D flip-flop with asynchronous reset in Verilog - Time & Space Complexity

Time Complexity: D flip-flop with asynchronous reset

O(n)

Understanding Time Complexity

We want to understand how the time needed to run a D flip-flop with asynchronous reset changes as inputs change.

Specifically, how does the circuit's behavior scale with input signals over time?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

module dff_async_reset (
  input wire clk,
  input wire rst_n,
  input wire d,
  output reg q
);

always @(posedge clk or negedge rst_n) begin
  if (!rst_n)
    q <= 1'b0;
  else
    q <= d;
end
endmodule

This code models a D flip-flop that updates its output on the clock's rising edge or resets immediately when reset is low.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The flip-flop updates output on every clock cycle or reset event.
• How many times: Once per clock cycle or reset signal change, repeating continuously as the clock runs.

How Execution Grows With Input

Each clock cycle triggers one update, so the number of operations grows directly with the number of clock cycles.

Input Size (n)	Approx. Operations
10 clock cycles	10 updates
100 clock cycles	100 updates
1000 clock cycles	1000 updates

Pattern observation: The operations increase linearly as the number of clock cycles increases.

Final Time Complexity

Time Complexity: O(n)

This means the work done grows in a straight line with the number of clock cycles or reset events.

Common Mistake

[X] Wrong: "The flip-flop updates multiple times per clock cycle, so complexity is higher."

[OK] Correct: The flip-flop only updates once per clock edge or reset event, so operations do not multiply within a single cycle.

Interview Connect

Understanding how hardware elements like flip-flops behave over time helps you reason about timing and performance in digital designs.

Self-Check

"What if we added a synchronous reset instead of asynchronous? How would the time complexity change?"

Practice

1.

What does the asynchronous reset in a D flip-flop do?

easy

A. It stores the input data only on the falling edge of the clock.

B. It clears the output immediately, ignoring the clock signal.

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

D. It toggles the output on every clock pulse.

Solution

1. Step 1: Understand asynchronous reset behavior

   Asynchronous reset forces the output to zero immediately, without waiting for the clock edge.

2. Step 2: Compare with other options

   Other options describe clock-dependent or toggling behavior, which is not asynchronous reset.

3. Final Answer:

   It clears the output immediately, ignoring the clock signal. -> Option B

4. Quick Check:

   Asynchronous reset = immediate clear [OK]

Hint: Async reset clears output instantly, no clock needed [OK]

Common Mistakes:

• Confusing asynchronous reset with synchronous reset
• Thinking reset waits for clock edge
• Mixing reset with toggle or clock enable

2.

Which of the following is the correct Verilog sensitivity list for a D flip-flop with asynchronous reset?

module dff(input clk, input rst_n, input d, output reg q);

easy

A. always @(rst_n or clk)

B. always @(posedge clk and negedge rst_n)

C. always @(posedge clk or negedge rst_n)

D. always @(posedge clk)

Solution

1. Step 1: Identify asynchronous reset sensitivity

   Asynchronous reset triggers on reset signal edge, so sensitivity list must include clock posedge and reset negedge.

2. Step 2: Check syntax correctness

   always @(posedge clk or negedge rst_n) uses 'or' to combine posedge clk and negedge rst_n, which is correct syntax.

3. Final Answer:

   always @(posedge clk or negedge rst_n) -> Option C

4. Quick Check:

   Sensitivity list uses 'or' for async reset [OK]

Hint: Use 'or' in sensitivity list for async reset [OK]

Common Mistakes:

• Using 'and' instead of 'or' in sensitivity list
• Omitting reset from sensitivity list
• Using only clock edge without reset

3.

Given the following Verilog code, what is the value of q after the reset is deasserted and one clock rising edge occurs with d = 1?

module dff(input clk, input rst_n, input d, output reg q);
  always @(posedge clk or negedge rst_n) begin
    if (!rst_n)
      q <= 0;
    else
      q <= d;
  end
endmodule

medium

A. q = 1

B. q retains previous value

C. q = 0

D. q is undefined

Solution

1. Step 1: Analyze reset effect

   When rst_n is low, q is set to 0 immediately.

2. Step 2: Analyze clock edge after reset

   After reset is released (rst_n=1), on next posedge clk, q takes value of d, which is 1.

3. Final Answer:

   q = 1 -> Option A

4. Quick Check:

   After reset, q follows d on clock edge [OK]

Hint: After reset, q updates to d on clock rising edge [OK]

Common Mistakes:

• Assuming q stays 0 after reset
• Ignoring clock edge for q update
• Thinking q is undefined after reset

4.

Identify the error in this D flip-flop with asynchronous reset code snippet:

always @(posedge clk or posedge rst_n) begin
  if (!rst_n)
    q <= 0;
  else
    q <= d;
end

medium

A. Reset edge should be negedge, not posedge.

B. Missing sensitivity to clock negedge.

C. The reset condition should check rst_n, not !rst_n.

D. The output q should not be declared as reg.

Solution

1. Step 1: Check reset polarity

   The reset is active low, so sensitivity should be on negedge rst_n, not posedge.

2. Step 2: Verify reset condition

   The condition uses !rst_n correctly for active low reset.

3. Final Answer:

   Reset edge should be negedge, not posedge. -> Option A

4. Quick Check:

   Active low reset uses negedge in sensitivity [OK]

Hint: Active low reset uses negedge, not posedge [OK]

Common Mistakes:

• Using posedge for active low reset
• Confusing reset polarity and condition
• Adding unnecessary clock edges

5.

You want to design a D flip-flop with asynchronous reset that sets q to 1 when reset is active (active low). Which code snippet correctly implements this behavior?

always @(posedge clk or negedge rst_n) begin
  if (!rst_n)
    q <= 1;
  else
    q <= d;
end

hard

A. The reset condition should be if (rst_n) instead of if (!rst_n).

B. Reset should set q to 0, not 1, so this code is incorrect.

C. Sensitivity list should use posedge rst_n for active low reset.

D. This code correctly sets q to 1 on reset and updates q on clock.

Solution

1. Step 1: Check reset polarity and action

   Reset is active low (negedge rst_n), and when active (!rst_n), q is set to 1 as required.

2. Step 2: Verify normal operation

   When reset is not active, q follows d on clock rising edge.

3. Final Answer:

   This code correctly sets q to 1 on reset and updates q on clock. -> Option D

4. Quick Check:

   Active low reset sets q=1, else q=d [OK]

Hint: Active low reset uses negedge and !rst_n condition [OK]

Common Mistakes:

• Assuming reset must clear q to 0
• Using posedge for active low reset
• Reversing reset condition logic