D flip-flop with rising edge in VHDL - Step-by-Step Execution

Concept Flow - D flip-flop with rising edge

Wait for rising clock edge

↓

Capture input D

↓

Update output Q with D

↓

Hold Q until next rising edge

The D flip-flop waits for a rising clock edge, then captures the input D and updates the output Q, holding it steady until the next rising edge.

Execution Sample

VHDL

process(clk)
begin
  if rising_edge(clk) then
    Q <= D;
  end if;
end process;

On each rising edge of the clock, the flip-flop copies the input D to the output Q.

Execution Table

Step	clk	D	rising_edge(clk)?	Action	Q (output)
1	0 -> 0	0	No	No change	Initial value (assume 0)
2	0 -> 1	1	Yes	Q <= D (1)	1
3	1 -> 1	0	No	No change	1
4	1 -> 0	0	No	No change	1
5	0 -> 1	0	Yes	Q <= D (0)	0
6	1 -> 1	1	No	No change	0
7	1 -> 0	1	No	No change	0
8	0 -> 1	1	Yes	Q <= D (1)	1

💡 Simulation ends after several clock cycles showing Q updates only on rising edges.

Variable Tracker

Variable	Start	After Step 2	After Step 5	After Step 8
clk	0	1	1	1
D	0	1	0	1
Q	0	1	0	1

Key Moments - 2 Insights

Why does Q only change on rising edges of clk?

Because the process uses 'if rising_edge(clk)' condition (see execution_table steps 2, 5, 8), Q updates only when clk changes from 0 to 1.

What happens if D changes but clk does not have a rising edge?

Q does not change until the next rising edge (see steps 3 and 6), because the flip-flop only samples D on rising edges.

Visual Quiz - 3 Questions

Look at the execution table, what is the value of Q after step 5?

A 1

B 0

C Undefined

D Same as D before step 5

💡 Hint

Check the Q column at step 5 in the execution_table.

At which step does the first rising edge of clk occur?

A Step 1

B Step 4

C Step 2

D Step 5

💡 Hint

Look at the 'rising_edge(clk)?' column in execution_table.

If D stayed at 1 all the time, what would Q be after step 8?

A 1

B 0

C Depends on clk

D No change from start

💡 Hint

Q copies D on rising edges, see variable_tracker and execution_table.

Concept Snapshot

D flip-flop in VHDL:
process(clk)
begin
  if rising_edge(clk) then
    Q <= D; -- capture input
  end if;
end process;
Q updates only on clk rising edge, holds value otherwise.

Full Transcript

This visual trace shows a D flip-flop in VHDL. The flip-flop waits for a rising clock edge, then copies the input D to output Q. The execution table tracks clock changes, input D, and output Q step-by-step. Q changes only on rising edges of clk, holding its value otherwise. The variable tracker summarizes key variable values at important steps. Key moments clarify why Q updates only on rising edges and what happens if D changes without a clock edge. The quiz tests understanding of Q's value after steps, identifying rising edges, and behavior if D stays constant. The snapshot summarizes the VHDL syntax and behavior of the D flip-flop.

Practice

1. What does a D flip-flop do on the rising edge of the clock signal?

easy

A. It stores the value of the D input into the output Q.

B. It resets the output Q to zero.

C. It toggles the output Q regardless of input.

D. It ignores the input and holds the previous state.

Solution

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

   The rising edge of the clock signal triggers the flip-flop to capture the input value.

2. Step 2: Identify what value is stored

   The flip-flop stores the current value of the D input into the output Q at the rising edge.

3. Final Answer:

   It stores the value of the D input into the output Q. -> Option A

4. Quick Check:

   D input stored on rising edge = It stores the value of the D input into the output Q. [OK]

Hint: Remember: rising_edge(clock) triggers storing D to Q [OK]

Common Mistakes:

• Confusing rising edge with falling edge
• Thinking output toggles automatically
• Assuming reset happens on rising edge

2. Which of the following is the correct VHDL syntax to detect a rising edge of a clock signal inside a process?

easy

A. if rising_edge(clock) then

B. if clock'event then

C. if clock = '1' then

D. if clock'event and clock = '0' then

Solution

1. Step 1: Recall the standard VHDL function for rising edge detection

   The function rising_edge(clock) is the recommended and clear way to detect a rising edge.

2. Step 2: Compare options

   if rising_edge(clock) then uses rising_edge(clock), which is correct and preferred over manual event checking.

3. Final Answer:

   if rising_edge(clock) then -> Option A

4. Quick Check:

   Use rising_edge(clock) for rising edge detection [OK]

Hint: Use rising_edge(clock) for clean rising edge detection [OK]

Common Mistakes:

• Using clock'event and clock = '0' (falling edge)
• Checking clock = '1' without event
• Confusing rising and falling edge syntax

3. Given the following VHDL process, what will be the value of Q after the rising edge of clock if D = '1' and Q was '0' before?

process(clock)
begin
  if rising_edge(clock) then
    Q <= D;
  end if;
end process;

medium

A. '0'

B. '1'

C. No change, remains previous value

D. Undefined behavior

Solution

1. Step 1: Understand the process behavior on rising edge

   On the rising edge of clock, Q is assigned the value of D.

2. Step 2: Apply given values

   D is '1' and Q was '0' before, so Q becomes '1' after the rising edge.

3. Final Answer:

   '1' -> Option B

4. Quick Check:

   Q follows D on rising edge = '1' [OK]

Hint: Q updates to D at rising edge of clock [OK]

Common Mistakes:

• Assuming Q stays the same without reset
• Confusing rising edge with falling edge
• Thinking Q toggles automatically

4. Identify the error in the following VHDL code for a D flip-flop:

process(clock)
begin
  if clock = '1' then
    Q <= D;
  end if;
end process;

medium

A. D input is not declared

B. Missing sensitivity list

C. Q should be assigned outside the process

D. Incorrect clock edge detection, should use rising_edge(clock)

Solution

1. Step 1: Check clock edge detection method

   The code checks if clock = '1' but does not detect the rising edge properly.

2. Step 2: Identify correct method

   Use rising_edge(clock) to detect the clock's rising edge correctly.

3. Final Answer:

   Incorrect clock edge detection, should use rising_edge(clock) -> Option D

4. Quick Check:

   Use rising_edge(clock) for proper edge detection [OK]

Hint: Always use rising_edge(clock) for clock edge detection [OK]

Common Mistakes:

• Checking clock = '1' without event
• Forgetting sensitivity list (though not shown here)
• Assigning signals outside process incorrectly

5. You want to design a D flip-flop in VHDL that asynchronously resets Q to '0' when reset = '1', but otherwise stores D on the rising edge of clock. Which code snippet correctly implements this behavior?

hard

A. process(reset)
begin
if reset = '1' then
Q <= '0';
end if;
end process;
process(clock)
begin
if rising_edge(clock) then
Q <= D;
end if;
end process;

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

C. process(clock, reset)
begin
if reset = '1' then
Q <= '0';
elsif rising_edge(clock) then
Q <= D;
end if;
end process;

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

Solution

1. Step 1: Understand asynchronous reset behavior

   An asynchronous reset acts immediately when reset = '1', regardless of clock.

2. Step 2: Check code for asynchronous reset and rising edge storage

   process(clock, reset) begin if reset = '1' then Q <= '0'; elsif rising_edge(clock) then Q <= D; end if; end process; checks reset first, outside rising_edge(clock), and includes both signals in sensitivity list, correctly implementing asynchronous reset.

3. Final Answer:

   process(clock, reset) begin if reset = '1' then Q <= '0'; elsif rising_edge(clock) then Q <= D; end if; end process; -> Option C

4. Quick Check:

   Reset outside rising_edge for async reset = process(clock, reset) begin if reset = '1' then Q <= '0'; elsif rising_edge(clock) then Q <= D; end if; end process; [OK]

Hint: Put async reset check before rising_edge(clock) in process [OK]

Common Mistakes:

• Placing reset check inside rising_edge (makes reset synchronous)
• Missing reset in sensitivity list
• Splitting reset and clock into separate processes incorrectly