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

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare the D input port.

VHDL

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

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

Aq

Brst

Cclk

Dd

Attempts:

3 left

💡 Hint

Common Mistakes

Confusing the clock or reset signals as the data input.

Using the output name 'q' as an input.

The D input is named d as per convention for D flip-flops.

2fill in blank

medium

Complete the process sensitivity list to include clock and asynchronous reset.

VHDL

architecture Behavioral of DFF is
begin
  process([1])
  begin
    -- process body
  end process;
end Behavioral;

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Arst, d

Bclk, rst

Cd, q

Dclk, d

Attempts:

3 left

💡 Hint

Common Mistakes

Leaving out the reset signal from the sensitivity list.

Including data or output signals in the sensitivity list.

The process must be sensitive to the clock and asynchronous reset signals.

3fill in blank

hard

Fix the error in the if statement to check asynchronous reset correctly.

VHDL

process(clk, rst)
begin
  if [1] = '1' then
    q <= '0';
  elsif rising_edge(clk) then
    q <= d;
  end if;
end process;

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

Arst

Bclk

Cd

Dq

Attempts:

3 left

💡 Hint

Common Mistakes

Checking clock signal instead of reset in the if condition.

Using data or output signals in the reset condition.

The asynchronous reset signal is rst, so the if condition must check rst = '1'.

4fill in blank

hard

Fill both blanks to assign the output and handle the clock edge.

VHDL

process(clk, rst)
begin
  if rst = '1' then
    [1] <= '0';
  elsif [2](clk) then
    q <= d;
  end if;
end process;

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Aq

Brising_edge

Cfalling_edge

Dd

Attempts:

3 left

💡 Hint

Common Mistakes

Assigning reset to the wrong signal.

Using falling_edge instead of rising_edge.

The output q is reset to '0'. The clock edge is detected with rising_edge(clk).

5fill in blank

hard

Fill all three blanks to complete the D flip-flop with asynchronous reset.

VHDL

library IEEE;
use IEEE.STD_LOGIC_1164.ALL;

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

architecture Behavioral of DFF is
begin
  process(clk, rst)
  begin
    if rst = '1' then
      [2] <= '0';
    elsif [3](clk) then
      q <= d;
    end if;
  end process;
end Behavioral;

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

Ad

Bq

Crising_edge

Dfalling_edge

Attempts:

3 left

💡 Hint

Common Mistakes

Mixing up input and output signal names.

Using falling_edge instead of rising_edge for clock.

The input is d, output is q, and clock edge detection uses rising_edge(clk).

Practice

1. What is the main function of the asynchronous reset in a D flip-flop?

easy

A. It stores data only on the rising edge of the clock.

B. It immediately clears the output regardless of the clock.

C. It delays the output change until the next clock edge.

D. It toggles the output on every clock cycle.

Solution

1. Step 1: Understand asynchronous reset behavior

   Asynchronous reset clears the output immediately, without waiting for the clock.

2. Step 2: Compare with clocked behavior

   Unlike clocked data storage, asynchronous reset acts instantly when active.

3. Final Answer:

   It immediately clears the output regardless of the clock. -> Option B

4. Quick Check:

   Asynchronous reset = immediate clear [OK]

Hint: Asynchronous reset acts instantly, ignoring clock [OK]

Common Mistakes:

• Confusing asynchronous reset with clocked data storage
• Thinking reset waits for clock edge
• Assuming reset toggles output

2. Which of the following VHDL code snippets correctly declares a D flip-flop with asynchronous reset?

easy

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

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

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

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

Solution

1. Step 1: Check sensitivity list for asynchronous reset

   Asynchronous reset requires reset signal in sensitivity list along with clock.

2. Step 2: Verify reset condition before clock edge

   Reset condition must be checked first, then clock edge for data update.

3. Final Answer:

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

4. Quick Check:

   Reset in sensitivity list + reset checked first = correct async reset [OK]

Hint: Async reset needs reset in sensitivity list and checked first [OK]

Common Mistakes:

• Omitting reset from sensitivity list
• Checking reset inside clock edge condition
• Using only reset process without clock

3. Given this VHDL process for a D flip-flop with asynchronous reset:

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

If reset is '0', d is '1', and a rising edge on clk occurs, what will be the value of q?

medium

A. Undefined

B. '0'

C. No change

D. '1'

Solution

1. Step 1: Check reset signal

   Reset is '0', so asynchronous reset does not clear output.

2. Step 2: Check clock edge and data input

   Rising edge on clk triggers q to take value of d, which is '1'.

3. Final Answer:

   '1' -> Option D

4. Quick Check:

   Reset '0' + rising clk + d='1' means q='1' [OK]

Hint: Reset '0' means normal clocked data update [OK]

Common Mistakes:

• Assuming reset clears output when '0'
• Ignoring clock edge
• Thinking output stays unchanged

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

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

medium

A. Reset signal is not in the sensitivity list for asynchronous reset.

B. The reset condition should be checked after the clock edge.

C. The process should not use rising_edge(clk) for a flip-flop.

D. The output q should be assigned using <= only inside clock edge.

Solution

1. Step 1: Check sensitivity list for asynchronous reset

   Asynchronous reset requires reset signal in sensitivity list to respond immediately.

2. Step 2: Analyze reset checking position

   Reset is checked before clock edge, which is correct, but missing reset in sensitivity list causes error.

3. Final Answer:

   Reset signal is not in the sensitivity list for asynchronous reset. -> Option A

4. Quick Check:

   Async reset needs reset in sensitivity list [OK]

Hint: Async reset must be in sensitivity list [OK]

Common Mistakes:

• Omitting reset from sensitivity list
• Checking reset after clock edge
• Misusing rising_edge for flip-flop

5. You want to design a D flip-flop with asynchronous reset that sets output q to '0' immediately when reset is '1', and otherwise updates q with d on the rising edge of clk. Which VHDL process correctly implements this behavior and avoids unintended latches?

hard

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

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

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

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

Solution

1. Step 1: Check sensitivity list for asynchronous reset

   Reset and clk must be in sensitivity list for asynchronous reset to work properly.

2. Step 2: Verify reset checked before clock edge

   Reset condition checked first ensures immediate clearing without waiting for clock.

3. Step 3: Confirm no latches created

   All outputs assigned in all branches, so no unintended latches occur.

4. Final Answer:

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

5. Quick Check:

   Reset in sensitivity + reset first + all assignments = correct async reset flip-flop [OK]

Hint: Reset in sensitivity list and checked first avoids latches [OK]

Common Mistakes:

• Checking reset inside clock edge (not async)
• Missing reset in sensitivity list
• Creating latches by incomplete assignments