Modulo-N counter in VHDL - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare a signal named 'count' of type integer.

VHDL

signal count : [1];

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

Abit_vector(3 downto 0)

Bstd_logic

Cinteger range 0 to N-1

Dboolean

Attempts:

3 left

💡 Hint

Common Mistakes

Using std_logic or bit_vector for the count signal instead of an integer range.

Not specifying the range limits for the integer.

The signal 'count' must be an integer within the range 0 to N-1 to represent the modulo-N counter value.

2fill in blank

medium

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

VHDL

process([1])
begin
  -- process body
end process;

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Aclk

Bclk, reset

Creset

Dclk or reset

Attempts:

3 left

💡 Hint

Common Mistakes

Including only the clock or only the reset signal in the sensitivity list.

Using 'or' instead of commas to separate signals.

The process must be sensitive to both 'clk' and 'reset' signals to react to clock edges and asynchronous reset.

3fill in blank

hard

Fix the error in the if statement checking for the rising edge of the clock.

VHDL

if [1] then
  -- increment count
end if;

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Arising_edge(clk)

Bclk = '0'

Cclk'event and clk = '1'

Dclk = '1'

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'clk = '1'' alone without event detection.

Using 'clk'event and clk = '1'' which is valid but less preferred than 'rising_edge'.

In VHDL, the correct way to check for a rising clock edge is using the 'rising_edge(clk)' function.

4fill in blank

hard

Fill both blanks to reset the count to zero and increment it modulo N.

VHDL

if reset = '1' then
  count <= [1];
else
  count <= (count + 1) [2] N;
end if;

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A0

B+

Cmod

D-

Attempts:

3 left

💡 Hint

Common Mistakes

Using '+' instead of 'mod' for modulo operation.

Setting reset value to 1 instead of 0.

On reset, count is set to 0. To increment modulo N, use the 'mod' operator to wrap around after N-1.

5fill in blank

hard

Fill all three blanks to complete the modulo-N counter process with asynchronous reset.

VHDL

process([1])
begin
  if [2] = '1' then
    count <= 0;
  elsif rising_edge([3]) then
    count <= (count + 1) mod N;
  end if;
end process;

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

Aclk, reset

Breset

Cclk

Denable

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'enable' instead of 'reset' in sensitivity list or condition.

Checking rising edge on 'reset' instead of 'clk'.

The process is sensitive to 'reset' and 'clk'. The reset condition checks if 'reset' is '1'. The rising edge is detected on 'clk'.

Practice

1. What does a Modulo-N counter do in VHDL?

easy

A. Counts from 0 up to N-1 and then restarts at 0

B. Counts from N down to 0 and stops

C. Counts only even numbers up to N

D. Counts from 1 to N without restarting

Solution

1. Step 1: Understand the counting range

   A Modulo-N counter counts starting at 0 and goes up to N-1.

2. Step 2: Recognize the restart behavior

   After reaching N-1, it restarts counting from 0 again.

3. Final Answer:

   Counts from 0 up to N-1 and then restarts at 0 -> Option A

4. Quick Check:

   Modulo-N counter = Counts 0 to N-1 then resets [OK]

Hint: Remember: Modulo means count resets after N-1 [OK]

Common Mistakes:

• Thinking it counts down instead of up
• Assuming it counts from 1 instead of 0
• Believing it stops after reaching N-1

2. Which of the following is the correct VHDL syntax to declare a signal for a 4-bit Modulo-16 counter?

easy

A. signal count : std_logic_vector(15 downto 0);

B. signal count : integer range 1 to 16;

C. signal count : integer range 0 to 15;

D. signal count : std_logic_vector(4 downto 0);

Solution

1. Step 1: Identify the correct data type and range

   For a Modulo-16 counter, integer range 0 to 15 covers all counts from 0 to 15.

2. Step 2: Check signal declaration syntax

   "signal count : integer range 0 to 15;" is valid VHDL syntax for this purpose.

3. Final Answer:

   signal count : integer range 0 to 15; -> Option C

4. Quick Check:

   Integer 0 to 15 fits 4-bit Modulo-16 counter [OK]

Hint: Use integer range 0 to N-1 for Modulo-N counters [OK]

Common Mistakes:

• Using std_logic_vector with wrong bit width
• Setting range from 1 to N instead of 0 to N-1
• Confusing bit width with integer range

3. Given the following VHDL process for a Modulo-4 counter, what is the value of count after 5 rising clock edges (assuming reset is inactive)?

process(clk, reset)
begin
  if reset = '1' then
    count <= 0;
  elsif rising_edge(clk) then
    if count = 3 then
      count <= 0;
    else
      count <= count + 1;
    end if;
  end if;
end process;

medium

A. 1

B. 2

C. 0

D. 3

Solution

1. Step 1: Trace count values on each clock edge

   Starting at 0, count increments each clock: 0->1->2->3->0->1 after 5 edges.

2. Step 2: Identify count after 5 edges

   After 5 rising edges (1st:1, 2nd:2, 3rd:3, 4th:0, 5th:1), count is 1.

3. Final Answer:

   1 -> Option A

4. Quick Check:

   Modulo-4 resets at 3, so after 5 edges count = 1 [OK]

Hint: Count resets after reaching N-1, track each clock edge [OK]

Common Mistakes:

• Counting from 1 instead of 0
• Forgetting the reset at count = 3
• Miscounting the number of clock edges

4. Identify the error in this VHDL code snippet for a Modulo-8 counter:

process(clk, reset)
begin
  if reset = '1' then
    count <= 0;
  elsif rising_edge(clk) then
    if count = 8 then
      count <= 0;
    else
      count <= count + 1;
    end if;
  end if;
end process;

medium

A. The reset condition should check for '0' instead of '1'

B. The count signal is not declared as std_logic_vector

C. The process is missing a sensitivity list

D. The count comparison should be 'count = 7' instead of 'count = 8'

Solution

1. Step 1: Check the counting limit for Modulo-8

   Modulo-8 counts from 0 to 7, so the reset should happen when count = 7.

2. Step 2: Identify the incorrect comparison

   The code compares count = 8, which is outside the valid range, causing incorrect behavior.

3. Final Answer:

   The count comparison should be 'count = 7' instead of 'count = 8' -> Option D

4. Quick Check:

   Modulo-8 max count = 7, not 8 [OK]

Hint: Check if count comparison matches N-1 for Modulo-N [OK]

Common Mistakes:

• Using count = N instead of count = N-1
• Confusing reset signal polarity
• Ignoring sensitivity list correctness

5. You want to design a Modulo-10 counter in VHDL that counts from 0 to 9 and then resets. Which of the following code snippets correctly implements this behavior inside a clocked process?

hard

A. if reset = '0' then
count <= 0;
elsif rising_edge(clk) then
if count = 10 then
count <= 0;
else
count <= count + 1;
end if;

B. if reset = '1' then
count <= 0;
elsif rising_edge(clk) then
if count = 9 then
count <= 0;
else
count <= count + 1;
end if;

C. if reset = '1' then
count <= 1;
elsif rising_edge(clk) then
if count = 10 then
count <= 1;
else
count <= count + 1;
end if;

D. if reset = '1' then
count <= 0;
elsif rising_edge(clk) then
if count = 10 then
count <= 1;
else
count <= count + 1;
end if;

Solution

1. Step 1: Check reset condition and initial count

   Reset should set count to 0, which is done correctly only in if reset = '1' then count <= 0; elsif rising_edge(clk) then if count = 9 then count <= 0; else count <= count + 1; end if;.

2. Step 2: Verify counting and reset limit

   Counting should reset when count = 9 (for Modulo-10), and then count restarts at 0, correctly shown in if reset = '1' then count <= 0; elsif rising_edge(clk) then if count = 9 then count <= 0; else count <= count + 1; end if;.

3. Final Answer:

   resets count at 9 and initializes at 0 -> Option B

4. Quick Check:

   Modulo-10 counts 0 to 9, reset at 9 [OK]

Hint: Reset at N-1 and start count at 0 for Modulo-N [OK]

Common Mistakes:

• Resetting count to 1 instead of 0
• Using count = N instead of count = N-1 for reset
• Incorrect reset signal polarity