ROM (Read-Only Memory) in Verilog - Step-by-Step Execution

Concept Flow - ROM (Read-Only Memory) in Verilog

Start

↓

Address Input

↓

ROM Lookup

↓

Output Data

↓

End

The ROM takes an address input, looks up the stored data at that address, and outputs the data without changing it.

Execution Sample

Verilog

module rom_example(
  input [1:0] addr,
  output reg [3:0] data
);
  always @(*) begin
    case(addr)
      2'b00: data = 4'hA;
      2'b01: data = 4'h3;
      2'b10: data = 4'hF;
      2'b11: data = 4'h0;
      default: data = 4'h0;
    endcase
  end
endmodule

This ROM module outputs a 4-bit value based on a 2-bit address input using a case statement.

Execution Table

Step	Input addr	Condition checked	Data assigned	Output data
1	2'b00	addr == 2'b00	4'hA	1010
2	2'b01	addr == 2'b01	4'h3	0011
3	2'b10	addr == 2'b10	4'hF	1111
4	2'b11	addr == 2'b11	4'h0	0000
5	2'b00	addr == 2'b00	4'hA	1010

💡 ROM outputs data based on address; no state changes or loops, so execution ends after output.

Variable Tracker

Variable	Start	Step 1	Step 2	Step 3	Step 4	Step 5
addr	undefined	2'b00	2'b01	2'b10	2'b11	2'b00
data	undefined	4'hA	4'h3	4'hF	4'h0	4'hA

Key Moments - 2 Insights

Why does the ROM output change immediately when the address changes?

Because the ROM uses a combinational always block (@(*)). It updates output 'data' instantly when 'addr' changes, as shown in execution_table steps 1 to 5.

Is the ROM storing data that can be changed during simulation?

No. The ROM data is fixed in the case statement and does not change during simulation, which is why 'data' only depends on 'addr' input.

Visual Quiz - 3 Questions

Look at the execution_table, what is the output data when addr is 2'b10?

A 4'h3

B 4'h0

C 4'hF

D 4'hA

💡 Hint

Check row 3 in execution_table where addr is 2'b10 and data assigned is 4'hF.

At which step does the ROM output data 4'h0?

A Step 2

B Step 4

C Step 3

D Step 5

💡 Hint

Look at execution_table row 4 where data assigned is 4'h0.

If the address input changes to 2'b01, what will be the output data?

A 4'h3

B 4'hA

C 4'hF

D 4'h0

💡 Hint

Refer to variable_tracker and execution_table rows where addr is 2'b01.

Concept Snapshot

ROM in Verilog stores fixed data accessed by address.
Use combinational always block with case statement.
Input: address; Output: stored data.
Data does not change during simulation.
Outputs update immediately on address change.

Full Transcript

This visual trace shows how a ROM module in Verilog works. The ROM takes a 2-bit address input and outputs a 4-bit data value. The data is fixed and defined in a case statement inside a combinational always block. As the address changes, the output data updates immediately without any delay or state change. The execution table shows each address tested and the corresponding output data. The variable tracker records the address and data values at each step. Key moments clarify why the output changes instantly and why ROM data is read-only. The quiz questions help check understanding of output values for given addresses. This trace helps beginners see how ROM lookup works step-by-step in Verilog.

Practice

1. What is the primary purpose of a ROM (Read-Only Memory) in Verilog?

easy

A. To store fixed data that cannot be changed during simulation or operation

B. To store data that can be written and read dynamically

C. To perform arithmetic operations

D. To generate clock signals

Solution

1. Step 1: Understand ROM functionality

   ROM is designed to hold data that is fixed and cannot be modified during operation.

2. Step 2: Compare with other memory types

   Unlike RAM, ROM data is read-only and does not support writing during runtime.

3. Final Answer:

   To store fixed data that cannot be changed during simulation or operation -> Option A

4. Quick Check:

   ROM = fixed data storage [OK]

Hint: ROM stores fixed data only, no writes allowed [OK]

Common Mistakes:

• Confusing ROM with RAM (read-write memory)
• Thinking ROM can perform operations
• Assuming ROM generates signals

2. Which of the following is the correct way to declare a ROM using an array and initial block in Verilog?

easy

A. integer rom_mem; always @(posedge clk) rom_mem = 8'hAA;

B. wire [7:0] rom_mem [0:3]; always @(posedge clk) rom_mem[0] <= 8'hAA;

C. reg [7:0] rom_mem; initial rom_mem = 8'hAA;

D. reg [7:0] rom_mem [0:3]; initial begin rom_mem[0] = 8'hAA; rom_mem[1] = 8'hBB; rom_mem[2] = 8'hCC; rom_mem[3] = 8'hDD; end

Solution

1. Step 1: Identify correct ROM declaration

   ROM is declared as a reg array with an initial block to assign fixed values.

2. Step 2: Check each option

   reg [7:0] rom_mem [0:3]; initial begin rom_mem[0] = 8'hAA; rom_mem[1] = 8'hBB; rom_mem[2] = 8'hCC; rom_mem[3] = 8'hDD; end uses reg array and initial block correctly. wire [7:0] rom_mem [0:3]; always @(posedge clk) rom_mem[0] <= 8'hAA; uses wire and always block incorrectly. Options C and D do not declare arrays properly.

3. Final Answer:

   reg [7:0] rom_mem [0:3]; initial begin rom_mem[0] = 8'hAA; rom_mem[1] = 8'hBB; rom_mem[2] = 8'hCC; rom_mem[3] = 8'hDD; end -> Option D

4. Quick Check:

   ROM array + initial block = reg [7:0] rom_mem [0:3]; initial begin rom_mem[0] = 8'hAA; rom_mem[1] = 8'hBB; rom_mem[2] = 8'hCC; rom_mem[3] = 8'hDD; end [OK]

Hint: Use reg array with initial block for ROM data [OK]

Common Mistakes:

• Using wire instead of reg for ROM array
• Assigning inside always block for ROM
• Not using an array for multiple ROM entries

3. Given the following Verilog ROM code, what is the output value of data_out when addr = 2'b10?

module rom_example(input [1:0] addr, output reg [7:0] data_out);
  always @(*) begin
    case(addr)
      2'b00: data_out = 8'h1A;
      2'b01: data_out = 8'h2B;
      2'b10: data_out = 8'h3C;
      2'b11: data_out = 8'h4D;
      default: data_out = 8'h00;
    endcase
  end
endmodule

medium

A. 8'h4D

B. 8'h1A

C. 8'h3C

D. 8'h00

Solution

1. Step 1: Locate the case for addr = 2'b10

   The case statement assigns data_out = 8'h3C when addr is 2'b10.

2. Step 2: Confirm output value

   Since addr matches 2'b10, data_out will be 8'h3C.

3. Final Answer:

   8'h3C -> Option C

4. Quick Check:

   addr 2'b10 -> data_out 8'h3C [OK]

Hint: Match addr in case to find data_out quickly [OK]

Common Mistakes:

• Choosing wrong case label
• Confusing hex values
• Ignoring default case

4. Identify the error in this ROM Verilog code snippet:

module rom_bug(input [1:0] addr, output reg [7:0] data_out);
  reg [7:0] rom_mem [0:3];
  initial begin
    rom_mem[0] = 8'hAA;
    rom_mem[1] = 8'hBB;
    rom_mem[2] = 8'hCC;
    rom_mem[3] = 8'hDD;
  end
  always @(*) begin
    data_out = rom_mem[addr];
  end
endmodule

medium

A. ROM array should be declared as wire, not reg

B. No error; code correctly implements ROM

C. Initial block cannot be used for ROM initialization

D. Using addr as index without conversion causes error

Solution

1. Step 1: Check ROM declaration and initialization

   ROM is declared as reg array and initialized in initial block, which is correct for simulation.

2. Step 2: Verify data output assignment

   Using addr as index to read rom_mem is valid and data_out is assigned in combinational always block.

3. Final Answer:

   No error; code correctly implements ROM -> Option B

4. Quick Check:

   ROM reg array + initial + combinational read = correct [OK]

Hint: ROM uses reg array + initial block + combinational read [OK]

Common Mistakes:

• Thinking initial block is illegal for ROM
• Confusing reg and wire for ROM arrays
• Assuming addr needs conversion

5. You want to implement a 4-word ROM in Verilog that outputs the square of the address value (0, 1, 4, 9). Which of the following code snippets correctly implements this ROM?

hard

A. always @(addr) begin case(addr) 2'b00: data_out=8'd0; 2'b01: data_out=8'd1; 2'b10: data_out=8'd4; 2'b11: data_out=8'd9; default: data_out=8'd0; endcase end

B. reg [7:0] rom [0:3]; initial begin rom[0]=8'd0; rom[1]=8'd1; rom[2]=8'd4; rom[3]=8'd9; end always data_out = rom[addr];

C. reg [7:0] rom [0:3]; assign rom[0]=0; assign rom[1]=1; assign rom[2]=4; assign rom[3]=9; assign data_out=rom[addr];

D. reg [7:0] rom [0:3]; always @(posedge clk) begin rom[0]<=0; rom[1]<=1; rom[2]<=4; rom[3]<=9; end assign data_out=rom[addr];

Solution

1. Step 1: Analyze each code snippet for ROM implementation

   reg [7:0] rom [0:3]; initial begin rom[0]=8'd0; rom[1]=8'd1; rom[2]=8'd4; rom[3]=8'd9; end always @(*) data_out = rom[addr]; lacks module and always block structure; data_out assignment is incomplete. always @(addr) begin case(addr) 2'b00: data_out=8'd0; 2'b01: data_out=8'd1; 2'b10: data_out=8'd4; 2'b11: data_out=8'd9; default: data_out=8'd0; endcase end uses a case statement to output fixed squares based on addr, which is a valid ROM implementation. reg [7:0] rom [0:3]; assign rom[0]=0; assign rom[1]=1; assign rom[2]=4; assign rom[3]=9; assign data_out=rom[addr]; tries to assign values to wire array elements, which is illegal. reg [7:0] rom [0:3]; always @(posedge clk) begin rom[0]<=0; rom[1]<=1; rom[2]<=4; rom[3]<=9; end assign data_out=rom[addr]; tries to assign ROM values inside a clocked always block, which is not suitable for ROM initialization.

2. Step 2: Confirm correct ROM behavior

   always @(addr) begin case(addr) 2'b00: data_out=8'd0; 2'b01: data_out=8'd1; 2'b10: data_out=8'd4; 2'b11: data_out=8'd9; default: data_out=8'd0; endcase end correctly outputs squares for each address using a combinational case statement, matching ROM behavior.

3. Final Answer:

   always @(addr) begin case(addr) 2'b00: data_out=8'd0; 2'b01: data_out=8'd1; 2'b10: data_out=8'd4; 2'b11: data_out=8'd9; default: data_out=8'd0; endcase end -> Option A

4. Quick Check:

   ROM as combinational case statement = always @(addr) begin case(addr) 2'b00: data_out=8'd0; 2'b01: data_out=8'd1; 2'b10: data_out=8'd4; 2'b11: data_out=8'd9; default: data_out=8'd0; endcase end [OK]

Hint: Use case statement for fixed ROM outputs [OK]

Common Mistakes:

• Assigning to wire arrays
• Initializing ROM inside clocked always block
• Not using combinational logic for ROM output