Circuit analysis

Discussion in 'Homework Help' started by safrinho, Feb 28, 2013.

  1. safrinho

    Thread Starter New Member

    Feb 27, 2013
    3
    0
    Dude im sorry for posting it on your topic, but i'm getting error 500 when i try to create a new topic, so im trying to get some help on yours

    I have to draw the waveform on R2, but i cant. The only thing i know is that in the positive semicicle i will get max voltage because the diode will be open. I cant figure out the exact moment when the diode open (the voltage on cathode is higher than anode). Is there a way to calculate this?

    Alson in the negative semicicle, i calculated the max voltage equals 40V. On multisim, it gives me around 32V. I really cant see what im missing.

    Thanks in advance


    https://www.dropbox.com/s/blvsrl3n5cx5odc/circuit.jpg
     
  2. WBahn

    Moderator

    Mar 31, 2012
    17,743
    4,790
    I'll ask the mods to split this off. So let's not post anything further to either topic until they've had a chance to do that.
     
  3. Georacer

    Moderator

    Nov 25, 2009
    5,142
    1,266
    Moved it to its own thread. If you need another title, say so.
    As a general rule, try to limit your new thread titles under 20 characters.
     
  4. WBahn

    Moderator

    Mar 31, 2012
    17,743
    4,790
    What diode model are you using for this problem?

    To find the point at which the diode switches, you need to determine the point at which whether the diode is on or off makes no difference. This will be when the diode is forward biased but doesn't actually have any current flowing in it. So impose those constraints and analyze the circuit with an unknown value of the voltage being output by the source. That will let you determine the point in the waveform that the circuit changes behavior due to the switching of the diode.
     
  5. safrinho

    Thread Starter New Member

    Feb 27, 2013
    3
    0
    IN4007

    I cant understand why the diode actually "open". Cause im my mind, all the voltage should be "consumed" by the resistences, and the diode would always be forward biased.

    I found now that when the AC signal is at 60 V. the diode is open, and remain open untill the voltage peak. I really cant understand why does this happen, since as what i said, all of the power should be at R1 and R3, then the voltage at cathode would be always zero.
     
  6. WBahn

    Moderator

    Mar 31, 2012
    17,743
    4,790
    I think you've got a lot of misconceptions going on. You need to show your work so that we can identify exactly where those misconceptions are rearing their ugly heads.
     
  7. safrinho

    Thread Starter New Member

    Feb 27, 2013
    3
    0
    well i managed to solve it. I set the voltage at cathode at 40V so there is no current flowing through it, then i could calculate the voltage at source as 60V. Im currently working on negative semicicle, but it seems to be easier =D
     
Loading...