MOSFET

Discussion in 'Homework Help' started by ihaveaquestion, May 6, 2009.

  1. ihaveaquestion

    Thread Starter Active Member

    May 1, 2009
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    First question is about this:

    http://img14.imageshack.us/img14/8672/13249405.jpg

    I'm only concerned with part a... no need for the truth table, I guess my question is regarding equation manipulation

    The equation I get, which was marked right for me was
    Z = NOT(A + (NOT(A)*B) + (B*C))

    Kind of hard to see from that, but its A + AB + BC with a line over the whole thing to not it, as well as a small not line over just the A in the AB term.

    However in the back of the book it is Z = NOT(A+B) which comes up with the same outputs as my equation in the truth table.

    Apparently some simplification was done using Demorgan's Theorem or something so I was wondering if someone could easily spot how to simplify it

    Second question
    I forgot how to do the following problem

    http://img14.imageshack.us/img14/8229/84474254.jpg

    I remember it has something to do with N amount of resistors in parallel R/N is the equivalent R, so as N goes to infinity the effective (total) resistance goes to 0.. and this is contradictory to what I might have thought, i.e. more resistors = more power consumed... so in this case more resistors = less power consumed? somethin' like that
     
    Last edited: May 6, 2009
  2. DonQ

    Active Member

    May 6, 2009
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    Homework problems...

    The answer is probably in the text.
     
  3. ihaveaquestion

    Thread Starter Active Member

    May 1, 2009
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    Actually they aren't, and even if they were... the book still doesn't explain things even half as well as the people on this forum
     
  4. vvkannan

    Active Member

    Aug 9, 2008
    138
    11
    For the first question

    (A + A'B + BC)' = A'.(A'B)'.(BC)' -Demorgan's law

    = A'.(A+B').(B'+C')

    = A'B' (B'+C') [A'A is 0]

    =A'B'+A'B'C'

    = A'B'(1+C') [1+C'=1]

    Z =A'B' = (A+B)'

    For the second question i think the equivalent of ON-state resistance of the Mosfets should equal R and hence maximum power is (Vs²/4 R) .Iam not sure though,lets wait together for some conformation
     
    Last edited: May 6, 2009
  5. ihaveaquestion

    Thread Starter Active Member

    May 1, 2009
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    Thanks vvkannan,

    Your Boolean Algebra makes sense...
     
  6. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    In the second problem, the minimum value of load resistance on Vss (maximum current) occurs when all inputs are high. Assuming R is internal to the gate, this is then the condition for maximum dissipation. You should be able to calculate dissipation as a function of R, Ron, and n. Come back with your answer and we can critique it.

    EDIT: If R is not internal to the gate, then the answer will be different.
     
  7. ihaveaquestion

    Thread Starter Active Member

    May 1, 2009
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    I'm still pretty lost..

    1) If my set of inputs are all 1, i.e. all the resistors in parallel are on:

    Rtotal = Ron/N where N -> infinity Ron -> 0 so I only have R

    2) If I have them all off, no current flows correct?

    I think maybe a fundamental example would help, this is one of my weaker areas (as if the others already aren't weak)
     
  8. steveb

    Senior Member

    Jul 3, 2008
    2,433
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    Remember the power formula P=V^2/Rt in the general case.

    In your case, V is constant and unchangable and equal to Vs, so maximum power will occur when the resistance is minimum.

    Rt=R+Ron/N

    Obviously Rt gets smaller as N gets larger.
     
  9. ihaveaquestion

    Thread Starter Active Member

    May 1, 2009
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    Thank you, this makes sense...
     
  10. vvkannan

    Active Member

    Aug 9, 2008
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    Sorry ihaveaquestion,I thought we need to find the maximum power absorbed by the Mosfets and hence applied max.power transfer theorem.steveb's reply cleared it up.
     
  11. ihaveaquestion

    Thread Starter Active Member

    May 1, 2009
    314
    0
    No problem vvkannan, I appreciate your input
     
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