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What is this Ro refering to?

  1. Aug 11, 2013 #1
    1. The problem statement, all variables and given/known data

    i) I am to find IB, IC and re
    ii) Determine Zin and Zout

    3. The attempt at a solution

    To find IB,


    I use the KVL, start from base

    V2 = IB R1 + VBE ,
    I assume my VBE = 0.7 V
    IB = 23.8462 μA

    then my IC = IB HFE
    IC = 2.3846 mA

    For re = 26m V / IC
    re = 10.9 Ω


    Zin = RB // Hie
    Hie = Hfe re = 1090 Ω
    Zin = 1.087 kΩ

    Zout = RC....I think this is where i need to use ro, but i have no idea wat is it refering to...
  2. jcsd
  3. Aug 11, 2013 #2
    I think Ro is 1/hoe.
  4. Aug 11, 2013 #3
    If it is 1/Hoe, then the Zout

    Zo = ro // RC
    Zo = 4.012 kΩ

    then i will proceed to the next part,
    iii) calculate Av and Ai

    Ib = Vin / Hie
    Vout = -hfe Ib (ro // RC)
    = -hfe ( Vin / Hie ) (ro // RC)

    Av = Vout / Vin
    =- hfe / hie (ro // RC)
    = -368.114 ...Awkward number...

    for Ai,

    Iin = Vin / 390k + Vin 1090
    = Vin (1/390K + 1/1090)

    Iout = IRC ...this is correct right?
    = -hfe Ib (ro//Rc)/Rc

    Ai =( -hfe/ hie)(ro//Rc)/Rc
    = -93.053

    Is all these working step correct?
    Last edited: Aug 11, 2013
  5. Aug 11, 2013 #4
    You have used a factor hfe in some places where strictly the proper factor should be (hfe+1). For example you have:

    Hie = Hfe re

    but it should actually be Hie = (Hfe+1) re

    Other than this, which leads to a very small error in some calculations (an error which is negligible if hfe is large), your answers are correct.
  6. Aug 12, 2013 #5

    rude man

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    Check your re. Do you have the correct formula?

    Not sure if Zo is meant to include C1 or not. Same for Zin and C2, come to think of it.

    EDIT: actually, this circuit is saturated: the collector current computes to 100*(9.3V/390K) = 2.3846 mA which x 4.3K = 10.25V which exceeds the collector supply.
    Last edited: Aug 12, 2013
  7. Aug 12, 2013 #6
    yeah i, but it still asking for amplification...

    and about the re, do u mean the Ebers–Moll equation? Well, i still don't know how to use it yet, so i wil syick with the 26mv/ic approximation first.
  8. Aug 12, 2013 #7

    rude man

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    Amplification is zero.

    You might have the right formula for re after all, I don't know your model.
  9. Aug 12, 2013 #8
    rude man, were you referring to the fact that the usual formula for re is re = 26m V / IE rather than re = 26mV / IC?

    Of course, if hfe is large, the difference is usually negligible, and somewhat variable since different people choose different values for the thermal voltage; sometimes 26 mV, sometimes 25 mV, etc.
  10. Aug 12, 2013 #9
    null void, tell your instructor that you noticed that the transistor is saturated, so you assumed that V1 is actually 12 volts! :biggrin:

    Or to really impress the instructor, tell him you analyzed a slight variation in the circuit, namely that V2 is gone and R1 is connected from collector to base.
  11. Aug 13, 2013 #10

    rude man

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    No, I had in mind re = 26mV/ib but that was probably wrong. As I said, i didn't know the OP's model.

    EDIT: 26 mV/ib is the input impedance to a decently-biased BJT (Vce between say 1V and 20V):

    Ic = Is exp(Vbe/Vt)
    ∂Ic/∂Vbe = Is (1/Vt) exp(Vbe/Vt) = Ic/Vt
    1/Zin = ∂Ib/∂Vbe by definition = (1/β) ∂Ic/∂Vbe = (1/β)(Ic/Vt)
    Zin = Vt/(Ic/β) = Vt/Ib = 26mV/Ib
    Last edited: Aug 13, 2013
  12. Aug 13, 2013 #11

    rude man

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    That would actually be a viable circuit! (the one with 12V Vcc is too dependent on beta).
  13. Aug 13, 2013 #12
    If i add a resistor at the emitter to the circuit and change the V1 and V2, is it possible that for the Vce to become negative value, is this also a case of saturation?
  14. Aug 13, 2013 #13


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    Staff: Mentor

    A negative value for VCE would mean the circuit bears no resemblance to an amplifier.

    An emitter resistor would allow a more stable design, but choose a lower value for V2.
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