# Engineering Shunt feedback amplifier circuit

#### favq

1. 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 $r_\pi=(\beta+1)r_e$.)

2. 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.

3. 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 iin:
$$i_{in}=\dfrac{i_b(R_B+r_\pi)+\dfrac{\beta i_bR_BR_C}{R_B+R_C}}{R_B}$$
$$i_{in}=i_b\left ( \frac{R_B+r_\pi}{R_B}+\dfrac{\beta R_C}{R_B+R_C} \right )$$
$$i_{in}=i_b\left ( \frac{(R_B+r_\pi)(R_B+R_C)+\beta R_BR_C}{R_B(R_B+R_C)} \right )$$
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.

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#### 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

Mentor
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

Gold Member
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

Mentor
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:
$$\dfrac{v_{out}}{i_{in}}=\dfrac{-R_B}{1+\dfrac{R_B+r_\pi}{(1+\beta)R_C}}$$

Mentor
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)/rπ.
(2) Vout/Vbe=Vout/(Ib*rπ)=-gRc which gives: To=Vout/Ib=-(β+1)Rc (because g=(β+1)/rπ).
(3) Transfer function with feedback: T=Vout/Iin=To/(1+To*k) with feedback factor k=If/Vout and If=(Vout-Vbe)/RBVout.
(4) Combining and solving for T gives the correct result.

Last edited:

#### favq

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

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