Miller effect

Discussion in 'Homework Help' started by mau80, Mar 13, 2011.

  1. mau80

    Thread Starter New Member

    Oct 24, 2010
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    Hi to all,

    I'd like to ask some help in computation of Miller equivalent capacitance in input. I've attached a simple EC amplifier with capacitance that I'd like to transform using miller theorem. I've write formula that I've take from text book.

    Could you suggest to me what is the value of Av? I've try and retry but I cannot find the correct result. I've to computate value of Ca and Cb, but before I've to find value for Av.

    Thank you very much for you help.

    I've attached the detail schema!

    Thank you very much for your help

    Regards
    Maurizio
     
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  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
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  3. Adjuster

    Well-Known Member

    Dec 26, 2010
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    What is the value of gm? You won't be able to find Av without knowing it.

    Is the value of Rc really 893kΩ, or is it 89.3kΩ? Does the circuit apply any loading on the collector, which would need to be included to find Av?
     
  4. mau80

    Thread Starter New Member

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

    The book doesn't give value for gm or Av.
    I'd like to know if that value could be computed in some way from other circuit parameter such as beta.

    Furthermore the book solve the exercise in first schema ( without applying Miller Effect ) finding transfer function in attachment ATT_1.
    Where Rp=(Rs^-1+Rbe^-1)^-1

    The book find out following results:

    Av(s) = 37.4 dB
    wp1 = 1.6E7 rad/s <- first pole
    wp2 = 8.0E8 rad/s <- second pole
    wz1 = 7.2E9 rad/s <- first zero
    Bandwidth 2.6Mhz

    I've spent many hours, trying to computate value of gm form ( Av reported above) and rewrite the function in MATLAB trying to find out correct result. My goal is determine Av without Miller Effect, after that apply Miller Effect and find out the same result ( just for exercise ).

    Usually the book use beta = 119 so I've gm = beta / Rbe =
    I've done following:

    s = tf('s');
    beta=119;Cbe=150E-12;Cbc=12E-12;Rbe=1.38E3;Rs=50;Rc=893E3;
    gm=0.08623;Rp=(Rs^-1+Rbe^-1)^-1;
    H=[(Rbe/(Rs+Rbe))*(s*Cbc*Rc-gm*Rc)]/[s^2*Cbc*Cbe*Rc*Rp+s*(Cbc*(Rp*(gm*Rc+1)+Rc)+Cbe*Rp)+1]

    If I computate: 20*log10(dcgain(H))

    I cannot find correct result, the same for wp1 and wp2, the only result that I can find is wz1.

    Could you give some help?

    Thank you very much for your help!

    Regards
    Maurizio
     
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  5. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    mau80
    Do you know Ic current ? Or maybe you know rbe ?

    gm = 1/re = Ic/Vt = Ic/26mV

    re = rbe/(β+1)


    hehe I see on the diagram rbe=1.38k

    re = 11.5Ω ---> Ic = 26mV/11.5Ω = 2.26mA

    gm = 87mS

    Av = Rc/re

    but I'm not sure the value of Rc
     
    Last edited: Mar 16, 2011
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