D flip-flop with synchronous reset in VHDL - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare the clock input port.

VHDL

entity DFF is
  Port (
    clk : [1];
    D : in std_logic;
    Q : out std_logic
  );
end DFF;

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Abuffer std_logic

Bout std_logic

Cin integer

Din std_logic

Attempts:

3 left

💡 Hint

Common Mistakes

Declaring clock as output or wrong data type.

The clock input must be declared as in std_logic to receive the clock signal.

2fill in blank

medium

Complete the process sensitivity list to trigger on clock's rising edge.

VHDL

architecture Behavioral of DFF is
begin
  process([1])
  begin
    if rising_edge(clk) then
      Q <= D;
    end if;
  end process;
end Behavioral;

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AD

Breset

Cclk

DQ

Attempts:

3 left

💡 Hint

Common Mistakes

Using data input or output signals in sensitivity list instead of clock.

The process must be sensitive to the clock signal clk to detect edges.

3fill in blank

hard

Fix the error in the synchronous reset condition inside the process.

VHDL

architecture Behavioral of DFF is
begin
  process(clk)
  begin
    if rising_edge(clk) then
      if [1] = '1' then
        Q <= '0';
      else
        Q <= D;
      end if;
    end if;
  end process;
end Behavioral;

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Areset

BD

Cclk

Drst_n

Attempts:

3 left

💡 Hint

Common Mistakes

Using clock or data signals instead of reset for reset condition.

The synchronous reset signal is usually named reset and checked inside the clock edge condition.

4fill in blank

hard

Fill both blanks to declare the reset input and add it to the sensitivity list.

VHDL

entity DFF is
  Port (
    clk : in std_logic;
    [1] : in std_logic;
    D : in std_logic;
    Q : out std_logic
  );
end DFF;

architecture Behavioral of DFF is
begin
  process([2], clk)
  begin
    if rising_edge(clk) then
      if reset = '1' then
        Q <= '0';
      else
        Q <= D;
      end if;
    end if;
  end process;
end Behavioral;

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Areset

Brst

Dclk

Attempts:

3 left

💡 Hint

Common Mistakes

Using different names for reset signal or missing it in sensitivity list.

The reset input is declared as reset and added to the sensitivity list to detect changes.

5fill in blank

hard

Fill all three blanks to complete the synchronous reset D flip-flop process.

VHDL

architecture Behavioral of DFF is
begin
  process([1], clk)
  begin
    if rising_edge(clk) then
      if [2] = '1' then
        Q <= [3];
      else
        Q <= D;
      end if;
    end if;
  end process;
end Behavioral;

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Areset

Bclk

C'0'

D'1'

Attempts:

3 left

💡 Hint

Common Mistakes

Mixing up reset and clock in sensitivity list or wrong reset value.

The process sensitivity list includes clk. The reset signal is checked for '1' and Q is set to '0' on reset.

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