Two stage amplifier help (help me to learn this on my own)

Discussion in 'Homework Help' started by gte, May 20, 2010.

  1. gte

    Thread Starter Active Member

    Sep 18, 2009
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    4
    Here is my circuit and my givens, correct me where I'm wrong.

    I have decided to split the BJT and FET for now in solving some of the things.

    BJT

    VC - 8.04vDC Vcc - (IC*RC) = 15 - (.012*580)
    VB - 2.2vDC Vcc *(R1/(R1+R2)) = 15 * (1100/7500)
    VE - 1.5vDC VB - VBE = 2.2 - ~.7

    IC - 12mA IC ~ IE
    IB - 2mA VB/R2 = 2.2/1100
    IE - 12mA VE/RE = 1.5v/125

    Rin(base) - 15000 Beta * RE = 120 * 125


    The BJT individual gain is Beta = IC/IB = 12mA/2mA = 6?






    FET

    Idss - 78.8mA (gm0 * VGSoff)/2 = (.035 * 4.5) /2

    Vg - ? (is it ground?)

    Vgs - -0.75v

    IS - 21.4mA VS/RS = .75/35

    ID - 21.4mA ID ~ IS

    VD - 7.296 VD = VDD - (ID * RD) = 15 - (21.4ma * 360) = 15 - 7.704

    gm - 29mS gm0(1 - Vgs/Vgs(off)) = 35mS(1 - (-.75/-4.5)) = 35mS(1-.1667) = .035 * .8333




    How am I doing? Then I have to factor in Rin(total) to get the actual gain? Thanks for reading!

    [​IMG]
     
  2. Audioguru

    New Member

    Dec 20, 2007
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    896
    Your circuit is not practical.
    It has a low input impedance and the FET has a very high input impedance that is not wanted.

    The transistor has no negative feedback so of course its output is very distorted.
     
  3. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,957
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    To be more precise to find Ic, Ib, Ie you need to find first Ib.
    So we start with replace R1, R2 with Thévenin's theorem equivalent circuit.
    Et = Vcc * R2/(R1+R2) = 15V * 1.1/7.5 = 2.2V
    Rt = R1||R2 = 938Ω
    So Ib = (Et - Vbe) / ( Rt + (β+1)*Re) = (2.2V - 0.7V)/ (983 +(120+1)*125) = 1.5V/16.108K = 93.121uA
    So if we know Ib now we can easily find Ic and Ie.
    Ic = Ib*β = 11.174mA and Ie = Ib + Ic = Ib*(β+1) = 11.267mA.
    Vc = Vcc - Ic*Rc = 8.519V
    Ve = Ie * Re = 1.408V
    Vb = Ve + Vbe = 2.108V





    If we know that Vs = 0.75V then
    Id = Is = 0.75V/Rs= 21.428mA
    Vg = 0V and Vd = 7.285V
    And gm = 29.166mS; Idss = 78m
    Not bad

    No, you need take in to account Rin of a JFET amplifier and RL to find the total gain
     
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  4. gte

    Thread Starter Active Member

    Sep 18, 2009
    347
    4
    Thank you for the response Jony

    Am I missing any other values based on the problem? I'm not 100% sure how to calculate RL to get the total gain?
     
  5. gte

    Thread Starter Active Member

    Sep 18, 2009
    347
    4
    Maybe this is close?


    I'm trying to figure out the RL that the FET puts onto the BJT. I know it's RC in combo with R3 (3M) and Rin of the JFET, and not RS because it has an AC filter, but I'm not sure how to calculate Rin of the JFET, as I don't have the Beta for the JFET, although I did calculate AV=gm*Rd or .0327*360 or 11.76, but I think it might be .0327 * (360 || 1000) or .0327 * 264.7 or 8.65569 gains.

    I think that the gain for the BJT is Av = RC || R3(3M) || Rin of Q2 /r'e but I'm having a hard time calculating the Rin? Is it negligible because it's very high? With what I know, it'd be 1/ ( 1/580 + 1/3,000,000 + 1/RinGate ) or 579.88, it appers it makes a very little difference? So the gain would be 579.88/2.083 or 278.39 gains?

    Then you multiply that times 8.6559 and you get 2409.6575391 gains? That just seems to high and doesn't seem correct?
     
  6. Audioguru

    New Member

    Dec 20, 2007
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    The Fet has an input resistance of almost infinity.
    The output of the transistor is loaded by the 3M resistor.
     
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  7. gte

    Thread Starter Active Member

    Sep 18, 2009
    347
    4
    Thanks guru! I was hoping that was the case, does that mean my calculations were correct?
     
  8. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Av_sf = Av1 * Av2

    Av1 = ( Rc|| R3 || 1/h22 )/re
    ≈ Rc/re ≈ 251 [V/V]

    AV2 = gm * (Rd||RL) ≈ 7.72 [V/V]

    Av_sf = 1.93K [V/V]

    And Rin has no effect on voltage gain as long Vin series resistance is equal 0Ω. BJT is driven from ideal voltage source.
     
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  9. gte

    Thread Starter Active Member

    Sep 18, 2009
    347
    4
    Just wanted to say thanks again to you two, I think I did well when I turned it in today
     
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