FPGA 实现LED闪烁10次数,可通过参数修改闪烁时间和闪烁次数

1 功能：

实现LED闪烁10次后熄灭

2 实现方法

         在原来定时计数器的基础上加入记录定时脉冲的计数器，每个定时脉冲加1，因此定时脉冲的计数器每计数两次LED闪烁一次，若需要使LED闪烁20次，则需要定时脉冲的计数器计数20次(0~19)。加入停止位stopFlag，当定时脉冲的计数器计数满后置为1，以后LED输出保持不变，这本程序为熄灭LED，也可以开启LED。

3 实现程序

module ledon_10times
#(
    parameter COUNTER_MAX = 25'd24_999_999,
    parameter LEDON_COUNTER = 5'd19
)
(
    input wire sys_clk,
    input wire sys_rst_n,

    output reg led_out
);

reg [24:0] counter;
reg [4:0] ledOnCounter;

reg stopFlag = 1'b0;
reg counter_flag;

// 设置counter
always @(posedge sys_clk or negedge sys_rst_n) begin
    if (sys_rst_n == 1'b0)
        counter <= 25'd0;
    else if(counter == COUNTER_MAX)
        counter <= 25'd0;
    else
        counter <= counter + 25'd1;
end

// 设置counter_flag
always @(posedge sys_clk or negedge sys_rst_n) begin
    if (sys_rst_n == 1'b0)
        counter_flag <= 1'b0;
    else if(counter == COUNTER_MAX - 25'd1)
        counter_flag <= 1'b1;
    else if(ledOnCounter == LEDON_COUNTER)
        counter_flag <= 1'b0;
    else
        counter_flag <= 1'b0;
end

// 设置ledout,前10s灯闪烁，10次后熄灭
// 不能在不同always块中对同一个信号进行赋值
always @(posedge sys_clk or negedge sys_rst_n) begin
    if (sys_rst_n == 1'b0)
        led_out <= 1'b0;
    else if(counter_flag == 1'b1 && stopFlag == 1'b0)
        led_out <= ~ led_out;
    else if(stopFlag == 1'b1)      // 10次后让灯熄灭
        led_out <= 1'b1;
    else
        led_out <= led_out;
end

// 设置ledOnCounter
always @(posedge sys_clk or negedge sys_rst_n) begin
    if (sys_rst_n == 1'b0)
        ledOnCounter <= 5'b0;
    else if(ledOnCounter == LEDON_COUNTER  && counter_flag == 1'b1)
        ledOnCounter <= 5'b0;
    else if( counter_flag == 1'b1)
        ledOnCounter <= ledOnCounter + 5'b1;
    else
        ledOnCounter <= ledOnCounter;
end

// 置位标志位
always @(posedge sys_clk or negedge sys_rst_n) begin
    if (sys_rst_n == 1'b0)
        stopFlag <= 1'b0;
    else if(ledOnCounter == LEDON_COUNTER-1  && counter_flag == 1'b1)
        stopFlag <= 1'b1;
    else 
        stopFlag <= stopFlag;
end
endmodule

•  板载FPGA 50MHz晶振，故一个时钟周期为 20 ns
• 要实现0.5s LED翻转一次则需计数25 x 10^6次(即：0~24_999_999)

3 仿真程序

`timescale 1 ns/ 1 ps
module tb_ledon_10times();

// test vector input registers
reg sys_clk;
reg sys_rst_n;
// wires                                               
wire led_out;

initial begin
	sys_clk = 1'b0;
	sys_rst_n = 1'b0;
	#20
	sys_rst_n = 1'b1;
end

always #20 sys_clk = ~sys_clk;

// assign statements (if any)                          
ledon_10times 
#(
    .COUNTER_MAX   (25'd24),
    .LEDON_COUNTER (5'd19)
)
ledon_10times_init1
(
// port map - connection between master ports and signals/registers   
	.led_out(led_out),
	.sys_clk(sys_clk),
	.sys_rst_n(sys_rst_n)
);
                                                      
endmodule

 仿真波形

 波形与设计的一致，仿真成功!