design common emitter amplifier with -20 gain, (working backwards)

Discussion in 'Homework Help' started by ninjaman, Jul 10, 2016.

  1. ninjaman

    Thread Starter Member

    May 18, 2013
    I have a question about designing a common emitter amplifier. I know the amp will invert the input so a gain of 20 is what I am looking for as whatever goes in, will come out inverted and multiplied by 20.
    Part of my coursework mentions that a rule-of-thumb is that the collector resistor (Rc) over the emitter resistor (Re) will give the voltage gain.
    So, example:
    Rc = 5.6k,
    Re = 1k,
    5.6k / 1k = 5.6 Av
    I am struggling with the coursework a little, it starts by selecting a supply voltage and choosing a quiescent current of 1mA. So, if the supply is 9 volts, the output would be half the supply, 4.5 volts. So, 4.5 / 0.001 = 4.5k or 4.7k (real value resistor). Then if the same current is going through the emitter and another rule of thumb is that the voltage across this resistor is 8% - 10% of the supply. So, 10% of 9 volts is 0.9 volts, find resistor by 0.9 / 0.001 = 900 or real value 910 ohms.
    with 4.7k over 910 = 5.16
    I want to work backwards. starting with 20 as a gain. if I keep the 9 volt supply and 1mA quiescent i will have 4.7k still. rearrange some stuff to get 20 / 4.7k to get 4.25 milliohms. thats a small resistor!!!?
    I was hoping there was a way to do this. The instructions in my course are really basic and only show this one way.
    any advice would be great!

  2. AlbertHall

    Well-Known Member

    Jun 4, 2014
    Try 4.7k / 20 => 220 ohms.
    ninjaman likes this.
  3. Bordodynov

    Active Member

    May 20, 2015
    Again=alfa*Requiv_load/(Re+re) Requiv_load=Rc||RLoad for RLoad=inf ==> Requiv_load=Rc
    Ic=1mA ==>re~25.8Ohm, alfa=Beta/(Beta+1), Beta=100 ==> alfa=0.99