two-stage JFET / BJT common-source / common-emitter amplifier

Discussion in 'Homework Help' started by stevy123, Mar 1, 2008.

  1. stevy123

    Thread Starter Active Member

    Nov 19, 2007
    61
    0
    Hi all

    Just wondering could anyone help me with this circuit.

    I am trying to work out values for Re and Rc but am confused.

    i have worked out the other resistor values by

    let Vg = Vcc/10 10/3 = 3.3v
    let Vgs = vp/2 -2/2=-1v

    Vs = 3.3 +1 =4.3v
    let Id=Is= 1 mA
    Rs= 4.3/1mA = 4.3K

    let Vd = 2Vcc/3 = 2x10/3 = 6.66v

    Rd= 10 -6.6/1mA = 3.4K

    R1 + R2 need to be as large as possible. R1 should be twice R2

    so let R1= 2M so R2 = 1M


    Also

    JFET
    Vp = -2 V
    IDSS = 4mA
    VA = 100 V


    BJT
    IC (rated) = 1 mA
    b = 150 (b = beta)
    VA = 100 V


    Could someone tell me if i have worked out the values correctly and how to go about working out Re and Rc

    Many thanks
    steve
     
  2. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
    657
    What does this mean?
     
  3. rwmoekoe

    Active Member

    Mar 1, 2007
    172
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    what is the supposed bandwidth of it?
    the gain?
    is the linearity critical at all?
    the circuit lacks negative feedback loop for linearity and stability.
    the gain will depend too much on frequency, temperature, beta, and all.
     
  4. stevy123

    Thread Starter Active Member

    Nov 19, 2007
    61
    0
    Hi Ron H

    With regards to what

    let Vg = Vcc/10 10/3 = 3.3v means

    It is the working out for Vg (the voltage at the gate)

    which is vcc/3 and equals 3.3v. Sorry typing mistake.

    Thanks
    Steve

    Can you advise on how to work out the values of Re and Rc
     
  5. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
    657
    JFETs have such sloppy specs, I don't see how you can have a predictable quiescent point for this circuit. Have you been assigned this circuit topology, and you have to pick component values, or did you choose the design yourself?
    Do you have part numbers for the transistors?
     
  6. rwmoekoe

    Active Member

    Mar 1, 2007
    172
    0
    you've already put the 100uf cap at the emitter. it's leaving the working re as internal rbe wich is very small and dependant on temperature, frequency, etc.
    let's say we must go on with that, we still have to decide the gain.
    if you want the gain to be 10, and the internal rbe at room temperature is around 60 ohms, then the rc should be 600 ohms and the re can be calculate to give the optimal dc level for the output. it depends on the dc level of the previous stage.
     
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