D flip-flop with synchronous reset in VHDL

Introduction

A D flip-flop stores a single bit of data on a clock edge. The synchronous reset lets you clear the stored data at the same time as the clock signal.

When you want to save a bit of information that changes only on a clock signal.

When you need to reset the stored data to zero at a specific clock time.

When designing digital circuits like counters or registers that need controlled resets.

When you want to avoid asynchronous glitches by resetting only on clock edges.

Syntax

VHDL

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

The process triggers on the rising edge of the clock.

The reset is checked inside the clock edge condition, making it synchronous.

Examples

Basic D flip-flop with synchronous reset inside a clocked process.

VHDL

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

This example shows the reset clearing the output only on the clock's rising edge.

VHDL

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

Sample Program

This is a full VHDL code for a D flip-flop with synchronous reset. The output q updates on the rising edge of clk. If reset is high at that moment, q is cleared to '0'. Otherwise, q takes the value of d.

VHDL

library ieee;
use ieee.std_logic_1164.all;

entity dff_sync_reset is
  port(
    clk   : in std_logic;
    reset : in std_logic;
    d     : in std_logic;
    q     : out std_logic
  );
end dff_sync_reset;

architecture behavioral of dff_sync_reset is
begin
  process(clk) is
  begin
    if rising_edge(clk) then
      if reset = '1' then
        q <= '0';
      else
        q <= d;
      end if;
    end if;
  end process;
end behavioral;

Important Notes

The reset only works when the clock rises, so it is synchronous.

This avoids unexpected resets caused by noise on the reset line.

Make sure the clock and reset signals are clean to avoid glitches.

Summary

A D flip-flop stores data on the clock's rising edge.

Synchronous reset clears the stored data only at the clock edge.

This design helps control when the reset happens and avoids glitches.

Practice

1.

What does a synchronous reset do in a D flip-flop?

easy

A. It disables the clock signal.

B. It clears the stored data immediately when reset is active.

C. It stores data continuously without any reset.

D. It clears the stored data only at the clock's rising edge.

Solution

1. Step 1: Understand synchronous reset behavior

   A synchronous reset affects the flip-flop only during the clock's active edge, not immediately.

2. Step 2: Compare with asynchronous reset

   Asynchronous reset clears data immediately, but synchronous reset waits for the clock edge.

3. Final Answer:

   It clears the stored data only at the clock's rising edge. -> Option D

4. Quick Check:

   Synchronous reset = clears data on clock edge [OK]

Hint: Synchronous reset acts only on clock edge, not instantly. [OK]

Common Mistakes:

• Confusing synchronous with asynchronous reset
• Thinking reset works immediately without clock
• Assuming reset disables clock

2.

Which of the following is the correct VHDL sensitivity list for a D flip-flop with synchronous reset?

process(???)

easy

A. clk, reset

B. clk

C. reset

D. clk, reset, d

Solution

1. Step 1: Identify signals triggering the process

   For synchronous reset, only the clock edge triggers the process, so only 'clk' is in the sensitivity list.

2. Step 2: Understand synchronous reset inside process

   The reset is checked inside the clock edge condition, so it does not go in sensitivity list.

3. Final Answer:

   clk -> Option B

4. Quick Check:

   Synchronous reset uses only clock in sensitivity list [OK]

Hint: Only clock in sensitivity list for synchronous reset. [OK]

Common Mistakes:

• Including reset in sensitivity list for synchronous reset
• Adding data input 'd' to sensitivity list
• Confusing asynchronous reset sensitivity

3.

Given this VHDL code snippet for a D flip-flop with synchronous reset, what is the output q after the rising edge of clk if reset = '1' and d = '0'?

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. '0'

B. '1'

C. No change

D. Undefined

Solution

1. Step 1: Check reset condition at clock edge

   At rising edge, reset is '1', so q is set to '0'.

2. Step 2: Ignore data input since reset is active

   Because reset is active, q does not take value from d.

3. Final Answer:

   '0' -> Option A

4. Quick Check:

   Reset '1' sets q to '0' on clock edge [OK]

Hint: Reset '1' overrides data input at clock edge. [OK]

Common Mistakes:

• Ignoring reset and taking d value
• Thinking output changes immediately without clock
• Confusing synchronous with asynchronous reset

4.

Identify the error in this VHDL code for a D flip-flop with synchronous reset:

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

medium

A. Reset should not be in the sensitivity list for synchronous reset.

B. Missing else statement for reset condition.

C. rising_edge(clk) should be replaced with falling_edge(clk).

D. Data input 'd' should be in sensitivity list.

Solution

1. Step 1: Check sensitivity list for synchronous reset

   Synchronous reset requires only clock in sensitivity list, not reset.

2. Step 2: Understand why reset is excluded

   Reset is checked inside clock edge condition, so including it causes unnecessary triggers.

3. Final Answer:

   Reset should not be in the sensitivity list for synchronous reset. -> Option A

4. Quick Check:

   Synchronous reset sensitivity list = clock only [OK]

Hint: Exclude reset from sensitivity list for synchronous reset. [OK]

Common Mistakes:

• Including reset in sensitivity list
• Changing clock edge without reason
• Adding data input to sensitivity list

5.

You want to design a D flip-flop with synchronous reset that sets output q to '1' when reset is active instead of '0'. Which code modification achieves this?

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

hard

A. Replace if reset = '1' with if reset = '0'

B. Add q <= '1'; before the if statement

C. Change q <= '0'; to q <= '1';

D. Move reset check outside the clock edge condition

Solution

1. Step 1: Identify reset output assignment

   The line q <= '0'; sets q to zero on reset.

2. Step 2: Modify reset output to '1'

   Changing this line to q <= '1'; sets q to one when reset is active.

3. Final Answer:

   Change q <= '0'; to q <= '1'; -> Option C

4. Quick Check:

   Reset output changed by modifying assignment line [OK]

Hint: Change reset assignment line to desired output value. [OK]

Common Mistakes:

• Adding assignment outside clock edge
• Reversing reset condition incorrectly
• Adding redundant assignments