computer. But when you try to turn that code into a real chip, it fails or behaves unexpectedly.

This happens because some parts of your code cannot be turned into hardware by the synthesis tool.

The Problem

Writing VHDL without knowing which parts can be synthesized is like writing a recipe with ingredients you can't find in the store. You waste time debugging, and your design might never work on real hardware.

Manual trial and error is slow and frustrating because simulation success does not guarantee hardware success.

The Solution

Understanding the difference between synthesizable and non-synthesizable constructs helps you write VHDL code that can be reliably turned into hardware.

This knowledge guides you to use only hardware-friendly code for synthesis, while keeping testbench or simulation-only code separate.

Before vs After

✗ Before

process
begin
  wait for 10 ns;
  signal <= '1';
end process;

✓ After

process(clk)
begin
  if rising_edge(clk) then
    signal <= '1';
  end if;
end process;

What It Enables

It enables you to create real, working digital circuits from your VHDL code without surprises.

Real Life Example

When designing a traffic light controller, using synthesizable constructs ensures the lights change correctly on the actual hardware, not just in simulation.

Key Takeaways

Not all VHDL code can be turned into hardware.

Knowing synthesizable constructs saves time and frustration.

It helps you build real circuits that work as expected.

Practice

1. Which of the following VHDL constructs is synthesizable and can be implemented in hardware?

easy

A. Use of assert statements for debugging

B. A process containing wait for 10 ns; statements

C. File I/O operations like reading from a text file

D. A process with only clocked assignments and no wait statements

Solution

1. Step 1: Identify synthesizable constructs

   Synthesizable code must describe hardware behavior, typically using clocked processes without delays or waits.

2. Step 2: Analyze each option

   A process with only clocked assignments and no wait statements uses clocked assignments without waits, so it is synthesizable. Options A, B, and D use waits, assertions, or file I/O, which are non-synthesizable.

3. Final Answer:

   A process with only clocked assignments and no wait statements -> Option D

4. Quick Check:

   Synthesizable = clocked process without wait [OK]

Hint: Synthesizable means no waits or file I/O, just clocked logic [OK]

Common Mistakes:

• Thinking wait statements are synthesizable
• Confusing assert for hardware logic
• Assuming file I/O can be synthesized

2. Which VHDL syntax is not synthesizable?

easy

A. A process with a wait until clk = '1'; statement

B. Concurrent signal assignment outside any process

C. Signal assignment inside a clocked process

D. Using rising_edge(clk) inside a process

Solution

1. Step 1: Understand wait statement restrictions

   Wait statements like wait until clk = '1'; are not synthesizable because hardware cannot implement arbitrary waits.

2. Step 2: Check other options

   Options A, B, and C are standard synthesizable constructs: clock edge detection, signal assignments in processes, and concurrent assignments.

3. Final Answer:

   A process with a wait until clk = '1'; statement -> Option A

4. Quick Check:

   Wait statements = non-synthesizable [OK]

Hint: Wait statements inside processes are never synthesizable [OK]

Common Mistakes:

• Assuming wait until is synthesizable
• Confusing sensitivity list with wait statements
• Thinking concurrent assignments are non-synthesizable

3. What is the output behavior of this VHDL process when synthesized?

process(clk)
begin
  if rising_edge(clk) then
    if reset = '1' then
      q <= '0';
    else
      q <= d;
    end if;
  end if;
end process;

medium

A. q synchronously resets to '0' on clock edge when reset is '1'

B. q asynchronously resets to '0' when reset is '1'

C. q changes immediately when reset is '1', ignoring clock

D. q never resets because reset is ignored

Solution

1. Step 1: Analyze reset behavior inside clocked process

   The reset check is inside the rising_edge(clk) condition, so reset is synchronous.

2. Step 2: Understand output assignment timing

   q is assigned '0' only on clock edge when reset is '1', otherwise q follows d on clock edges.

3. Final Answer:

   q synchronously resets to '0' on clock edge when reset is '1' -> Option A

4. Quick Check:

   Synchronous reset inside clock edge = q synchronously resets to '0' on clock edge when reset is '1' [OK]

Hint: Reset inside rising_edge means synchronous reset [OK]

Common Mistakes:

• Thinking reset is asynchronous here
• Assuming q changes immediately without clock
• Ignoring the rising_edge condition

4. Identify the error in this VHDL code snippet that prevents synthesis:

process(clk)
begin
  wait until clk = '1';
  q <= d;
end process;

medium

A. Signal assignment outside process

B. Missing sensitivity list in process

C. Using wait statement inside a clocked process

D. Incorrect use of <= operator

Solution

1. Step 1: Check for wait statement usage

   Wait statements inside processes are not synthesizable and cause synthesis errors.

2. Step 2: Verify other syntax elements

   The process has a sensitivity list (clk), signal assignment is inside process, and <= is correctly used.

3. Final Answer:

   Using wait statement inside a clocked process -> Option C

4. Quick Check:

   Wait inside process = synthesis error [OK]

Hint: Wait statements inside processes cause synthesis errors [OK]

Common Mistakes:

• Ignoring that wait is not synthesizable
• Confusing sensitivity list with wait usage
• Thinking <= operator is wrong here

5. You want to write a testbench to verify a VHDL design. Which construct is non-synthesizable but useful for simulation only?

hard

A. A clocked process with if rising_edge(clk) condition

B. A process with wait for 20 ns; to generate clock signals

C. Concurrent signal assignment for output signals

D. Signal assignment inside a synchronous reset process

Solution

1. Step 1: Identify testbench constructs

   Testbenches often use wait for statements to simulate timing delays, which are non-synthesizable.

2. Step 2: Compare with synthesizable constructs

   Options A, C, and D describe synthesizable hardware behavior, not simulation-only code.

3. Final Answer:

   A process with wait for 20 ns; to generate clock signals -> Option B

4. Quick Check:

   Wait for delays = simulation only [OK]

Hint: Use wait for delays only in testbenches, not hardware [OK]

Common Mistakes:

• Using wait for in synthesizable design code
• Confusing testbench and design code constructs
• Assuming clocked processes are non-synthesizable