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Digital to Analog converter concept problem

  1. Jan 19, 2014 #1
    1. The problem statement, all variables and given/known data
    Hello PF :smile:
    i am stuck with a concept of weighted resistor D/A converter.
    For a 4-bit resistive ladder,

    VA={V0/R0+V1/(R0/2)+V2/(R0/4)+V3/(R0/8)}/{1/R0+1/(R0/2)+1/(R0/4)+1/(R0/8)}

    attachment.php?attachmentid=65838&stc=1&d=1390196087.png

    I want to find out how the equation is derived but have no clue.

    Thanks.
     

    Attached Files:

  2. jcsd
  3. Jan 20, 2014 #2

    maajdl

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    Write the equations of this circuit and solve.
    For example:
    V0 = R0 i0 + RL iL
    ...
    iL=i0+i1+i2+i3
     
  4. Jan 20, 2014 #3

    rude man

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    You'll have a hard time doing that because your expression omits RL which is clearly necessary.
     
  5. Jan 20, 2014 #4
    finally i have ended up with this:


    VA/RL=V0/(R0+RL) + 2V1/(R0+2RL) +4V2/(R0+4RL) + 8V3/(R0+8RL)

    Any way to get rid of RL??
     
  6. Jan 20, 2014 #5

    rude man

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    No. Assume RL=0. What would VA be then?
     
  7. Jan 20, 2014 #6
    thanks for the reply :smile:

    if RL is zero then VA would be grounded and hence it will be zero.
     
  8. Jan 21, 2014 #7

    rude man

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    Right. So VA depends a lot on RL, does it not.
    So your given answer can't be right.

    So now let's do it right:
    Sum currents at VA to zero: (V3-VA)/(R0/8) + ... = VA/RL.
    Fill in the blanks. and solve for VA.
     
  9. Jan 21, 2014 #8
    the equation --VA={V0/R0+V1/(R0/2)+V2/(R0/4)+V3/(R0/8)}/{1/R0+1/(R0/2)+1/(R0/4)+1/(R0/8)} is right from my text book.

    i have got that equation on Post No. 4
     
  10. Jan 21, 2014 #9
    here is the equation:

    (V3-VA)/(R0/8) + (V2-VA)/(R0/4)+(V1-VA)/(R0/2)+(V0-VA)/(R0)=VA/RL
     
    Last edited: Jan 21, 2014
  11. Jan 21, 2014 #10

    rude man

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    Well, your textbook is right wrong. See my next post.
     
  12. Jan 21, 2014 #11

    rude man

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    That is wonderful!
    Oops, you forgot the divide signs for several terms on the left ....
     
    Last edited: Jan 21, 2014
  13. Jan 21, 2014 #12
    My text book says """ Resistance RL represents load to which divider is connected and RL is considered to be large enough so that it does not load the divider network.


    For a 4-bit resistive ladder,

    VA={V0/R0+V1/(R0/2)+V2/(R0/4)+V3/(R0/8)}/{1/R0+1/(R0/2)+1/(R0/4)+1/(R0/8)} """
     
  14. Jan 21, 2014 #13
    so does RL>>R0 affects the equation?
     
  15. Jan 21, 2014 #14

    rude man

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    If RL >> R0 then RL drops out, approximately.
    So then, compute VA for RL >> R0 ...

    I have the answer if you want to check yours ...

    PS R0 drops out too ... which is obvious if you look at your given solution. Multiply numerator and denominator by R0.
     
    Last edited: Jan 21, 2014
  16. Jan 21, 2014 #15
    Can you explain me how RL drops out?
    And what does it mean by ""RL is considered to be large enough so that it does not load the divider network.""
     
  17. Jan 21, 2014 #16

    rude man

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    Solve the equation in your post #9 for VA. Then let RL >> R0.
    That will answer both your questions.

    (The divider network is R0, R0/2, R0/4 and R0/8.)
     
  18. Jan 21, 2014 #17
    solving that i get
    (8V3+4V2+2V1+V0)/15=VA
     
    Last edited: Jan 21, 2014
  19. Jan 21, 2014 #18
    thanks a lot!!!:smile:
    I got it.:thumbs:



    (My text book should have stated that it is appox.)
     
  20. Jan 21, 2014 #19

    rude man

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    Good work!
    Yes, it should have.
    A finite RL reduces the output voltage by a constant: R0/(15R0 + RL) instead of just 1/15.

    The idea here is that in a real application there will be some finite value of RL; this computation shows that it should be >> R0 or the output voltage will be a function of RL which makes the d/a converter inaccurate since RL is usually not constant.
     
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