题解 | #使用3-8译码器①实现逻辑函数#

`timescale 1ns/1ns

module decoder_38(
   input             E1_n   ,
   input             E2_n   ,
   input             E3     ,
   input             A0     ,
   input             A1     ,
   input             A2     ,
   
   output wire       Y0_n   ,  
   output wire       Y1_n   , 
   output wire       Y2_n   , 
   output wire       Y3_n   , 
   output wire       Y4_n   , 
   output wire       Y5_n   , 
   output wire       Y6_n   , 
   output wire       Y7_n   
);
wire E ;
assign E = E3 & ~E2_n & ~E1_n;
assign  Y0_n = ~(E & ~A2 & ~A1 & ~A0);
assign  Y1_n = ~(E & ~A2 & ~A1 &  A0);
assign  Y2_n = ~(E & ~A2 &  A1 & ~A0);
assign  Y3_n = ~(E & ~A2 &  A1 &  A0);
assign  Y4_n = ~(E &  A2 & ~A1 & ~A0);
assign  Y5_n = ~(E &  A2 & ~A1 &  A0);
assign  Y6_n = ~(E &  A2 &  A1 & ~A0);
assign  Y7_n = ~(E &  A2 &  A1 &  A0);
     
endmodule

module decoder0(
   input             A     ,
   input             B     ,
   input             C     ,
   
   output wire       L
);
wire Y0_n,Y1_n,Y2_n,Y3_n,Y4_n,Y5_n,Y6_n,Y7_n;
decoder_38 u_1(0,0,1,A,B,C,Y0_n,Y1_n,Y2_n,Y3_n,Y4_n,Y5_n,Y6_n,Y7_n);
assign L = ~(Y3_n & Y4_n & Y6_n & Y7_n);
endmodule

需要注意的是该38译码器是输出0有效，所以以ABC为输入化为最小项以后，得到的Y7,YC,Y6,Y4相与之后需要再取反