UART连接ESP8266的verilog代码
时间: 2023-12-08 13:03:51 浏览: 87
以下是一个简单的Verilog代码,用于实现UART连接ESP8266:
```verilog
module uart_esp8266(
input CLK,
input RST,
input UART_RX,
output reg UART_TX
);
reg [7:0] uart_data;
reg [3:0] uart_state;
localparam IDLE = 4'd0;
localparam START = 4'd1;
localparam RECEIVE = 4'd2;
localparam TRANSMIT = 4'd3;
localparam STOP = 4'd4;
always @(posedge CLK) begin
if (RST) begin
uart_data <= 8'b0;
uart_state <= IDLE;
UART_TX <= 1'b1;
end else begin
case(uart_state)
IDLE: begin
if (!UART_RX) begin
uart_state <= START;
end
end
START: begin
uart_data <= 8'b0;
uart_state <= RECEIVE;
end
RECEIVE: begin
uart_data <= {uart_data[6:0], UART_RX};
uart_state <= (uart_state == TRANSMIT) ? STOP : RECEIVE;
end
TRANSMIT: begin
UART_TX <= uart_data[0];
uart_data <= {1'b0, uart_data[7:1]};
uart_state <= (uart_state == STOP) ? IDLE : TRANSMIT;
end
STOP: begin
UART_TX <= 1'b1;
uart_state <= IDLE;
end
endcase
end
end
endmodule
```
这个模块在时钟上升沿时工作,包含一个有限状态机,用于处理UART通信。它接收UART_RX信号并将其存储在一个8位寄存器中,然后将数据发送到UART_TX输出端口。在发送每个字节之前,它会向UART_TX发送起始位,然后向UART_TX发送一个停止位。
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