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Mid-Band Voltage Gain Equations

  1. Feb 22, 2014 #1
    1. The problem statement, all variables and given/known data
    Problem picture is attached also some of my work is attached.
    The problem states derive midband voltage gain v0/vi expression in terms of circuit parameters. From the expression you just derived calculate numerically the midband gain value?


    2. Relevant equations



    3. The attempt at a solution
    I really only need help with part b finding the mid band voltage gain. So far I have V0=gmvpi(Rc||Rl) and vpi=-Vs((Rpi/(Ri+(Rb||Rib))). Then Vo/Vi = voltage gain but my professor told me I was wrong how should I proceed? I have also uploaded my work for a and shown some of b
     

    Attached Files:

  2. jcsd
  3. Feb 25, 2014 #2

    rude man

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    Your high-pass corner frequency calculation is correct: 3.2 Hz.

    The mid-band gain is when you assume the input capacitor → ∞ but without the 50 pF collector capacitor hooked up.

    You won't ever get the rolloff frequency if you don't incorporate the collector 50 pF capacitor in your equivalent circuit.

    I dn't have time to figure out where you went wrong in your computation of Vout/Vin. Maybe tomorrow. Offhand your Vπ calculation looks strange.
     
  4. Feb 25, 2014 #3
    Ok I recalculated the gain V0=gmVpi(Rc||Rl) and Vi=-Vpi((Ri+Rb||rpi)/(Rb||rpi)) so V0/Vi = -gm(Rb||rpi)(Rc||Rl)/(Ri+Rb||rpi) = -39.531
     
  5. Feb 25, 2014 #4

    rude man

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    If that's your mid-band gain it's way off.

    To get an approximate idea of the gain (the input capacitor shorted and the collector capacitor gone), imagine a small voltage change ΔVi. Now, ΔVi is almost equal to ΔVb and ΔVb is almost equal to ΔVe. So the emitter current change is ~ (0-ΔVi)/RE ~ collector current change which is Δvc/RL + Δvc/Rc. Now solve for Δvc/Δvi & what do you get?

    Sorry, I don't have the time to check your math. Maybe someone else will.
     
  6. Feb 25, 2014 #5
    Ok so I know you don't have much time but does this sound ok for my approach. Ic=beta(Ib) and Ie=(1+beta)Ib so Ie=.906 mA, Ic=0.9 mA, and Ib=.006 mA. Then deltaVi=-9.06 V and delta Vc=1.285 V so Vc/Vi=-0.1418 this seems off?
     
  7. Feb 25, 2014 #6

    rude man

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    That number is right on the money. Good going!
     
  8. Feb 25, 2014 #7

    rude man

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    If you ever get into this stuff professionally you will almost never use equivalent circuits. You will follow the sort of argument I offered. Basically one assumes beta goes to infinity, ib = 0 etc. A good circuit will never depend on the value of beta since it can vary all over the place even for the same type number, like 2N2222.

    The exception might be integrated circuit designers who can assume well-matched transistor pairs due to proximity, common process, etc.
     
  9. Feb 25, 2014 #8
    Ok man cool yah I don't plan too well if I do it will be a mixture of this and power maybe electric vehicles or something
     
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