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Help with 16F876

Discussion in '8bit Microcontrollers' started by CJunk, Jun 1, 2004.

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  1. CJunk

    CJunk Guest

    Hi there. Just was wondering if anyone can lead me to a site or give me
    info that will show me basic operation of the ADC in the PIC16F876. I have
    tried getting it to work but cannot seem to do it. I have checked out
    www.microchip.com and the info there but it dont help me. Would like a
    simple schematic and software guide if possible.thx
     
  2. CJunk

    CJunk Guest

    Dont worry I figured it out.

    thx
     
  3. david

    david Guest

    was it a simple mistake why tell us how you fixed so quickly it might help
    someone else.
    David
     
  4. CJunk

    CJunk Guest

    ok.... Initially I was connecting the 5k pot to the analogue input (RA0),
    but didnt put the connection to the center of it and + to the one pot pin
    and Gnd to other pot pin. I was just using two of the pins on the pot. I
    hope what I typed came out all right. ...and hopefully it will help some
    beginners out there.
     
  5. Howard Long

    Howard Long Guest

    The other 'gotha' I've encountered on the 16F876 ADC is the relatively low
    impedance of the ADC input.

    Kind Regards, Howard
     
  6. CJunk

    CJunk Guest

    Could you please let me know what 'impedance' affects? thx
     
  7. Eric Bohlman

    Eric Bohlman Guest

    It means that the ADC will draw a non-trivial amount of current from the
    source of the voltage that it's measuring. If the voltage source has a
    high impedance, then drawing too much current from it will actually lower
    the voltage it's supplying, and the reading will be wrong.

    Consider a hypothetical ADC input with an input impedance of 5K. If you
    were to connect it directly to a 5-volt supply, you'd accurately read 5
    volts. But now consider connecting it to the 5-volt supply through a 1K
    resistor. You'd read only 4.17 volts, because the impedance of the ADC
    input forms a voltage divider with the resistor. If, OTOH, the input
    impedance were 50K, you'd read 4.9 volts; you'd still have a voltage
    divider, but the drop across the source resistor would be much smaller.
     
  8. Howard Long

    Howard Long Guest

    .... I think that about covers it!

    Regards, Howard
     
  9. CJunk

    CJunk Guest

    thx
     
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