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Physical Design/PDK methodology Engineer

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Restoring And Non Restoring Division Algorithms Using Verilog HDL

 Hello Dear Readers,

Today In this post I have implemented Restoring Division Algorithm Using Verilog HDL. So I have followed this One video from Tutorial Point Youtube Channel So go through it before Verilog code.

start means the start of the division; busy indicates that the divider is busy (cannot start a new division); ready indicates that the quotient and remainder are available, and the count is the output of a counter that is used to control the iterations of the division.

Verilog Code(Restoring):

module divider_32(a,b,start,clk,reset,q,r,busy,ready,count);
input [31:0] a; // dividend
input [15:0] b; // divisor
input start; // start
input clk,reset; // clk,reset
output [31:0] q; // quotient
output [15:0] r; // remainder
output reg busy; // busy
output reg ready; // ready
output [4:0] count; // counter
reg [31:0] reg_q;
reg [15:0] reg_r;
reg [15:0] reg_b;
reg [4:0] count;
wire [16:0] sub_out = {reg_r,reg_q[31]} - {1'b0,reg_b}; // sub
wire [15:0] mux_out = sub_out[16]? // restoring logic
{reg_r[14:0],reg_q[31]} : sub_out[15:0]; // or not
assign q = reg_q;
assign r = reg_r;
always @ (posedge clk or negedge reset) begin
if (!reset) begin
busy <= 0;
ready <= 0;
end else begin
if (start) begin
reg_q <= a; // load a
reg_b <= b; // load b
reg_r <= 0;
busy <= 1;
ready <= 0;
count <= 31;
end else if (busy) begin
reg_q <= {reg_q[30:0],~sub_out[16]}; // << 1
reg_r <= mux_out;
count <= count - 5'b1; // counter++
  if (count == 5'h0) begin // finished
    busy <= 0;
    ready <= 1; // q,r ready
      end
    end
  end
end
endmodule

Verilog Code(Non Restoring):

module nonrestore_32(a,b,start,clk,reset,q,r,busy,ready,count);
   input [31:0] a; // dividend
   input [15:0] b; // divisor
input start; // start
   input clk,reset; // clk,reset
   output [31:0] q; // quotient
   output [15:0] r; // remainder
   output reg busy; // busy
   output reg ready; // ready
   output [4:0] count; // count
   reg [31:0] reg_q;
   reg [15:0] reg_r;
   reg [15:0] reg_b;
   reg [4:0] count;
wire [16:0] sub_add = reg_r[15]?
{reg_r,reg_q[31]} + {1'b0,reg_b} : // + b
{reg_r,reg_q[31]} - {1'b0,reg_b}; // - b
   assign q = reg_q;
   assign r = reg_r[15]? reg_r + reg_b : reg_r; // adjust r
always @ (posedge clk or negedge reset) begin
       if (!reset) begin
           busy <= 0;
           ready <= 0;
        end 
        else begin
           if (start) begin
              reg_q <= a; // load a
              reg_b <= b; // load b
              reg_r <= 0;
              busy <= 1;
              ready <= 0;
              count <= 0;
        end else if (busy) begin
              reg_q <= {reg_q[30:0],~sub_add[16]}; // << 1
              reg_r <= sub_add[15:0];
              count <= count + 5'b1; // count++
         if (count == 5'h1f) begin // finish
               busy <= 0;
               ready <= 1; // q,r ready
           end
       end
    end
end
endmodule


Testbench:

module tb();
   reg [31:0] a; // dividend
reg [15:0] b; // divisor
reg start; // start
reg clk,reset; // clk,reset
wire [31:0] q; // quotient
wire [15:0] r; // remainder
wire busy; // busy
wire ready; // ready
wire [4:0] count; // counter

//Device Under Test(DUT)
 divider_32 test(a,b,start,clk,reset,q,r,busy,ready,count);
 always
  #2 clk=~clk; 
  
initial begin
   reset=0;
clk=0;
start=0;
a=32'h4c7f228a;
b=32'h6a0e;
#2
reset=1;
start=1;
#1
start=0;

end

endmodule

Timing Parameters:

  Minimum period: 5.325ns (Maximum Frequency: 187.786MHz)
 Minimum input arrival time before clock: 4.389ns
 Maximum output required time after clock: 4.252ns

Simulational Results:







Comments

  1. Thanks for your effort for giving test bench also and you are only one who provide complete expected code.

    ReplyDelete
  2. Great post ☺️

    ReplyDelete
  3. Wow Non restore also added

    ReplyDelete

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