超前进位加法器

1、说明文档4到8 LED编码器4位超前进位加法器abcincout8段LED显示器8段LED显示器1.结构图：2.八段LED显示器的编码：假设8段LED显示器是共阴极的，DPGFEDCBA正好组成一个8位的数8seg，给这个数赋值，即可以显示相应的数字。编码：3.编码器子模块4到8位LED编码器module coder(in,out);input3:0 in;output7:0 out;wire3:0 in;reg7:0 out;always(in)begincase(in)4'h0: out=8'h3f;4'h1: out=8'h06;4'h2: out

2、=8'h5b;4'h3: out=8'h4f;4'h4: out=8'h66;4'h5: out=8'h6d;4'h6: out=8'h7d;4'h7: out=8'h07;4'h8: out=8'h7f;4'h9: out=8'h6f;4'ha: out=8'h77;4'hb: out=8'h7c;4'hc: out=8'h39;4'hd: out=8'h5e;4'he: out=8'h79;4

3、'hf: out=8'h71;endcaseendendmodule4. 超前进位加法器子模块对一个4位全加器，第i位的两加数分别是Ai和Bi,进位输入信号是Ci-1,这个位产生的进位输出是 Ci，则可以有得到：Ci=AiBi（AiBi）Ci-1令Gi= AiBi(进位产生函数) Pi= AiBi(进位传递函数)则：Ci=Gi+ PiCi-1所以有：C0=G0+P0C-1 C1=G1+P1C0 C2=G2+P2C1 C3=G3+P3C2将上式逐次代入则可以得到如下表达式（C-1是最低位的进位输入）COG0+P0C-1C1=G1+P1C0=G1+P1G0+P1P0C-1C2=G2

4、+P2C1=G2+P2G1+P2P1C0=G2+P2G1+P2P1G0+P2P1P0C-1C3=G3+P3C2 =G3+P3G2+P3P2C1 =G3+P3P2P3P2C1 =G3+P3P2P3P2G1P3P2P1C0 =G3+P3P2P3P2G1P3P2P1G0P3P2P1P0C-1上述表达式说明只要各位数据和最低位进位同时输入，各位之间的进位信号与和就能同时产生。要据这样的原理可以实现超前进位加法器，这种加法器运算速度显著提高，但是复杂度也提高了。module add4_head( a, b, cin, sum,cout);input3:0 a;input3:0 b;input cin;

5、output3:0 sum;output cout;wire3:0 p;wire3:0 g;wire3:0 t;wire2:0 c;assign p0 = a0 | b0;assign p1 = a1 | b1;assign p2 = a2 | b2;assign p3 = a3 | b3;assign t0 = a0 b0;assign t1 = a1 b1;assign t2 = a2 b2;assign t3 = a3 b3;assign g0 = a0 & b0;assign g1 = a1 & b1;assign g2 = a2 & b2;assign g3

6、= a3 & b3;assign c0 = (p0 & cin) | g0;assign c1 = (p1 & c0) | g1;assign c2 = (p2 & c1) | g2;assign cout = g3 | (p3 & (g2 | p2 & (g1 | p1 & (g0|(p0&cin);assign sum0 = t0 cin;assign sum1 = t1 c0;assign sum2 = t2 c1;assign sum3 = t3 c2; endmodule5.例化两个子模块把超前进位加法器和4到8位LED

7、编码器 两个模块例化放一起，完成本设计module add_display(a,b,cin,cout,out);input3:0 a;input3:0 b;input cin;output cout;output7:0 out;wire3:0 temp;add4_head add4_head(.a(a), .b(b), .cin(cin), .sum(temp), .cout(cout) );coder coder ( .in(temp), .out(out);endmodule6.对本设计进行测试/timescale 1ns / 1psmodule testbench;/ Inputsreg 3:0 a;reg 3:0 b;reg cin;/ Outputswire 7:0 out;wire cout; add_display add_display(.a(a),.b(b),.cin(cin),.cout(cout),.out(out);integer m,n,l;initial begin a = 4'b0000; b = 4'b0000;cin= 0;/ Add stimul