Multi-stage BJT Amplifiers Input Resistance

Discussion in 'Homework Help' started by h_ngm_n, Aug 16, 2014.

  1. h_ngm_n

    Thread Starter New Member

    Aug 16, 2014
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    Hi all,

    I am working on analyzing a common emitter amp feeding into an emitter follower, and there is one point I am stuck on. Is the input resistance effectively the input resistance of stage 1 only (i.e. the common emitter only) or do I need to take into account the parameters of the second stage as well? I understand the input resistance of the CE amp to be R1||R2||hie.

    Any help greatly appreciated, thanks.
     
    Last edited: Aug 16, 2014
  2. to3metalcan

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    Jul 20, 2014
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    Schematic? Is there negative feedback or an emitter resistor ir anything?
     
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  3. Jony130

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    Feb 17, 2009
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    No, you don't need to take into account the second stage.
    The second stage will almost have no effect on Rin of the first stage.
     
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  4. h_ngm_n

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    Aug 16, 2014
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    Ok thank you I suspected this but have a few confusing examples in my study materials so just wanted to make sure
     
  5. h_ngm_n

    Thread Starter New Member

    Aug 16, 2014
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    Sorry I didn't think to upload a schematic as I was asking in more a generic sense, I'll try to post one anyway as a courtesy and practice.
     
  6. to3metalcan

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    Jul 20, 2014
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    There are definitely situations where the second stage can change the impedance of the first, through bootstrapping or negative feedback. But not here! :) Just R1 || R2 || β x R4
     
  7. anhnha

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    Apr 19, 2012
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    Is the exact value of input resistance of the common emitter R1 || R2 || (β*re + (β +1)R4) ?
     
  8. Jony130

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    Are you sure about this ?
     
  9. The Electrician

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    It's R1 || R2 || ((β+1)re + (β+1)R4)

    And, even without feedback through external components, the transistor's hre parameter, if not zero, will cause the input resistance to be affected by the second stage.
     
  10. anhnha

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    Apr 19, 2012
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    I think I am right.
    The input resistance of only BJT with R4 at emitter is β*re + (β +1)R4.
     
  11. anhnha

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    Apr 19, 2012
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    I am wondering why it is (β+1)re. According to this model, shouldn't it be β*re?

    [​IMG]
     
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  12. The Electrician

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    If there is no R4, then the input resistance is either β*re or (β+1)*re. If β is zero, then in the first case the input resistance would be zero, and this is clearly not correct. The (+1) part is there because the emitter current is the sum of the base current and the collector current, and is not just equal to the collector current..
     
  13. anhnha

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    Apr 19, 2012
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    I can't figure out why it can be (β+1)*re.
    Are you talking about β + 1 in (β+1)*R4?
    I am confused about this?
     
  14. Jony130

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    Well

    re = vbe/ie = Vt/Ie

    So vbe = ib*r_pi = ie*re

    thus

    r_pi =vbe/ib = (Ie/Ib)*re = (β + 1)*re

    and

    Ie = Ib + Ic = Ib + β*Ib = (β + 1)*ib

    so Ie/Ib =
    (β + 1)
     
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  15. to3metalcan

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    Jul 20, 2014
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    It's β+1 in both terms because you input a small current (Ib) and the transistor adds a larger collector current (Ib x β). Both of these flow through the emitter together, so the total current is Ie = Ib + (Ib x β) = Ib x (β + 1).

    But base current and re are not relevant in most CE analysis because β is hugely more than one and is also not predictable for a given device...variations in β are going to render the other two numbers mostly irrelevant. re becomes more important if R4 is small or zero ohms.
     
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  16. The Electrician

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  17. MrAl

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    Jun 17, 2014
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    Hi,

    Example:
    Re=1 ohms (sum of re and RE, re internal, RE external)
    Ib=1 amps
    1amps*1ohm=1v (voltage drop across Re due to base current alone)
    10amps*1ohm=10v (volage drop across Re due to Beta=10 times base current)
    Total Re voltage drop=11v
    11v/1a=11 ohms input resistance Rin
    Rin=(Beta+1)*Re=(10+1)*1 ohm=11 ohms input resistance

    A single common emitter stage has Rin dependence on Ic because re is dependent on Ic:
    Rin=(Beta+1)*Re=(Beta+1)*(re+RE)=(Beta+1)*(re(Ic)+RE)
    which shows the dependence of re on Ic by re(Ic).
    Note this shows that even without a second stage the input resistance will be dependent partly on the load resistance because that changes Ic (RL shunts some current to ground or increases current from the power supply source).

    It should be noted however that this dependence will be minimal with well chosen RE where RE swamps re, but with purposefully added feedback it will usually change Rin by a much greater factor. Thus it can be said that a stage with added feedback will change Rin more than without added feedback.
     
  18. to3metalcan

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    If you think about it, anhnha, re is exactly the same (for purposes of this analysis) as adding a small resistor in series with R4...there's no reason why the two of them would be calculated differently!
     
  19. Veracohr

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    Jan 3, 2011
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    Think about it analytically:

    You know how to calculate the input resistance of the CE amp, you know that the output of the CE amp is the collector, and you'll notice there's nothing in the input resistance formula about the collector.

    On the other hand, if you swapped the two stages, so that the emitter follower was first and the CE amp second, the input resistance of the second stage would factor in. In that case, the input resistance of the first stage (emitter follower) is determined (in part) by the resistance connected to the emitter, which now includes the input resistance of the second stage.
     
  20. MrAl

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    Jun 17, 2014
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    Hi,

    You must have not read the posts before yours or you would have seen that the input resistance depends partly on the collector current Ic. In most cases this can be minimized with proper selection of RE the external emitter resistance, but it is worth studying the reverse effect anyway.
     
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