AVR Freaks

Swap these two around...

 

    MIDI_UBRRL = (unsigned char) baudrate;
    MIDI_UBRRH = (unsigned char) (baudrate >> 8);

...to give...

 

    MIDI_UBRRH = (unsigned char) (baudrate >> 8);
    MIDI_UBRRL = (unsigned char) baudrate;

See the datasheet for why.

 

I've just tried your code on a real 1284P sitting here in a customers unit and I get data out of UART0.

 

One thought...what are your MIDI_UBRRL and MIDI_UBRRH defined as?

 

Now to put the customers unit back to doing what is should!

 

 

This is my code which is sending data out on PD1 (pin15 on a DIP package).

 

(note, I use codevision as a compiler so my includes look a bit different)

 

#include <io.h>
#include <stdio.h>
#include <stdint.h>

#define F_CPU 11059200UL
#define MIDI_BAUDRATEVAL 31250


void sendMidiData(uint8_t midiVal){
    while((UCSR0A & (1<<UDRE0)) == 0);
    UDR0 = midiVal;
}

void midiInit(){

    unsigned int baudrate;
    //set up USART

    UCSR0B =  (1<<TXEN0);
    UCSR0C = (1<<UCSZ00) | (1<<UCSZ01); //8 bit, 1 stop, no parity
    // Set the baud rate

    baudrate = (((F_CPU / (MIDI_BAUDRATEVAL * 16UL))) - 1);

    UBRR0H = (unsigned char) (baudrate >> 8);
    UBRR0L = (unsigned char) baudrate;

}

void main(void)
{
	midiInit();

	while (1)
	{
		sendMidiData(0xE0); //controller midi ch0
		sendMidiData(0xE5); //random controller number
        sendMidiData(0xE7);  //random value
	}
}

 

Actually it doesn't matter.  Although it is a 16-bit register, UBRRn doesn't have a temporary register for the high byte in UBRRnH (like the 16-bit timer registers).

 

The only restriction:

The USART Baud Rate Register (UBRRn) and the down-counter connected to it function as a programmable prescaler or baud rate generator. The down-counter, running at system clock (fosc), is loaded with the UBRRn value each time the counter has counted down to zero or when the UBRRLn Register is written. A clock is generated each time the counter reaches zero.

The worst case is that the first clock period runs at the previous UBRRn value.  Since there at 8 or 16 clock periods per bit time, it's unlikely that even the first bit of the first frame would be adversely affected.

 

In any event, why write to UBRRnL and UBRRnH separately?  The toolchain provides 16-bit access via UBRRn.

@oleoleo2

 

Since it's clear that something odd is going on here I suggest you strip it back and start again.

 

1) Knock up a bit of code that toggles the pin at 1Hz.

2) Take the minimal code that is needed to send data out of the USART; for example my code (with adjustments for gcc) above.

Have we established that your MPU is actually running at 20MHz?