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BJT DC Analysis

  1. Apr 12, 2015 #1

    Zondrina

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    Homework Helper

    1. The problem statement, all variables and given/known data

    Screen Shot 2015-04-12 at 10.50.51 AM.png

    For the circuit above, draw the DC equivalent circuit.

    Then compute ##I_B, V_C, V_{CE}##.

    2. Relevant equations


    3. The attempt at a solution

    So the DC equivalent circuit was easy to draw. My concern is about the final answer. The answer given does not match the answer I've obtained and I'm wondering who is correct.

    After drawing the DC circuit, I obtained the following relationships:

    $$I_C = \frac{V_{CC} - V_C}{R_C} = \frac{21 - V_C}{2k}$$
    $$I_B = \frac{V_C - V_{BE}}{R_1 + R_2} = \frac{V_C - 0.7}{81k}$$
    $$I_E = \frac{V_E}{R_E} = \frac{V_E}{1k}$$

    Now we know ##I_C = \beta I_B##, so the first and second equations can be related:

    $$40I_B = \frac{21 - V_C}{2k}$$
    $$I_B = \frac{V_C - 0.7}{81k}$$

    Simplifying we get:

    $$I_B = \frac{21 - V_C}{80k}$$
    $$I_B = \frac{V_C - 0.7}{81k}$$

    Solving we get:

    $$\frac{21 - V_C}{80k} = \frac{V_C - 0.7}{81k}$$
    $$V_C = 10.9V$$

    Going back to the original equations, we see:

    $$I_C = \frac{21 - V_C}{2k} = \frac{21 - 10.9}{2k} = 5.05 mA$$
    $$I_B = \frac{V_C - 0.7}{81k} = \frac{10.9 - 0.7}{81k} = 0. 126 mA$$

    Then we know:

    $$I_E = I_C + I_B = 5.05 mA + 0.126 mA = 5.176 mA$$

    From which we find:

    $$I_E = \frac{V_E}{1k}$$
    $$V_E = (5.176 mA)(1k) = 5.176 V$$

    Hence we know:

    $$V_{CE} = V_C - V_E = 10.9V - 5.176V = 5.72V$$

    Now all of my analysis seems reasonable to me, but for some reason the answer is given:

    $$I_B = 0.0995 mA$$
    $$V_C = 12.8 V$$
    $$V_{CE} = 8.76 V$$

    This answer does not match mine whatsoever. Have I done something wrong?

    EDIT: Here is a picture of the analysis given in the answer:

    Screen Shot 2015-04-12 at 11.17.22 AM.png
     
  2. jcsd
  3. Apr 12, 2015 #2
    This doesn't look right:
    I didn't check the rest. I think the solution you were given is correct.

    You have the two equations in three unknowns ##I_C = \beta I_B,I_C + I_B = I_E##, so find a third using KVL and your knowledge of ##V_\mathrm{BE}##.
     
  4. Apr 12, 2015 #3

    gneill

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    Staff: Mentor

    When you calculated IB you did not take into account the fact that the emitter is not at ground potential, but has some potential VE due to the emitter resistor carrying current (IB and IC).


    Edit: Oops! Beaten to the punch by milesyoung!
     
  5. Apr 12, 2015 #4

    Zondrina

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    So I have the two equations:

    $$I_C = \beta I_B$$
    $$I_E = I_C + I_B$$

    That connection from the collector to the base is what's causing a little bit of confusion. Why do they write ##I_C + I_B = I_E## in the spot where usually they write ##I_C##? In other words, why are they saying ##I_E## flows through ##R_C##? If I understood this, then everything else is clear.
     
  6. Apr 12, 2015 #5
    Consider where the base current comes from in this circuit compared to the ones you're used to.

    Maybe think of it this way: The emitter current flows down through ##R_C##, splits into ##I_B## and ##I_C##, and then forms up as ##I_E## again through ##R_E##.
     
  7. Apr 12, 2015 #6

    gneill

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    Staff: Mentor

    Maybe a picture to help?

    Fig1.gif
     
  8. Apr 12, 2015 #7

    Zondrina

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    I see how I'm supposed to think about the current "breaking up" now. Pictures are worth a million words.

    Thank you so much to both of you. That one connection threw me off my usual thinking. So:

    $$-21V + I_ER_C + I_B(R_1 + R_2) + 0.7V + I_ER_E = 0$$
    $$I_ER_C + I_B(R_1 + R_2) + I_ER_E = 20.3V$$

    Then using ##I_E = (\beta + 1)I_B##:

    $$I_B(R_1 + R_2) + (\beta + 1)I_B(R_C + R_E) = 20.3V$$

    The rest of the analysis follows.
     
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