Settling time of ADC with changed bit range

Discussion in 'Homework Help' started by elmaggan, Jul 15, 2015.

  1. elmaggan

    Thread Starter New Member

    Jul 15, 2015
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    0
    Hi! I need help with part of an assignment. It is about a 12-bit AD Converter.

    First with given Vref=3.3 V and digital output i were to calculate Vin at that time => Vin=780.879 mV. I also had to calculate the LSB=805.664*10^-6 V.

    So now an analog multiplexer is connected, with inner resistance Ri=7000 ohm and generates 0-Vref V. It now looks like an RC circuit. Given values are the R in the RC circuit is R=3000 ohm and C=12 pF.

    First task was to find the time of the capacitor (which is initially uncharged) to reach 1/4 LSB accuracy in the worst case scenario.

    I found a formula which i used that stated: t=-ln(1/(2^N*4))*RC where i used R=R+Ri, given C and N=12.

    Next task is the one i need help with! Now i should with same conditions as above calculate the time of the capacitor to reach 1/4LSB accuracy but only when using the 6 most significant bits of the result.
    How do i do that? I tried the same formula but with N=6 but it gave apparently wrong answer. I also divided first calculated time in 2 because i thought it would help, but it didn't.

    Maybe i need to use another formula or something, im clueless.

    Thanks!
     
    Last edited: Jul 15, 2015
  2. crutschow

    Expert

    Mar 14, 2008
    13,009
    3,233
    Instead of using a canned formula that you found and don't understand you should work with the basic equations.
    You calculate the step size of one LSB based upon the number of bits.
    Then you use the basic RC exponential equation Vo/Vin = [1 - e^(t/τ)] (t is time and τ is the RC time-constant) to determine how long it takes for that 1 LSB level to change to within 1/4 LSB of the desired value.
    You should be able to use basic Algebra to convert that equation into the form you need (solve for t).
     
  3. elmaggan

    Thread Starter New Member

    Jul 15, 2015
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    0
    Okey. So i calculated the LSB by using the bit like:
    LSB=Vref/2^N=3.3/2^12=805.664*10^-6 V

    I recognize the formula, so the way i see it Vin is the final value the capacitor gets. So i think that should be 3.3 then. But i have no clue what to put in Vo, what is that? The desired value? I don't understand the sentence "determine how long it takes for that 1 LSB level to change to within 1/4 LSB of the desired value" Do you mean that Vo=3.3-1/4LSB or Vo=3.3/4 or something?

    Or have i gotten it all wrong?
     
    Last edited: Jul 15, 2015
  4. crutschow

    Expert

    Mar 14, 2008
    13,009
    3,233
    Vo is the desired voltage value (in this case to within 3/4 of an LSB value plus the initial voltage) after it settles for the desired time.
    Vin is the step voltage change (due to the bit change) plus the initial voltage value.
     
  5. elmaggan

    Thread Starter New Member

    Jul 15, 2015
    5
    0
    Okey, so the initial voltage is zero right? What does it mean by step voltage change, is that the step size aka. 1 LSB?
     
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