Effective load of a common emitter amplifier

Discussion in 'General Electronics Chat' started by hrs, Jun 14, 2016.

  1. hrs

    Thread Starter Member

    Jun 13, 2014
    82
    7
    Hi,

    I'm trying to go through the very basics using the Talking electronics book. Chapter 8 discusses transistors and the performance of various amplifiers. It says that the effective load is R_C // R_O and since R_O is much larger this reduces to R_C (see first attachment). But how do you arrive at R_C // R_O?

    Suppose we replace the collector emitter junction with a resistor R_CE (second attachment), I would understand that the load is R_C + R_CE // R_O. Since R_CE << R_O this would reduce to R_C + R_CE. Perhaps R_CE << R_C so you would arrive at R_C too. But that's not what the book says.

    Can you help me understand this?
     
  2. dl324

    Distinguished Member

    Mar 30, 2015
    3,250
    626
    Supplies are treated as ground for AC analysis.
     
    hrs likes this.
  3. AlbertHall

    Well-Known Member

    Jun 4, 2014
    1,964
    387
    Yes. Look at it like this. Suppose you have a signal on the collector of the transistor. There would be (ideally) none of that signal on the supply line and none of it on the ground line, so the collector signal voltage is across both R_C and R_O, i.e. effectively they are in parallel.
     
    hrs likes this.
  4. hrs

    Thread Starter Member

    Jun 13, 2014
    82
    7
    I think I can see how that would make them parallel. Thank you both!
     
  5. dannyf

    Well-Known Member

    Sep 13, 2015
    1,828
    365
    From thee collector looking out, you see two paths to ground, one through RO and another through RC in serial with Vcc (zero impedance) to ground.

    Thus your answer.
     
Loading...