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find o/p voltage

Discussion in 'Electronics Homework Help' started by strameshkumar, Feb 23, 2012.

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  1. strameshkumar

    strameshkumar

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    Feb 23, 2012
    find the Vo of the circuit and explain it?
     

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  2. davenn

    davenn Moderator

    13,865
    1,956
    Sep 5, 2009
    ok so what have you so far considered / worked out?

    break the circuit down into stages and go from there
    initially work out the voltage at the common point of the 2 x 10k resistors.
    That is a standard voltage divider.

    the resistors are the same value so that should tell you something about the
    voltage measured at that point

    tell me what you think that voltage should be and why you think so :)

    cheers
    Dave
     
  3. electr0Dave

    electr0Dave

    3
    0
    Feb 23, 2012
    Hi

    Beware, then there is another resistor that forces greater current flow.

    Consider the two resistors as one. Make an equivalent resistor.

    This'll make you have a value nearest to the real.

    Vo is about 1.33 V.

    ;)
     
  4. jackorocko

    jackorocko

    1,284
    1
    Apr 4, 2010
    you sure about that? Besides it really against the policy to just give answers in the homework section.
     
    Last edited: Feb 23, 2012
  5. strameshkumar

    strameshkumar

    25
    0
    Feb 23, 2012
    hi
    when i simulate in Multisim its 1.44V, but divider ckt theroritically 1.66v across diode + load resistor.
    if i minus 0.7v of diode 0.96 is Vo. there is too much of difference
     
  6. jackorocko

    jackorocko

    1,284
    1
    Apr 4, 2010
    I am with you, in that the output voltage is 1.67V minus the diode voltage drop, but proteus and multisim both seem to disagree that there is a 0.7 volt drop across the diode. Not sure why either to be honest. proteus has the output voltage at 1.3 with an ideal battery. *shrugs*
     
  7. electr0Dave

    electr0Dave

    3
    0
    Feb 23, 2012
    I only simulates this in proteus and this is the value it gave.

    For me, this value is right. Of course, this value can be different of real.
     
  8. davenn

    davenn Moderator

    13,865
    1,956
    Sep 5, 2009
    Please dont give answers electr0Dave

    in the homework section we help people to work out the answers for then selves
    instead of some one coming along and spilling the beans

    The Op didnt even get a chance to respond to my original question
    before you gave final answers

    I wanted the OP to learn about resistors used as a voltage divider and move through the circuit from there

    Dave
     
  9. electr0Dave

    electr0Dave

    3
    0
    Feb 23, 2012
    Oh ... OK, sorry.

    But this question also made me think, because I can not solve this with analysis of simple circuits.

    Now I just read this topic and wait for the right moment to give a clue.

    ;);)
     
  10. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,497
    2,839
    Jan 21, 2010
    But the question is, how would you calculate this value? (Or I presume that is the question) "Because I simulated it" seems to be a very unsatisfactory explanation of how you arrived at the figure.

    edit: the "answers" you've got are a good way of determining if your calculations result in a correct value.
     
  11. Laplace

    Laplace

    1,252
    184
    Apr 4, 2010
    As shown in the attached diagram there are several ways to look at the circuit for analysis. I prefer the one where the diode offsets some of the Thevenin equivalent voltage from the input source, and the remainder goes through a two-thirds voltage divider to the output.

    If the diode had no forward voltage drop, then the current would be 167 uA. A typical diode with a forward current of 100 uA will have a voltage drop of 0.5 V which will rise to 0.7 V at 5,000 uA. So Vo will be somewhere between two-thirds of (2.5 - 0.5) and (2.5 - 0.7). Given the simulation result for Vo can you calculate what diode voltage drop the simulator used?
     

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