I/O Port Operations in AVR


AVR Series

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Hello friends! In this post, we will discuss about the port operations in AVR. Before going further, I suggest that you read my previous post regarding AVR Basics. The examples discussed here are in accordance with ATMEGA16/32 MCU. However, the concepts are equally good for any AVR MCU.

Register

Okay, now I hope you are familiar with the term register. If not, then you must have heard of it. Basically, a processor register is a memory space within the CPU itself so that they can be accessed very frequently and fast. These registers are linked with the operation of the MCU. Let’s consider the following memory space.

Register Memory Space

Register Memory Space

Here, you can see that I have represented 8 bits together to form a memory of 1 byte. Note the sequence in which the bits are numbered. They are numbered as 7, 6, 5, … , 1, 0. This is because the bits are numbered from the Least Significant Bit (LSB) to the Most Significant Bit (MSB). From the knowledge of digital logic, we know that the last bit is the LSB whereas the first bit is the MSB. Hence, Bit 0 = LSB and Bit 7 = MSB.

Register Concept Explained

Let me explain you why LSB is the last bit. Let’s take an example. Please note that 0b stands for binary and 0x stands for hexadecimal. If nothing is prefixed, it means that it is in decimal number system.

A = 0b 0110 0111 = 103

Now here, let’s change the value of the last bit (orange colour bit) from 1 to 0. This makes

B = 0b 0110 0110 = 102

Now, once again in A, let’s change the first bit (magenta colour bit) from 0 to 1. This makes

C = 0b 1110 0111 = 231

We see that by changing the last bit, the result (B) is very close to the original data (A), whereas by changing the first bit, the result (C) varies quite significantly from the original data (A). Hence, the last bit is the LSB (as the data doesn’t change significantly) whereas the first bit is the MSB (as the data changes significantly).

Now, we also know that 1 nibble = 4 bits. Hence, bits 0,1,2,3 are called lower nibble whereas bits 4,5,6,7 are called upper nibble. So basically, a register is a memory allocated in the CPU, usually having a size of 1 byte (8 bits).

Next, every register has a name, and every bit of it also has a name. Take the following example.

ADMUX

ADMUX Register

Here, the name of the register is ADMUX (don’t worry about the name, we will discuss it later). Also, note that each bit also has a name and an initial value. Each bit of the register can have a value of either 0 or 1 (initially it is 0). Now suppose, I write

ADMUX = 0b01000111;

This means that the ADMUX register has been updated as follows:

ADMUX Register after setting values

ADMUX Register after setting values

This can also be achieved by the following codes:

ADMUX = (1<<REFS0)|(1<<MUX2)|(1<<MUX1)|(1<<MUX0);
ADMUX = 0x47;    //Hex form

So, got an idea of what registers are and how are they defined/initialized? We will discuss about various registers one by one as required. Right now, we are concerned with only three registers, DDR, PIN and PORT.

Port Operation Registers

The following registers are related to the various port operations that we can perform with the GPIO pins.

  • DDRx – Data Direction Register
  • PORTx – Pin Output Register
  • PINx – Pin Input Register

where x = GPIO port name (A, B, C or D)

Different Port Operations

Different Port Operations

DDRx Register

As I mentioned earlier, the GPIO pins are the digital I/O pins i.e. they can act as both input and output. Now, how do we know that the pin is an output pin or input? The DDRx (Data Direction Register) helps in it.

DDRx initializes the port. Have a look at it’s bit structure.

DDRx Register

DDRx Register

The ‘x’ in DDRx is just for representation. It is replaced by the corresponding port i.e. x = A, B, C, D. Say for the example shown in the diagram above

DDRC = (1<<DDC0)|(1<<DDC4)|(1<<DDC5)|(1<<DDC7);

This is equivalent to

DDRC = 0b10110001;

and as well as

DDRC = 0xB1;

and even

DDRC = (1<<0)|(1<<4)|(1<<5)|(1<<7);

So, did you get how to declare it? Suppose I would like to initialize my port B, then I would have written

DDRB = (1<<DDB0)|(1<<DDB4)|(1<<DDB5)|(1<<DDB7);
DDRC Example

DDRC Example

All right, now that we are done with the declaration, let me explain you what it does. Always remember, in the case of DDRx, 1 stands for output and 0 stands for input. In the following statement (given below), port C is initialized such that the pins PC0, PC4, PC5 and PC7 are output pins whereas pins PC1, PC2, PC3 and PC6 are input pins.

This is represented in the adjoining figure. The pins marked as input now has the capability to read the voltage level at that pin and then treat it as HIGH if it is above the threshold level, or else it will treat it as LOW. Generally, the threshold level is half the VCC.

Similarily, the pins marked as output have the capability to either become HIGH (voltage = VCC) or LOW (voltage = zero) as directed/written in the code.

DDRC = (1<<DDC0)|(1<<DDC4)|(1<<DDC5)|(1<<DDC7);

PORTx Register

The PORTx register determines whether the output should be HIGH or LOW of the output pins. In simplified terms, once the DDRx register has defined the output pins, we need to set them to give an output of HIGH or LOW. The PORTx register goes as follows.

PORTx Register

PORTx Register

The register declaration is similar to the DDRx register, except that we change the names, that’s all! One such example is given above in the diagram. The following declarations are one and the same.

PORTD = (1 << PD0)|(1 << PD3)|(1 << PD6);
PORTD = (1 << 0)|(1 << 3)|(1 << 6);
PORTD = 0b01001001;
PORTD = 0x49;

Now, let’s consider the following statements:

DDRC   = 0b10110001;
PORTC  = 0b10010001;
OUTPUT = 0b10010001;  /*This is not a C executable line, this line is just for explanation*/

The port C is initialized using the DDRx register. The highlighted bits correspond to the output pins. Now, just concentrate on the highlighted bits only. Since they are output pins, wherever I state ‘1’ in PORTC, that pin goes HIGH (1), giving an output voltage of VCC at that pin.

Now consider the following set of statements:

DDRC   = 0b10110001;
PORTC  = 0b10010101;
OUTPUT = 0b10010001;  /*This is not a C executable line, this line is just for explanation*/

Once again, the bits highlighted in orange correspond to the output pins. So, whatever value (0 or 1) desired in the orange region is reflected in the output. Now, look at the magenta highlighted bit. Inspite of being set as HIGH in the PORTx register, the output is LOW. This is because that pin is initialized as an input pin by DDRx. Hence, PORTx cannot change the properties of that pin. Hence, in general, PORTx cannot modify the properties of a pin which is initialized as input by DDRx.

PINx Register

The PINx register gets the reading from the input pins of the MCU. The register goes as follows:

PINx Register

PINx Register

The register declaration procedure stands the same. Also note that the names of the bits of PORTx and PINx registers are same.

Now, let’s consider the following statements:

DDRC   = 0b10110001;
PINC   = 0b01001011;
INPUT  = 0b01001011;  /*This is not a C executable line, this line is just for explanation*/

Here, the highlighted bits correspond to the pins that are initialized as input by the DDRx. In the second line, the PINx register is defined. Well, this line is just to explain the concept, practically, we always use PINx as a condition (like in IF or in WHILE loop). As per the second statement, the PINx command reads the values only at the input pins.

Now, consider the following set of statements:

DDRC   = 0b10110001;
PINC   = 0b01011010;
INPUT  = 0b01001010;  /*This is not a C executable line, this line is just for explanation*/

Here, you can compare it with the example I gave for PORTx. Since the magenta-highlighted bit is an output pin, PINx cannot change it’s properties. Hence, in general, PINx cannot modify the properties of a pin which is initialized as output by DDRx and vice versa.

Example Code Snippet

Let’s demonstrate the use of the DDRx, PORTx and PINx registers from the following code snippet:

DDRC = 0x0F;
PORTC = 0x0C;

// lets assume a 4V supply comes to PORTC.6 and Vcc = 5V
if (PINC == 0b01000000)
PORTC = 0x0B;
else
PORTC = 0x00;

Code Explained:

  • DDRC = 0x0F; is equivalent to DDRC = 0b00001111; This means that the pins PC0…PC3 are output pins (can be manipulated using PORTC) and pins PC4…PC7 are input pins (whose levels determine the value of PINC).
  • PORTC = 0x0C; is equivalent to PORTC = 0b00001100; This means that the pins PC2 and PC3 have a HIGH voltage (Vcc = 5V) and pins PC0 and PC1 have LOW voltage (0V). The other pins have low voltage by default.
  • if (PINC = 0b01000000) checks the input voltage at pin PC6. Since it is mentioned in the comment that a 4V is supplied to PORTC.6 (same as pin PC6), this condition is true (as 4 > 2.5, where 2.5V is the threshold, 5/2 = 2.5).
  • Since the if condition is true, PORTC = 0x0B; is executed.
  • If the if condition is not satisfied, PORTC = 0x00; will be executed.

We can also put it inside a while loop to run it continuously i.e. it always checks for the voltage at pin PC6, and the outputs at PC0, PC1 and PC3 go high only if the voltage at PC6 is greater than 4V.

DDRC = 0x0F;

while(1)
{
    // a condition of 4V supply to PORTC.6 and Vcc = 5V
    if (PINC == 0b01000000)
        PORTC = 0x0B;
    else 
        PORTC = 0x00;
}

To learn how to code and simulate using AVR Studio 5, visit this page.

So, here we end up with the port operations that can be performed with an AVR MCU. Though the concept has been explained using ATMEGA16/32, the concepts are equally good for any AVR MCU! Just go through the datasheet in order to get an idea of the registers.

Thank you for reading this! The comments’ box is down below! 😉

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128 responses to “I/O Port Operations in AVR

  1. Hi,

    I need your suggestion in avr coding.

    I have 3 different frequency values that I have as input. A corresponding conversion factor for calculation of intensity is : Light intensity (i)=frequncy (i)/100 for i=1,2,3

    Now I perform my calculation using 3 inputs LI(1), LI (2) and LI(3).

    Ad(1) = Log10(2.383/LI(1))
    Ad(2) = Log10(2.383/LI(2))
    Ad(3) = Log10(2.383/LI(3))

    net absorbance= avg. of all 3

    How do I implement these calculation in atmega32. What should be the avrcode corresponding to identifying the inputs and doing this calculation to display the output?

    • You can include the math.h header file where you can have functions for log operations. Define LI and Ad as float. Rest should be fine. Btw, where are you taking the input for the frequency – from ADC?

  2. heyy Mayank…i really appreciate the way u explain…i am making a breathlyzer using the butterfly board and want to get the input from the breathlyzer into that. how do i do that…a simple explaination would help

    • Hello Jagroop
      Which breathlyzer are you using? If possible please send me the link of the datasheet of the breathlyzer you are using.

  3. Hi Mayank,

    Can you please help me out with this easy but very commonly used coding style ? Below you will find the code.

    I already have read out “simulator using topic” written by you. After that I am trying to debug my code using the simulator but it is not working properly. Will you please help me out? I think this will be also very helpful to other newbies like me.

    Problem: On the simulator when I am selecting bits on the PINB in the following way,
    PINB=00000011 or PINB = 0xFF
    I cannt see my output on PORTD in the following way,
    PORTD = 0xFB;
    rather I am having different output.

    Thanks in advance. I will be waiting for your reply.

    Code: http://pastebin.com/BpeKa57x

    • Hi Tonoy

      I have simulated the same code and it was working fine. There are small things that need to be taken care off.

      1. Please check that you have selected the proper simulation device, it should not be Atmega8 in the simulator, in the code it is mentioned clearly Atmega48PA, for ease you can use Atmega16 or 32.

      2. Once the while loop starts executing, the very first if statement, which ever is true is considered and the output is of that case.

      3. In this code, the if statement outcome is TRUE for any non zero value, it results FALSE only if there is zero.

      4. While performing bitwise AND operation, if the resultant is non zero, it is executed else it goes to next condition

      I would suggest you to do a manual calculation for the given input and you will find that for input 0b00000011 both the Case 2 and 3 are true but because of the else if structure only Case 2 will be executed.

      If you delete the Case 2 and again execute then for the input 0b00000011 you will get the previously Case 3 i.e. 0xFC as output.

      Now for the input 0xFF, the very first case if true, so it will not check any other case because of else if structure and the output will be 0xFE.

      Now if you replace all the “else if ” with ” if ” statement then, the last True condition will fetch you the corresponding output.

      You can easily watch this (in Atmel Studio) by enabling the Debugging mode and then Right click on driveBuffer variable and select “Add Watch”.

      Hope this will solve your doubts. 😀
      If you still find any difficulty, feel free to contact us….

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  5. hi…what can i use only one bit in a register in my codes,for example whats the true form for this cod : while (!(ADCSRA.4)) {…}

    • Hi Mr Gi!
      You need to use bitwise operators. If you want to access 5th bit of ADCSRA register, then you can write: (ADCSRA & 0b00010000)
      I would suggest you to read the Programming 101 tutorial to know more about bitwise operators.

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  9. I’m having an issue where I have LEDs for output on port d, and also inputs – one of which operates on a threshold type level.
    The problem is that as I monitor pin 7 to see if it has turned to a 1 (threshold reached/exceeded) I want to flip an LED on to show that this has happened – the detection works fine unless I alter the port d state to turn on this LED, and then suddenly the threshold changes.
    I know it’s a bit hard to explain, but I am unsure how to put it better.
    I am only changing the one bit on the port d to flip the LED on or off. If I leave the LED alone, the threshold detection works fine. Once the LED is on, POOF, the detector pin no longer detects the input threshold correctly. Turn off the LED, and POOF, all is OK again. I’m wondering if turning on the LED is switching on an internal pull-up resistor or something…

    • Hi Norman,
      How do you know that without the LED your threshold detection works fine? I mean, how are you checking (without the LED) that your threshold detection works?

      To add to it, I would ask you how have you powered up your microcontroller – i.e. the source of power – is it a battery, or through your computer, or using a power adapter? Because sometimes it might happen that if the power supply isn’t good or is inconsistent, then connecting the LED might draw a little more current, which would eventually reduce your effective Vcc, thus reducing your threshold level.
      -Max

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  12. DDR and PORT registers may be both written to, and read.
    so i have question-
    What is the use of reading these two registers؟

    • Masam,
      You read from these two registers if you wanna know the previous value written to it. This condition usually doesn’t arise however. Makes sense?

    • Initial value of DDR is zero.
      Initial value of PORT is also zero.
      But initial value of PIN is not defined. Whenever you try to read in using PIN, it takes in the real-world values at that instant.

  13. while(1)
    {
    // a condition of 4V supply to PORTC.6 and Vcc = 5V
    if (PINC == 0b01000000)
    PORTC = 0x0B;
    else
    PORTC = 0x00;
    }
    i don’t understand-“a condition of 4v supply to PORTC and Vcc=5v”
    when this code will work…. and what is threashold voltage?

    • Alamin,
      Threshold voltage is the voltage above which a HIGH signal is considered HIGH.

      For example, I am giving a chain of input pulses to the microcontroller with HIGH = 2.8v and LOW = 0.1v. Assuming that threshold voltage (Vth) = 2.5v, the HIGH input will be considered 1 whereas LOW input will be considered 0. If instead of that, I give input with HIGH = 2.2v and LOW = 0.1v, keeping the Vth same, both HIGH and LOW will be considered as 0 by the microcontroller. Makes sense?

  14. Hi,
    for my final year project i’m using Atmega 16 interfacing lcd. My purpose here is to display an analog input coming in to any one of the atmega adc inputs which is then to be displayed on the lcd. As i’m weak in coding can you please help me code this. I’m trying really hard since a week but to no success. your help will really be appreciated..
    my email id is sidmody92@gmail.com

    • Hey Siddharth,
      Sorry, but we can’t. We don’t solve people’s homework here. If you are weak in coding, I don’t understand why did you choose this project in the first place! My suggestion would be to go and learn C, and then visit my site again. Things will make more sense to you then. Thanks for stopping by! Good luck!

    • Your best bet is maybe to switch to the Arduino (ATmega328P) as it has an LCD communication library already which makes projects like yours extremely simple.

  15. Hi there – just thought I would drop an update as to what has happened with the project since I last came here looking for advice.

    Yes, I am actually still using D6 for my audio trigger input.
    If you want to skip the blah-de-blah-blah and go right to the audio input part of the demo, go to about 4:30 in the video.

  16. Hai, for the past ten years I was using only PIC Microcontrollers. Now I have started learning AVR. I use Atmel Visulal studio 6. In PIC I find very to ease to toggle a pin. like
    while(1){
    RC0=1;
    delay1s()l
    RC0=0;
    delay1s();
    }

    Is there any easy method to toggle a pin in ATmega 16

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