Electronic – Bitwise operation for UART, AVR

avrcmicrocontrolleruart

I'm trying to work on UART for AVR. After setting the baud rate registers UBRRnX, I'm trying to configure the UCSRnA register.

I'm trying to set the 2nd bit, U2XO. Which is the right way to do it?
UCSR0A = (1 << U2X0)

UCSR0A = UCSR0A | (1 << U2X0) (Notice the OR operator in the 2nd statement)

Best Answer

In this case:

UCSR0A = (1 << U2X0);

1<<U2XO = 00000010, 1 is shifted to left to U2XO position so when you give this value to UCSR0A the register's content will be:

UCSR0A = 00000010

This operation will set all bits to 0 but the U2X0, for example if RXC0 would have been set to 1, then this operation would clear it to 0. If you are not aware of this that could cause some headache later.

In this one:

UCSR0A = UCSR0A | (1 << U2X0);

which is equal to this:

UCSR0A |= (1<<U2X0);

Here you set only the second bit to one and leave the rest unchanged which is preferable, because you won't change something accidentally. The content will be:

// UCSRA0 = xxxxxxxx               before operation something is in the register
UCSR0A = UCSR0A | (1 << U2X0);     // OR current with 00000010
// UCSRA0 = xxxxxx1x               something OR 1 will be 1

where x is the unchanged/previous value of the bit. If x is 1, 1 OR 0 will remain 1 and if x is 0, 0 OR 0 will remain 0.

So I suggest the second way, because that way it is easier to track which bit is set and which not and you won't change previously applied configuration by mistake.

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Electronic – UART signal distortion with AVR

Your problem is likely in using the wrong baudrate: 112500 is most likely a mistake, the standard rate in that range is 115200 (a search of the manufacturer's website for this number finds many hits, but none for 112500). You could also not be producing the programmed baud rate on one end or the other due to divider granularity; sometimes changing the oversampling of the UART can help. EDIT: Specifically, a value of 8 (divisor of 9) will get you 111111 baud from 8MHz if you set the "double speed" mode bit.

Secondarily, you have a problem such as lack of a common ground between the boards, or haven't grounded the scope to them, thus resulting in the distorted waveform. It's not clear yet if that is what the receiver sees, or if it's merely a measurement mistake in applying the scope.

Additionally, have you verified that the un-named external board also runs without a serial level translator? Most modular inter-board serial communications is at RS232 levels and logically inverted from the logic-level signals, though there are exceptions.

Electronic – UART on the atmega328. Dataregister on the USART (UCSR0A) status never set on

I don't see where you put the correct F_CPU definition, which is mandatory for setbaud.h.

You also have to define the BAUD macro to use UBRR(L|H)_VALUE, like this:

#define F_CPU 1e6       // 1Mhz clock
#include <avr/io.h>
#include <util/delay.h>

void initUSART(void) {
 #define BAUD 9600
 #include <util/setbaud.h> 
 UBRR0H = UBRRH_VALUE;
 UBRR0L = UBRRL_VALUE;
// ...
}

You can also calculate the UBRRnx values with this site. The values depend on your clock speed and the prescaler setting. At the default configuration it is simply one, so it is not used.

And don't forget the pull-up resistors (about 10K ohm) at the TX/RX lines.

I hope this helps.

Best Answer

Related Solutions

Electronic – UART signal distortion with AVR

Electronic – UART on the atmega328. Dataregister on the USART (UCSR0A) status never set on

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