CE Amplifier Design question

Discussion in 'Homework Help' started by ECE101, Nov 25, 2011.

  1. ECE101

    Thread Starter New Member

    Nov 25, 2011
    29
    0
    Hi,

    I'm currently studying electronics and came across this diagram for the design of CE amplifiers.

    I'm just trying to figure out why Rc has been chosen to be 20k?

    In order for Ic to be 0.5mA shouldn't Rc be 40k? ( R = v/i = 20/0.5*10^-3 )

    Also, how was the value for Re chosen?


    Any help would be greatly appreciated!


    [​IMG]
     
  2. hobbyist

    Distinguished Member

    Aug 10, 2008
    764
    56
    For a CE amp the voltage at the collector needs to be around 1/2 of VCC.

    So 10v. / 0.5mA will be the 20K ohm resistance.

    The DC bias voltage at the emitter is 1v. so 1v/ 0.5mA will give the 2K value.

    R3 is not part of the biasing voltage at the emitter, it is used for gain calculations.
    Due to the bypass cap.
     
    ECE101 likes this.
  3. ECE101

    Thread Starter New Member

    Nov 25, 2011
    29
    0
    Ahhh it's obvious now! Thanks hobbyist, that has cleared everything up :)
     
Loading...