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Need help!! electrical stuff -diodes

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  • WoAiMeiMei
    WoAiMeiMei's Avatar
    180 posts since Jul '07
  • justinfoo
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    18 posts since Oct '02
  • WoAiMeiMei
    WoAiMeiMei's Avatar
    180 posts since Jul '07
    • 01 May `08, 6:24AM

      will there be any exceptions? or it's good?

       

      ps: will check it in the afternoon.. got test soon D:

      Edited by WoAiMeiMei 01 May `08, 6:48AM
  • WoAiMeiMei
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    180 posts since Jul '07
  • Moderator
    eagle
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    16,362 posts since Aug '01
    • 05 May `08, 11:12AM

      I presume Q1 has been done...

       

      Q3)
      a)
      Vout = Vz = 8V
      IR2 = 8/880 = 9.1 mA
      Iz is just positive, can be assumed to be zero/neglible

      therefore, current through Vs is 9.1 mA (Kirchoff's current law)

      Then voltage drop across R1 = 0.0091 * 750 = 6.82V
      Voltage drop across diode = 0.64 V

      Thus, Vs = sum of voltage drops = 15.46 V (Kirchoff's voltage law)



      b) Vz is still 8V
      So Vout is still 8V, and IR2 = 9.1 mA

      So voltage drop across R1 = 41 - 8V - 0.64 V = 32.36V (Kirchoff's voltage law)
      so IR1 = 32.36 / 750 = 43.147 mA

      Thus, Iz = IR1 - IR2 = 34 mA (Kirchoff's current law)

      Edited by eagle 05 May `08, 11:13AM
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    eagle
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    16,362 posts since Aug '01
    • 05 May `08, 11:15AM

      Q2)

      0 to 5 sec
      Both off, so Vo = 0V

      5 to 8 sec
      Only V1 is on, so voltage drop of 0.6V, means Vo = 3.4V

      8 to 13 sec
      Can use law of superposition
      From V1, it is 3.4V
      From V2, it is 4.4V
      So add up, you get 7.8V

      13 to 15 sec
      Both off, so Vo = 0V

  • WoAiMeiMei
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    180 posts since Jul '07
    • 05 May `08, 11:34AM

      thanks, will go home and try it out cos the questions always change. 1 question for 1) though.. the negative values... i plugged that in the equation but i got a very small number. should it be 0?

      i notice the two negative values give u the same answer

  • WoAiMeiMei
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    180 posts since Jul '07
    • 05 May `08, 8:30PM

      edit: ignore above..

       

      i managed to solve it on the second round :D

      now i understand how to do them great....! however...

      the 3rd part in the diode graph is wrong... i tried it twice n two different graphs and the superposition method doesnt apply... D:

  • WoAiMeiMei
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    180 posts since Jul '07
    • 05 May `08, 8:38PM

      ahh how do i do this one D:

      (using same time intervals)

      they seem to cross.. so i guess average? :s

  • Moderator
    eagle
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    16,362 posts since Aug '01
    • 05 May `08, 9:54PM

      no not superposition... mistake there...

      I think the diode for the smaller voltage will not turn on at all. Means you consider only the bigger voltage source.

  • WoAiMeiMei
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    180 posts since Jul '07
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