Chapter 6, Problem 56E

To determine

Find the current $I_L$ in the circuit.

Answer to Problem 56E

The current $I_L$ is $\frac{1}{5000}{V}_1$ in op amp.

Explanation of Solution

Given data:

Value of voltage $V_2$ is $0\text{V}$ and

Value of resistances $R_1=2R_2=1\text{k}\Omega$, $R_1=R_3$, $R_2=R_4$, and $R_L=100\Omega$.

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 $R_4$ is 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 the equation (2).

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

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

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

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

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

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

$v_{\text{out}}=\left[\frac{R_4{R}_L\left(R_1+R_2\right)}{R_1{R}_4{R}_L+R_1{R}_3{R}_4-R_3{R}_L{R}_2}\right]V_1-\left[\frac{R_2{R}_L{R}_4+R_2{R}_3{R}_L+R_2{R}_3{R}_4}{R_1{R}_4{R}_L+R_1{R}_3{R}_4-R_3{R}_L{R}_2}\right]V_2$ (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 the 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\end{array}$$I_L=\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$ (9)

Substitute $0\text{V}$ for $V_2$, $1\text{ k}\Omega$ for $R_1$, $1\text{k}\Omega$ for $R_3$, $0.5\text{}\text{k}\Omega$ for $R_4$, $0.5\text{k}\Omega$ for $R_2$, and $100\Omega$ for $R_L$ in equation (9).

$\begin{array}{c}{I}_L=\left[\begin{array}{l}\left\{\frac{100\Omega \times 1\text{k}\Omega }{\left(1\text{k}\Omega \times 0.5\text{k}\Omega \times 100\Omega \right)+1\text{k}\Omega \times 1\text{k}\Omega \times 0.5\text{k}\Omega -1\text{k}\Omega \times 100\Omega \times 0.5\text{k}\Omega }\right\}{V}_1\\ -\left\{\frac{0.5\text{k}\Omega \times 1\text{k}\Omega }{1\text{k}\Omega \times 0.5\text{k}\Omega \times 100\Omega +1\text{k}\Omega \times 1\text{k}\Omega \times 0.5\text{k}\Omega -1\text{k}\Omega \times 100\Omega \times 0.5\text{k}\Omega }\right\}\times 0\text{V}\end{array}\right]\\ =\left[\begin{array}{l}\left\{\frac{100\text{k}\Omega }{500\text{M}\Omega }\right\}{V}_1\\ -\times 0\text{V}\end{array}\right]\\ =\frac{1}{5000}{V}_1\end{array}$

$I_L=\frac{1}{5000}{V}_1$

Conclusion:

Thus, the current $I_L$ is $\frac{1}{5000}{V}_1$ in op amp.