Question

Write in verilog, code to implement a 6 bit multiplier. Cascade a 1 bit multiplier to implement this.

Answer 1

The 1 bit multiplier verilog code:

module multiplier_1bit(a,b,p);
input a,b; //1 bit inputs
output wire p; //1 bit output
assign p=a&b; //product assignment
endmodule

code snippet:

6_bit multiplier by cascading 1 bit multiplier:

Verilog code:

module multiplier_6bit(a,b,prod);
input [5:0] a,b; //6 bit input a,b
output wire [11:0] prod; //the product is 12 bit stored in prod
wire [5:0]p0,p1,p2,p3,p4,p5; //which stores the 1 bit multiplier outputs
wire [9:0] cy; //stores the carry value while adding 1 bit products
//The below 1 bit module instantiation stores p0=B*A0
multiplier_1bit m1(a[0],b[0],p0[0]);
multiplier_1bit m2(a[0],b[1],p0[1]);
multiplier_1bit m3(a[0],b[2],p0[2]);
multiplier_1bit m4(a[0],b[3],p0[3]);
multiplier_1bit m5(a[0],b[4],p0[4]);
multiplier_1bit m6(a[0],b[5],p0[5]);
//The below 1 bit module instantiation stores p1=B*A1
multiplier_1bit mm1(a[1],b[0],p1[0]);
multiplier_1bit mm2(a[1],b[1],p1[1]);
multiplier_1bit mm3(a[1],b[2],p1[2]);
multiplier_1bit mm4(a[1],b[3],p1[3]);
multiplier_1bit mm5(a[1],b[4],p1[4]);
multiplier_1bit mm6(a[1],b[5],p1[5]);
//The below 1 bit module instantiation stores p2=B*A2
multiplier_1bit mmm1(a[2],b[0],p2[0]);
multiplier_1bit mmm2(a[2],b[1],p2[1]);
multiplier_1bit mmm3(a[2],b[2],p2[2]);
multiplier_1bit mmm4(a[2],b[3],p2[3]);
multiplier_1bit mmm5(a[2],b[4],p2[4]);
multiplier_1bit mmm6(a[2],b[5],p2[5]);
//The below 1 bit module instantiation stores p3=B*A3
multiplier_1bit mmm11(a[3],b[0],p3[0]);
multiplier_1bit mmm22(a[3],b[1],p3[1]);
multiplier_1bit mmm33(a[3],b[2],p3[2]);
multiplier_1bit mmm44(a[3],b[3],p3[3]);
multiplier_1bit mmm55(a[3],b[4],p3[4]);
multiplier_1bit mmm66(a[3],b[5],p3[5]);
//The below 1 bit module instantiation stores p4=B*A4
multiplier_1bit mm11(a[4],b[0],p4[0]);
multiplier_1bit mm22(a[4],b[1],p4[1]);
multiplier_1bit mm33(a[4],b[2],p4[2]);
multiplier_1bit mm44(a[4],b[3],p4[3]);
multiplier_1bit mm55(a[4],b[4],p4[4]);
multiplier_1bit mm66(a[4],b[5],p4[5]);
//The below 1 bit module instantiation stores p5=B*A5
multiplier_1bit m11(a[5],b[0],p5[0]);
multiplier_1bit m22(a[5],b[1],p5[1]);
multiplier_1bit m33(a[5],b[2],p5[2]);
multiplier_1bit m44(a[5],b[3],p5[3]);
multiplier_1bit m55(a[5],b[4],p5[4]);
multiplier_1bit m66(a[5],b[5],p5[5]);
//The below assignment operations used to assign the final product.
assign prod[0]=p0[0];
assign {cy[0],prod[1]}=p0[1]+p1[0];
assign {cy[1],prod[2]}=p0[2]+p1[1]+p2[0]+cy[0];
assign {cy[2],prod[3]}=p0[3]+p1[2]+p2[1]+p3[0]+cy[1];
assign {cy[3],prod[4]}=p0[4]+p1[3]+p2[2]+p3[1]+p4[0]+cy[2];
assign {cy[4],prod[5]}=p0[5]+p1[4]+p2[3]+p3[2]+p4[1]+p5[0]+cy[3];
assign {cy[5],prod[6]}=p1[5]+p2[4]+p3[3]+p4[2]+p5[1]+cy[4];
assign {cy[6],prod[7]}=p2[5]+p3[4]+p4[3]+p5[2]+cy[5];
assign {cy[7],prod[8]}=p3[5]+p4[4]+p5[3]+cy[6];
assign {cy[8],prod[9]}=p4[5]+p5[4]+cy[7];
assign {cy[9],prod[10]}=p5[5]+cy[8];
assign prod[11]=cy[9];
endmodule

Code snippets:

Simulation results:

Here i just multiplied the each single bits of A with B.This is like

A0*B=(B5A0)(B4A0)(B3A0)(B2A0)(B1A0)(B0A0)

Simillarly for A1*B;A2*B;A3*B;A4*B;A5*B

By using this i assigned the product values.Please see the below figure for better understanding.By using this picture i assigned the sum values to the product.

For prod[1]=p0[1]+p1[0];

prod[2]=p0[2]+p1[1]+p2[0]+cy[0];

simillary for remaining.

Here i used vivado tool for simulation of the code.

In simulation results I convert a,b,prod radix to unsigned decimal.

Please upvote this..