Electronic – Using 10 bit ADC for only 8 bits on a PIC MCU

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I have a PIC16f1824, which has a 10bit built-in ADC, but I only want 8 bit precision. I don't just want an 8 bit value, but actual 8 bit conversion i.e. higher samples per second at the cost of bits. Here I am stuck because the ADRESH register get loaded only once whole 10 bit conversion is performed.

Best Answer

This thread: http://www.edaboard.com/thread155430.html should help. Basically, you'll have to do some converting on your own, but configuring the ADC to left justify will give you bits 9-2 in the ADRESH byte, which you can then use as your 8-bit value.

I don't think you can configure the chip to give you 8-bit precision at the cost of higher samples/sec. That's a design design made when you order the part. I could be wrong though and I would consult with the appropriate data sheet.

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You should be using a bitwise AND. The expression in the first while loop evaluates to zero, which causes the compiler to remove the loop altogether.

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You need to re-read the ADC status register inside the loop; otherwise, you're just re-testing the bit you read the first time. Also, you need to reverse the sense of the test — you want to repeat the test if the bit is "1" and drop out of the loop once it switches to "0". It may also be necessary to toggle chip select to the ADC on each read, as shown in the comments below.

unsigned char CheckStatus(void)
{
  unsigned char adcStatus;
  do {
    /* TBD: assert chip select here? */
    spi(0x40);
    adcStatus = spi(0xFF);
    /* TBD: negate chip select here? */
  } while ((adcStatus & 0x80) == 0x80);                         
  return adcStatus;
}

Best Answer

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