Chapter 6, Problem 55E

To determine

Derive the expression for $v_{\text{out}}$ and $I_L$ in the op amp circuit.

Answer to Problem 55E

The derived expression for $v_{\text{out}}$ is

$\left[\frac{R_4{R}_L\left(R_1+R_2\right)}{R_1{R}_4{R}_L+R_1{R}_3{R}_4-R_3{R}_L{R}_2}\right]V_1-\left[\frac{R_2{R}_L{R}_4+R_2{R}_3{R}_L+R_2{R}_3{R}_4}{R_1{R}_4{R}_L+R_1{R}_3{R}_4-R_3{R}_L{R}_2}\right]V_2$ and for $I_L$ is

$\left(\frac{R_4{R}_1}{R_1{R}_4{R}_L+R_1{R}_3{R}_4-R_3{R}_L{R}_2}\right)V_1-\left[\frac{R_2{R}_3}{R_1{R}_4{R}_L+R_1{R}_3{R}_4-R_3{R}_L{R}_2}\right]V_2$.

Explanation of Solution

Calculation:

The redrawn circuit is shown in Figure 1 as follows.

Refer to the Figure 1.

The expression for nodal analysis at node voltage $V_a$ is,

$\left(\frac{V_a-V_2}{R_1}\right)+\left(\frac{V_a-v_{\text{out}}}{R_2}\right)=0$ (1)

Here,

$V_2$ is the input voltage of op amp,

$v_{\text{out}}$ is the output voltage of op amp and

$R_1$ and $R_2$ are the resistances of op amp.

The expression for nodal analysis at node voltage $V_b$ is,

$\left(\frac{V_b}{R_L}\right)+\left(\frac{V_b-V_1}{R_3}\right)+\left(\frac{V_b-v_{\text{out}}}{R_4}\right)=0$ (2)

Here,

$V_1$ is the input voltage of op amp and

$R_L$, $R_3$ and $R_4$ are the resistance of op amp.

The expression for the virtual ground concept is as follows.

$V_a=V_b$ (3)

The expression for current $I_L$ in the circuit is,

$I_L=\frac{V_b}{R_L}$ (4)

Here,

$I_L$ is the current for resistance $R_L$ in circuit.

Simplify equation (1) for $V_a$.

$\begin{array}{c}{V}_a\left(\frac{1}{R_1}+\frac{1}{R_2}\right)=\frac{V_2}{R_1}+\frac{v_{\text{out}}}{R_2}\\ V_a\left(\frac{R_1+R_2}{R_1{R}_2}\right)=\frac{V_2{R}_2+v_{\text{out}}{R}_1}{R_1{R}_2}\end{array}$

$V_a=\frac{V_2{R}_2+v_{\text{out}}{R}_1}{\left(R_1+R_2\right)}$ (5)

Simplify equation (2) as follows.

$V_a\left(\frac{1}{R_3}+\frac{1}{R_4}+\frac{1}{R_L}\right)=\frac{V_1}{R_3}+\frac{v_{\text{out}}}{R_4}$ (6)

Substitute $\frac{V_2{R}_2+v_{\text{out}}{R}_1}{\left(R_1+R_2\right)}$ for $V_a$ in the equation (6).

$\begin{array}{l}\left(\frac{V_2{R}_2+v_{\text{out}}{R}_1}{R_1+R_2}\right)\left(\frac{R_4{R}_L+R_3{R}_L+R_3{R}_4}{R_3{R}_4{R}_L}\right)=\frac{V_1{R}_4+v_{\text{out}}{R}_3}{R_3{R}_4}\\ \left(\frac{V_2{R}_2}{R_1+R_2}\right)\left(\frac{R_4{R}_L+R_3{R}_L+R_3{R}_4}{R_L}\right)+\left(\frac{v_{\text{out}}{R}_1}{R_1+R_2}\right)\left(\frac{R_4{R}_L+R_3{R}_L+R_3{R}_4}{R_L}\right)=V_1{R}_4+v_{\text{out}}{R}_3\end{array}$

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

$\begin{array}{c}{v}_{\text{out}}\left(\frac{R_1{R}_4{R}_L+R_1{R}_3{R}_4+R_1+R_3+R_L}{R_L\left(R_1+R_2\right)}-R_3\right)=V_1{R}_4-\left(\frac{R_2{R}_4{R}_L+R_3{R}_2{R}_L+R_2{R}_3{R}_4}{R_L\left(R_1+R_2\right)}\right)V_2\\ v_{\text{out}}\left(\begin{array}{l}{R}_1{R}_4{R}_L+R_1{R}_3{R}_L+R_1{R}_3{R}_4\\ -R_3{R}_L{R}_1-R_3{R}_L{R}_2\end{array}\right)=\left[\begin{array}{l}{V}_1{R}_4{R}_L\left(R_1+R_2\right)\\ -\left(R_2{R}_4{R}_L+R_2{R}_3{R}_L+R_2{R}_3{R}_4\right)V_2\end{array}\right]\end{array}$

Simplify for $v_{\text{out}}$.

$v_{\text{out}}\left(R_1{R}_4{R}_L+R_1{R}_3{R}_4-R_3{R}_L{R}_2\right)=V_1{R}_4{R}_L\left(R_1+R_2\right)-\left(R_2{R}_4{R}_L+R_2{R}_3{R}_L+R_2{R}_3{R}_4\right)V_2$

$v_{\text{out}}=\left[\frac{R_4{R}_L\left(R_1+R_2\right)}{R_1{R}_4{R}_L+R_1{R}_3{R}_4-R_3{R}_L{R}_2}\right]V_1-\left[\frac{R_2{R}_L{R}_4+R_2{R}_3{R}_L+R_2{R}_3{R}_4}{R_1{R}_4{R}_L+R_1{R}_3{R}_4-R_3{R}_L{R}_2}\right]V_2$ (7)

Substitute $\frac{V_2{R}_2+v_{\text{out}}{R}_1}{\left(R_1+R_2\right)}$ for $V_b$ in equation (4).

$I_L=\frac{V_2{R}_2+v_{\text{out}}{R}_1}{R_L\left(R_1+R_2\right)}$ (8)

Substitute value of $v_{\text{out}}$ from equation (7) in equation (8).

$\begin{array}{c}{I}_L=\frac{V_2{R}_2}{R_L\left(R_1+R_2\right)}+\frac{R_1}{R_L\left(R_1+R_2\right)}\left\{\begin{array}{l}\left[\frac{R_4{R}_L\left(R_1+R_2\right)}{R_1{R}_4{R}_{{}_L}+R_1{R}_3{R}_4-R_3{R}_L{R}_2}\right]V_1\\ -\left[\frac{R_2{R}_4{R}_L+R_L{R}_2{R}_3+R_2{R}_3{R}_4}{R_1{R}_4{R}_L+R_1{R}_3{R}_4-R_3{R}_L{R}_2}\right]V_2\end{array}\right\}\\ =\left(\frac{R_4{R}_1}{R_1{R}_4{R}_L+R_1{R}_3{R}_4-R_3{R}_L{R}_2}\right)V_1+\left[\begin{array}{l}\frac{R_2}{R_L\left(R_1+R_2\right)}\\ -\frac{R_1\left(R_2{R}_4{R}_L+R_2{R}_3{R}_L+R_2{R}_3{R}_4\right)}{R_L\left(R_1+R_2\right)\left(R_1{R}_4{R}_L+R_1{R}_3{R}_4-R_3{R}_2{R}_L\right)}\end{array}\right]V_2\\ =\left(\frac{R_4{R}_1}{R_1{R}_4{R}_L+R_1{R}_3{R}_4-R_3{R}_L{R}_2}\right)V_1-\left[\frac{R_3{R}_2{R}_L\left(R_1+R_2\right)}{R_L\left(R_1+R_2\right)\left(R_1{R}_4{R}_L+R_4{R}_3{R}_1-R_3{R}_L{R}_2\right)}\right]V_2\\ =\left(\frac{R_4{R}_1}{R_1{R}_4{R}_L+R_1{R}_3{R}_4-R_3{R}_L{R}_2}\right)V_1-\left[\frac{R_2{R}_3}{R_1{R}_4{R}_L+R_1{R}_3{R}_4-R_3{R}_L{R}_2}\right]V_2\end{array}$

Conclusion:

Thus, the derived expression for $v_{\text{out}}$ is

$\left[\frac{R_4{R}_L\left(R_1+R_2\right)}{R_1{R}_4{R}_L+R_1{R}_3{R}_4-R_3{R}_L{R}_2}\right]V_1-\left[\frac{R_2{R}_L{R}_4+R_2{R}_3{R}_L+R_2{R}_3{R}_4}{R_1{R}_4{R}_L+R_1{R}_3{R}_4-R_3{R}_L{R}_2}\right]V_2$ and for $I_L$ is

$\left(\frac{R_4{R}_1}{R_1{R}_4{R}_L+R_1{R}_3{R}_4-R_3{R}_L{R}_2}\right)V_1-\left[\frac{R_2{R}_3}{R_1{R}_4{R}_L+R_1{R}_3{R}_4-R_3{R}_L{R}_2}\right]V_2$.