Emitter-Follower Question..

Discussion in 'Homework Help' started by Permia, May 6, 2010.

  1. Permia

    Thread Starter New Member

    May 6, 2010
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    0
    [​IMG]


    Assuming the transistor is conducting :. VVE = 0.7V,

    And that the transistor is operating in its active region :. IB is porportional to IC..


    IC = β x IB

    IE = IB + IC

    and so

    IE = IB(1+β)
    = 101 x Ib

    80x10^3xIB + 1.2x10^3xIE + 0.7 = 20

    and after working that out I get:

    201200IB = 19.3

    Then using the above equations:

    IB = 95.92μA
    IE = 9.69mA
    IC = 9.59mA

    VE = 11.628

    but

    VC = 26V ?

    I've been over it and over it but cant figure out what I've done wrong. Can anyone explain why I get this?
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    This equation is not correct. You are assuming that the only current flowing in the 80K is the base current - which won't be the case. Keep in mind that the 80K and 8K in combination with the 20V supply are forming a voltage divider from which the base current is drawn.
     
  3. Ethan

    New Member

    Feb 7, 2009
    6
    1
    You need to use: Ie=(1+B)Ib. You will have: -Vth +IbRth +Vbe +ReIe then you can solve for the collector current and Vc.
     
  4. Permia

    Thread Starter New Member

    May 6, 2010
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    I follow what you guys are saying.

    I've calculated Vth as 1.81

    and Rth as 7.27

    and so I get

    -1.81 + 7.27xIB + 0.7 + 1.2x10^3x101xIB = 20

    am I right so far?

    This gives IB as 174.16μA
     
  5. Ethan

    New Member

    Feb 7, 2009
    6
    1
    You are applying KVL in the Base-Emitter-Source loop where did you get the =20 from?

    -Vth +IbRth +Vbe +Re(1+B)Ib=0
     
  6. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    No ...

    -1.81 + 7.27x10^3xIB + 0.7 + 1.2x10^3x101xIB = 0
     
  7. Permia

    Thread Starter New Member

    May 6, 2010
    8
    0
    yeah figured that after I typed it out. Once I wrote it down on paper I realised I was being an idiot.

    Time for bed methinks..

    IB = 8.64μA?


    ..can get the rest from there.

    Thanks guys.
     
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