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## Homework Statement

The attached figure shows a "shunt feedback amplifier" circuit, and its AC equivalent model.

Verify that:

[tex]r_T = \dfrac{v_{out}}{i_{in}}=\dfrac{-R_B}{1+\dfrac{R_B+r_\pi}{(1+\beta)R_C}}[/tex]

(assuming [itex]R_B\gg r_e[/itex], where [itex]r_\pi=(\beta+1)r_e[/itex].)

## Homework Equations

In the transistor's AC model, the b-e junction resistance [itex]r_\pi[/itex] is defined as [itex]r_\pi=(\beta+1)r_e[/itex], where [itex]r_e[/itex] is a very small resistance.

## The Attempt at a Solution

__Equation for Node 1:__[tex]i_{in}=i_b+\dfrac{v_{be}-v_{out}}{R_B}[/tex]

Since [itex]v_{be} = i_br_\pi[/itex]:

[tex]i_{in}=i_b+\dfrac{i_br_\pi-v_{out}}{R_B}[/tex]

[tex]i_{in}=\dfrac{i_b(R_B+r_\pi)-v_{out}}{R_B}[/tex]

__Equation for Node 2:__[tex]\dfrac{i_br_\pi-v_{out}}{R_B}=\beta i_b + \dfrac{v_{out}}{R_C}[/tex]

[tex]v_{out} \left ( \dfrac{1}{R_C} + \dfrac{1}{R_B} \right ) = i_b \left ( \dfrac{r_\pi}{R_B} - \beta \right )[/tex]

Since [itex]r_\pi=(\beta+1)r_e[/itex]:

[tex]v_{out} \left ( \dfrac{1}{R_C} + \dfrac{1}{R_B} \right ) = i_b \left ( \dfrac{(\beta+1)r_e}{R_B} - \beta \right )[/tex]

[tex]v_{out} \left ( \dfrac{1}{R_C} + \dfrac{1}{R_B} \right ) = i_b(\beta+1) \left ( \dfrac{r_e}{R_B} - \dfrac{\beta}{\beta+1} \right )[/tex]

Since [itex]R_B\gg r_e[/itex], I tried to neglect [itex]\dfrac{r_e}{R_B}[/itex], making it 0:

[tex]v_{out} \left ( \dfrac{1}{R_C} + \dfrac{1}{R_B} \right ) = i_b(\beta+1) \left ( 0 - \dfrac{\beta}{\beta+1} \right )[/tex]

[tex]v_{out} \left ( \dfrac{1}{R_C} + \dfrac{1}{R_B} \right ) = - \beta i_b[/tex]

[tex]v_{out} = -\dfrac{\beta i_bR_BR_C}{R_B+R_C}[/tex]

__Plugging this back into the expression for__i_{in}__:__[tex]i_{in}=\dfrac{i_b(R_B+r_\pi)+\dfrac{\beta i_bR_BR_C}{R_B+R_C}}{R_B}[/tex]

[tex]i_{in}=i_b\left ( \frac{R_B+r_\pi}{R_B}+\dfrac{\beta R_C}{R_B+R_C} \right )[/tex]

[tex]i_{in}=i_b\left ( \frac{(R_B+r_\pi)(R_B+R_C)+\beta R_BR_C}{R_B(R_B+R_C)} \right )[/tex]

__[itex]\dfrac{v_{out}}{i_{in}}[/itex]__

*Calculating*

*:*[tex]\dfrac{v_{out}}{i_{in}}=\left (-\dfrac{\beta R_BR_C}{R_B+R_C} \right )\left(\frac{R_B(R_B+R_C)}{(R_B+r_\pi)(R_B+R_C)+\beta R_BR_C} \right )[/tex]

[tex]\dfrac{v_{out}}{i_{in}}=-\frac{R_B(\beta R_BR_C)}{(R_B+r_\pi)(R_B+R_C)+\beta R_BR_C}[/tex]

[tex]\dfrac{v_{out}}{i_{in}}=-\frac{R_B(\beta R_BR_C)}{R_B^2+R_BR_C+R_Br_\pi+R_Cr_\pi+\beta R_BR_C}[/tex]

[tex]\dfrac{v_{out}}{i_{in}}=-\frac{R_B}{\dfrac{R_B}{\beta R_C}+ \dfrac{1}{\beta}+\dfrac{r_\pi}{\beta R_C}+\dfrac{r_\pi}{\beta R_B} +1}[/tex]

[tex]\dfrac{v_{out}}{i_{in}}=-\frac{R_B}{\dfrac{R_B+r_\pi}{\beta R_C}+ \dfrac{1}{\beta}+\dfrac{r_\pi}{\beta R_B} +1}[/tex]

If I neglect [itex]\dfrac{1}{\beta}[/itex] as 0, and neglect [itex]\dfrac{r_\pi}{\beta R_B} = \dfrac{(\beta+1)r_e}{\beta R_B}[/itex] (since [itex]R_B\gg r_e[/itex]), I get:

[tex]\dfrac{v_{out}}{i_{in}}=-\frac{R_B}{\dfrac{R_B+r_\pi}{\beta R_C} +1}[/tex]

This is very close to the desired result, but not equal. Am I missing something?

Thank you in advance.