the maximum symmetrical swing in output voltage

Discussion in 'Homework Help' started by ADCapacitor, Apr 24, 2016.

  1. ADCapacitor

    Thread Starter Member

    Feb 3, 2016
    38
    0
    Hello
    I don't quite understand what does the question mean
    So I ask my teacher and he tells me draw the load line first
    I try to write down the dc and ac load line equation.
    dc:
    Id=-Vds/(Rd+Rs) + Vdd/(Rd+Rs)
    Id=-Vds/5000 +12/5000
    ac:
    Vds=Vd-Vs
    Vds=-id*(Rd//RL)-id*Rs
    id=-Vds/3500

    I put Vds=0 and Id=0 to dc load line equation
    then I get the x-intercept and y-intercept
    But I put Vds=0 and Id=0 to ac load line equation
    I get both 0 at LHS and RHS
    How to find the x and y intercept and plot the ac load line?
     
  2. WBahn

    Moderator

    Mar 31, 2012
    17,720
    4,788
    The question is simply asking how large the amplitude of the output signal can be, about the quiescent point, without clipping.
     
  3. ADCapacitor

    Thread Starter Member

    Feb 3, 2016
    38
    0
    Do you mean the amplitude of Vds?
     
  4. WBahn

    Moderator

    Mar 31, 2012
    17,720
    4,788
    No. Vds is not the output of the amplifier -- Vout is (so Vout relative to ground).
     
    ADCapacitor likes this.
  5. ADCapacitor

    Thread Starter Member

    Feb 3, 2016
    38
    0
    thank you
    I also have another problem
    From the above ac load line equation, I only get the slope of ac load line
    How to find the rest part of ac load line ? (x-intercept/y-intercept)
    Or is there another method to plot the ac load line?
     
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