Clock divider design in Verilog - Step-by-Step Execution

Concept Flow - Clock divider design

Input Clock

↓

Counter increments

↓

Check if counter reached N

↓

Toggle output

↓

Reset counter

↩Back to Counter increments

The input clock drives a counter that increments each cycle. When the counter reaches a set value, the output clock toggles and the counter resets, effectively dividing the clock frequency.

Execution Sample

Verilog

module clk_divider(
  input clk_in,
  output reg clk_out
);
  reg [2:0] count = 0;
  always @(posedge clk_in) begin
    if (count == 3) begin
      clk_out <= ~clk_out;
      count <= 0;
    end else begin
      count <= count + 1;
    end
  end
endmodule

This code divides the input clock by 8 by toggling the output clock every 4 input clock cycles.

Execution Table

Step	clk_in posedge	count before	Condition (count==3)	Action	count after	clk_out after	Output clock toggle
1	Yes	0	No	count = count + 1	1	0	No
2	Yes	1	No	count = count + 1	2	0	No
3	Yes	2	No	count = count + 1	3	0	No
4	Yes	3	Yes	Toggle clk_out, count=0	0	1	Yes
5	Yes	0	No	count = count + 1	1	1	No
6	Yes	1	No	count = count + 1	2	1	No
7	Yes	2	No	count = count + 1	3	1	No
8	Yes	3	Yes	Toggle clk_out, count=0	0	0	Yes
Exit	No more posedges	-	-	-	-	-	Simulation ends

💡 No more positive edges on clk_in, simulation ends

Variable Tracker

Variable	Start	After 1	After 2	After 3	After 4	After 5	After 6	After 7	After 8	Final
count	0	1	2	3	0	1	2	3	0	0
clk_out	0	0	0	0	1	1	1	1	0	0

Key Moments - 3 Insights

Why does clk_out toggle only when count reaches 3?

Because the condition 'count == 3' triggers the toggle and counter reset, as shown in execution_table rows 4 and 8.

Why does count reset to 0 after toggling clk_out?

Resetting count to 0 restarts the counting cycle to divide the clock correctly, as seen in the 'Action' column in rows 4 and 8.

Why does clk_out start at 0 and toggle to 1 at step 4?

clk_out is initialized to 0 and toggles (inverts) when count reaches 3, changing from 0 to 1 at step 4.

Visual Quiz - 3 Questions

Look at the execution_table at step 4, what is the value of clk_out after the action?

A 1

B 0

C 3

D Undefined

💡 Hint

Check the 'clk_out after' column at step 4 in the execution_table.

At which step does the count reset to 0 for the first time?

A Step 3

B Step 1

C Step 4

D Step 8

💡 Hint

Look at the 'count after' column in execution_table rows to find when count becomes 0.

If we change the condition to 'count == 1', how would the output clock toggle frequency change?

A It would toggle half as fast

B It would toggle twice as fast

C It would not toggle at all

D It would toggle at the same rate

💡 Hint

Consider how often the toggle happens based on the count value in the execution_table.

Concept Snapshot

Clock divider uses a counter to reduce input clock frequency.
On each input clock rising edge, counter increments.
When counter reaches preset value N, output clock toggles and counter resets.
Output clock frequency = input frequency / (2*(N+1)).
This design is synchronous and simple to implement.

Full Transcript

This visual execution traces a Verilog clock divider module. The input clock triggers a counter that increments on each rising edge. When the counter reaches 3, the output clock toggles and the counter resets to zero. This process repeats, effectively dividing the input clock frequency by 8. The execution table shows each step with variable values and actions. Key moments clarify why toggling and resetting happen at count 3. The quiz tests understanding of variable states and frequency changes. The snapshot summarizes the clock divider logic and formula.

Practice

1.

What is the main purpose of a clock divider in Verilog?

easy

A. To reset the clock signal

B. To increase the clock frequency

C. To slow down a fast clock signal by dividing its frequency

D. To generate random clock pulses

Solution

1. Step 1: Understand clock divider function

   A clock divider reduces the frequency of an input clock signal by toggling the output clock after a set number of input cycles.

2. Step 2: Identify correct purpose

   Among the options, only slowing down the clock frequency matches the clock divider's role.

3. Final Answer:

   To slow down a fast clock signal by dividing its frequency -> Option C

4. Quick Check:

   Clock divider = slows clock [OK]

Hint: Clock divider reduces frequency, not increases it [OK]

Common Mistakes:

• Confusing clock divider with clock multiplier
• Thinking clock divider resets the clock
• Assuming it generates random pulses

2.

Which of the following is the correct way to declare a 4-bit counter register in Verilog?

easy

A. reg [3:0] counter;

B. wire [4:0] counter;

C. reg counter[3:0];

D. integer counter;

Solution

1. Step 1: Understand register declaration

   In Verilog, a 4-bit register is declared as reg [3:0] name; where 3:0 means 4 bits from 3 down to 0.

2. Step 2: Check options

   reg [3:0] counter; uses reg and correct bit range. wire [4:0] counter; uses wire and wrong bit width. reg counter[3:0]; syntax is incorrect for vector registers. integer counter; declares integer, not a 4-bit register.

3. Final Answer:

   reg [3:0] counter; -> Option A

4. Quick Check:

   4-bit reg = reg [3:0] [OK]

Hint: Use reg [3:0] for 4-bit registers in Verilog [OK]

Common Mistakes:

• Using wire instead of reg for counters
• Incorrect bit range syntax
• Confusing integer with fixed bit registers

3.

Consider this Verilog snippet for a clock divider:

reg [2:0] count = 0;
reg clk_out = 0;
always @(posedge clk_in or posedge reset) begin
  if (reset) begin
    count <= 0;
    clk_out <= 0;
  end else if (count == 3) begin
    count <= 0;
    clk_out <= ~clk_out;
  end else begin
    count <= count + 1;
  end
end

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

medium

A. 4

B. 6

C. 2

D. 8

Solution

1. Step 1: Analyze counter behavior

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

2. Step 2: Calculate frequency division

   clk_out toggles every 4 cycles, so one full period of clk_out is 8 input cycles (4 cycles high, 4 cycles low). Thus, frequency is divided by 8.

3. Final Answer:

   8 -> Option D

4. Quick Check:

   Toggle every 4 cycles -> divide by 8 [OK]

Hint: Frequency division = 2 x (counter max + 1) [OK]

Common Mistakes:

• Counting only toggle cycles, not full period
• Misreading counter max value
• Ignoring clk_out toggling logic

4.

Identify the error in this clock divider code snippet:

reg [3:0] count;
reg clk_out;
always @(posedge clk_in) begin
  if (count == 7) begin
    count = 0;
    clk_out = ~clk_out;
  end else begin
    count = count + 1;
  end
end

medium

A. Missing reset condition for count and clk_out

B. Using blocking assignments (=) instead of non-blocking (<=) inside always block

C. Counter max value should be 15, not 7

D. clk_out should not toggle inside always block

Solution

1. Step 1: Check assignment types in sequential logic

   Inside an always block triggered by clock edge, non-blocking assignments (<=) should be used to avoid race conditions.

2. Step 2: Identify error in code

   The code uses blocking assignments (=) which can cause incorrect behavior in sequential logic.

3. Final Answer:

   Using blocking assignments (=) instead of non-blocking (<=) inside always block -> Option B

4. Quick Check:

   Sequential always block needs <= [OK]

Hint: Use <= for registers inside clocked always blocks [OK]

Common Mistakes:

• Using = instead of <= in clocked always blocks
• Forgetting reset logic (but not main error here)
• Incorrect max count value (not error here)

5.

You want to design a clock divider that divides the input clock frequency by 10 using a 4-bit counter. Which approach correctly achieves this?

always @(posedge clk_in or posedge reset) begin
  if (reset) begin
    count <= 0;
    clk_out <= 0;
  end else if (count == 4) begin
    count <= 0;
    clk_out <= ~clk_out;
  end else begin
    count <= count + 1;
  end
end

Choose the correct explanation.

hard

A. This code divides frequency by 10 because clk_out toggles every 5 cycles, full period 10

B. This code divides frequency by 5 because clk_out toggles every 5 cycles, full period 5

C. This code divides frequency by 8 because count max is 4, toggling twice per 8 cycles

D. This code divides frequency by 4 because count resets at 4

Solution

1. Step 1: Understand toggle and count max

   The counter counts from 0 to 4 (5 cycles). clk_out toggles after 5 input cycles.

2. Step 2: Calculate frequency division

   Since clk_out toggles every 5 cycles, one full period is 10 cycles (5 high + 5 low), so frequency divides by 10.

3. Final Answer:

   This code divides frequency by 10 because clk_out toggles every 5 cycles, full period 10 -> Option A

4. Quick Check:

   Toggle every 5 cycles -> divide by 10 [OK]

Hint: Frequency division = 2 x (count max + 1) [OK]

Common Mistakes:

• Confusing toggle count with full period
• Assuming count max is frequency division directly
• Ignoring clk_out toggling effect