Can I make 8-bit ADC using PIC16F876 ?

[aljamri]

Hi,

With valuable help from ETO fellows, I completed my first 10bit ADC using PIC16F876 (see this Thread)
Now I’m trying to implement the knowledge I gained there, to make another ADC, but with 8bit. Is it possible using the same PIC 16F876 ?
Thanks.

 

[Misterbenn]

Well if you can make a 10 bit one then yes an eight bit is easy - just dont use the two lowest bits....

 

[hexreader]

To make 8 bits even easier......

During initialisation, make sure bit ADFM is cleared to zero in register ADCON1 (left justified). You can simply read register ADRESH to get the 8 bits that you want.

Hope this helps

 

[aljamri]

Well if you can make a 10 bit one then yes an eight bit is easy - just dont use the two lowest bits....

Hi Misterbenn;

But thats going to make 11111111 is = 1.25V if my +ref=5V if I ignore two MSB.

Thanks.

 

[aljamri]

To make 8 bits even easier......

During initialisation, make sure bit ADFM is cleared to zero in register ADCON1 (left justified). You can simply read register ADRESH to get the 8 bits that you want.

Hope this helps

Hi hexreader;

Again my Vref=5V, I'll get maximum Digital Output by applying 1.25V only.

Thnaks

 

[ericgibbs]

Hi hexreader;

Again my Vref=5V, I'll get maximum Digital Output by applying 1.25V only.

Thnaks

hi aljamri,
Its doesn't work like that.

If you have a 5Vref and set the adc for 10 bit resolution you will get for a 5V adc input 5V/1023 ie approx 4.88mV /bit

If you set the adc for 8bit resolution, for a 5V adc input you will get 19.6mV/bit.

Its the resolution that changes not the voltage range...OK.?

 

[be80be]

Your looking at this wrong

10 bit is 1023 = 5 volts b1111111111

127.5 x 0.0048875855327468230694037145650049 = .623 volts

8 bit is 255 = 5 volts b11111111

127.5 x 0.01960784313725490196078431372549 = 2.5 volts at 8 bit

Sorry ericgibbs you posted the same thing I'll leave it you hadn't posted when I was there

 

[Misterbenn]

Just an additional note to the comments made above-

The main point is that you would get rid of the two LOWEST bits, the LSBs NOT the two Highest bits, the MSBs.

This way you only lose resolution, not range. hope that helps

 

[aljamri]

Thanks for all.

If I want to send my ADC output to computer using parallel port, and I'll send it in form of 8-bit to read it in the computer as 5Volts, will the following work?

void main() {
  ADCON0 = 0b11000000;
  ADCON1 = 0b00001111;
  TRISA  = 0xFF;
  TRISC  = 0;
  PORTC  = 0;
  do {
  
      PORTC = ADC_Read(0);
     
  } while(1);
       }

 

[aljamri]

hi aljamri,

If you set the adc for 8bit resolution,

That's exactly what I'm looking for, how to set ADC to 8-bit instead of 10 bit.

 

[aljamri]

Your looking at this wrong

10 bit is 1023 = 5 volts b1111111111

127.5 x 0.0048875855327468230694037145650049 = .623 volts

8 bit is 255 = 5 volts b11111111

127.5 x 0.01960784313725490196078431372549 = 2.5 volts at 8 bit

Do you mean it can be done just by ignoring two bits ?! or I have to set the ADC differently to get 8-bit output

 

[ericgibbs]

That's exactly what I'm looking for, how to set ADC to 8-bit instead of 10 bit.

Morning aljamri,
Corrected.!
When set for 8bit ADC, read the ADRESH register after the conversion, value will be 0 thru 255

 

[Mr RB]

(edit) removed.

Sorry I was saying the same thing as Eric, as he was posting it.

 

[aljamri]

Now it is clear for me.

Thanks for all, I like ETO