Question in relation to CE amplifier voltage gain.

Discussion in 'General Electronics Chat' started by ECE101, Dec 9, 2011.

  1. ECE101

    Thread Starter New Member

    Nov 25, 2011

    Just wondering if anyone could explain the answer to this question?
  2. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
    Do you have a slightly bigger version of the image?

    I'm reading it as the voltage is applied at the output (after the capacitor), is that correct?
    ECE101 likes this.
  3. Audioguru

    AAC Fanatic!

    Dec 20, 2007
    The gm and re of the transistor changes when the transistor's current changes.
    When the collector voltage increases because the transistor's current is reducing then the gm is also reducing which reduces the voltage gain.

    If the transistor is amplifying a sine-wave then the top portion is compressed and the botton portion is expanded causing severe distortion like this simulation:
    ECE101 likes this.
  4. ECE101

    Thread Starter New Member

    Nov 25, 2011
    Vin is to the left, before the input coupling cap C1, V out is to the right after C2.

    Ah I understand, thanks for the explanation!
  5. crutschow


    Mar 14, 2008
    The relative effect of the change in re (and gm which is determined by re) with collector current is determined by the value of re as compared to the emitter impedance at the frequency being measured. If the emitter impedance is much larger than re, then the change in re will have little effect on the circuit gain.
    ECE101 likes this.
  6. ECE101

    Thread Starter New Member

    Nov 25, 2011
    From a mathematical point of view:

    Gain = -gm*Rc

    gm = Ic1/Vt
    Rc = 8k

    When Vo = +15

    AV = -((15/8*10^3)/25*10-3) * 8*10^3
    = -600

    When Vo = -10

    AV = ((10/8*10^3)/25*10-3) * 8*10^3
    = 400

    Is this correct?