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First experience with an NPN BJT

  1. Mar 8, 2015 #1

    Zondrina

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

    I had a few questions about the npn BJT with regards to designing a few circuits. The questions and my attempts are below.

    For the calculations, assume ##\beta = 140## and ##V_{BE} = 0.7V## for npn active mode.

    1. Fixed bias circuit: Design the resistor values required to bias the circuit in the active region, such that ##I_C = 1 mA##, ##V_{CE} = 3V## (calculate ##R_C, I_B, R_B##):

    Screen Shot 2015-03-08 at 9.33.56 AM.png

    I don't know why they are asking for ##R_C## in this circuit when they gave ##R_C = 12 k##. I think it may be a typo, but I'm not certain. I will solve for ##I_B## and ##R_B## assuming ##R_C## was a typo.

    2. Self bias circuit: Design the self bias circuit for the BJT CE amplifier shown in question 3:

    Screen Shot 2015-03-08 at 9.34.10 AM.png

    I assume they want me to find ##R_1## and ##R_2##.

    3. BJT CE Amplifier: Design the CE amplifier (as per the specifications given). Derive the input impedance ##Z_i## as well as the open circuit voltage gain ##\frac{v_o}{v_b}##:

    Screen Shot 2015-03-08 at 9.33.25 AM.png

    What will happen to the gain under loaded conditions?
    What will happen to the gain if ##R_E## is bypassed?
    What will happen to the input impedance if ##R_E## is bypassed?

    2. Relevant equations


    3. The attempt at a solution

    1. Just to prove ##R_C## was a typo:

    $$I_C = \frac{V_{CC} - V_{CE}}{R_C} \Rightarrow R_C = \frac{V_{CC} - V_{CE}}{I_C} = \frac{15V - 3V}{1 \times 10^{-3} A} = 12k \Omega$$

    Now for active mode operation we know:

    $$I_C = \beta I_B \Rightarrow I_B = \frac{I_C}{\beta} = \frac{1 \times 10^{-3} A}{140} = 7.143 \mu A$$

    Now that we know the base current we can write:

    $$I_B = \frac{V_{CC} - V_{BE}}{R_B} \Rightarrow R_B = \frac{V_{CC} - V_{BE}}{I_B} = \frac{15V - 0.7V}{7.143 \times 10^{-6} A} = 2.002 M \Omega$$

    The circuit will be in active mode with the given specifications.

    2. First, use Thevenin's theorem to simplify the base of the BJT. Short out ##V_{CC}## to find ##R_{th}##:

    $$V_{th} = \frac{R_1}{R_1 + R_2} V_{CC}$$
    $$R_{th} = R_1 || R_2 = \frac{R_1R_2}{R_1 + R_2}$$

    So we will have ##V_{th}## and ##R_{th}## in the base of the BJT. Applying KVL to the BE loop now:

    $$V_{th} = R_{th} I_B + V_{BE} + R_E I_E = \frac{R_{th}}{1 + \beta} I_E + V_{BE} + R_E I_E = \left(\frac{R_{th}}{1 + \beta} + R_E \right) I_E + V_{BE} $$

    Now we must find ##R_{th}## and ##I_E##. We apply the stability criterion to select ##R_{th}##, namely we choose:

    $$R_{th} = 0.1 \times (1 + \beta) R_E = 0.1 \times (1 + 140) (3.3 k \Omega) = 46.53 k \Omega$$

    Now we must find the bias point so we can determine ##I_E##. Let the load line be given by:

    $$V_{CE} = - (R_C + R_E)I_C + V_{CC}$$

    The bias point should be in the middle of the load line, so:

    $$V_{CE,Q} = \frac{V_{CC}}{2} = \frac{15V}{2} = 7.5V$$
    $$I_{C,Q} = \frac{V_{CC}}{2(R_C + R_E)} = \frac{15V}{2(8.2k \Omega + 3.3k \Omega)} = 0.652 mA$$

    Now going back to the previous KVL equation for ##V_{th}##, we find:

    $$V_{th} = \left(\frac{R_{th}}{1 + \beta} + R_E \right) I_E + V_{BE} = \left(\frac{46.53 k \Omega}{141} + 3.3 k \Omega \right) (0.652 mA) + 0.7V = 3.067 V$$

    This gives us two equations in the two unknowns ##R_1## and ##R_2##:

    $$3.067 V = \frac{R_1}{R_1 + R_2} (15V)$$
    $$46.53 k \Omega = \frac{R_1R_2}{R_1 + R_2}$$

    The first equation gives:

    $$0.204 = \frac{R_1}{R_1 + R_2}$$
    $$R_2 = \frac{R_1}{0.204} - R_1 = 3.9R_1$$

    Solving the second equation then yields:

    $$46.53 k \Omega = \frac{R_1 (3.9R_1)}{R_1 + 3.9R_1} = \frac{3.9 R_1^2}{4.9R_1} = 1.256 R_1$$
    $$R_1 = \frac{46.53 k \Omega}{1.256} = 37.05 k \Omega$$

    Subbing back into the first equation then gives:

    $$R_2 = 3.9R_1 = 3.9(37.05 k \Omega) = 144.5 k \Omega$$

    Phew, that was a lot of work. I hope my answers to questions 1 and 2 look okay. If they do then I will try question 3.

    If someone could let me know if my calculations look okay, then I will proceed to #3.

    Thank you in advance.
     
  2. jcsd
  3. Mar 8, 2015 #2

    Zondrina

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    I thought I would at least give question #3 a try while I wait to see if I had any clue as to what is happening.

    We know ##R_1, R_2, R_C## and ##R_E## from question #2. We wish to find the gain ##\frac{v_o}{v_b}## and the input impedance ##Z_i##. A small signal model would be appropriate I believe.

    From prior we know ##I_C = 0.652 mA## and so we can compute:

    $$g_m = \frac{I_C}{V_T} = \frac{0.652 mA}{25 mV} = 26 mS$$

    Hence we can find ##r_{\pi}##:

    $$r_{\pi} = \frac{\beta}{g_m} = \frac{140}{26 mS} = 5.385 k \Omega$$

    We can calculate the gain:

    $$A_v = \frac{v_o}{v_b} = - \frac{g_m R_c}{1 + g_m R_E} = - \frac{(26 \times 10^{-3} S) (8.2 \times 10^3 \Omega)}{1 + (26 \times 10^{-3} S) (3.3 \times 10^3 \Omega)} = - 2.46 \frac{V}{V}$$

    Now we wish to find the input impedance taking the source resistance into account:

    $$Z_i = R_1 || R_2 || (r_{\pi} + R_S) = 37.05 k \Omega || 144.5 k \Omega || (5.385 k \Omega + 10 k \Omega) = 10.11 k \Omega$$

    I'm not entirely sure this is correct, but hopefully I have some idea.
     
    Last edited: Mar 8, 2015
  4. Mar 8, 2015 #3

    LvW

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    I must admit that I didn`t check all your calculations.
    However, the last equation for Zi seems to be incorrect.

    In the task desription you are asked "What will happen to the input impedance if RE is bypassed?"
    However, RE does not appear in the formula at all.

    Without the source resistance Rs the input impedance at the base node is

    Z=R1||R2||r,in.

    Two cases for the dynamic input resustance of the BJT:

    (1) Without a capacitor CE: r,in=rπ + β*RE (because negative feedback caused by RE);

    (2) With CE: r,in=rπ
     
  5. Mar 8, 2015 #4

    Zondrina

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    Okay so the bypass capacitor is not connected in the emitter branch with ##R_E##. So taking the source resistance into account, what you're telling me is:

    $$Z_i = R_1 || R_2 || (r_{\pi} + (1 + \beta)R_E + R_S) = 37.05 k \Omega || 144.5 k \Omega || (5.385 k \Omega + (141)(3.3 k \Omega) + 10 k \Omega)$$

    ?

    Sorry for the many other lengthy calculations.
     
  6. Mar 8, 2015 #5

    LvW

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    No - why do you think that RS is - together with the two other parts in parallel to R1 and R2?

    The resistance Z as given in my post is referenced to the base node.
    That means, if your signal source is connected to the BJT via an external resitot RS the resulting input resistance as seen by the signal source is (RS+Z).
     
  7. Mar 8, 2015 #6

    Zondrina

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    Yes sorry, ##R_{in}## is looking into the terminals of ##v_i##. So it would be:

    $$Z_i = R_1 || R_2 || (r_{\pi} + (1 + \beta)R_E) + R_S = 37.05 k \Omega || 144.5 k \Omega || (5.385 k \Omega + (141)(3.3 k \Omega)) + 10 k \Omega$$

    I'm guessing the rest of the calculations look okay since no one has mentioned anything.
     
    Last edited: Mar 8, 2015
  8. Mar 9, 2015 #7

    LvW

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    Yes - looks OK to me.
     
  9. Mar 10, 2015 #8
    In second question you calculated Vth using voltage division right ? Then doesn't it should be R2/(R2+R1)
     
    Last edited: Mar 10, 2015
  10. Mar 10, 2015 #9
    You calculated Vth using voltage division right ? Shouldn't it be R2/(R1+R2)?
     
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