Pole Frequency - Small signal Circuit

Discussion in 'Homework Help' started by mau80, May 18, 2011.

  1. mau80

    Thread Starter New Member

    Oct 24, 2010
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    0
    Hi to all,

    I've to find out a pole frequency in a simple common base amplifier.
    I've spent many hours, but I cannot reach the correct result, I'd like to ask some help about.

    In attach there is a pdf with the circuit, and thre track that I've followed.

    Thank you very much for your help!

    Regards
    Maurizio
     
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  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    782
    Looking into the emitter leg at point Vi you see a total emitter side series resistance

    RG+re=50+26/.3=137Ω

    Plus ......

    The transferred parallel values referred from the base side will be Ceq=β*C=10uF

    and (RB1||RB2)/β=2kΩ

    The effective resistance which determines the pole will then be

    Req=2kΩ||137Ω=128.2Ω

    The pole frequency f=1/(2πReqCeq)=124Hz

    EDIT:

    Rather than using just β one should more accurately use (1+β). But it's only a small difference in result.
     
    Last edited: May 18, 2011
  3. mau80

    Thread Starter New Member

    Oct 24, 2010
    21
    0
    Thank you very much for your answer!

    I'd like to ask some more detail about this circuit:

    Related your solution I'd like to ask following:

    * why you're interested in total emitter side series resistance

    * transferred parallel values referred from the base side

    I'd like to know why my computation using small signal circuit doesn't give to me correct results, I've try to recalculate all but I cannot find correct result. My textbook alway perform small signal analysis for computate Vo/Vi.

    Thank you very very very much for your time!

    Bye
    Maurizio
     
  4. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    Hi mau80,

    I could post a complete worked solution for the problem but that's a lot of work for me.

    In summary the gain is given by

    A_v=\frac{V_o}{V_i}=\frac{\beta R_C}{(1+\beta)} [\frac{1}{r_e+R_G+\frac{Z_B}{(1+\beta)}}]

    Z_B=R_{B1}||R_{B2}||\frac{1}{j\omega C}

    re is the dynamic emitter resistance typically approximated as

    r_e=\frac{26}{I_E (mA) }

    It's that part of the equation within the square braces

    r_e+R_G+\frac{Z_B}{(1+\beta)}

    that leads you to the pole. You'll note that the gain equation "shows" that the base side total parallel impedance ZB is divided by (1+β).
     
  5. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    While I haven't shown the derivation of the amplifier gain, I have derived the pole frequency condition. As you can see it's a substantial piece of work of itself. The analysis focuses on that part of the gain function Av inside the square braces - as indicated in my earlier post.

    In the end, the more formally derived result confirms the rather more simple approach outlined in post #2. From the emitter side of the circuit, the referred base terminal appears as having a shunt impedance ZB/(1+β). So "looking into" the input terminal (at Vi) one sees a series part (re+RG) up to the referred base node. At the base node there is a shunt impedance ZB/(1+β). ZB can be split into two parallel elements, namely a resistor RB=RB1||RB2 and capacitor C. The emitter side referred equivalent of RB is then RB/(1+β) and the referred equivalent capacitance is (1+β)C.

    The attached analysis may also be used to note that a frequency zero also occurs at ωz=1/(RBC)=50 rads/sec=7.96Hz where the gain Av begins to roll off at the lower 3dB point.
     
    mau80 likes this.
  6. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    Mau80, I looked at your work. Your procedure is rather abbreviated and I found it hard to follow. It's more work to figure what you did wrong than to show you a correct way to do it.

    On the second page you have an equation:

    (I2-I1)*Rg + Vx + I2*Rc - V1 = 0

    What happened to Vx? I don't see where you used it in later calculations.

    Furthermore, when you have a current source (-gm*Vbe) in a branch of a loop, not shared by any other loop, the equation for that loop becomes a constraint equation. All that equation should be is I2 = -gm*Vbe. If you'll substitute -gm*Vbe for I2 in your first equation, you should be able to get an expression for Vo/Vi.

    If all you want is the pole frequency, you could also work from the base side of your circuit. Reflect the impedances at the emitter to the base side and get the time constant of the pole.

    On the emitter side you have Rg and re in series. Reflected to the base, these become (1+β)*(Rg+re). Parallel that with RB1, RB2 and sC and you should get about 125 Hz for the pole frequency.

    A full derivation of Av can be done using the method of Shekel.

    See this thread:
    http://forum.allaboutcircuits.com/showthread.php?t=26710

    starting at post #43 et seq.

    Here's an attachment showing the derivation for your circuit. This method is just a systematized nodal analysis method.
     
    mau80 likes this.
  7. mau80

    Thread Starter New Member

    Oct 24, 2010
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    Dear guys,

    what you've done for me is magnificent great!!!!!

    I've not word for thank you for your works and for your precius time that you've spent for hel me with this exercise!!

    I've anylized your computation and website proposed, and I've been able to get the result!

    Thank you again for your great works and you help!

    Have a nice day!!!!

    Thanks Thanks Thanks

    Maurizio
     
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