Transistor hFE equation question

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The discussion focuses on deriving the equation Z_{out} = Z_{in}/(h_F_E + 1) from the relationship ΔI_E = ΔV_B/R. The original poster is confused about the transition from ΔI_E to ΔI_B in the textbook "The Art of Electronics." A participant clarifies that the derivation involves substituting ΔI_E into the equation for ΔI_B, leading to the correct form. They also note that Z represents a complex version of resistance, suggesting that starting with complex impedances would be more appropriate. The conversation emphasizes the importance of clearly defining input and output impedances in the context of BJTs.
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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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