Hybrid-pi model (low frequency..?)

Discussion in 'Homework Help' started by Steve|, Nov 6, 2008.

  1. Steve|

    Thread Starter Member

    Nov 4, 2008
    17
    0
    [​IMG]

    I've found the values of gm,g∏,go,gμ and they satisfy:
    gm>>g∏>>go>>gμ
    So from that I can use the low frequency hybrid pi model.. (I'm not 100% sure on this)

    So from the full hybrid-pi model, I can ignore rμ and open circuit the capacitors (since dc analysis) and also ignore rb since its so small.
    In my tutorial solutions they drew this:

    [​IMG]

    But I can't actually get/understand this...
    (ground is meant to be under Vs right?)
    Instead I have gm.Vbe where B.ib is. I don't understand how they changed the dependent source gm.Vbe to independent source B.ib=Ic

    And where is Vo actually meant to be? I'm assuming it replaces the whole BJT? (So the new Vs is where Q1 is now.)
    I also noticed in the rest of the solution (not linked), they don't consider IEE - I don't actually know what this is either other then something to do with the emitter current.

    Thanks.
     
  2. mik3

    Senior Member

    Feb 4, 2008
    4,846
    63
    This is not a correct pi-model. Also, you can change gm*Vpi with b*Ib because Ib is not constant but it varies with Vs and other factors.
     
  3. Steve|

    Thread Starter Member

    Nov 4, 2008
    17
    0
    Oh ok, is this then a mistake on their behalf or am I approaching the question the wrong way?
     
  4. mik3

    Senior Member

    Feb 4, 2008
    4,846
    63
    Its their mistake! They omitted rπ and they connected the input signal to the collector, this is wrong unless they had made other calculations before and simplify the circuit to this but i dont think so.
     
  5. Steve|

    Thread Starter Member

    Nov 4, 2008
    17
    0
    Oh I'm so sorry! I should have stated what the resistor values stood for:
    rpi = 38.9 kohms
    ro = 1.3 Mohms
    Rs = 50 kohms

    Sorry again, I omitted these and the rest of the solution because I wanted to try it myself, but the only calculations before that simplified circuit was for gm, ro and rpi.
     
  6. mik3

    Senior Member

    Feb 4, 2008
    4,846
    63
    But its still wrong. Is it only for the first transistor?
     
  7. Steve|

    Thread Starter Member

    Nov 4, 2008
    17
    0
    Yeah, for the next part of the solutions they have Stage 2, where they drew a similar circuit with the voltage source being equal to Vo of the first stage (also with a resister in series with it - Ro1).
    To save any hassle I'll just link the full solution (bar the final expansion for Ri):

    [​IMG]
     
  8. mik3

    Senior Member

    Feb 4, 2008
    4,846
    63
    With all i know i dont agree with the first stage, if you are going to do a serious analysis. If you assume than Vπ is very small and thus gmVπ is small, then you can remove the gmVπ current source of the equivalent pi-model. Maybe that what they did, i am not sure.
     
  9. Steve|

    Thread Starter Member

    Nov 4, 2008
    17
    0
    Ok, thanks a lot for the for help! I'll try ask around and see if I can get to the bottom of this, if I do find out I'll post it here :)
     
  10. The Electrician

    AAC Fanatic!

    Oct 9, 2007
    2,281
    326
    Do you get a result for Ri near the given value?

    I only get something like 5.8 meg.
     
  11. Steve|

    Thread Starter Member

    Nov 4, 2008
    17
    0
    They had 9.56M ohms
    Ri = rpi + (B+1)(1.3M||(rb2+rbpi2)
    = 38.96kohms + (151)(37.92kohms)
    = 9.56Mohms
     
  12. The Electrician

    AAC Fanatic!

    Oct 9, 2007
    2,281
    326
    Looks like a plain old arithmetic error to me.

    38.96kohms + (151)(37.92kohms)

    = 38.96kohms + 5725.92kohms
    = 5764.88kohms
    = 5.76Mohms
     
Loading...