C.E. Amp. w/ Emitter Resistance (Quick/Simple)

Discussion in 'Homework Help' started by jegues, Apr 24, 2011.

  1. jegues

    Thread Starter Well-Known Member

    Sep 13, 2010
    735
    43
    I've got quick question about the input resistance for this particular circuit.

    The input resistance is looking in from the white node to the right of Rsig right?

    Why doesn't his calculations for Rin include that 0.1MΩ resistor as well? Is there a mistake?

    Can someone clarify?
     
  2. wmodavis

    Well-Known Member

    Oct 23, 2010
    737
    150
    I do not see any 0.1M Ohm resistor.
     
  3. designnut

    Member

    Apr 21, 2011
    33
    1
    The .1 ma refers to an unknown current source. Did you mistake it for a resistor?
     
  4. jegues

    Thread Starter Well-Known Member

    Sep 13, 2010
    735
    43
    No, the 1M ohm resistor to the right first of the capacitor that couples the signal voltage.
     
  5. jegues

    Thread Starter Well-Known Member

    Sep 13, 2010
    735
    43
    Still looking for some clarification on this one. My test is soon!:confused:
     
  6. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    Strictly speaking the 1MΩ should be included - maybe the person who wrote the solution thought it wouldn't change things by that much.

    With the 1MΩ included Rin=50.5k||1M=48.07k which is about a 5% change and probably shouldn't be disregarded.
     
  7. jegues

    Thread Starter Well-Known Member

    Sep 13, 2010
    735
    43
    Ah so that's the best way to deal with the 1MΩ being in there, find the input resistance to the right of it (say Rib), and combine that in parallel with the 1MΩ to get the actual input resistance(Rin).

    Where, Rin = (Rib//1MΩ)

    Thank you for clearing this up!
     
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