Prescaler for clock division in VHDL - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare a clock input signal.

VHDL

signal clk : std_logic := [1];

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

A'X'

B'1'

C'Z'

D'0'

Attempts:

3 left

💡 Hint

Common Mistakes

Using '1' as initial clock value may cause unexpected behavior.

The clock signal is usually initialized to '0' to start from a known state.

2fill in blank

medium

Complete the code to define a 4-bit counter signal.

VHDL

signal counter : std_logic_vector([1] downto 0) := (others => '0');

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

A7

B4

C3

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using 4 downto 0 creates 5 bits, which is incorrect.

A 4-bit vector counts from 3 downto 0 (4 bits total).

3fill in blank

hard

Fix the error in the process sensitivity list to include the clock signal.

VHDL

process([1])
begin
  if rising_edge(clk) then
    -- counter logic
  end if;
end process;

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

Aclk

Breset

Ccounter

Dclk, reset

Attempts:

3 left

💡 Hint

Common Mistakes

Including signals other than clk can cause unintended triggers.

The process must be sensitive to the clock signal to detect edges.

4fill in blank

hard

Fill both blanks to complete the counter increment and reset logic.

VHDL

if rising_edge(clk) then
  if [1] = '1' then
    counter <= (others => '0');
  else
    counter <= std_logic_vector(unsigned(counter) [2] 1);
  end if;
end if;

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

Areset

B+

C-

Denable

Attempts:

3 left

💡 Hint

Common Mistakes

Using '-' decrements instead of increments the counter.

Reset clears the counter; on clock edge, counter increments by 1.

5fill in blank

hard

Fill all three blanks to complete the prescaler output assignment with condition.

VHDL

if counter = [1] then
  prescaled_clk <= not prescaled_clk;
  counter <= [2];
end if;

[3] <= prescaled_clk;

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

A"1111"

B(others => '0')

Coutput_clk

Dclk_div

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong max count or forgetting to reset counter.

When counter reaches max (1111), toggle prescaled_clk, reset counter, and assign output.

Practice

1. What is the main purpose of a prescaler in VHDL clock division?

easy

A. To generate random clock pulses

B. To multiply the input clock frequency

C. To reset the clock signal

D. To divide a fast input clock into a slower output clock

Solution

1. Step 1: Understand the function of a prescaler

   A prescaler reduces the frequency of a fast clock by dividing it to create a slower clock signal.

2. Step 2: Compare options with the prescaler function

   Only To divide a fast input clock into a slower output clock correctly describes this division function; others describe unrelated actions.

3. Final Answer:

   To divide a fast input clock into a slower output clock -> Option D

4. Quick Check:

   Prescaler divides clock = To divide clock [OK]

Hint: Prescaler always divides clock frequency, never multiplies [OK]

Common Mistakes:

• Thinking prescaler multiplies clock frequency
• Confusing prescaler with reset function
• Assuming prescaler generates random pulses

2. Which VHDL code snippet correctly declares a counter signal for a prescaler?

easy

A. signal counter : integer range 0 to 99 := 0;

B. signal counter : std_logic_vector(7 downto 0) := '0';

C. signal counter : boolean := false;

D. signal counter : integer := '0';

Solution

1. Step 1: Identify proper signal type for counting

   Counting requires an integer with a defined range and initial value, as in signal counter : integer range 0 to 99 := 0;.

2. Step 2: Check other options for correctness

   signal counter : std_logic_vector(7 downto 0) := '0'; uses std_logic_vector but initializes incorrectly with a single quote; signal counter : boolean := false; uses boolean which can't count; signal counter : integer := '0'; assigns a character to integer, which is invalid.

3. Final Answer:

   signal counter : integer range 0 to 99 := 0; -> Option A

4. Quick Check:

   Integer counter with range and init = Correct declaration [OK]

Hint: Use integer with range and initial value for counters [OK]

Common Mistakes:

• Using std_logic_vector without proper initialization
• Using boolean type for counting
• Assigning character literals to integer signals

3. Given this VHDL process for a prescaler with a max count of 3:

process(clk, reset)
begin
  if reset = '1' then
    counter <= 0;
    clk_out <= '0';
  elsif rising_edge(clk) then
    if counter = 3 then
      counter <= 0;
      clk_out <= not clk_out;
    else
      counter <= counter + 1;
    end if;
  end if;
end process;

What is the frequency division factor of clk_out relative to clk?

medium

A. 4

B. 8

C. 6

D. 3

Solution

1. Step 1: Understand counter behavior

   The counter counts from 0 to 3, so it counts 4 clock cycles before toggling clk_out.

2. Step 2: Calculate total division factor

   Since clk_out toggles every 4 input clocks, one full cycle of clk_out is 8 input clocks (4 up, 4 down).

3. Final Answer:

   8 -> Option B

4. Quick Check:

   Toggle every 4 clocks means full period 8 clocks [OK]

Hint: Multiply count by 2 for full output clock period [OK]

Common Mistakes:

• Using counter max value as division factor directly
• Forgetting clk_out toggles, doubling period
• Counting only half the clock cycles

4. Identify the error in this prescaler process code:

process(clk, reset)
begin
  if reset = '1' then
    counter <= 0;
    clk_out <= '0';
  elsif rising_edge(clk) then
    if counter = 4 then
      counter <= 0;
      clk_out <= not clk_out;
    else
      counter <= counter + 1;
    end if;
  end if;
end process;

Assuming counter is declared as integer range 0 to 3, what will happen?

medium

A. Counter never reaches 4, so clk_out never toggles

B. Counter resets correctly at 4

C. Runtime error due to counter overflow

D. clk_out toggles every clock cycle

Solution

1. Step 1: Check counter range and comparison

   The counter is declared with range 0 to 3, but the code checks if counter = 4, which is outside the range.

2. Step 2: Understand effect of out-of-range value

   Since counter can only count up to 3, it will never equal 4, so the condition is never true and clk_out never toggles.

3. Final Answer:

   Counter never reaches 4, so clk_out never toggles -> Option A

4. Quick Check:

   Counter max 3 but checks 4 = Condition never met [OK]

Hint: Check counter max matches comparison value [OK]

Common Mistakes:

• Ignoring counter range limits
• Assuming counter can reach out-of-range values
• Thinking clk_out toggles without counter reset

5. You want to design a prescaler in VHDL that divides a 50 MHz clock down to 1 kHz. Which counter range is closest to the correct choice?
Use the formula: division factor = 50,000,000 / 1,000 = 50,000.
Remember the output clock toggles every time the counter reaches max and resets.

hard

A. signal counter : integer range 0 to 99999 := 0;

B. signal counter : integer range 0 to 49999 := 0;

C. signal counter : integer range 0 to 24999 := 0;

D. signal counter : integer range 0 to 12499 := 0;

Solution

1. Step 1: Calculate division factor for toggling

   The full division factor is 50,000, but clk_out toggles every time counter reaches max, so counter max = 50,000 / 2 = 25,000.

2. Step 2: Choose closest counter range

   Since counting starts at 0, range should be 0 to 24,999 to count 25,000 cycles.

3. Final Answer:

   signal counter : integer range 0 to 24999 := 0; -> Option C

4. Quick Check:

   Divide by 2 for toggle count = 24999 max [OK]

Hint: Divide total count by 2 for toggle max count [OK]

Common Mistakes:

• Using full division factor as counter max
• Not dividing by 2 for toggle behavior
• Choosing incorrect range limits