Common Base CB BJT

Discussion in 'Homework Help' started by Vicros, Jan 25, 2012.

  1. Vicros

    Thread Starter New Member

    Jan 22, 2012
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    0
    Hi all,

    Back again with another BJT configuration, this time it is a Common base question.

    I am asked to final suitable values for V_{cc} and R_E with a quiescent point I_{CQ} = 1mA

    I have tried doing Thévenin on each of the terminals but end up with both V_{cc} and R_E in the equation. As well as using my regular method of

    V_{cc} = (R_C + R_E)I_C + V_{CE}

    Im having a mental block and cant seem to find V_{cc} on its own first.

    Please find attached the question and my attempt at solving it
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    Given Vb is 0V then Ve will be -0.7V.

    With Vceq=5V then Vc=+4.3V. This will lead you to Vcc since you are given Ic=1mA & Rc=2k2.
     
  3. Vicros

    Thread Starter New Member

    Jan 22, 2012
    13
    0
    Brilliant, thank you

    I think i did it right, I ended up with Vcc as 6.5V when i did:

    Vcc = Ic*Rc + Vc

    I assumed that it was correct, so I moved on and continued to do most of the questions. Turns out that the common base AC analysis is quite similar to the common emitter i think.

    I did what i believe is an alpha model of the question...
     
  4. Vicros

    Thread Starter New Member

    Jan 22, 2012
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    Ok, I have finished off the question now

    The two major things that i would like some one to look over is the AC analysis circuit that i drew and the last section where it ask to calculate the transresistance gain v_o / i_i

    What i did was replaced i_i with v_i / R_i so that the equation came down to A_V * R_i

    but my answer is very large which doesnt look right
     
  5. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    I would dispute both your values of voltage gain and input resistance.

    To find Av one would use Rc||Rload=2.2k||4.7k=1.5k

    Also keep in mind that Vo and Vi will be in phase for the CB configuration - at midband frequencies.

    Also the effective input resistance at the emitter is dominated by re which is quite small. RE and re are in parallel rather than in series as your analysis suggests.

    Re-check your values.
     
  6. Vicros

    Thread Starter New Member

    Jan 22, 2012
    13
    0
    thanks for pointing that out, i can now see that i did forget to include the load resistance.

    I changed my values. i now have smaller values for both voltage and current gain.
     
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