Up-down counter with direction control in Verilog - Mini Project: Build & Apply

Up-down counter with direction control

📖 Scenario: You are designing a simple digital circuit that counts numbers up or down based on a control signal. This is useful in many electronics projects like timers, clocks, or simple user interfaces.

🎯 Goal: Create a Verilog module for an up-down counter that changes its count direction based on a control input signal.

📋 What You'll Learn

Create a 4-bit register to hold the count value

Use a clock input to update the count on each rising edge

Use a direction input signal to decide whether to count up or down

Reset the count to zero when reset signal is active

💡 Why This Matters

🌍 Real World

Up-down counters are used in digital clocks, volume controls, and event counters where counting direction changes dynamically.

💼 Career

Understanding counters and direction control 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_down_counter with inputs clk, reset, and dir, and a 4-bit output count. Declare count as a reg type.

Verilog

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

Hint

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

2

Initialize the count register

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

Verilog

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

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

endmodule

Hint

Use an asynchronous reset inside the always block to set count to zero.

3

Add counting logic based on direction

Inside the always block, add an else branch after the reset condition. Use an if statement to check if dir is 1. If yes, increment count by 1. Otherwise, decrement count by 1.

Verilog

module up_down_counter(input clk, input reset, input dir, 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

Use nested if-else inside the always block to update count based on dir.

4

Test the counter output

Add a initial block to display the value of count every time it changes using $monitor. This will help you see the counter output during simulation.

Verilog

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

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

// Your code here
endmodule

Hint

Use $monitor inside an initial block to print count changes during simulation.

Practice

1. What does the direction signal control in an up-down counter in Verilog?

easy

A. The reset state of the counter

B. The speed of the clock signal

C. Whether the counter counts up or down

D. The width of the counter output

Solution

1. Step 1: Understand the role of the direction signal

   The direction signal determines if the counter increments or decrements its value.

2. Step 2: Relate direction to counting behavior

   If direction is high, the counter counts up; if low, it counts down.

3. Final Answer:

   Whether the counter counts up or down -> Option C

4. Quick Check:

   Direction controls count direction = C [OK]

Hint: Direction signal decides count up or down [OK]

Common Mistakes:

• Confusing direction with clock speed
• Thinking reset controls counting direction
• Assuming direction changes output width

2. Which of the following is the correct Verilog syntax to declare a 4-bit up-down counter with direction input?

easy

A. wire [4:0] count; input direction; always @(negedge clk) begin if(reset) count = 0; else if(direction) count = count + 1; else count = count - 1; end

B. reg [3:0] count; input dir; always @(posedge clk) begin if(reset) count <= 0; else if(dir) count <= count + 1; else count <= count - 1; end

C. reg [3:0] count; input dir; always @(posedge clk) begin if(reset) count = 0; else if(dir) count = count + 1; else count = count - 1; end

D. reg [3:0] count; input dir; always @(posedge clk) begin if(reset) count <= 0; else if(dir) count <= count - 1; else count <= count + 1; end

Solution

1. Step 1: Check variable declarations and types

   Count should be a 4-bit reg, direction input is 'dir'.

2. Step 2: Verify always block and assignments

   Use posedge clk, non-blocking assignments '<=' for sequential logic, reset sets count to 0, dir controls increment or decrement.

3. Final Answer:

   Correct syntax with reg, posedge clk, non-blocking assignments, and direction control -> Option B

4. Quick Check:

   Proper reg and non-blocking assignments = A [OK]

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

Common Mistakes:

• Using wire instead of reg for count
• Using blocking '=' in sequential always block
• Incorrect clock edge sensitivity
• Reversing increment and decrement logic

3. Given this Verilog code snippet for a 3-bit up-down counter with direction control:

reg [2:0] count = 3'b000;
always @(posedge clk) begin
  if(reset) count <= 3'b000;
  else if(dir) count <= count + 1;
  else count <= count - 1;
end

If reset=0, dir=1, and the clock pulses 5 times starting from 0, what is the value of count after 5 pulses?

medium

A. 3'b101

B. 3'b111

C. 3'b100

D. 3'b110

Solution

1. Step 1: Understand initial conditions and direction

   Count starts at 0, reset is inactive, dir=1 means count up.

2. Step 2: Calculate count after 5 increments

   Counting up 5 times from 0: 0->1->2->3->4->5, which is 3'b101 in binary.

3. Final Answer:

   3'b101 -> Option A

4. Quick Check:

   0 + 5 increments = 5 (binary 101) [OK]

Hint: Count up adds 1 each clock pulse [OK]

Common Mistakes:

• Forgetting initial count is zero
• Confusing binary values
• Counting down instead of up

4. Identify the error in this Verilog code for an up-down counter:

reg [3:0] count;
always @(posedge clk) begin
  if(reset) count = 0;
  else if(dir) count <= count + 1;
  else count <= count - 1;
end

medium

A. Using blocking assignment '=' for reset inside always block

B. Missing sensitivity to negedge clk

C. Count should be declared as wire, not reg

D. Direction signal 'dir' is not declared

Solution

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

   Sequential logic requires non-blocking assignments '<=' for all assignments inside always @(posedge clk).

2. Step 2: Identify inconsistent assignment usage

   Reset uses blocking '=' which can cause simulation mismatches; it should use '<=' like other assignments.

3. Final Answer:

   Using blocking assignment '=' for reset inside always block -> Option A

4. Quick Check:

   Use '<=' for all sequential assignments [OK]

Hint: Use non-blocking '<=' in sequential always blocks [OK]

Common Mistakes:

• Mixing blocking and non-blocking assignments
• Ignoring declaration of signals
• Wrong sensitivity list for clock

5. You want to design a 4-bit up-down counter with direction control and synchronous reset in Verilog. The counter should wrap around when it reaches max or min values (i.e., counting up from 15 goes to 0, counting down from 0 goes to 15). Which code snippet correctly implements this behavior?

hard

A. always @(posedge clk) begin
if(reset) count <= 0;
else if(dir) count <= count + 1;
else count <= count - 1;
end

B. always @(posedge clk) begin
if(reset) count <= 0;
else if(dir) count <= (count == 4'b0000) ? 4'b1111 : count - 1;
else count <= (count == 4'b1111) ? 0 : count + 1;
end

C. always @(posedge clk) begin
if(reset) count <= 0;
else if(dir) count <= count - 1;
else count <= count + 1;
end

D. always @(posedge clk) begin
if(reset) count <= 0;
else if(dir) count <= (count == 4'b1111) ? 0 : count + 1;
else count <= (count == 4'b0000) ? 4'b1111 : count - 1;
end

Solution

1. Step 1: Understand wrap-around conditions

   When counting up at max (15), reset to 0; when counting down at 0, reset to 15.

2. Step 2: Check code logic for direction and wrap-around

   always @(posedge clk) begin if(reset) count <= 0; else if(dir) count <= (count == 4'b1111) ? 0 : count + 1; else count <= (count == 4'b0000) ? 4'b1111 : count - 1; end correctly uses conditional operator to wrap count on max and min values based on direction.

3. Final Answer:

   Code with conditional wrap-around for max and min counts -> Option D

4. Quick Check:

   Wrap-around logic matches direction and limits [OK]

Hint: Use conditional operator to wrap count at limits [OK]

Common Mistakes:

• Not wrapping count at max/min values
• Reversing increment and decrement logic
• Using incorrect boundary values