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Transistor DC Circuit Analysis

  1. Mar 18, 2007 #1


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    I have an amplifier circuit as shown in this picture,
    http://www.fileden.com/files/2006/8/19/175303/4.wmf" [Broken]
    And I need to do a DC analysis on it to find the current into the base lead (I_B) and the voltage over the collector-emitter (V_CE).

    Realizing this is DC and that the capacitor will block any input signal, I can re-draw the circuit as,
    http://www.fileden.com/files/2006/8/19/175303/5.wmf" [Broken]
    And now re-arrange it to be,
    http://www.fileden.com/files/2006/8/19/175303/6.wmf" [Broken]
    Which can be further simplified to,
    http://www.fileden.com/files/2006/8/19/175303/7.wmf" [Broken]
    By using the thevenin equivalent of the circuit to the left of the transistor.

    I now need to write KVL expressions around the loops (in the last picture),

    V_BB = I_B * R_th + .7 volts + R4 * (I_B + I_C)
    Assuming we are in the active region, I_C = β * I_B
    V_BB = I_B * (R_th + R4 (1 + β))

    Solving for I_B,
    I_B = (V_BB - .7 volts) / (R_th + R4(1 + β))
    Where the .7 volts is arising due to the nature of the silicon transistor being used in the circuit, I_B is the current into the base lead, I_C is the current into the collector, R_th is the thevenin equivalent resistance of R1 in parallel with R2, and V_BB is the thevenin voltage equal to the voltage drop across R2 in the circuit.

    Now to find the voltage drop over the collector-emitter,
    It should be,
    V_CE = V_CC – I_C * (R3 + R4)
    Where, again, I_C = β * I_B since we assume we are in the active region of the transistor
    The external power supply used to amplify the input signal, V_CC must drop down to zero over 3 segments…there is one drop over R3, one over R4, so the rest must drop over the collector-emitter.

    V_CE = V_CC – β * (R3 + R4) * I_B,
    This should be the load line of the circuit.

    To help find the end points of the load line,

    Set I_C = 0, so V_CE = V_CC
    Set V_CE = 0, so I_C = V_CC / (R3 + R4)

    I now have (in theory) all the information I desire.
    But I am not at all very confident in my work and if I did everything correctly.

    Also, I am unsure of myself when graphing the I_C curves (as a function of V_CE) to confirm that I am actually operating in the active region of the transistor when I compare my operating point on the load line with the I_C curve.
    I have,
    I_C = β * I_B,
    So if I just choose I_B = some value, and then get a corresponding I_C value, I should be WELL within the active region (which is what it should be), even for small values of I_B.

    So how does all of this look? I am I doing it correctly?
    You-all don’t need to worry so much about the numbers, I am just trying to solve everything symbolically right now, I can go back and plug in values later.
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Mar 19, 2007 #2


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    I'm not sure if your redrawing step for the Thevenin equivalent is valid, because the base-emitter voltage is constrained to be 0.7V (assuming the transistor is active) so the circuit is nonlinear.

    The practical way to go about this sort of problem is to look at the circuit and (from experience) see that it's a sensible looking common-emitter amplifier design.

    So ignoring the base current for now, from the potential divider, Vb = 12*3430/37000 = 1.11V
    Ve = 1.11 - 0.7 = 0.41V
    Ie = 0.41/.372 = = 1.10 mA

    The base current is Ie/beta = less than 0.01mA (assuming beta > 110)

    The current in the base potential divider is 12/37.0 = 0.32mA, so the error in ignoring the base current is too small to bother about, for practical component tolerances.

    Finally, if beta is large, Ic = Ie approximately, and so Vc = 12 - 1.1*4.67 = 6.86V.

    OTOH if you are required to do an "exact" analysis, this approach may not be good enough for your prof - though it's certainly good enough to design amplifiers that work!
    Last edited: Mar 19, 2007
  4. Mar 19, 2007 #3
    unless i'm missing something your last anaylsis doesn't account for negitive signal swing, and don't forget the output cap if your running single ended.
  5. Mar 19, 2007 #4


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    That is the part I am most concerned about getting, findnig the voltage between the collector and the emitter.

    If I use the formula I have come up with in my own analysis,
    V_CE = V_CC – β * (R3 + R4) * I_B
    I get some very large, negative, number which cannot be correct.
    If V_CC = 15 volts, β = 179, (R3 + R4) = 4744 Ohms, and my value of I_B from the other equation of 194 microAmps,
    plugging this all in gets me a Collector-Emitter voltage of about -150 volts, which obviously is wrong, it should have a max of +15 volts and a minimum of zero volts. I am not certain where exactly the problem lies, either in the calculuation of I_B, or in figuring out which resistors to use to calculate the voltage drop, or both.

    How did you get,
    Vc = 12 - 1.1*4.67 = 6.86V

    In calculating V_BB, I think I might be wrong with the (1 + β) term multiplied by R4.
    V_BB = I_B * (R_th + R4 (1 + β))
    I have seen it done somewhere else where it is just,
    V_BB = I_B * (R_th + R4 *β)
    Even if this is true (not "1+" needed, numerically, it really doesnt change the answer much (only by +/- the value of R4, which is already considerably smaller than R_th.

    I have a fair bit of confidence that my Thevenin equivilent circuit is valid since I have seen it done before in my text book and in class.

    How should I account for the negitive signal swing?
  6. Mar 19, 2007 #5


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    First off, I misread your diagram and thought it said Vcc = 12V not 15V. So that threw all the other numbers off. Sorry about that.

    Your equation is OK, but your value of I_B is a lot bigger than mine was.

    If R3+R4 = 4744 ohms (??? I'm seeing R3 = 4672 and R4 = 372 on your first picture so I get 5044 ohms) then because there must be some of the 15V across the transistor, Ic and Ie can't be more than 15/4.744 = 3.2mA

    So Ib must be less than about 3.2/179 = 18 microamps. My estimate (with the wrong power supply voltage and the wrong value of beta!) was about 10 microamps.

    Ohms law across R4, assuming the power supply was 12V and Ic = Ie = 1.1mA

    All this suggests to me your Thevenin calculations are wrong somewhere - but that's a black hole in my memory of how to do circuit analysis, so I would be guessing to try and find the mistake.
  7. Mar 19, 2007 #6


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    Once you have got the operating point sorted out, the limit on negative swing is that the emitter voltage can't go below 0 (or Ib can't go below 0, which amounts to the same thing).

    The limit on positive swing is that V_CE must be greater than 0.

    For a "properly" designed amplifier, you would expect the limits of +ve and -ve swing to be about the same size.

    To do a proper AC analysis you need to include the re from the transistor itself, as well as the external emitter resistance.

    But your first post said the question was a DC analysis, so this isn't relevant to the original question.
    Last edited: Mar 19, 2007
  8. Mar 19, 2007 #7


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    Right you are, I must have added 4672 + 172 instead of 3[72 ohms by mistake.

    Somehow my I_B is coming out about two orders of magnitude too larger. The only reason I can see for this would be possibly some error in the thevenin equivalent since everything else look correct to me.
    If we assume that I_B is about 1/100 of what I am calculating, then I get a reasonable for V_CE.
  9. Mar 19, 2007 #8
    i prototyped one in about half an hour to check if i was thinking right, in your (using that as a referance) first circuit v+ to base is 78k with 12k from base to ground, v+ has 100 ohms going to the collector, the emitier is at ground, both collector and base go through 47uf caps i had laying around out to an 32 ohm speaker, the input was a 32 ohm speaker, actually a pair of headphones. speak in one side amplified out the other, the transister was a pn2222a, i followed the curves on the datasheet and hfe was 300, v+ was a 9 volt battery. sounded quite good for a 5 piece amp and it really doesn't get any simpler. if you prototype you can measure what should be against what is. try something like that with another type of transistor you may have laying around, a tip120 would be good and i'd bet you'd get a high mark if you brought it to class. imho there is no substute for prototyping.
    Last edited: Mar 19, 2007
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