JK触发器与模12计数器

JK触发器

JK触发器具有保持，置0，置1和翻转四个功能。

则可得出次态方程： $Q_{n + 1} = J Q_{n}^{'} + K^{'} Q_{n}$

Design

`timescale 1ns / 1ps
module jk(clk,Reset,Set,J,K,Q);
input clk;
input Reset;
input Set;
input J;
input K;

output Q;

reg q;

always@(posedge clk)
    if(!Reset)            
        begin
            q <= 1'b0;
        end
    else if(!Set)    
        begin
            q <= 1'b1;
        end
    else
        begin
            case({J,K})
                2'b00 : q <= q;
                2'b01 : q <= 0;
                2'b10 : q <= 1;
                default : q <= ~q;
            endcase
         end
     assign Q=q;
endmodule

Simulation

`timescale 1ns / 1ps
module JK_sim();
reg Reset, Set, clk;
wire Q;
reg J, K;

jk uu(clk, Reset, Set, J, K, Q);
initial
    begin
    Reset = 0;
    Set = 0;
    clk = 1;
    J = 0;
    K = 0;
    #40 Reset = 1;
    #40 Set = 1;
    end
always #20{J, K} = {J, K} + 1;
always #10 clk = ~clk;
endmodule

波形图

模12计数器

需要使用4个JK触发器，卡诺图化简后得到触发器控制函数的最简与或式如下：

$J_{1} = K_{1} = 1$

$J_{2} = K_{2} = Y_{1}^{n}$

$J_{3} = (Y_{4}^{n})^{'} Y_{2}^{n} Y_{1}^{n}$

$K_{3} = Y_{2}^{n} Y_{1}^{n}$

$J_{4} = Y_{3}^{n} Y_{2}^{n} Y_{1}^{n}$

$K_{4} = Y_{2}^{n} Y_{1}^{n}$

Design

`timescale 1ns / 1ps
module count(Reset,Set,clk,Y);
input Reset;
input Set;
input clk;
output [3:0]Y;

wire j1,j2,j3,j4,k1,k2,k3,k4;

assign j1 = 1;
assign k1 = 1;
assign j2 = Y[0];
assign k2 = Y[0];
assign j3 = (~Y[3])&Y[1]&Y[0];
assign k3 = Y[0]&Y[1];
assign j4 = Y[0]&Y[1]&Y[2];
assign k4 = Y[0]&Y[1];
jk_0 u1 (clk,Reset,Set,j1,k1,Y[0]);
jk_0 u2 (clk,Reset,Set,j2,k2,Y[1]);
jk_0 u3 (clk,Reset,Set,j3,k3,Y[2]);
jk_0 u4 (clk,Reset,Set,j4,k4,Y[3]);

endmodule

Simulation

`timescale 1ns / 1ps
module sim_dev1();
reg Reset,Set,clk;
wire [3:0]Y;

initial
    begin
        Reset = 0;
        Set = 0;
        clk = 1;
        #20 Reset = 1;
        #20 Set = 1;
    end
always#20 clk = ~clk;
CNT12 u(Reset,Set,clk,Y);
endmodule