Transistor hFE equation question

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Discussion Overview

The discussion revolves around deriving the output impedance equation for a transistor circuit, specifically how to transition from the equation involving the emitter current to the output impedance expression. The context is homework-related, focusing on the application of transistor models in circuit analysis.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • TFM seeks clarification on how to derive the equation Z_{out} = Z_{in}/(h_F_E + 1) from the relationship ΔI_E = ΔV_B/R.
  • TFM notes a perceived jump in logic in the textbook "The Art of Electronics" from ΔI_E = ΔV_B/R to ΔI_B = (1/(h_{FE} + 1))ΔI_E.
  • Another participant suggests that the transition involves substituting ΔI_E into the equation relating ΔI_E and ΔI_B, indicating that the book uses h_{FE} instead of β.
  • TFM provides a step-by-step substitution leading to the conclusion that R_{output} = R_{input}/(h_{FE} + 1), asserting that Z is a complex version of R.
  • A later reply suggests that TFM should clarify the definitions of input and output impedances and recommends starting with complex impedances rather than switching between R and Z.

Areas of Agreement / Disagreement

Participants generally agree on the substitution process but express differing views on the clarity and justification of the definitions used in the context of the problem. The discussion remains unresolved regarding the best approach to present the derivation.

Contextual Notes

There are limitations regarding the assumptions made about input and output impedances, as well as the transition between resistance and impedance without clear definitions. The mathematical steps leading to the final expressions are not fully resolved.

TFM
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Homework Statement



Showing how to get:

Z_{out} = \frac{Z_{in}}{h_F_E + 1}

from

\Delta I_E = \Delta V_B/R

Homework Equations



\Delta I_E = \Delta V_B/R

Z_{out} = \frac{Z_{in}}{h_F_E + 1}

\Delta V_E = \Delta V_B

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:

\Delta I_E = \Delta V_B/R - (1)

straight to

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

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

Thanks,

TFM
 
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TFM said:

Homework Statement



Showing how to get:

Z_{out} = \frac{Z_{in}}{h_F_E + 1}

from

\Delta I_E = \Delta V_B/R

Homework Equations



\Delta I_E = \Delta V_B/R

Z_{out} = \frac{Z_{in}}{h_F_E + 1}

\Delta V_E = \Delta V_B

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:

\Delta I_E = \Delta V_B/R - (1)

straight to

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

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

Thanks,

TFM

They don't. They substitute (1) into the equation relating \Delta I_E and \Delta I_B (the first half of (2)). I believe AoE uses h_{FE} in place of \beta
 
Okay so:

\Delta I_E = \Delta V_B/R

substitute into:

\Delta I_B = \frac{1}{h_{FE} + 1}\Delta I_E

gives:

\Delta I_B = \frac{1}{h_{FE} + 1}\frac{\Delta V_B}{R}

\Delta I_B = \frac{\Delta V_B}{(h_{FE} + 1)R}

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

R\Delta I_B = \frac{\Delta V_B}{(h_{FE} + 1)}

divide by I_B

R = \frac{\Delta V_B}{(h_{FE} + 1)\Delta I_B}

V = IR
R = V/I

thus:

R_{output} = \frac{R_{input}}{(h_{FE} + 1)}


Z_{out} = \frac{Z_{in}}{h_F_E + 1}

is also the same as:

R_{out} = \frac{R_{in}}{h_F_E + 1}

Z is just a complex version of R

Does this look correct?

TFM
 
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.
 
Okay, Thanks for all your assistance :smile:

Thanks,

TFM
 

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