ADC value displaying to LCD as voltage

[thannara123]

I am trying to Displaying ADC value to LCD as voltage ie 00.00Volt.

May i get a tutorial program for it .

I just write the following code .please help me

void value_coverter(unsigned int adc_value)
{   unsigned char temp[8];
    vout = adc_value * 0.00488;
    int_value = vout;   // only decimal part 
    vout = vout - int_value; // for only fractional part 
    temp[0] = int_value /10; // saving first decimal digit
    temp[1]  = int_value % 10; // saving second digit
    temp[2]  = '.';  // point
    vout = vout * 100; // making fractional part to decimal 
    temp[3] = vout /10;
    //temp[4] = vout % 10;
    for(int i=0; i<4; i++)    
    lcd_data(temp[i]+0x38);    
}

I searched the forum a lot of but I didnt get better one .I want to learn it

 

[ec_nisarg]

"0.00488" seems to be a scaling factor. it might be possible that this code is for different circuit. you have to find your scaling factor.
also
"vout" should be float and "int_out" should be integer type.

 

[paulfjujo]

hello,


with this coeff of 0.0048 wich represente 5V/1024
to get maxi= 5.00 , better to scale with 0.488 to get 500.0

you must use floating point variable

float Vout;
.....

Vout= (float) Adc_value * 0.488;
int_Vout=(unsigned int) Vout;
.. etc ..

 

[Easyrider83]

Just for example, like I'm doing this kind of measure

float POS9V_Measure (void)
{
	return ((float)ADC_Get_Result(POS9V_MES) * ((R9 + R10) / R10) * VDD) / ADC_RESOLUTION;
}

Where R9 and R10 - value of resistors in devider (1k and 10k for example)
VDD = 3.3V or 5.0V
ADC_RESOLUTION is 4096 for 12 bit, 1024 for 10 bit and 256 for 8 bit
POS9V_MES - adc channel, where devider is connected
Returned value - voltage in volts.

 

[thannara123]

Vout is declared a global variable ie float vout .

Easyrider83 is float POS9V_Measure (void) function return a correct float value . Can i give it directly give to lcd to display ?

 

[Gorgon]

You need to use a conversion function like sprintf(...) to convert the float value to an ascii characterstring in a buffer. You can then write the contents of the buffer to the LCD.

 

[thannara123]

Whenever I try sprint() it shows only a question mark .

 

[Gorgon]

It says 'int sprintf(char *s, const char *format, ..... );' in my copy of 'The Standard C Library'. It's in <stdio.h>.

 

[thannara123]

sorry didn't get ,can you elaborate ?

 

[paulfjujo]

hello

Simplest is the best,

i suppose you have a function to write a string to your LCD !
convert 0 -> 1023 to 0.00 to 5.00

 float vout;
 unsigned int adc_value;
 unsigned int value;

  void value_converter(unsigned int adc_value)
{   unsigned char temp[8];
    vout = (float) adc_value * 0.488;
    value = (unsigned int) vout;   // only decimal part
       temp[0] =  value/100+48;   //Add 48 to get the ASCII character
       temp[1]='.' ;
       temp[2] = ( value/10)%10+48;
       temp[3] =  value % 10 +48;
       LCD_Write_Text(1,1,temp) ;
     
}


//the test 
  
   LCD_CLS();
   for (i=0;i<1023;i=i+16)
   {
   value_converter(i);
   Delay_ms(500);
   }
   while(1);