Half Adder HDL Verilog Code
I am supposed to create 4 bit full adder verilog code in vivado.But when I try to test in the simulation.It give me z and x output.Which part of code I have to change to get an output in simulation. How to solve 4 bit full adder verilog. This is the one bit full adder verilog code.
This page of verilog sourcecode covers HDL code for half adder, half substractor, full substractor using verilog. Mst209 mathematical methods and modeling madness 2017.
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The half adder truth table and schematic (fig-1) is mentioned below. The boolean expressions are:
S= A (EXOR) B
C=A.B
S= A (EXOR) B
C=A.B
Half Adder Schematic
Half Adder Verilog code
module ha ( a, b, s, c)
input a, b;
output s, c;
assign s= a ^ b;
assign c= a & b;
end module
input a, b;
output s, c;
assign s= a ^ b;
assign c= a & b;
end module
Half Substractor
The half substractor truth table and schematic (fig-2) is mentioned below. The boolean expressions are:
D= A (EXOR) B
Br=A'.B
D= A (EXOR) B
Br=A'.B
Half substractor Schematic
Half Substractor Verilog code
![Verilog Verilog](/uploads/1/2/4/7/124730792/463835054.jpg)
module hs ( a, b, d, br)
input a, b;
output d, br;
assign d= a ^ b;
assign br= ~a & b;
end module
input a, b;
output d, br;
assign d= a ^ b;
assign br= ~a & b;
end module
Verilog Full Adder Module
Full Substractor
The full substractor truth table and schematic (fig-3) is mentioned below. The boolean expressions are:
D= A (EXOR) B (EXOR) C
Br=A'.B + B.Cin + A'.Cin
D= A (EXOR) B (EXOR) C
Br=A'.B + B.Cin + A'.Cin
Input-A | Input-B | Input-Cin | Output-D | Output-Br |
---|---|---|---|---|
0 | 0 | 0 | 0 | 0 |
0 | 0 | 1 | 1 | 1 |
0 | 1 | 0 | 1 | 1 |
0 | 1 | 1 | 0 | 1 |
1 | 0 | 0 | 1 | 0 |
1 | 0 | 1 | 0 | 0 |
1 | 1 | 0 | 0 | 0 |
1 | 1 | 1 | 1 | 1 |
Full substractor Schematic
Full Substractor Verilog code
module fs ( a, b, c, d, br)
input a, b, c;
output d, br;
assign d= a ^ b ^ c;
assign br=(( ~a)& (b ^ c)) | (b & c);
end module
input a, b, c;
output d, br;
assign d= a ^ b ^ c;
assign br=(( ~a)& (b ^ c)) | (b & c);
end module