Chapter 5, Problem 45E

To determine

Find the Thevenin equivalent of the circuit.

Answer to Problem 45E

The Thevenin voltage is $\frac{R_i\left(R_o-A{R}_f\right)}{R_i{R}_f+R_1{R}_f+R_1{R}_i+R_i{R}_o+R_1{R}_o+A{R}_1{R}_i}{v}_{in}$ and the Thevenin equivalent resistance is $\frac{R_o\left(R_i{R}_f+R_1{R}_f+R_1{R}_i\right)}{R_i{R}_f+R_1{R}_f+R_1{R}_i+R_i{R}_o+R_1{R}_o+A{R}_1{R}_i}$ and the equivalent circuit is drawn as shown in Figure 3.

Explanation of Solution

Calculation:

The redrawn circuit diagram is given in Figure 1,

Refer to Figure 1,

Apply Kirchhoff’s current law at node $a$,

$\frac{-v_d-v_{\text{in}}}{R_1}+\frac{-v_d}{R_i}+\frac{-v_d-v_{\text{out}}}{R_f}=0\text{A}$ (1)

Here,

$R_1$, $R_i$ and $R_f$ are the resistances of an operational amplifier,

$R_o$ is the load resistance,

$v_d$ is the voltage across the $R_i$ resistor,

$v_{\text{in}}$ is the input voltage of an operational amplifier and

$v_{\text{out}}$ is the output voltage of an operational amplifier.

Rearrange equation (1),

$\begin{array}{l}\frac{v_d+v_{\text{in}}}{R_1}+\frac{v_d}{R_i}+\frac{v_d+v_{\text{out}}}{R_f}=0\text{A}\\ \frac{R_i{R}_f{v}_d+R_i{R}_f{v}_{\text{in}}+R_1{R}_f{v}_d+R_1{R}_i{v}_d+R_1{R}_i{v}_{\text{out}}}{R_1{R}_i{R}_f}=0\text{A}\\ R_i{R}_f{v}_d+R_i{R}_f{v}_{\text{in}}+R_1{R}_f{v}_d+R_1{R}_i{v}_d+R_1{R}_i{v}_{\text{out}}=0\text{V}\end{array}$

$v_d\left(R_i{R}_f+R_1{R}_f+R_1{R}_i\right)+R_i{R}_f{v}_{\text{in}}+R_1{R}_i{v}_{\text{out}}=0\text{V}$ (2)

Apply Kirchhoff’s current law at node $b$,

$\frac{v_{\text{out}}-v_o}{R_o}+\frac{v_{\text{out}}+v_d}{R_f}=0\text{A}$ (3)

Here,

$v_o$ is the dependent voltage, which is equal to $A{v}_d$.

Substitute $A{v}_d$ for $v_o$ in equation (2),

$\begin{array}{l}\frac{v_{\text{out}}-A{v}_d}{R_o}+\frac{v_{\text{out}}+v_d}{R_f}=0\text{A}\\ \frac{R_f{v}_{\text{out}}-A{R}_f{v}_d+R_o{v}_{\text{out}}+R_o{v}_d}{R_o{R}_f}=0\text{A}\end{array}$

Rearrange for $v_d$,

$\begin{array}{l}{R}_f{v}_{\text{out}}-A{R}_f{v}_d+R_o{v}_{\text{out}}+R_o{v}_d=0\text{V}\\ v_d\left(R_o-A{R}_f\right)+v_{\text{out}}\left(R_f+R_o\right)=0\text{V}\\ v_d\left(A{R}_f-R_o\right)=v_{\text{out}}\left(R_f+R_o\right)\end{array}$

$v_d=\frac{\left(R_f+R_o\right)}{\left(A{R}_f-R_o\right)}{v}_{\text{out}}$ (4)

Substitute $\frac{\left(R_f+R_o\right)}{\left(A{R}_f-R_o\right)}{v}_{\text{out}}$ for $v_d$ in equation (2),

$\frac{\left(R_f+R_o\right)}{\left(A{R}_f-R_o\right)}{v}_{\text{out}}\left(R_i{R}_f+R_1{R}_f+R_1{R}_i\right)+R_i{R}_f{v}_{\text{in}}+R_1{R}_i{v}_{\text{out}}=0\text{V}$

Rearrange the equation for $v_{\text{out}}$,

$\begin{array}{l}\left(\frac{\left(R_f+R_o\right)\left(R_i{R}_f+R_1{R}_f+R_1{R}_i\right)}{\left(A{R}_f-R_o\right)}+R_1{R}_i\right)v_{\text{out}}=-R_i{R}_f{v}_{\text{in}}\\ \left(\frac{\left(R_f+R_o\right)\left(R_i{R}_f+R_1{R}_f+R_1{R}_i\right)+R_1{R}_i\left(A{R}_f-R_o\right)}{\left(A{R}_f-R_o\right)}\right)v_{\text{out}}=-R_i{R}_f{v}_{\text{in}}\\ \left(\frac{R_i{\left(R_f\right)}^2+R_1{\left(R_f\right)}^2+R_1{R}_i{R}_f+R_i{R}_f{R}_o+R_1{R}_f{R}_o+R_1{R}_i{R}_o+A{R}_1{R}_i{R}_f-R_1{R}_i{R}_o}{\left(A{R}_f-R_o\right)}\right)v_{\text{out}}=-R_i{R}_f{v}_{\text{in}}\\ \left(\frac{R_i{R}_f+R_1{R}_f+R_1{R}_i+R_i{R}_o+R_1{R}_o+A{R}_1{R}_i}{\left(A{R}_f-R_o\right)}\right)R_f{v}_{\text{out}}=-R_i{R}_f{v}_{\text{in}}\end{array}$

Modify the equation as,

$\left(\frac{R_i{R}_f+R_1{R}_f+R_1{R}_i+R_i{R}_o+R_1{R}_o+A{R}_1{R}_i}{\left(A{R}_f-R_o\right)}\right)v_{\text{out}}=-R_i{v}_{\text{in}}$

Write the equation to find voltage $v_{\text{out}}$ as follows.

$\begin{array}{l}{v}_{\text{out}}=-\frac{R_i\left(A{R}_f-R_o\right)}{R_i{R}_f+R_1{R}_f+R_1{R}_i+R_i{R}_o+R_1{R}_o+A{R}_1{R}_i}{v}_{\text{in}}\\ v_{\text{out}}=\frac{R_i\left(R_o-A{R}_f\right)}{R_i{R}_f+R_1{R}_f+R_1{R}_i+R_i{R}_o+R_1{R}_o+A{R}_1{R}_i}{v}_{\text{in}}\end{array}$

So, the Thevenin voltage is $\frac{R_i\left(R_o-A{R}_f\right)}{R_i{R}_f+R_1{R}_f+R_1{R}_i+R_i{R}_o+R_1{R}_o+A{R}_1{R}_i}{v}_{\text{in}}$

To find equivalent resistance of a circuit the independent voltage source replaced by short circuit and load resistor is disconnected and $1\text{A}$ test source is connected across open circuit terminal.

The redrawn circuit diagram is given in Figure 2,

Refer to the redrawn Figure 2,

Apply Kirchhoff’s current law at node $a$,

$\frac{-v_d}{R_1}+\frac{-v_d}{R_i}+\frac{-v_d-v_{\text{out}}}{R_f}=0\text{A}$ (5)

Rearrange for $v_d$,

$\begin{array}{l}\frac{v_d}{R_1}+\frac{v_d}{R_i}+\frac{v_d+v_{\text{out}}}{R_f}=0\text{A}\\ \frac{R_i{R}_f{v}_d+R_1{R}_f{v}_d+R_1{R}_i{v}_d+R_1{R}_i{v}_{\text{out}}}{R_1{R}_i{R}_f}=0\text{A}\\ R_i{R}_f{v}_d+R_1{R}_f{v}_d+R_1{R}_i{v}_d+R_1{R}_i{v}_{\text{out}}=0\text{V}\\ v_d\left(R_i{R}_f+R_1{R}_f+R_1{R}_i\right)=-R_1{R}_i{v}_{\text{out}}\end{array}$

$v_d=-\frac{R_1{R}_i}{R_i{R}_f+R_1{R}_f+R_1{R}_i}{v}_{\text{out}}$ (6)

Apply Kirchhoff’s current law at node $b$,

$\frac{v_{\text{out}}-v_o}{R_o}+\frac{v_{\text{out}}+v_d}{R_f}=i_o$ (7)

Substitute $A{v}_d$ for $v_o$ and $1\text{A}$ for $i_o$ in equation (7),

$\begin{array}{l}\frac{v_{\text{out}}-A{v}_d}{R_o}+\frac{v_{\text{out}}+v_d}{R_f}=1\text{A}\\ \frac{R_f{v}_{\text{out}}-A{R}_f{v}_d+R_o{v}_{\text{out}}+R_o{v}_d}{R_o{R}_f}=1\text{A}\end{array}$

$\frac{v_{\text{out}}\left(R_f+R_o\right)+v_d\left(R_o-A{R}_f\right)}{R_o{R}_f}=1\text{A}$ (8)

Substitute $-\frac{R_1{R}_i}{R_i{R}_f+R_1{R}_f+R_1{R}_i}{v}_{\text{out}}$ for $v_d$ in equation (8),

$\begin{array}{l}\frac{v_{\text{out}}\left(R_f+R_o\right)+\left(-\frac{R_1{R}_i}{R_i{R}_f+R_1{R}_f+R_1{R}_i}{v}_{\text{out}}\right)\left(R_o-A{R}_f\right)}{R_o{R}_f}=1\text{A}\\ \frac{v_{\text{out}}\left(R_f+R_o\right)+\left(-\frac{R_1{R}_i}{R_i{R}_f+R_1{R}_f+R_1{R}_i}{v}_{\text{out}}\right)\left(R_o-A{R}_f\right)}{R_o{R}_f}=1\text{A}\\ \frac{v_{\text{out}}\left(\left(R_f+R_o\right)-\frac{R_1{R}_i\left(R_o-A{R}_f\right)}{R_i{R}_f+R_1{R}_f+R_1{R}_i}\right)}{R_o{R}_f}=1\text{A}\end{array}$

Rearrange for $v_{\text{out}}$,

$\begin{array}{l}{v}_{\text{out}}\left(\left(R_f+R_o\right)-\frac{R_1{R}_i\left(R_o-A{R}_f\right)}{R_i{R}_f+R_1{R}_f+R_1{R}_i}\right)=R_o{R}_f\\ v_{\text{out}}\left(\frac{\left(R_f+R_o\right)\left(R_i{R}_f+R_1{R}_f+R_1{R}_i\right)-R_1{R}_i\left(R_o-A{R}_f\right)}{R_i{R}_f+R_1{R}_f+R_1{R}_i}\right)=R_o{R}_f\\ v_{\text{out}}\left(\frac{R_i{\left(R_f\right)}^2+R_1{\left(R_f\right)}^2+R_1{R}_i{R}_f+R_i{R}_f{R}_o+R_1{R}_f{R}_o+R_1{R}_i{R}_o-R_1{R}_i{R}_o+A{R}_1{R}_i{R}_f}{R_i{R}_f+R_1{R}_f+R_1{R}_i}\right)=R_o{R}_f\\ v_{\text{out}}\left(\frac{R_i{R}_f+R_1{R}_f+R_1{R}_i+R_i{R}_o+R_1{R}_o+A{R}_1{R}_i}{R_i{R}_f+R_1{R}_f+R_1{R}_i}\right)=R_o\end{array}$

$v_{\text{out}}=\frac{R_o\left(R_i{R}_f+R_1{R}_f+R_1{R}_i\right)}{R_i{R}_f+R_1{R}_f+R_1{R}_i+R_i{R}_o+R_1{R}_o+A{R}_1{R}_i}$

The expression for the Thevenin equivalent resistance is as follows,

$R_{TH}=\frac{v_{\text{out}}}{i_o}$ (9)

Substitute $\frac{R_o\left(R_i{R}_f+R_1{R}_f+R_1{R}_i\right)}{R_i{R}_f+R_1{R}_f+R_1{R}_i+R_i{R}_o+R_1{R}_o+A{R}_1{R}_i}$ for $v_{\text{out}}$ and $1\text{A}$ for $i_o$ in equation (9),

$\begin{array}{c}{R}_{TH}=\frac{\frac{R_o\left(R_i{R}_f+R_1{R}_f+R_1{R}_i\right)}{R_i{R}_f+R_1{R}_f+R_1{R}_i+R_i{R}_o+R_1{R}_o+A{R}_1{R}_i}}{1\text{A}}\\ =\frac{R_o\left(R_i{R}_f+R_1{R}_f+R_1{R}_i\right)}{R_i{R}_f+R_1{R}_f+R_1{R}_i+R_i{R}_o+R_1{R}_o+A{R}_1{R}_i}\end{array}$

So, the Thevenin equivalent resistance is $\frac{R_o\left(R_i{R}_f+R_1{R}_f+R_1{R}_i\right)}{R_i{R}_f+R_1{R}_f+R_1{R}_i+R_i{R}_o+R_1{R}_o+A{R}_1{R}_i}$.

The Thevenin equivalent circuit is drawn as shown in Figure 3,

Conclusion:

Thus, the Thevenin voltage is $\frac{R_i\left(R_o-A{R}_f\right)}{R_i{R}_f+R_1{R}_f+R_1{R}_i+R_i{R}_o+R_1{R}_o+A{R}_1{R}_i}{v}_{\text{in}}$ and the Thevenin equivalent resistance is $\frac{R_o\left(R_i{R}_f+R_1{R}_f+R_1{R}_i\right)}{R_i{R}_f+R_1{R}_f+R_1{R}_i+R_i{R}_o+R_1{R}_o+A{R}_1{R}_i}$ and the equivalent circuit is drawn as shown in Figure 3.