【iCore4 双核心板_FPGA】例程二:GPIO输入实验——识别按键输入
实验现象:
按键每按下一次,三色LED切换一次状态。
核心源代码:
module key_ctrl( input clk_25m, input rst_n, input key, output fpga_ledr, output fpga_ledg, output fpga_ledb ); //--------------------key_in--------------------------------// parameter ms_10 = 250000; reg key_r; reg [17:0]low_cnt; reg [17:0]hig_cnt; always @(posedge clk_25m or negedge rst_n) //按键消抖动,提取按键状态 if (!rst_n) begin key_r <= 1'd0; low_cnt <= 18'd0; hig_cnt <= 18'd0; end else if(key) //检测按键状态为高时,延时10ms,把按键状态提取出来。 begin low_cnt <= 18'd0; if (hig_cnt == ms_10) begin key_r <= key; hig_cnt <= hig_cnt; end else hig_cnt <= hig_cnt + 1'd1; end else //按键状态为低时,延时10ms,把按键状态提取出来。 begin hig_cnt <= 18'd0; if (low_cnt == ms_10) begin key_r <= key; low_cnt <= low_cnt; end else low_cnt <= low_cnt + 1'd1; end wire key_state = key_r; //--------------------led_ctrl-----------------------------// reg [1:0]led_cnt; reg ledr,ledg,ledb; always@(negedge key_state or negedge rst_n) //按键下降沿控制led状态切换 if (!rst_n) begin led_cnt <= 2'd0; end else if (led_cnt == 2'd2) begin led_cnt <= 2'd0; end else led_cnt <= led_cnt + 1'd1; always@(posedge clk_25m or negedge rst_n) //led状态切换的状态机 if (!rst_n) begin ledr <= 1'd1; ledg <= 1'd1; ledb <= 1'd1; end else case(led_cnt) 2'd0: //红灯亮 begin ledr <= 1'd0; ledg <= 1'd1; ledb <= 1'd1; end 2'd1: //绿灯亮 begin ledr <= 1'd1; ledg <= 1'd0; ledb <= 1'd1; end 2'd2: //蓝灯亮 begin ledr <= 1'd1; ledg <= 1'd1; ledb <= 1'd0; end default: //都不亮 begin ledr <= 1'd1; ledg <= 1'd1; ledb <= 1'd1; end endcase assign {fpga_ledr,fpga_ledg,fpga_ledb} = {ledr,ledg,ledb}; //--------------------endmodule-----------------------------// endmodule
源代码下载链接:
链接:http://pan.baidu.com/s/1kVkOCgJ 密码:u1gd
iCore4链接: