1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Transistor hFE equation question

  1. Dec 8, 2008 #1

    TFM

    User Avatar

    1. The problem statement, all variables and given/known data

    Showing how to get:

    [tex] Z_{out} = \frac{Z_{in}}{h_F_E + 1} [/tex]

    from

    [tex] \Delta I_E = \Delta V_B/R [/tex]

    2. Relevant equations

    [tex] \Delta I_E = \Delta V_B/R [/tex]

    [tex] Z_{out} = \frac{Z_{in}}{h_F_E + 1} [/tex]

    [tex] \Delta V_E = \Delta V_B [/tex]

    3. The attempt at a solution

    I am trying to prove the above, but the book makes quite a large jump (again, "The Art of Electronics")

    It goes from:

    [tex] \Delta I_E = \Delta V_B/R [/tex] - (1)

    straight to

    [tex] \Delta I_B = \frac{1}{h_{FE} + 1}\Delta I_E = \frac{\Delta V_B}{R(h_{FE} + 1)} [/tex] -(2)

    Can any one help show how they've gone from (1) to (2)?

    Thanks,

    TFM
     
  2. jcsd
  3. Dec 9, 2008 #2

    MATLABdude

    User Avatar
    Science Advisor

    They don't. They substitute (1) into the equation relating [tex]\Delta I_E[/tex] and [tex]\Delta I_B[/tex] (the first half of (2)). I believe AoE uses [tex]h_{FE}[/tex] in place of [tex]\beta[/tex]
     
  4. Dec 9, 2008 #3

    TFM

    User Avatar

    Okay so:

    [tex] \Delta I_E = \Delta V_B/R [/tex]

    substitute into:

    [tex] \Delta I_B = \frac{1}{h_{FE} + 1}\Delta I_E [/tex]

    gives:

    [tex] \Delta I_B = \frac{1}{h_{FE} + 1}\frac{\Delta V_B}{R} [/tex]

    [tex] \Delta I_B = \frac{\Delta V_B}{(h_{FE} + 1)R} [/tex]

    R is the load, so I am assuming that this is the R output. Multiply it out:

    [tex] R\Delta I_B = \frac{\Delta V_B}{(h_{FE} + 1)} [/tex]

    divide by I_B

    [tex] R = \frac{\Delta V_B}{(h_{FE} + 1)\Delta I_B} [/tex]

    V = IR
    R = V/I

    thus:

    [tex] R_{output} = \frac{R_{input}}{(h_{FE} + 1)} [/tex]


    [tex] Z_{out} = \frac{Z_{in}}{h_F_E + 1} [/tex]

    is also the same as:

    [tex] R_{out} = \frac{R_{in}}{h_F_E + 1} [/tex]

    Z is just a complex version of R

    Does this look correct?

    TFM
     
  5. Dec 9, 2008 #4

    MATLABdude

    User Avatar
    Science Advisor

    I think that's okay. Though you might want to make some justifications as to what the input and output (and input and output impedances) are of the BJT (unless this was part of the setup for the question). And you should probably start with complex impedances (Z=V/I) instead of changing from R to Z mid-way through.

    Just my 2c.
     
  6. Dec 10, 2008 #5

    TFM

    User Avatar

    Okay, Thanks for all your assistance :smile:

    Thanks,

    TFM
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Transistor hFE equation question
  1. Transistor question (Replies: 3)

Loading...