D flip-flop with synchronous reset in VHDL - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of a D flip-flop with synchronous reset

What is the output Q after the rising edge of the clock if D = '1' and reset = '1' in this D flip-flop with synchronous reset?

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)
    begin
        if rising_edge(clk) then
            if reset = '1' then
                Q <= '0';
            else
                Q <= D;
            end if;
        end if;
    end process;
end Behavioral;

A Q retains previous value

B Q = '1'

C Q = '0'

D Q = 'Z' (high impedance)

💡 Hint

Remember synchronous reset means reset affects output only on clock edge.

Since reset is '1' at the rising clock edge, the flip-flop sets Q to '0' regardless of D.

❓ Predict Output

intermediate

Behavior when reset is '0' and D changes

Given the same D flip-flop with synchronous reset, what will be the output Q after the rising edge of the clock if reset = '0' and D = '1'?

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)
    begin
        if rising_edge(clk) then
            if reset = '1' then
                Q <= '0';
            else
                Q <= D;
            end if;
        end if;
    end process;
end Behavioral;

A Q = '1'

B Q = '0'

C Q retains previous value

D Q = 'X' (unknown)

💡 Hint

When reset is '0', output follows D at clock edge.

Since reset is '0', the flip-flop captures D value '1' at rising clock edge, so Q becomes '1'.

🔧 Debug

advanced

Identify the error in this D flip-flop with synchronous reset code

Which option correctly identifies the error in this VHDL code for a D flip-flop with synchronous reset?

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)
    begin
        if clk'event and clk = '1' then
            if reset = '1' then
                Q <= '0';
            else
                Q <= D;
            end if;
        end if;
    end process;
end Behavioral;

A Using clk'event and clk = '1' is outdated; should use rising_edge(clk)

B Q should be declared as signal, not port

C Reset should be asynchronous, not synchronous

D Missing sensitivity list in process

💡 Hint

Check the recommended way to detect clock edges in modern VHDL.

Using rising_edge(clk) is the modern, recommended way to detect clock edges instead of clk'event and clk = '1'.

📝 Syntax

advanced

Syntax error in D flip-flop with synchronous reset

Which option contains a syntax error in the process block of this D flip-flop with synchronous reset?

VHDL

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

A Q should be declared as variable, not signal

B Missing semicolon after Q <= '0'

C Missing sensitivity list in process

D Missing end if for the first if statement

💡 Hint

Check each statement for proper ending punctuation.

In VHDL, each assignment statement must end with a semicolon. The line Q <= '0' is missing a semicolon.

🚀 Application

expert

Number of clock cycles to reset and set output

Consider a D flip-flop with synchronous reset initialized with Q = '1'. The reset signal is held high for 2 clock cycles, then low for 3 clock cycles with D = '1' during those 3 cycles. How many clock cycles after the first rising edge will Q be '1' again?

A 6 clock cycles

B 3 clock cycles

C 5 clock cycles

D 4 clock cycles

💡 Hint

Remember synchronous reset sets Q to '0' only at clock edges when reset is '1'. Q updates to D only when reset is '0' at clock edge.

At first clock edge with reset='1', Q becomes '0'. It stays '0' at second clock edge (reset still '1'). At third clock edge, reset='0' and D='1', so Q becomes '1'. Thus, 4 clock edges after start.

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