[SOLVED] problem with Floating point display on LCD

[vibodha]

I have made a small program to measure voltage(0-5) on AN2. Its works fine but have a little problem. When the voltage less than 1v (ie. 0.99) it shows wrong voltage on LCD.
ie when voltage 0.568V , Lcd shows 5.68v
0.0213v -----> 0.213v

Can anyone correct me..
here is the program (mikroc)

// LCD module connections
sbit LCD_RS at RB4_bit;
sbit LCD_EN at RB5_bit;
sbit LCD_D4 at RB0_bit;
sbit LCD_D5 at RB1_bit;
sbit LCD_D6 at RB2_bit;
sbit LCD_D7 at RB3_bit;

sbit LCD_RS_Direction at TRISB4_bit;
sbit LCD_EN_Direction at TRISB5_bit;
sbit LCD_D4_Direction at TRISB0_bit;
sbit LCD_D5_Direction at TRISB1_bit;
sbit LCD_D6_Direction at TRISB2_bit;
sbit LCD_D7_Direction at TRISB3_bit;
// End LCD module connections
char txt1[] = "ADC Example";
char txt[2];
char txt2[5];
unsigned int temp_res;
float data2;

void main() {
ansel=4;  // Configure AN2 pin as analog
  anselh=0;
  c1on_bit=0;
  c2on_bit=0;
   ////////////////////////////////////////////////////////////////
  TRISA  = 0xFF;              // PORTA is input
  Lcd_Init();
  Lcd_Cmd(_LCD_CLEAR);               // Clear display
  Lcd_Cmd(_LCD_CURSOR_OFF);          // Cursor off
  Lcd_Out(1,1,txt1);
  adc_init();
  do {

    temp_res = ADC_read(2);   // Get 10-bit results of AD conversion
   

    data2 =  (temp_res*5)/1023.0;
    FloatToStr(data2,txt2);
    txt2[5]=0;
    Lcd_Out(2,1,txt2);
    Delay_ms(1000);
    }
    while(1) ;


}

 

[jayanth.devarayanadurga]

char txt2[5]; should be char txt2[17];

Remove this line txt2[5]=0;.

 

[vibodha]

char txt2[5]; should be char txt2[17];

Remove this line txt2[5]=0;.

Tnx Mr.Jayanth..
but Still having the same prb.. now its shows more floating points.. but have the same problem..

 

[jayanth.devarayanadurga]

Which PIC are you using. See if you have to configure ADCON1 register.

 

[vibodha]

I'm Using PIC16F887 .. I have used ADC library in mikroC

 

[jayanth.devarayanadurga]

Try this code.

---

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// LCD module connections
sbit LCD_RS at RB4_bit;
sbit LCD_EN at RB5_bit;
sbit LCD_D4 at RB0_bit;
sbit LCD_D5 at RB1_bit;
sbit LCD_D6 at RB2_bit;
sbit LCD_D7 at RB3_bit;
 
sbit LCD_RS_Direction at TRISB4_bit;
sbit LCD_EN_Direction at TRISB5_bit;
sbit LCD_D4_Direction at TRISB0_bit;
sbit LCD_D5_Direction at TRISB1_bit;
sbit LCD_D6_Direction at TRISB2_bit;
sbit LCD_D7_Direction at TRISB3_bit;
// End LCD module connections
 
char txt1[] = "ADC Example";
char txt[23];
float data2;
 
void main() {
 
    TRISA  = 0xFF;              // PORTA is input
    TRISB  = 0x00;              // PORTA is input
    PORTA  = 0x00;              // PORTA is input
    PORTB  = 0x00;              // PORTA is input
    ANSEL  = 0x04;              // Configure AN2 pin as analog
    ANSELH = 0x00;
    ADCON1 = 0x80;
    c1on_bit=0;
    c2on_bit=0;
        
    Lcd_Init();
    Lcd_Cmd(_LCD_CLEAR);        // Clear display
    Lcd_Cmd(_LCD_CURSOR_OFF);   // Cursor off
    Lcd_Out(1,1,txt1);
  
    do {
 
        data2 = ADC_read(2);    // Get 10-bit results of AD conversion
        data2 =  data2 * 5.0 / 1023.0;
        FloatToStr(data2,txt2);
        Lcd_Out(2,1,txt2);
        Delay_ms(1000);
    }
    while(1) ;
 
 
}

---

 

[embpic]

what reason to increase size of array?
is it like
float = 4byte
so

5*4 = 20;?

 

[vibodha]

Mr. jayanth I tried the way but the problem is same... but somehow I found another method here is the changed code.

unsigned long ADC_Value, value, ADC_Value1, value1;
char *volts = "00.00 V";


/////////// Inside the main Function //////////////
ADC_Value = ADC_read(2);
    value = (ADC_Value * 500)/1023;
   volts[0] = value/1000+ 48 ;
   volts[1] = (value/100)%10 + 48;
   volts[3] = (value/10)%10 + 48;
   volts[4] = (value)%10 + 48;
   Lcd_Out(2,1,volts )   ;

I got this from internet searching and of that I cannot understand some parts:shock::shock: such as the meaning of "48" of the following code

 volts[0] = value/1000+ 48 ;

Thank You very much for your contribution

 

[jayanth.devarayanadurga]

float is 4 bytes. FloatToStr() conversion function needs array size to be minimum 17 bytes.

 

[jayanth.devarayanadurga]

48 in decimal is equal to character '0' and hex 0x30. By adding 48 you are converting the integer value to character. 0 + 48 = 48 = '0' = 0x30.

 

[vibodha]

Mr.jayanth.devarayanadurga Thank you very much for the help