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4 bit ADC and DAC

  1. Sep 13, 2007 #1
    Here is my problem:
    There is a 4 bit ADC with an input range of -5 to +5 Volts.

    I would like someone to check my answers.

    Q. How many states does the ADC have?
    A. The ADC has 2^4 = 16 states.

    Q. What voltage does each step represent?
    A. range = 10 V
    no. of steps = 16-1 = 15
    10/15 = .6667 V

    Q. Vout = -5 + (digital count*.6667)
    What is the largest number an 8 bit binary number can represent? Determine Vout for
    this number. If it does not equal 5V, explain.
    A. largest 8 bit binary number can represent is 255
    Vout = -5 + (255 *.6667)
    Vout = 165 V
    largest number 4 bits can represent is 15
    255 exceeds capability of 4 bit ADC

    Q. What is the dynamic range of an 8 bit DAC in dB?
    A. range of 8 bit DAC is 0 to 255
    20 log (255) = 48.13 dB
  2. jcsd
  3. Sep 13, 2007 #2


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    Your answers are all correct. However, make sure your convention for expressing dynamic range really involves 20 log and not 10 log.

    - Warren
  4. Sep 13, 2007 #3
    I'm a little confused on how dynamic range for the DAC is defined. I don't think there was an explicit definition given in class, but I found the log formula in the lab book. Can you shed light on this idea?

  5. Sep 13, 2007 #4


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    The dynamic range is just the ratio of the largest and smallest output signals. Since the ADC has 255 steps, the largest possible output is 255 times larger than the smallest possible non-zero output (code one). When expressed in decibels, this ratio (255:1) is about 24 dB (using 10*log). I don't know why you would be expected to use 20*log.

    - Warren
  6. Sep 13, 2007 #5


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    Staff: Mentor

    Maybe because it's the ratio of voltages, and not powers. Along the same lines as the gain in a Bode plot.
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