Can anybody solve attached questions

Discussion in 'General Electronics Chat' started by mishra87, May 9, 2016.

  1. mishra87

    Thread Starter Member

    Jan 17, 2016
    170
    2
    Could any body help me out to solve attached circuit of diode and transistor !!!
     
  2. ElectronicMotor

    Member

    May 1, 2016
    53
    6
    Question 1:
    (1)
    Loop1:
    6V = Vr + Vd1 + Vd2
    = I1.1k + 2.Vf
    I1 = (6 - 1.4) / 1k
    = 4m6 A
    Loop2:
    Vd2 = Vr
    = I2.1k
    I2 = 0.7 / 1k
    = 0m7 A
    The current through D1 is I1 = 4m6 A.
    The current through D2 is I1 - I2 = 3m9 A.

    (2)
    Loop1:
    10V = Vr + Vd1 + Vd2
    = I1.1k + 2.Vf
    I1 = (10 - 1.4) / 1k
    = 8m6 A
    Loop2:
    Vd1 = Vr
    = I2.1k
    I2 = 0.7 / 1k
    = 0m7 A
    Loop3:
    Vd2 = Vr
    = I3.500
    I3 = 0.7 / 500
    = 1m4 A
    The current through D1 is I1 - I2 = 7m9 A.
    The current through D2 is I1 - I3 = 7m2 A.

    Question 2:
    Ib = Vbb / Rb
    = 5 / 47 k
    = 106 uA
    Ic = ∞
    Of course, the C-E junction has just avalanched, shorting the B-E junction, and an infinite current is flowing through the collector. Why do they teach kids these inoperable and useless circuits ?
    :D
    Note: Theoretically, a current source can be put in series with a voltage source, so that the voltage source maintains a voltage across the current source, and the ideal current source maintains a current through the ideal voltage source.
    :rolleyes:
     
    Last edited: May 9, 2016
    Shaira Rose likes this.
  3. dl324

    Distinguished Member

    Mar 30, 2015
    3,250
    626
    Is this homework?
     
  4. kubeek

    AAC Fanatic!

    Sep 20, 2005
    4,670
    804
    Why do you think so? I would say that Ib=(Vbb-0.7)/Rb, and Ic=Ib*hFE and I can´t see any reason for CE to avalanche.
     
  5. AnalogKid

    Distinguished Member

    Aug 1, 2013
    4,546
    1,252
    0 for 2.

    Ib = (Vbb - Vbe) / Rb = 4.3 / 47 k = 91.5 uA
    Ic = Ib x hfe = 91.5 uA x 80 = 7.32 mA

    And by the way...

    P = E x I = Vce x Ic = 9 x 7.32 mA = 66 mW

    Depending on its package, the transistor probably is not overheating.

    ak
     
  6. ElectronicMotor

    Member

    May 1, 2016
    53
    6
    You got me. Rash. But I have never seen a common emitter circuit without a collector resistor, but then again, even after years of uni, I don't think I understand that circuit. :rolleyes:
    After thought, I kind of knew this was an exam question, and decided to 'throw a spanner in the works'. As I do ?
    ;)
     
    KJ6EAD likes this.
  7. AnalogKid

    Distinguished Member

    Aug 1, 2013
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    Dangle-biasing always is the very first way taught to bias a transistor, and (hopefully) the first way used as a reason to do it some other way.

    ak
     
  8. ElectronicMotor

    Member

    May 1, 2016
    53
    6
    Common Collector.jpg
    I agree that understanding the fundamentals is good, but mixing ideal voltage sources with BJT's that are kind of ideal current sources, is dumb.
     
  9. JohnInTX

    Moderator

    Jun 26, 2012
    2,348
    1,029
    How about you don't do that?
     
  10. ElectronicMotor

    Member

    May 1, 2016
    53
    6
    How about I do ?
     
  11. JohnInTX

    Moderator

    Jun 26, 2012
    2,348
    1,029
    Because you were asked nicely not to?
     
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