Verilog 帮我写一个uart接受代码
时间: 2024-05-06 08:19:19 浏览: 13
好的,以下是一个简单的Verilog UART接收代码:
```
module uart_rx(
input clk,
input rst,
input rx,
output reg [7:0] data,
output reg valid
);
localparam idle = 2'd0;
localparam start_bit = 2'd1;
localparam data_bits = 2'd2;
localparam stop_bit = 2'd3;
reg [3:0] state;
always @(posedge clk or posedge rst) begin
if (rst) begin
state <= idle;
data <= 8'b0;
valid <= 1'b0;
end else begin
case (state)
idle: begin
if (!rx) begin
state <= start_bit;
end
end
start_bit: begin
state <= data_bits;
end
data_bits: begin
data <= {data[6:0], rx};
state <= state + 1;
end
stop_bit: begin
valid <= 1'b1;
state <= idle;
end
default: begin
state <= idle;
end
endcase
end
end
endmodule
```
这段代码实现了一个简单的8位UART接收器。它在时钟和重置信号作用下,通过状态机控制UART接收的过程,将接收到的数据存储在`data`寄存器中,并将`valid`信号置为1表示数据接收完毕。注意修改时钟和重置信号的名称以适应你的设计。