(问题)关于一个分频电路方式的书写
(问题)关于一个分频电路的书写
1、学习Verilog有一段时间后,有点对于基本电路的书写不再上心了。可慢慢的发觉对于基本电路学习是很重要的,在最近的FPGA设计中更多的是在基本电路上得到提示的方法。所以就坚持学习过去分析过的代码,
2、这是在国外一个网站看到了一种写法,很困惑他的实现方式,就发出来和大家研究研究。
3、他的控制方式不像网上流行的的那种,不明白这样的电路对生成那样的电路
//-----------------------------------------------------
// Design Name : clk_div_45
// File Name : clk_div_45.v
// Function : Divide by 4.5
// Coder : Deepak Kumar Tala
//-----------------------------------------------------
module clk_div_45 (
clk_in, // Input Clock
enable, // Enable is sync with falling edge of clk_in
clk_out // Output Clock
);
// --------------Port Declaration-----------------------
input clk_in ;
input enable ;
output clk_out ;
//--------------Port data type declaration-------------
wire clk_in ;
wire enable ;
wire clk_out ;
//--------------Internal Registers----------------------
reg [3:0] counter1 ;
reg [3:0] counter2 ;
reg toggle1 ;
reg toggle2 ;
//--------------Code Starts Here-----------------------
always @ (posedge clk_in)
if (enable == 1'b0)
begin
counter1 <= 4'b0;
toggle1 <= 0;
end
else
if ((counter1 == 3 && toggle2) || (~toggle1 && counter1 == 4))
begin
counter1 <= 4'b0;
toggle1 <= ~toggle1;
end
else
begin
counter1 <= counter1 + 1;
end
always @ (negedge clk_in)
if (enable == 1'b0)
begin
counter2 <= 4'b0;
toggle2 <= 0;
end
else
if ((counter2 == 3 && ~toggle2) || (toggle2 && counter2 == 4))
begin
counter2 <= 4'b0;
toggle2 <= ~toggle2;
end
else
begin
counter2 <= counter2 + 1;
end
assign clk_out = (counter1 <3 && counter2 < 3) & enable;
// File Name : clk_div_45.v
// Function : Divide by 4.5
// Coder : Deepak Kumar Tala
//-----------------------------------------------------
module clk_div_45 (
clk_in, // Input Clock
enable, // Enable is sync with falling edge of clk_in
clk_out // Output Clock
);
// --------------Port Declaration-----------------------
input clk_in ;
input enable ;
output clk_out ;
//--------------Port data type declaration-------------
wire clk_in ;
wire enable ;
wire clk_out ;
//--------------Internal Registers----------------------
reg [3:0] counter1 ;
reg [3:0] counter2 ;
reg toggle1 ;
reg toggle2 ;
//--------------Code Starts Here-----------------------
always @ (posedge clk_in)
if (enable == 1'b0)
begin
counter1 <= 4'b0;
toggle1 <= 0;
end
else
if ((counter1 == 3 && toggle2) || (~toggle1 && counter1 == 4))
begin
counter1 <= 4'b0;
toggle1 <= ~toggle1;
end
else
begin
counter1 <= counter1 + 1;
end
always @ (negedge clk_in)
if (enable == 1'b0)
begin
counter2 <= 4'b0;
toggle2 <= 0;
end
else
if ((counter2 == 3 && ~toggle2) || (toggle2 && counter2 == 4))
begin
counter2 <= 4'b0;
toggle2 <= ~toggle2;
end
else
begin
counter2 <= counter2 + 1;
end
assign clk_out = (counter1 <3 && counter2 < 3) & enable;
endmodule