Large and small signal analysis

Discussion in 'Homework Help' started by Mjollnir, Apr 22, 2004.

  1. Mjollnir

    Thread Starter Member

    Apr 22, 2004
    27
    0
    Hi there ppl, first time posting on this board, came across it by accident :)

    [​IMG]

    Vcc = 5V, RA = 1K, RC= 3.9K, Is (reverse saturation) = 2.682nA, emission factor = 2, and Vt = 0.026V

    The large mode analysis part is relatively easy, I can just find out IDQ using

    Vcc - 0.7 - IDQ (RA +RC) = 0 (where 0.7 is the diode 'turn on' voltage) and then sub IDQ into shockley to find VDQ, or use loadline analysis

    Is this correct??

    [​IMG]

    Now this part confuses me, as far as i know, I can find out the value of rd for small signal analysis by rd = nVT/IDQ, for the particular operating point

    but after that what I do? use superposition and kill off Vcc? (making Ra connected to ground)? and just do everything in AC steady state?? or is there some other trick to it all?

    http://members.iinet.net.au/~zzhang/431-222%20ASSY.pdf (entire assignment)
     
  2. Mjollnir

    Thread Starter Member

    Apr 22, 2004
    27
    0
    come on ppl, this is not hard stuff :D
     
  3. djwhiplash2001

    New Member

    Apr 29, 2004
    2
    0
    Yes, you superimpose - ground the resistor, short the caps, and you're good to go.
     
  4. Battousai

    Senior Member

    Nov 14, 2003
    141
    44
    In Analog Circuits, people talk about large signal and small signal. In school you learn that in large signal you bias everything around a linear point and then you apply a "small" signal input such that you hope you don't swing off the linear point that you are biased at.


    For large signal:
    The impedance a capacitor presents is Z=1/jwC. So at low frequencies, the capacitor is basically a fancy wire (a short). So ignore all caps at DC and set up a nodal equation: which looks like you've done it correctly.

    For small signal:
    If the diode is simply a diode connected BJT, then the small signal impedance will be 1/gm = Vt/Ic. The supply voltage, Vcc is a small signal ground, so replace Vcc with a ground symbol. You know the frequency of the input signal, so you can determine the impedance of each capacitor. Now just treat each cap just like a resistor (with a imaginary impedance) and write nodal equations to solve for vo/vi.

    You'll get a result like:

    vo/vi = A/[(1-s*p1)*(1-s*p2)]

    where A is some real constant (low frequency gain), s=j*w, and p1 and p2 are your pole frequencies.

    To get the transient response you have to use the inverse laplace transform...
     
  5. Mjollnir

    Thread Starter Member

    Apr 22, 2004
    27
    0
    The 2nd part of the assignment we had to design a value of Rc so we get a gain of 0.7

    http://members.iinet.net.au/~zzhang/500MV.JPG

    Here's the Pspice simulation for Vo/Vi for 500mV small signal/2KHz with 10uF caps, and Rc = 33kohm, Rl = 1k and Ra = 1k

    I understand the gain for a 'large' small signal is different coz the diode can no longer be assumed to be linear, but why do u get the clipping on the negative wave?
     
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