Calculating mid-band voltage gain of multistage BJT

Discussion in 'Homework Help' started by Whyregister, Sep 12, 2012.

  1. Whyregister

    Thread Starter New Member

    Sep 12, 2012
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    So basically this is the initial circuit:

    [​IMG]

    For one of the parts of the question, I have to calculate the mid-band voltage gain. rb = 0, rμ = ∞, Va = ∞

    I am not entirely sure if my small-signal model is correct:

    [​IMG]

    If it is, then I'm thinking Vo = -Rcβ(ib1 + ib2)

    What I am confused about here is that since ib1 is neglected, I cannot use that. I was thinking perhaps doing KCL at the E1B2 node labelled, I can say that ib1 = ib2 - ic1?

    I am at a loss to find parameters in terms of Vs. I first tried doing KVL:

    Vs = is*Rs + ib1*r∏1 + ib2*r∏2 + ie2*re2

    The problem with this is that I don't have is.

    Any help appreciated!
     
  2. fila

    Member

    Feb 14, 2011
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    Why are you using small signal model if you have to find the DC values? Use the Thevenin theorem to find the base voltage of Q1 and base current Ib1.

    EDIT: Sorry, my bad, didn't see the red text! :)

    Why don't you represent Darlington with just one current source?

    βdarlington = β1 * β2 = 100 * 100 = 10 000.
     
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  3. Whyregister

    Thread Starter New Member

    Sep 12, 2012
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    I am sorry, I am not sure what a Darlington is, we haven't learnt that. Although I will look into that now.
     
  4. Audioguru

    New Member

    Dec 20, 2007
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    Rs feeding RB1 and RB2 form an attenuator with an output of 0.833 times the input level.
    Re and Rc have the same value so the transistors have a voltage gain of 1.
    Then the output level is 0.833 times the input level.
     
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  5. Whyregister

    Thread Starter New Member

    Sep 12, 2012
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    I have to do this using the small-signal model, so it won't be acceptable to use short-cuts like that. Also, that's interesting as to why you divided the resistances of Re/Rc ? Is that like a feedback loop or something?

    I reading over Darlington pair, I am not entirely sure but I assume using the Darlington pair model means there's a lot of assumptions and simplifications about some of the currents?

    I am hesitant to use methods not taught as there's a marking scheme that they follow.
     
  6. #12

    Expert

    Nov 30, 2010
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    Consider that Re has a DC current through it. Rc is in series with Re, so it must have the same current.

    Consider that Re has an ac voltage imposed on it, and thus an AC current. It is in series with RC, so Rc must have the same current in both DC and AC.

    This is the principle that Audioguru used to declare that the gain is one.
     
  7. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    The problem says to neglect base currents where appropriate. It doesn't say that you are compelled to neglect Ib1. If you need to use a particular base current to obtain a solution, then use it.

    Since you have dependent current sources, KCL would be appropriate. I see 4 nodes (you don't need to deal explicitly with the node between Vs and Rs), so you will have 4 equations to set up and solve.
     
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  8. Whyregister

    Thread Starter New Member

    Sep 12, 2012
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    It is now solved, although it did take a page of ugly algebraic manipulation. I however didn't really set up 4 equations with 4 unknowns.

    I got ib2 and ie2 in terms of ib1. Using voltage division and KCL at the first node, I got is in terms of Vs, and then substituted here and there and solved.
     
  9. Audioguru

    New Member

    Dec 20, 2007
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    What is your complicated result? Is the voltage gain of the transistor about 0.99? 0.98?
    Isn't that almost 1?

    Does the input attenuator have an output of about 0.833 which is calculated with simple arithmatic?
     
  10. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    I'm also curious. What values did you use for r∏1 and r∏2, and what was your final result?
     
  11. Whyregister

    Thread Starter New Member

    Sep 12, 2012
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    r∏1 was 165,837 ohms
    r∏2 was 1642 ohms

    The final gain value I calculated was -0.806
     
  12. Audioguru

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    Dec 20, 2007
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    I used simple arithmatic to calculate the input attenuation is 0.833. Then I knew that the transistor had a voltage gain a little less than 1, about 0.98. My total gain is 0.816 which is very close (1 percent) to your complicated calculation.
     
  13. Whyregister

    Thread Starter New Member

    Sep 12, 2012
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    I have no doubt, but unfortunately we are meant to use the small-signal model to calculate it, which involves a few equations and something a bit complicated than what you presented. My final formula came down to a fraction upon a fraction of several variables which was very annoying to manipulate.

    In a way this is good because I've so far come across at least 2 different methods which is more simplistic than what we have been taught. Thanks for your responses. If I were to ever do this again in my own research and experimentations, I would definitely use what you have presented.

    DO they use "small-signal" model in industry? What are the practical applications of using this?
     
  14. Audioguru

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    Dec 20, 2007
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    I had a 35 years career designing audio circuits. I have NEVER used your teacher's complicated calculations.

    As I showed, my simple arithmatic and my experienced guess gave a result that is within 1% of your complicated calculation. Most resistor values are not 1%, they are 5% so who cares (maybe your teacher cares).

    Many people today simulate a circuit using a software program to see if it works.
    I never simulated a circuit during my career. I carefully designed them and they all worked perfectly.
    I simulate circuits today and post the results in forums.
     
  15. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    We can find voltage gain quite simply. Because we are having here a Darlington pare we can ignore Ib current and find Ie current directly form the diagram.
    Voltage at base Q1 is equal Vb = 3V and Ve = 3V - 2Vbe = 1.6V
    So Ie = 1.6V/1K = 1.6mA.
    The BJT is connect as a CE amplifier so the gain is equal Av ≈ Rc/(Re + re)
    But we have a Darlington so gain is equal to Av ≈ Rc/(Re + 2re)
    re ≈ 26mV/Ic = 26mV/1.6mA ≈ 16.25Ω
    So the gain Av = 1K/( 1K + 32.5Ω) = 0.968V/V
    But we also have a Rs resistor which form a voltage divider with Rin resistance.
    And this decreases the overall voltage gain to:

    Aov = Rin/(Rin + Rs) * Av ≈ 50K/(50K + 10K) * 0.968 0.833 * 0.968 0.806V/V

    And this is how you can find the answer without a small signal analysis.
    Of course you need to know some facts about BJT. And this method is shown in a book "the art of electronics".
     
    Last edited: Sep 16, 2012
  16. Whyregister

    Thread Starter New Member

    Sep 12, 2012
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    If only we were taught proper design and simplification techniques rather than bulldozing through weird models.
     
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