I believe I've understood the concept of interrupt and how to initialize it, but I've seen in various places where they would first AND (select) the NVIC_PR registers against F bits before (ORing) the priority bits. For example in TM4C123 GPIO Port Interrupt Programming there was this line towards the end:
NVIC_PRI7_R = (NVIC_PRI7_R & 0xFF00FFFF) | 0x00A00000
The purpose is to set Port F interrupt a priority of 5 (by setting the top 3 bits [23:21] to the value; hence .1010. or 0xA to represent the a value of 5). So, why can't I just do this instead?
NVIC_PRI7_R |= 0x00A00000
What is & 0xFF00FFFF doing here? Why do I want to clear bits [23:16] before OR-ing priority bits [23:21]? The bits between [20:16] are not used anyway.
If bits [23:21] were all already set to 1, then NVIC_PRI7_R |= 0x00A00000 does nothing, it doesn't set them to the value you want, they all stay as 1s, that's why you clear them before applying your mask. The & with 0xFF00FFFF ensures that only the bits [23:16] are modified, the rest are left as they were.
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I'm a begginer in programming AVR microcontroler and I get a lot of headacke sometimes from reading the datasheets.
I'm trying to make a communication between my AVR and PC just to send some caracters and receive it on my computer.
There are two lines I don't understand from the whole program and that is:
void USART_init(void)
{
UBRRH = (uint8_t)(BAUD_PRESCALLER>>8); <---- this one!
UBRRL = (uint8_t)(BAUD_PRESCALLER); <--- and this one
UCSRB = (1<<RXEN)|(1<<TXEN);
UCSRC = (1<<UCSZ0)|(1<<UCSZ1)|(1<<URSEL);
}
Datasheet
Why do I have to shift BAUD_PRESCALLER with 8? If BAUD_PRESCALLER is a number and shifting that number with 8 doesn't mean the result will be zero?(Because we are shifting it too many times)
From the datasheet I understand that UBRRH contains the four most significant bits and the UBRRL contains the eight least signicant bits of the USART baut rate.(Note:UBBR is a 12-bit register)
So how actually we put all the required numbers in the UBBR register?
You have to shift it right 8 bits because the result of BAUD_PRESCALLER is larger than 8 bits. Shifting it right 8 bits gives you the most significant byte of a 16-bit value.
For example, if the value of BAUD_PRESCALAR is 0x123 - then 0x1 would be assigned to UBRRH and 0x23 would be assigned to UBRRL.
If the library was smart it could also perform sanity checking on the BAUD_PRESCALAR to make sure it fits in 16bits. If it can't, that means you cannot achieve the baud rate you want given the clock you are using. If you're UBRRx is truly 12bits, the sanity check would look something like this:
#if BAUD_PRESCALAR > 0xFFF
#error Invalid prescalar
#endif
I'm using an ATmega328. I am currently doing several measurements using the 10-bit ADC. I would like to store the values it converts in variables in order to be able to operate with them. For example:
int a;
(...)
ADMUX = 0b01000011; //Vref = 5V, ADC3
ADCSRA |= (1<<ADSC); //Starts conversion
while(!(ADCSRA & (1<<ADIF))); //Wait until it finishes
ADCSRA |= (1<<ADIF); //Clear flag
Suppose that the ADC stored the value 576 in ADCH:ADCL. Is it possible to achieve, somehow, the variable a to take that same value? (i.e. a=576;).
The full 16-bit result register should be accessible as such:
a = ADC;
But if you want to read both parts manually, then
a = ADCL;
a |= ADCH << 8;
That has to be done in two separate statements to force ADCH to be read last. The I/O modules have a temporary register to hold the high byte, preventing the module itself from corrupting the read value if it changes the value of the register. (i.e. if the ADC finishes another conversion and stores the new value.)
If you have interrupts that access the ADC (or need to use the value at a), you'll need to disable them for the duration of the access (that also goes for a = ADC, since it also compiles into multiple 8-bit reads).
The answer above is close, still comes up a little short based on the recommendations from the chip manufacturer. To be safe you should follow the instructions in the authoritative reference linked below.
Simply turn off interrupts just before and restore them after:
unsigned int a; // 16-bit word
// other code
cli();
a = ADCL;
a |= ADCH << 8;
sei();
Every example in this reference from the chip manufacturer follows this pattern for atomic 16-bit reads and writes. Reference: AVR Application Note 072.
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I am trying to read ADC(ADS8320) values using SPI communication. I am using IAR embedded workbench for ARM and SM470R1B1 controller. In the data sheet it says first 6 bits are dummy read and next 16 bits is the actval data. I am trying to read 24 bits and ignore first 6 bits and last 2 bits. I am getting wrong values when i try with the below code.
unsigned int readADC8320(void)
{
value = AD8320_16(0xFFFFFF); // read registe
return value;
}
unsigned int AD8320_16(unsigned int value)
{
unsigned int data;
AD8320_CS_Status(0);
SW_Delay(DELAY_10US);
while(GIODOUTA&X2);
data = spi2(value >> 16); //read high 8 bits
data = (data << 6)| spi2(value >> 8); //read next 8 bits but 6+8 =14
data = (data << 8)| spi2(value >> 2); //add last two bits only
SW_Delay(DELAY_10US);
AD8320_CS_Status(1);
return data;
}
unsigned char spi2(unsigned char data)
{
// Write byte to SPI2DAT1 register
SPI2DAT1 = data;
while(!(SPI2CTRL3 & 0x01)){} // Wait for RxFlag to get set
return (SPI2BUF & 0x000000FF); // Read SPIBUF
}
Can anyone suggest me where i am doing wrong. I am quite poor with shift operations.
Your code looks wonky. For example, what purpose does value have?
The datasheet you linked to indicates that the 22 clock cycles suffice (and that 24 clock cycles are okay), no extra delays are needed, assuming the SPI clock is within limits (at most 2.4 MHz, I believe). So, I'd expect your code to look more like
void AD8320_setup(void)
{
/* TODO:
* - Set SPI2 clock frequency, if programmable (max. 2.4 MHz)
* - Set SPI2 transfer size, if programmable (8 bits per byte)
* - Set SPI2 to latch data on the falling edge of clock pulse
* - Set SPI2 to pull clock high when inactive
* - Set AD8320 chip select pin as an output
* - Set AD8320 chip select pin high (it is active low)
* - Ensure AD8320 is powered
*/
}
uint16_t AD8320_16(void)
{
uint16_t result;
/* TODO: Set AD8320 chip select pin 0/LOW (to select it)
*/
/* TODO: Clear SPI2 FIFO if it has one,
* so that we get actual wire data,
* not old cached data.
*/
result = ((uint16_t)(spi2_read() & 3U)) << 14;
result |= ((uint16_t)spi2_read()) << 6;
result |= ((uint16_t)spi2_read()) >> 2;
/* TODO: Set AD8320 chip select pin 1/HIGH (to deselect it)
*/
return result;
}
The prototype for spi2_read is typically unsigned char spi2_read(void);, but it might be of any integer type that can represent all eight bit values correctly (0 to 255, inclusive). Above, I cast the result -- after masking out any irrelevant bits, if any, with a binary AND operation -- to uint16_t before shifting it into the correct position in the result, so that regardless of the spi2_read() return type, we can use it to construct our 16-bit value.
The first spi2_read() reads the first 8 bits. I am assuming your device SPI bus sends and receives the most significant bit first; this means that the 6 dummy bits are most significant bits, and the 2 least significant bits are the most significant bits of the result. That's why we only keep the two least significant bits of the first byte, and transfer it 14 places left -- that way we get them in the most significant position.
The second spi2_read() reads the next eight bits. These are bits 13..6 of the data; that's why we shift this byte left by 6 places, and OR with the existing data.
The last spi2_read() reads the next eight bits. The last two bits are to be discarded, because only the first (most significant) 6 bits contain the rest of the data we're interested in. So, we shift this down by 2 places, and OR with the existing data.
Preparing spi2 involves stuff like word size (8 bits), data rate (if programmable), and so on.
The chip select for the AD8320 is just a generic output pin, it's just active low. That's why it's described as ^CS/SHDN in the datasheet: chip selected when low, shutdown when high.
The chip is a micropower one, so it requires less than 2mA, so you could conceivably power it from an output pin on many microcontrollers. If you were to do so, you'd also have to remember to make that pin an output and 1/high in _setup() function.
The code below is used for programming microcontrollers. I want to know what the code below is doing. I know that '|' is OR and '&' AND but what is the whole line doing?
lcd_port = (((dat >> 4) & 0x0F)|LCD_EN|LCD_RS);
It's hard to put into context since we don't know what dat contains, but we can see that:
The data is right-shifted by 4 bits, so 11111111 becomes 00001111, for instance.
That value is AND'ed with 0x0F. This is a common trick to remove unwanted bits, since b & 1 = 1 and b & 0 = 0. Think of your number as a sequence of bits, here's a 2-byte example :
0011010100111010
&
0000000000001111
0000000000001010
Now the LCD_EN and LCD_RS flags are OR'ed. Again, this is a common binary trick, since b | 1 = 1 and b | 0 = b, so you can add flag but not remove them. So, if say LCD_EN = 0x01 and LCD_RS = 0x02,
0000000000001010
|
0000000000000011
0000000000001011
Hope that's clearer for you.
Some guesses, as you'll probably need to find chip datasheets to confirm this:-
lcd_port is probably a variable that directly maps to a piece of memory-mapped hardware - likely an alphanumeric LCD display.
The display probably takes data as four-bit 'nibbles' (hence the shift/and operations) and the higher four bits of the port are control signals.
LCD_EN is probably an abbreviation for LCD ENABLE - a control line used on the port.
LCD_RS is probably an abbreviation for LCD READ STROBE (or LCD REGISTER SELECT) - another control line used on the port. Setting these bits while writing to the port probably tells the port the kind of operation to perform.
I wouldn't be at all surprised if the hardware in use was a Hitachi HD44780 or some derivative.
It appears to be setting some data and flags on the lcd_port. The first part applies the mask 0x0F to (dat >> 4) (shift dat right 4) which is followed by applying the LCD_EN flag and then LCD_RS flag.
It is shifting the variable data four bits to the right, then masking the value with the value 15. This results in a value ranging from 0-15 (four left-most bits). This result is binary ORd with the LCD_EN and LCD_RS flags.
This code is shifting the bits of dat 4 bits to the right and then using & 0x0F to ensure it gets only those 4 least significant bits. It's then using OR to find which bits exist in that value OR LCD_EN OR LCD_RS and assigning that value to lcd_port.
I am using an IC, DS1620 to read 1 bit serial data coming on a single line. I need to read this data using one of the ports of ARM microcontroller (LPC2378). ARM ports are 32 bit. How do I get this value into a 1 bit variable?
Edit: In other words I need direct reference to a port pin.
there are no 1 bit variables, but you could isolate a particular bit for example:
uint32_t original_value = whatever();
uint32_t bit15 = (original_value >> 15) & 1; /*bit15 now contains either a 1 or a 0 representing the 15th bit */
Note: I don't know if you were counting bit numbers starting at 0 or 1, so the >> 15 may be off by one, but you get the idea.
The other option is to use bit fields, but that gets messy and IMO is not worth it unless every bit in the value is useful in some way. If you just want one or two bits, shifting and masking is the way to go.
Overall, this article may be of use to you.
For your CPU the answer from Evan Teran should be used. I just wanted to mention the bit-band feature of some other ARM CPU's like the Cortex-M3. For some region of RAM/Peripherals all bits are mapped to a separate address for easy access.
See http://infocenter.arm.com/help/index.jsp?topic=/com.arm.doc.ddi0337e/Behcjiic.html for more information.
It's simple if you can access the port register directly (I don't have any experience with ARM), just bitwise AND it with the binary mask that corresponds to the bit you want:
var = (PORT_REGISTER & 0x00008000);
Now var contains either 0 if the 15th bit is '0' or 0x00008000 if the 15th bit is '1'.
Also, you can shift it if you want to have either '0' or '1':
var = ((PORT_REGISTER & 0x00008000) >> 15);
The header file(s) which come with your compiler will contains declarations for all of the microcontroller's registers, and the bits in those registers.
In this specific article, let's pretend that the port input register is called PORTA, and the bit you want has a mask defined for it called PORTA15.
Then to read the state of that pin:
PinIsSet = (PORTA & PORTA15) == PORTA15;
Or equivalently, using the ternary operator:
PinIsSet = (PORTA & PORTA15) ? 1 : 0;
As a general point, refer to the reference manual for what all the registers and bits do. Also, look at some examples. (This page on the Keil website contains both, and there are plenty of other examples on the web.)
In LPC2378 ( as the other LPC2xxxx microcontroller family ), I/O ports are in system memory, so you need to declare some variables like this:
#define DALLAS_PIN (*(volatile unsigned long int *)(0xE0028000)) /* Port 0 data register */
#define DALLAS_DDR (*(volatile unsigned long int *)(0xE0028008)) /* Port 0 data direction reg */
#define DALLAS_PIN (1<<15)
Please note that 0xE0028000 is the address for the data register of port0, and 0xE0028008 is the data direction register address for port0. You need to modify this according to the port and bit used in your app.
After that, in your code function, the code or macros for write 1, write 0 and read must be something like this:
#define set_dqout() (DALLAS_DDR&=~DALLAS_PIN) /* Let the pull-up force one, putting I/O pin in input mode */
#define reset_dqout() (DALLAS_DDR|=DALLAS_PIN,DALLAS_PORT&=~DALLAS_PIN) /* force zero putting the I/O in output mode and writing zero on it */
#define read_dqin() (DALLAS_DDR&=~DALLAS_PIN,((DALLAS_PORT & DALLAS_PIN)!= 0)) /* put i/o in input mode and test the state of the i/o pin */
I hope this can help.
Regards!
If testing for bits, it is good to keep in mind the operator evaluation order of C, e.g.
if( port & 0x80 && whatever() )
can result in unexpected behaviour, as you just wrote
if( port & (0x80 && whatever()) )
but probably ment
if( (port & 0x80) && whatever() )