Common base emitter junction with resistor at the emitter

Discussion in 'Homework Help' started by hitmen, Dec 5, 2008.

  1. hitmen

    Thread Starter Active Member

    Sep 21, 2008
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    This exam question has stumped me too. A resistor is placed at the emitter of the common base-emitter junction instead of the collector. I was stumped.

    For NPN transistor assume: Vbe = 0.6V when just on
    Vbe = 0.7V Bf = 100 in active region
    Vbe=0.75, Vce = 0.2V

    Question : When Vin = 5V in fig Q4a, transistor is in active region
    (i) Write equation to show relationship between Ib, Ic , Ie
    (ii) Express Ic and Ie in terms of Ib
    (iii) Hence determine Vo.

    Here is my answer:

    Ib + Ic = Ie

    Ic=100Ib

    Vo = 10-0.2 = 9.8V (obviously wrong)

    Can someone guide me through?
    Thanks.
     
    Last edited: Dec 5, 2008
  2. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    Your attachment is blurry, and the lower part of the schematic is clipped off.
     
  3. hitmen

    Thread Starter Active Member

    Sep 21, 2008
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    I have already posted an better auto adjusted image. I have already typed in the question and my answer
     
  4. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    Auto adjusted, huh? I guess I need new glasses.:rolleyes:
    What is the value of the base resistor?
    You said Vin=5V. Your attachment appears to show Vb=0.5V.:confused:
     
  5. hitmen

    Thread Starter Active Member

    Sep 21, 2008
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    Ignore the statements in the attachment just looking at the diagrams will do:) I know my photography skill sucks. For Vb = 0.5V, I knew how to do that part of the question.

    It is the next part that is the problem. In the attachment, Vcc = 10V, Vb = 5V, Rb = 10kΩ, Re = 5kΩ. Find V o (located just above Re).

    Thanks.
     
  6. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    It's simple algebra.

    Vbase=5-IbRb
    Vo=Vbase-0.7
    Ib=Ie/(β+1)
    Ie=Vo/Re

    Solve for Vo.:)
     
  7. hitmen

    Thread Starter Active Member

    Sep 21, 2008
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    There are too many unknowns. Do we need Vcc to come into the picture? How are we gonna calculate Ic?
     
  8. hitmen

    Thread Starter Active Member

    Sep 21, 2008
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    I got it. I got it. 4 equation and 4 unknowns. Thanks!:)
     
  9. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    Rb is known.
    Re is known.
    β is known.

    There are not too many unknowns. I calculated the answer. You can too.
    Vcc is irrelevant as long as it is higher than +5V, and it is. I think you already know that Ic=β*Ib.
    I'll give you some more hints.
    You can start with Vo=Vbase-0.7, and substitute the above equation for Vbase. For Vbase, substitute its equation. For Ib, substitute its equation. For Ie, substitute its equation. If you do all that, you can calculate Vo.

    EDIT: I was composing while you posted your most recent entry.
     
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