Up counter design in Verilog - Mini Project: Build & Apply

Up Counter Design in Verilog

📖 Scenario: You are designing a simple digital up counter for a small device. This counter will count from 0 upwards each time a clock pulse arrives.

🎯 Goal: Build a Verilog module that counts up by 1 on each clock pulse and resets to 0 when a reset signal is active.

📋 What You'll Learn

Create a 4-bit register to hold the count

Use a clock and reset input

Increment the count on the rising edge of the clock

Reset the count to 0 when reset is high

Display the count value

💡 Why This Matters

🌍 Real World

Up counters are used in digital clocks, timers, and event counters in many electronic devices.

💼 Career

Understanding counters is fundamental for hardware design engineers working with digital circuits and FPGA programming.

1

Create the module and declare inputs and outputs

Write a Verilog module named up_counter with inputs clk and reset, and a 4-bit output count. Declare count as a reg type.

Verilog

module up_counter(input clk, input reset, output reg [3:0] count);
// Your code here
endmodule

Hint

Start by defining the module and its ports exactly as described.

2

Initialize the count register

Inside the up_counter module, declare an always block triggered on the rising edge of clk or when reset is high. Inside it, set count to 0 if reset is high.

Verilog

module up_counter(input clk, input reset, output reg [3:0] count);

always @(posedge clk or posedge reset) begin
    // Your code here
end

endmodule

Hint

Use a non-blocking assignment to set count to zero when reset is high.

3

Add the counting logic

In the same always block, add an else clause that increments count by 1 on each rising clock edge.

Verilog

module up_counter(input clk, input reset, output reg [3:0] count);

always @(posedge clk or posedge reset) begin
    if (reset) begin
        count <= 4'b0000;
    end else begin
        // Your code here
    end
end

endmodule

Hint

Increment count by 1 using a non-blocking assignment inside the else block.

4

Test and display the count value

Add a testbench module named testbench that instantiates up_counter. Generate a clock signal that toggles every 5 time units. Apply a reset pulse for 10 time units at the start. Run the simulation for 50 time units and use $display to print the count value at each positive clock edge.

Verilog

module up_counter(input clk, input reset, output reg [3:0] count);

always @(posedge clk or posedge reset) begin
    if (reset) begin
        count <= 4'b0000;
    end else begin
        count <= count + 1;
    end
end

endmodule

// Your testbench code here

Hint

Create a testbench that toggles the clock every 5 time units, applies reset for 10 time units, and prints the count on each clock rising edge.

Practice

1. What does an up counter do in Verilog?

easy

A. It holds the value constant regardless of clock.

B. It decreases its value by 1 on each clock pulse.

C. It resets the value to zero on each clock pulse.

D. It increases its value by 1 on each clock pulse.

Solution

1. Step 1: Understand the function of an up counter

   An up counter increments its stored value by 1 every time the clock signal pulses.

2. Step 2: Compare options with the definition

   Only It increases its value by 1 on each clock pulse. describes increasing the value by 1 on each clock pulse, which matches the up counter behavior.

3. Final Answer:

   It increases its value by 1 on each clock pulse. -> Option D

4. Quick Check:

   Up counter = increments by 1 [OK]

Hint: Up counter always counts up by 1 each clock pulse [OK]

Common Mistakes:

• Confusing up counter with down counter
• Thinking reset happens every clock pulse
• Assuming counter holds value without change

2. Which of the following is the correct Verilog syntax to define a 4-bit up counter with synchronous reset?

easy

A. always @(negedge clk) if (reset) count = 0; else count = count + 1;

B. always @(posedge clk) if (reset) count <= 0; else count <= count + 1;

C. always @(posedge clk) if (reset) count = 0; else count = count - 1;

D. always @(posedge clk) if (reset) count <= 1; else count <= count + 1;

Solution

1. Step 1: Identify correct clock edge and reset behavior

   Up counters usually update on the positive edge of the clock and reset synchronously to zero.

2. Step 2: Check assignment operators and counting direction

   Non-blocking assignment '<=' is preferred in sequential logic. Counting up means adding 1, and reset sets count to 0.

3. Final Answer:

   always @(posedge clk) if (reset) count <= 0; else count <= count + 1; -> Option B

4. Quick Check:

   posedge + non-blocking + reset to 0 + count up = A [OK]

Hint: Use posedge clk and non-blocking <= for counters [OK]

Common Mistakes:

• Using negedge instead of posedge
• Using blocking '=' in sequential logic
• Resetting count to 1 instead of 0
• Counting down instead of up

3. What will be the output sequence of a 3-bit up counter starting from 0 after 5 clock pulses (assuming reset is inactive)?

module counter(input clk, input reset, output reg [2:0] count);
  always @(posedge clk) begin
    if (reset) count <= 0;
    else count <= count + 1;
  end
endmodule

medium

A. [0, 1, 2, 3, 4]

B. [1, 2, 3, 4, 5]

C. [1, 2, 3, 4, 0]

D. [0, 1, 2, 3, 4, 5]

Solution

1. Step 1: Understand initial value and counting

   The counter starts at 0 after reset. On first clock pulse, count increments to 1.

2. Step 2: List values after each clock pulse

   After 1st pulse: 1, 2nd: 2, 3rd: 3, 4th: 4, 5th: 5.

3. Step 3: Consider 3-bit counter wrap-around

   Since it's a 3-bit counter, it counts from 0 to 7 and then wraps around. After 5 pulses starting from 0, the sequence is [1, 2, 3, 4, 5]. However, the question asks for the output sequence starting from 0 after 5 pulses, which includes the initial 0 and the next 4 increments: [0, 1, 2, 3, 4].

4. Final Answer:

   [0, 1, 2, 3, 4] -> Option A

5. Quick Check:

   Count starts at 0, increments each pulse = [0..4] [OK]

Hint: Count increments after clock pulse, starts from 0 [OK]

Common Mistakes:

• Including initial 0 before first pulse
• Counting 6 values instead of 5
• Confusing count start value

4. Identify the error in this Verilog code for a 4-bit up counter:

module counter(input clk, input reset, output reg [3:0] count);
  always @(posedge clk) begin
    if (reset == 1) count = 0;
    else count = count + 1;
  end
endmodule

medium

A. Output 'count' should not be declared as reg.

B. Reset condition should be 'if (reset != 1)'.

C. Using blocking assignments '=' instead of non-blocking '<=' inside always block.

D. Clock edge sensitivity should be negedge instead of posedge.

Solution

1. Step 1: Check assignment type in sequential always block

   Inside always @(posedge clk), non-blocking assignments '<=' should be used to avoid race conditions.

2. Step 2: Verify other parts of the code

   Reset condition and reg declaration are correct. Clock edge is correctly posedge.

3. Final Answer:

   Using blocking assignments '=' instead of non-blocking '<=' inside always block. -> Option C

4. Quick Check:

   Sequential logic uses '<=' not '=' [OK]

Hint: Use '<=' for sequential logic assignments [OK]

Common Mistakes:

• Using '=' in always @(posedge clk)
• Misunderstanding reset condition
• Thinking reg is wrong for output
• Changing clock edge unnecessarily

5. How can you modify a 4-bit up counter to count only when an enable signal is high?

module counter(input clk, input reset, output reg [3:0] count);
  always @(posedge clk) begin
    if (reset) count <= 0;
    else count <= count + 1;
  end
endmodule

hard

A. Add 'if (enable) count <= count + 1;' inside the always block after reset check.

B. Add 'if (enable) count = count + 1;' outside the always block.

C. Replace 'count <= count + 1;' with 'count <= count - 1;' when enable is high.

D. Add 'enable' to the sensitivity list of the always block.

Solution

1. Step 1: Understand how to control counting with enable

   Counting should happen only when enable is high, so add a condition inside the always block after reset.

2. Step 2: Implement enable check correctly

   Use 'if (enable) count <= count + 1;' with non-blocking assignment inside the always block.

3. Final Answer:

   Add 'if (enable) count <= count + 1;' inside the always block after reset check. -> Option A

4. Quick Check:

   Enable controls counting inside always block [OK]

Hint: Check enable inside always block after reset [OK]

Common Mistakes:

• Using blocking assignment outside always block
• Changing count direction instead of gating
• Adding enable to sensitivity list (not needed)