Help me calculate output impedance

Discussion in 'Circuit Help' started by GhostLoveScore, Jan 11, 2017.

  1. GhostLoveScore

    GhostLoveScore

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    I've been reading on amplifier's output impedance. It's common collector amplifier. Take a look

    circuit2.png

    Let me first warn you that circuit in red frame has not been calculated, meaning, I didn't calculate base resistor divider for some specific bias. I just put them there because for what I want to know they doesn't matter.

    I want to know what is output impedance of circuit on the right, in red frame, at emitter. I read that it's Rsource/beta

    [​IMG]

    But in this case what is Rsource? Is it output impedance of circuit on the left? If that's the case then Rsource is 330 ohms? Then if beta for 2n2222 is 100, output impedance of circuit on the right should be 3.3 ohms?

    On the other hand, R1 could be interpreted as a load resistor on circuit on the left and then output impedance of circuit on the left would be Rc||Rl(R1)?
     
    GhostLoveScore, Jan 11, 2017
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  2. GhostLoveScore

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    (*steve*), Jan 11, 2017
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  3. GhostLoveScore

    GhostLoveScore

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    believe it or not, almost every link I already visited, but couldn't find an answer.
     
    GhostLoveScore, Jan 11, 2017
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  4. GhostLoveScore

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    I scrolled through the first one I saw and got the answer.
     
    (*steve*), Jan 11, 2017
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  5. GhostLoveScore

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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  6. GhostLoveScore

    LvW

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    At first, using approximations one should know if such a simplification is allowed or not. The explaining text (right from the transistor) gives an expression fo the output impedance which is a rough approximation.
    The correct formula is:
    r(out)=RE||[1/gm + Rsource/(1+beta)]
    With: transconductance gm=Ic/Vt (Ic: collector current, Vt: temperature voltage).
    The dynamic (differential) signal resistance 1/gm at the E node will be much larger than the 3.3 ohms you have calculated (dependent on the DC current).
    It is correct that Rsource is the parallel combination of R1 and R2 and the signal output resistance of the previous stage.

    Remark: In the document - linked by Steve - appears the base-emitter diff. resistance rbe. Please note that this value is in most cases unknown. However, the ratio rbe/beta is identical to the transconductance 1/gm as contained in the formula given above. This value can easily found using the known DC current Ic.
     
    Last edited: Jan 11, 2017
    LvW, Jan 11, 2017
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  7. GhostLoveScore

    GhostLoveScore

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    GhostLoveScore, Jan 11, 2017
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  8. GhostLoveScore

    GhostLoveScore

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    Ah, thanks, that's around 320ohms.
     
    GhostLoveScore, Jan 11, 2017
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  9. GhostLoveScore

    Ratch

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    First of all, you should read this thread and learn about the PIT or GIT https://www.electronicspoint.com/threads/output-resistance-of-equivalent-op-amp-circuit.278977/ .

    For your circuit, rs = source resistance, and rb = r1||r2 .

    Applying the PIT below, rout is calculated.

    GhostLoveScore.JPG
    Note that rout does not include the value of re.

    Ratch
     
    Ratch, Jan 11, 2017
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  10. GhostLoveScore

    LvW

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    Ratch - I think, your equations are not easy to read.(and to understand)
    * Why don`t you distinguish between static (R) and differential (r) resistances ?
    * In the transfer function, what is rs (with respect to the given circuit)?
    * T=1 for rs=0 ?
    * I doubt if your expression for rout is correct (re=RE ?).
     
    Last edited: Jan 12, 2017
    LvW, Jan 12, 2017
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  11. GhostLoveScore

    GhostLoveScore

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    Can you explain where does this formula comes from? I've seen few variations on my formula, but I've never seen this one.
     
    GhostLoveScore, Jan 12, 2017
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  12. GhostLoveScore

    LvW

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    In the link which was given by Steve (post #5), you will find the following expression:

    [​IMG]

    This is the output resistance without the external ohmic resistor RE.
    This can be writte as
    rout=rbe/beta +Rs/beta.

    We knbow that rbe/beta=1/gm=Vt/Ic.
    Hence, rout=1/gm +Rs/beta.

    (I have explained this already at the end of my post#6)

    It is very advantageous to use this last form because we do not know the value of the base-emitter resistance rbe.
    Also the value of beta is known only with very large tolerances. But the RATIO rbe/beta is known because it is the slope of the transfer function Ic=f(Vbe). This slope is the transconductance gm which can easily be found because of the exponential function. Remember that the BJT is - physically spoken - a voltage controlled device.

    The approximation error (beta+1) instead of "beta" can be neglected.
    OK?
     
    LvW, Jan 12, 2017
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  13. GhostLoveScore

    Ratch

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    You have a point. I should have used capital letters to designate the external base resistors (Rb), the external emitter resistor (Re), and source resistance (Rs). My analysis disregards the internal resistances of the transistor diodes (rb and re) because they are usually small compared to the external circuit resistances.

    Ratch
     
    Ratch, Jan 12, 2017
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  14. GhostLoveScore

    Ratch

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    I believe that equation will have to be modified if an external base resistance (Rb) is present.

    Ratch
     
    Ratch, Jan 12, 2017
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  15. GhostLoveScore

    LvW

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    I think, everything is OK (except "beta+1" instead of "beta") if we consider Rs to be the effective (resulting) source resistance at the base node.
     
    LvW, Jan 12, 2017
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