D flip-flop with enable in VHDL - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare the clock signal in the entity.

VHDL

entity DFF_Enable is
  Port (
    clk : [1];
    d : in std_logic;
    en : in std_logic;
    q : out std_logic
  );
end DFF_Enable;

Drag options to blanks, or click blank then click option'

Ain std_logic

Bout std_logic

Cin integer

Dbuffer std_logic

Attempts:

3 left

💡 Hint

Common Mistakes

Declaring clock as output instead of input

Using incorrect data type like integer

The clock signal must be an input of type std_logic to receive the clock pulses.

2fill in blank

medium

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

VHDL

process([1])
begin
  if rising_edge(clk) then
    if en = '1' then
      q <= d;
    end if;
  end if;
end process;

Drag options to blanks, or click blank then click option'

Ad

Bq

Cen

Dclk

Attempts:

3 left

💡 Hint

Common Mistakes

Including data or enable signals in sensitivity list unnecessarily

Leaving sensitivity list empty

The process should be sensitive to the clock signal to detect its rising edge.

3fill in blank

hard

Fix the error in the assignment inside the process to correctly update output q.

VHDL

process(clk)
begin
  if rising_edge(clk) then
    if en = '1' then
      q <= [1];
    end if;
  end if;
end process;

Drag options to blanks, or click blank then click option'

Ad

Bq

Cen

Dclk

Attempts:

3 left

💡 Hint

Common Mistakes

Assigning q to itself causing no change

Assigning clock or enable signals instead of data input

The output q should be assigned the input d when enabled.

4fill in blank

hard

Fill both blanks to complete the entity port declaration for data input and output.

VHDL

entity DFF_Enable is
  Port (
    d : [1];
    q : [2]
  );
end DFF_Enable;

Drag options to blanks, or click blank then click option'

Ain std_logic

Bout std_logic

Cin integer

Dbuffer std_logic

Attempts:

3 left

💡 Hint

Common Mistakes

Using incorrect directions for ports

Using wrong data types like integer

Data input d is an input signal, and output q is an output signal, both of type std_logic.

5fill in blank

hard

Fill all three blanks to complete the process sensitivity list and the enable condition.

VHDL

process([1])
begin
  if rising_edge(clk) then
    if en = '[2]' then
      q <= [3];
    end if;
  end if;
end process;

Drag options to blanks, or click blank then click option'

Ad

Bclk

C1

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong sensitivity list signals

Checking enable for wrong value

Assigning wrong signals to output

The process is sensitive to clk. The enable signal is checked for '1', and q is assigned d.

Practice

1. What is the main function of the enable signal in a D flip-flop with enable?

easy

A. It acts as the clock signal.

B. It controls when the flip-flop updates its stored data.

C. It resets the flip-flop to zero.

D. It outputs the stored data.

Solution

1. Step 1: Understand the role of enable in a D flip-flop

   The enable signal allows the flip-flop to update its stored data only when it is active (usually high).

2. Step 2: Differentiate enable from other signals

   The clock triggers the flip-flop, but enable decides if the data input is accepted or ignored.

3. Final Answer:

   It controls when the flip-flop updates its stored data. -> Option B

4. Quick Check:

   Enable controls data update = C [OK]

Hint: Enable lets data change only when active [OK]

Common Mistakes:

• Confusing enable with clock signal
• Thinking enable resets the flip-flop
• Assuming enable outputs data

2. Which of the following VHDL code snippets correctly describes a D flip-flop with enable inside a process triggered by a rising clock edge?

easy

A. if enable = '1' then
if falling_edge(clk) then
q <= d;
end if;
end if;

B. if rising_edge(clk) then
q <= d;
end if;

C. if rising_edge(clk) then
q <= d when enable = '1' else '0';
end if;

D. if rising_edge(clk) then
if enable = '1' then
q <= d;
end if;
end if;

Solution

1. Step 1: Check the correct order of conditions

   The clock edge check must come first to synchronize the flip-flop operation.

2. Step 2: Verify enable condition inside clock check

   Enable must be checked inside the clock edge condition to control data update.

3. Final Answer:

   if rising_edge(clk) then if enable = '1' then q <= d; end if; end if; -> Option D

4. Quick Check:

   Clock edge first, then enable check = B [OK]

Hint: Clock edge check always comes before enable inside process [OK]

Common Mistakes:

• Placing enable check before clock edge
• Using conditional assignment incorrectly inside process
• Omitting enable check

3. Given the following VHDL code for a D flip-flop with enable, what will be the value of q after the rising edge of clk if d = '1' and enable = '0'?

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

medium

A. Previous value of q (no change)

B. '0'

C. Undefined

D. '1'

Solution

1. Step 1: Analyze enable signal value

   Enable is '0', so the condition to update q is false.

2. Step 2: Determine q behavior on clock edge

   Since enable is not '1', q retains its previous value (no update).

3. Final Answer:

   Previous value of q (no change) -> Option A

4. Quick Check:

   Enable '0' means no update = D [OK]

Hint: Enable '0' means q keeps old value [OK]

Common Mistakes:

• Assuming q updates regardless of enable
• Confusing enable with clock
• Thinking q resets automatically

4. Identify the error in this VHDL process for a D flip-flop with enable:

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

medium

A. The assignment to q should be outside both conditions.

B. The process should be sensitive to enable signal.

C. The enable check should be inside the rising_edge(clk) condition.

D. The clock edge check should be replaced with falling_edge.

Solution

1. Step 1: Review the order of conditions in a flip-flop process

   The clock edge check must come first to ensure synchronous operation.

2. Step 2: Identify the misplaced enable check

   Enable should be checked inside the clock edge condition, not before it.

3. Final Answer:

   The enable check should be inside the rising_edge(clk) condition. -> Option C

4. Quick Check:

   Clock edge first, then enable check = A [OK]

Hint: Clock edge check always comes first inside process [OK]

Common Mistakes:

• Checking enable before clock edge
• Adding enable to sensitivity list unnecessarily
• Using falling_edge instead of rising_edge without reason

5. You want to design a D flip-flop with enable that also asynchronously resets q to '0' when reset is '1'. Which VHDL process correctly implements this behavior?

hard

A. process(clk, reset)
begin
if reset = '1' then
q <= '0';
elsif rising_edge(clk) then
if enable = '1' then
q <= d;
end if;
end if;
end process;

B. process(clk)
begin
if rising_edge(clk) then
if reset = '1' then
q <= '0';
elsif enable = '1' then
q <= d;
end if;
end if;
end process;

C. process(clk, reset)
begin
if rising_edge(clk) then
if reset = '1' then
q <= '0';
elsif enable = '1' then
q <= d;
end if;
end if;
end process;

D. process(reset)
begin
if reset = '1' then
q <= '0';
end if;
end process;

Solution

1. Step 1: Identify asynchronous reset implementation

   Asynchronous reset requires reset in sensitivity list and immediate check before clock edge.

2. Step 2: Confirm enable and clock edge behavior

   Data updates only on rising clock edge when enable is '1', after reset is checked.

3. Final Answer:

   process(clk, reset) begin if reset = '1' then q <= '0'; elsif rising_edge(clk) then if enable = '1' then q <= d; end if; end if; end process; -> Option A

4. Quick Check:

   Reset async first, then clock and enable = A [OK]

Hint: Async reset checked before clock edge in sensitivity list [OK]

Common Mistakes:

• Placing reset check inside clock edge (making it synchronous)
• Omitting reset from sensitivity list
• Using reset process without clock for flip-flop