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beginner with C
Guest
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| I look for C cod for optical encoder, thks |
 Posted: Wed Apr 21, 2004 1:35 am |
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I'm begginer with CCS and with C at all. And I need from C code for optical encoder, whitch determine directions.
Thks in advance. |
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Ttelmah
Guest
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| Re: I look for C cod for optical encoder, thks |
| Posted: Wed Apr 21, 2004 2:13 am |
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| beginner with C wrote: | I'm begginer with CCS and with C at all. And I need from C code for optical encoder, whitch determine directions.
Thks in advance. |
There are lots of ways of doing this, with the simplest/best, depending on the actual speed that the pulses are likely to arrive. This is a version designed to use the portb 'changed' interrupt, with minimum hardware needed to handle reasonable speeds. This requires that the two pulses are coming into pins b4, and b5, and that these are setup as inputs, and the interrupt enabled. You can go significantly faster, by using a logic gate to decode the direction, and using the CCP pins to count the movement. Details of this are given in a couple of Microchip application notes, on 'servos'.
| Code: |
signed int32 position;
#INT_RB
void quad(void) {
static int old;
static int new;
static int value;
//Here I have an edge on one of the quadrature inputs
new=portb;
/*Now I have to decode the quadrature changes. There are four
possibilities:
I can have a rising or falling edge, on each of the two inputs. I have to
look at the state of the other bit, and increment/decrement according to
this. /*
value=new^old;
//'value', now has the bit set, which has changed
if (value & 0x10) {
//Here the low bit has changed
if (new & 0x10) {
//Here a rising edge on A
if (new & 0x20) --position;
else ++position;
}
else {
//Here a falling edge on A
if (new & 0x20) ++position;
else --position;
}
}
else {
//Here the high bit (B) must have changed
if (new & 0x20) {
//Here a rising edge on B
if (new & 0x10) ++position;
else --position;
}
else {
//Here a falling edge on B
if (new & 0x10) --position;
else ++position;
}
}
old=new;
}
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This code just updates the value 'position', which is a 32bit number, increasing it for every edge in one direction, and decreasing it in the other. This decodes all four of the possible quadrature states (giving 2000 changes on a 500 'line' encoder), and despite looking fairly long, executes quickly, with just three integer 'tests', and a single increment/decrement, whichever edge has changed. The limiting factor on speed here, is th interrupt latency of the processor. Typically on a 20MHz 16F877 for example, the code will handle up to about 60000 edges/second (giving a maximum rotation rate of about 1800rpm, on a 500 line encoder).
Best Wishes |
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guest
Guest
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| optical encoder |
| Posted: Wed Apr 21, 2004 5:28 am |
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Hi
I could look on the example from ccs.
You can use an IC LS7083 from
http://www.lsicsi.com/
I found it on internet.
Quadrature Encoder ISR
by David Kott
struct RotaryEncoderStruct {
int unused:4; // RB[0:3]
int A:1; // RB4
int B:1; // RB5
int unused2:2; // RB6-7
};
struct RotaryEncoderStruct RotaryEncoder;
#pragma BYTE PORTB = 0x06
#pragma BYTE RotaryEncoder = 0x06 // Place structure right over PORTB at
location 0x06
USHORT g_usEncoderCount = 0;
#pragma INT_RB
void
PortBInterrupt (void)
{
static BYTE OldPortB;
BYTE byTemp;
static struct RotaryEncoderStruct OldEncoder;
if (OldPortB != (byTemp=(PORTB & 0x30))) {
OldPortB = byTemp;
if (RotaryEncoder.A == OldEncoder.A) {
if (RotaryEncoder.B == RotaryEncoder.A)
--g_usEncoderCount;
else
++g_usEncoderCount;
} else {
OldEncoder.A = RotaryEncoder.A;
if (RotaryEncoder.B == RotaryEncoder.A)
++g_usEncoderCount;
else
--g_usEncoderCount;
}
}
// Handle more PORTB stuff here, like a button, or something.
} |
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beginner with C
Guest
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| Posted: Thu Apr 22, 2004 5:33 am |
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Thanks all
I succssed. |
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