Shunt feedback amplifier circuit

[favq]

Homework Statement

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

Verify that:
r_T = \dfrac{v_{out}}{i_{in}}=\dfrac{-R_B}{1+\dfrac{R_B+r_\pi}{(1+\beta)R_C}}
(assuming R_B\gg r_e, where r_\pi=(\beta+1)r_e.)

Homework Equations

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

The Attempt at a Solution

Equation for Node 1:
i_{in}=i_b+\dfrac{v_{be}-v_{out}}{R_B}
Since v_{be} = i_br_\pi:
i_{in}=i_b+\dfrac{i_br_\pi-v_{out}}{R_B}
i_{in}=\dfrac{i_b(R_B+r_\pi)-v_{out}}{R_B}

Equation for Node 2:
\dfrac{i_br_\pi-v_{out}}{R_B}=\beta i_b + \dfrac{v_{out}}{R_C}
v_{out} \left ( \dfrac{1}{R_C} + \dfrac{1}{R_B} \right ) = i_b \left ( \dfrac{r_\pi}{R_B} - \beta \right )
Since r_\pi=(\beta+1)r_e:
v_{out} \left ( \dfrac{1}{R_C} + \dfrac{1}{R_B} \right ) = i_b \left ( \dfrac{(\beta+1)r_e}{R_B} - \beta \right )
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 )
Since R_B\gg r_e, I tried to neglect \dfrac{r_e}{R_B}, making it 0:
v_{out} \left ( \dfrac{1}{R_C} + \dfrac{1}{R_B} \right ) = i_b(\beta+1) \left ( 0 - \dfrac{\beta}{\beta+1} \right )
v_{out} \left ( \dfrac{1}{R_C} + \dfrac{1}{R_B} \right ) = - \beta i_b
v_{out} = -\dfrac{\beta i_bR_BR_C}{R_B+R_C}

Plugging this back into the expression for i_in:
i_{in}=\dfrac{i_b(R_B+r_\pi)+\dfrac{\beta i_bR_BR_C}{R_B+R_C}}{R_B}
i_{in}=i_b\left ( \frac{R_B+r_\pi}{R_B}+\dfrac{\beta R_C}{R_B+R_C} \right )
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 )
Calculating \dfrac{v_{out}}{i_{in}}:
\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 )
\dfrac{v_{out}}{i_{in}}=-\frac{R_B(\beta R_BR_C)}{(R_B+r_\pi)(R_B+R_C)+\beta R_BR_C}
\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}
\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}
\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}
If I neglect \dfrac{1}{\beta} as 0, and neglect \dfrac{r_\pi}{\beta R_B} = \dfrac{(\beta+1)r_e}{\beta R_B} (since R_B\gg r_e), I get:
\dfrac{v_{out}}{i_{in}}=-\frac{R_B}{\dfrac{R_B+r_\pi}{\beta R_C} +1}
This is very close to the desired result, but not equal. Am I missing something?
Thank you in advance.

 

[LvW]

Two remarks:
(1) What about Rin ? Did you forget the influence of Rin?
(EDIT: OK, only now I have realized that you are asking for Vout/Iin).
(2) The "very small resistance re" is identical to re=1/gm (with transconductance gm=Ic/Vt ; Vt: temeprature voltage).

 

[favq]

Thank you for the reply.
I didn't include R_{IN} in the calculations because, in the KCL equation for Node 1, I think I can just use the quantity i_{in} as the current going into Node 1 from the input source.
I could have written \dfrac{v_{in}-v_{be}}{R_{IN}} instead of i_{in} as the current going into Node 1 from the input source, but, since the result I'm trying to verify is in terms of just i_{in} (not v_{in} and R_{IN}), I just wrote i_{in}.

 

[LvW]

OK - I see.
But, as far as I can see, the difference between both results is just in the denominator: β vs. (β+1), correct?
This difference is really negligible. Don`t forget that β is (a) very large and (b) has very large tolerances.

 

[gneill]

Hi favq,

Welcome to Physics Forums!

I think I would hold off using approximations until a bit later in the derivation. Find an expression for $v_{out}/i_{in}$ first from the node equations. There will be a lot of algebra, but when the smoke clears you should get something like:

$\frac{v_{out}}{v_{in}} = \frac{(r_\pi - \beta R_B) R_C}{\beta R_C + R_C +R_B + r_\pi}$

This way you'll rule out any missing any multipliers that might have made the early approximation problematical.
You should be able to work from there towards your goal.

 

[The Electrician]

Hi favq,

Welcome to Physics Forums!

I think I would hold off using approximations until a bit later in the derivation. Find an expression for $v_{out}/i_{in}$ first from the node equations. There will be a lot of algebra, but when the smoke clears you should get something like:

$\frac{v_{out}}{v_{in}} = \frac{(r_\pi - \beta R_B) R_C}{\beta R_C + R_C +R_B + r_\pi}$

This way you'll rule out any missing any multipliers that might have made the early approximation problematical.
You should be able to work from there towards your goal.

Don't you mean: $\frac{v_{out}}{i_{in}} = \frac{(r_\pi - \beta R_B) R_C}{\beta R_C + R_C +R_B + r_\pi}$

 

[gneill]

Don't you mean: $\frac{v_{out}}{i_{in}} = \frac{(r_\pi - \beta R_B) R_C}{\beta R_C + R_C +R_B + r_\pi}$

Yes indeed. That's a typo on my part. Thanks for catching that.

 

[favq]

Thank you all for the feedback.
Following gneill's suggestion, I redid this problem delaying the approximations, and I got the expected result.
For reference, here is my new solution:
I'm going to start again from this point (which is right before I applied the R_B\gg r_e approximation):
v_{out} \left ( \dfrac{1}{R_C} + \dfrac{1}{R_B} \right ) = i_b \left ( \dfrac{r_\pi}{R_B} - \beta \right )
v_{out} = i_b \dfrac{(r_\pi - \beta R_B)R_C}{R_B+R_C}
Plugging this v_{out} expression back into the i_{in} expression:
i_{in}=i_b\dfrac{(R_B+r_\pi)- \dfrac{(r_\pi - \beta R_B)R_C}{R_B+R_C} }{R_B}
i_{in}=i_b\dfrac{(R_B+r_\pi)(R_B+R_C)-(r_\pi - \beta R_B)R_C}{R_B(R_B+R_C)}
Calculating \dfrac{v_{out}}{i_{in}}:
\dfrac{v_{out}}{i_{in}}= \dfrac{(r_\pi - \beta R_B)R_C}{R_B+R_C} \dfrac{R_B(R_B+R_C)}{(R_B+r_\pi)(R_B+R_C)-(r_\pi - \beta R_B)R_C}
\dfrac{v_{out}}{i_{in}}= \dfrac{(r_\pi - \beta R_B)R_CR_B}{(R_B+r_\pi)(R_B+R_C)-(r_\pi - \beta R_B)R_C}
\frac{v_{out}}{i_{in}} = \frac{(r_\pi - \beta R_B) R_C}{\beta R_C + R_C +R_B + r_\pi} = \frac{(r_\pi - \beta R_B) R_C}{(\beta+1)R_C +R_B + r_\pi}
\frac{v_{out}}{i_{in}} = \frac{((\beta+1))r_e - \beta R_B) R_C}{(\beta + 1)R_C +R_B + r_\pi}
Dividing the numerator and denominator by (\beta+1)R_C:
\frac{v_{out}}{i_{in}} = \frac{r_e - \left (\dfrac{\beta}{\beta+1} \right ) R_B}{1 + \dfrac{R_B+r_\pi}{(1+\beta)R_C}}
Now, I will make \dfrac{\beta}{\beta+1} \approx 1 and, since R_B\gg r_e, neglect r_e in comparison with R_B. This yields:
\dfrac{v_{out}}{i_{in}}=\dfrac{-R_B}{1+\dfrac{R_B+r_\pi}{(1+\beta)R_C}}

 

[gneill]

Nice

 

[LvW]

Hi favq, perhaps you are also interested in an alternative calculation - based on feedback theory?
Here is my approach:
(1) Voltage gain without feedback (open-loop): Ao=Vout/Vbe=-gRc with transconductance g=d(Ie)/d(Vbe)=(β+1)/r_π.
(2) Vout/Vbe=Vout/(Ib*r_π)=-gRc which gives: To=Vout/Ib=-(β+1)Rc (because g=(β+1)/r_π).
(3) Transfer function with feedback: T=Vout/Iin=To/(1+To*k) with feedback factor k=I_f/Vout and I_f=(Vout-Vbe)/R_BVout.
(4) Combining and solving for T gives the correct result.

 

[favq]

LvW,
Thank you for the alternative calculation. I will look into it to deepen my understanding.