Common Base problem

Discussion in 'Homework Help' started by Kubbe, Nov 14, 2006.

  1. Kubbe

    Thread Starter New Member

    Nov 14, 2006
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    Hello,

    I am trying to calculate the working point of the transistor used in this common base circuit. Please see the this image: http://c213-100-21-126.swipnet.se/cb_part.jpg. The circuit is larger than just the common base part, but that's where I'm stuck at the moment. I've marked it with a red box.

    I am trying to make a "direct current scheme" of that part and then "split" it by the base and emitter part (at least I think that's what I should do). I don't really know how to treat the input signal when creating the direct current scheme, though. Could anyone help me with this part and maybe give tips on how to calculate the working point of the transistor?

    The symbols are European and the instruction text is in Swedish, but please ignore that :) R1, R2 etc are resistances, e(t) is the input signal, which is an ideal voltage source (sinus shaped AC).


    Thanks in advance!
     
  2. hgmjr

    Moderator

    Jan 28, 2005
    9,030
    214
    I would pick as my starting point, the establishment of the DC voltage being applied to the base of the PNP transistor. It is a simple voltage divider so that should be a straightforward exercise.

    With the value of the base voltage in hand, I would imagine that the emitter was at ground and compute the voltage across the emitter resistor under this specific condition.

    I would then multiply that current times alpha (based on a beta of 100) which though I don't read or speak Swedish I believe that is what B=100 indicates.

    You would then have an estimate of the current that would be flowing in the collector.

    Since you know that one end of the collector is connected to -E or -12V and you now the currenct flowing in the collector load resistor, you have all of the information you need to establish the dc setpoint of the collector of the PNP transistor.

    You should be able then to evaluate the effect on that dc setpoint as the input signal swings positive from ground.

    hgmjr
     
  3. Kubbe

    Thread Starter New Member

    Nov 14, 2006
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    0
    Thank you for your reply!

    I'm not sure what assumptions I can make when calculating the voltage at the base. When simply

    calculating between -E and 0 V and over only R2 and R3, the voltage falls by 1.2 Volts, so it

    would be -1.2V at the base. Is this correct?

    I wasn't sure how to calculate the current though the emitter resistor (this is R1, right?), but I tried

    a slightly different path of solving the problem;
    I identified two circuits:
    R1 - R2 - (UBE) - R1 and
    R4 - (UBC) - R3 - R4.
    I then used KVL to form two equations. Unfortunately, there were three unknown variables in

    these, so I had to make the assumption that UBE = 0.7 V. Then I actually got a dc setpoint with

    UCEQ = -2.8V , ICQ = 1.2 mA.
    Is the above a correct path to go as well? Are these circuits correct and enough to get an

    answer?

    Thanks!
     
  4. hgmjr

    Moderator

    Jan 28, 2005
    9,030
    214
    -1.2V is correct. Good start.
    The emitter resistor is R1 as you have surmised.
    Things get a little murky here. I am not clear on how you arrived at your two equations.

    What you need to do is determine the current that is flowing in the emitter resistor. Since you have correctly identified the base voltage at -1.2V and you can assume that the input signal end of R1 is at ground when the signal passes through 0V, then you have the two pieces of information that you need to compute the voltage across R1.

    HINT: Don't forget to account for the emitter base forward voltage of 0.7V in calculating the voltage drop across R1.

    Once you have the current flowing in R1, you can multiply it by alpha to obtain the collector current. With the collector current, you can then establish the quiescent DC bias point for collector of the transisitor.

    There is still a little bit of work to do before we can declare total victory.

    hgmjr
     
  5. Kubbe

    Thread Starter New Member

    Nov 14, 2006
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    Cheers! I calculated the current flowing through R1 into the transistor to be 5 mA (voltage at emitter would be -1.2+0.7 = -0.5 V). I don't know what you mean by alpha, but with B=100 and IE = IC + IB and IC = 100*IB, I get the following expression: IC = IE*(100/101) = 4.95 mA.

    With this collector current, I get that the voltage at the collector is -12 + 4.95mA*R4 = -1.1 V. Giving me that UCE = -1.1 - (-0.5) = - 0.6V. Did I get it right this time? :)
    Thanks
     
  6. hgmjr

    Moderator

    Jan 28, 2005
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    214
    That looks very good. Great job.

    The equation for alpha is:

    alpha = beta/(beta+1) or in the case of beta = 100 that would be 100/101.

    You actually used alpha correctly in your calculation of Ic.

    Congratulation, You are well on your way to becoming familiar with transistors.

    hgmjr
     
  7. Kubbe

    Thread Starter New Member

    Nov 14, 2006
    4
    0
    Great! Thanks for your help!
    There's also a second part to the assignment, but I think I just might manage that on my own.
     
  8. hgmjr

    Moderator

    Jan 28, 2005
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    214
    You are very welcome..

    You appear to be grasping the finer points of transistor analysis so I am quite confident that you can tackle the second part without too much difficulty.

    hgmjr
     
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