Spanning multiple ports in C

[Brian Hoskins]

Hi there all,

I've presented myself with what I believe to be a small problem but at the moment I'm banging my head against the desk trying to come up with a good method of solving it.

Basically, I have 10 LED displays that I want to select from within my C program. Only 1 display will be selected at a time (I'm multiplexing them). I've done this before with 4 LED displays and to do it I created an array as follows;

const char select [4] = {1, 2, 4, 8} ;

And then when I wanted to select one of the four displays I just wrote something like;

PORTA = select[*insert display number here*]

And then PORTA would output the required bit pattern to enable the intended display. That all works fine. Here's my problem;

I now want to select from 10 displays instead, using the same idea. But that required more than 8 bits so it's larger than PORT A!

To remedy this I want to use all 8 bits of PORT A, and also two bits from PORT E to make a 10 bit port. I want to define this somehow, so that I can say something like

SELECTPORT = select[*insert display number here*]

where SELECTPORT refers to all of PORTA and also the required two bits of PORTE.

And this is the basis of my question - how do I go about defining this? I'm racking my brains and I can't think how but I'm certain there's going to be an easy way of doing it.

Suggestions?

Cheers!

Brian

 

[3v0]

Make two arrays which contain the LATA values need to turn on each led. The 0th element of each array is associated with led 0.

in c

rom const char lataVal[10]={0x??,0x??,..};
rom const char latcVal[10]={0x??,0x??,..};

void turnOnLed(char led)
{
    LATA=LATC=0x00; // all off to prevent ghosting
    LATA=lataVal[led];
    LATC=latcVal[led];
}

If you need to use the other 6 bits of porte for output you need to use masking when writting that port.

 

[Brian Hoskins]

Hmmmm so there isn't an easy way of "defining" your own I/O port which is made up of bits from more than one I/O port then?

I read through your solution but I'm not certain I understand it. Well, if I can't solve it by combining two ports perhaps I can come up with another method. Back to the drawing board!

Thanks for the response,

Bri

 

[Brian Hoskins]

Actually, thinking about it I might be able to do this using pointers....

 

[3v0]

This is the easy way but I made it look hard.

I redid the example give it another look.

char selectPortA[10]={0x??,0x??,..};  // 10 PORTA values
char selectPortE[10]={0x??,0x??,..};   // 10 PORTE values

void turnOnLed(char n)
{
    PORTA=selectPortA[n];
    PORTE=selectPortE[n];
}
// instead of
//   PORTA = select[*insert display number here*]
// use
turnOnLed(*insert display number here*);

You could use a macro but it is clearer with the function.

If you still do not understand ask about it.

3v0

 

[Brian Hoskins]

Yeah I see what you're getting at actually.

So the 'walking 1' values would be inserted for the first 8 values of the selectPortA array and the remaining two values of the selectPortA array would = 0.
For the remaining values of the array (9 and 10) the walking 1 would hop over to PortE where the first 8 values of the selectPortE array would = 0 and the walking 1 values would be inserted for the remaining 2 selectPortE array values (9 & 10)

So I guess it would look like this;

const char selectPortA [10] = {1, 2, 4, 8, 16, 32, 64, 128, 0, 0} ;
const char selectPortE [10] = {0, 0, 0, 0, 0, 0, 0, 0, 1, 2};

Then to select a display I can use a value n as you suggested. A bit wasteful, but certainly functional!

Have I understood correctly? I'm going to try that - but tomorrow now because it's getting late here

Bri

 

[3v0]

The array method is robust in that it works for more complicated multiplexing where more then one bit is on at at time.

For a simple walking 1's try this.

PORTA=PORTE=0x00;  // turn off last selection

if (n<8)
  PORTA = 1<<n; // 1 shifted n times is correct bit
else
  PORTE = 1<<(n-8);

3v0

 

[Pommie]

Or, even simpler,

  PORTA = 1<<n;
  PORTE = ((int)1<<n)/256;

Mike.