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Common Emitter with emitter resistance small signal model

  1. Sep 10, 2015 #1
    [Mentor's Note: Thread Moved from EE Forum]

    This link has the picture, for some reason it wasnt showing in this post. http://i.imgur.com/TYlb32t.jpg
    Here is my problem, I am supposed to use this circuit and solve for Av and Rin. My problem is that I have solved for Av but I cant solve for Rin no matter what I use. Its very frustrating. I need to use the pi model.

    Here is my work for the Av gain.

    i2yiMT6.jpg

    Please let me now if this is correct and then a tip on how I can solve for Rin which should be Rin = RB|| r_pi (R_E(1+B)). Is Vbe only across r_pi or r_pi and R_E?

    Av = -B(R_C||R_L)/(r_pi) which is what I have in my work in the upper right middle
     
    Last edited by a moderator: Sep 10, 2015
  2. jcsd
  3. Sep 10, 2015 #2

    Baluncore

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    Science Advisor

    Welcome to PF.
    Is this either coursework or homework ?

    small_circuit.png
     
    Last edited: Sep 10, 2015
  4. Sep 10, 2015 #3
    Its homework, do you think there is a typo in the Rin given? Or is the emitter resistance in the wrong place?
     
  5. Sep 10, 2015 #4

    Baluncore

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    I shrank and posted the circuit.
    Rin is shown as an indeterminate arrow, Is it measured at Vs of at the base terminal ?
    Rin is frequency dependent because of the series C in the base circuit.
    Re will appear higher resistance to Rin because it is seen through the base current gain, beta.
     
  6. Sep 11, 2015 #5

    NascentOxygen

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    Staff: Mentor

    Vbe is always the voltage across the active b-e junction; it doesn't embrace voltage across externally placed resistors.

    Your formula for Rin looks awkward. The expressions on each side of the "||" symbol should have units of Ohms.
     
  7. Sep 11, 2015 #6
    Should Rin be Rin = RB || (rpi + RE*(beta+1)) ?
     
  8. Sep 11, 2015 #7

    NascentOxygen

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    That looks better.
     
  9. Sep 11, 2015 #8

    Baluncore

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    If R_pi has dimension ohms then your Rin = RB || r_pi (R_E(1+B)) has dimensions of ohm || ohm2
    Where has your R_pi come from ?

    See schematic in post #2
    Do you think Rbase = R1 || R2 ?

    How is beta defined ?
    Maybe; Rin = Rbase || (Re * beta)
    or; Rin = Rbase || (Re * (1+beta) )
     
  10. Sep 11, 2015 #9
    Baluncore, I think the reason I was having trouble is because the answer they provided was missing the plus sign, a previous professor of mine provided me a pdf file explaining how they get Rin and it definitely shows that it should be Rin = RB || ( rpi + RE(beta +1)).
     
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