Chapter 6, Problem 24E

(a)

To determine

Find the output voltage of the circuit.

Answer to Problem 24E

The output voltage of combined circuit is $-36.5\text{V}$.

Explanation of Solution

Given data:

Combine the two circuits by eliminating the $1.5\text{V}$ source of FIGURE 6.50,

Connect the output of circuit shown in FIGURE 6.49 to left-hand terminal of $500\Omega$ resistor,

Value of resistance $R_4$ is $2\text{k}\Omega$.

Value of input voltage of 1^st op amp $v_{\text{in}}$ is $2\text{V}$ and

Value of input voltage of 4^th op amp $v_1$ is $1\text{V}$.

Calculation:

The redrawn circuit from given data is shown in Figure 1 as follows.

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

$\frac{v_a}{R_1}+\frac{v_a-v_x}{R_{f_1}}=0$ (1)

Here,

$v_a$ is the voltage node across op amp,

$v_x$ is the output voltage of 1^st op amp and

$R_1$ and $R_{f_1}$ are the resistances of op amp

The expression for the virtual ground concept across 1^st op amp is as follows.

$v_a=v_b$

$v_a=v_{\text{in}}\text{V}$ (2)

Substitute $v_{\text{in}}$ for $v_a$ from equation (2) in equation (1).

$\begin{array}{l}\frac{v_{\text{in}}}{R_1}+\frac{v_{\text{in}}-v_x}{R_{f_1}}=0\\ \frac{v_{\text{in}}}{R_1}+\frac{v_{\text{in}}}{R_{f_1}}-\frac{v_x}{R_{f_1}}=0\end{array}$

Rearrange for $v_x$.

$\begin{array}{l}\frac{v_{\text{in}}}{R_1}+\frac{v_{\text{in}}}{R_{f_1}}=\frac{v_x}{R_{f_1}}\\ v_{\text{in}}\left(\frac{1}{R_1}+\frac{1}{R_{f_1}}\right)=\frac{v_x}{R_{f_1}}\\ v_{\text{in}}\left(\frac{R_1+R_{f_1}}{R_1{R}_{f_1}}\right)=\frac{v_x}{R_{f_1}}\end{array}$

Rearrange for $v_x$.

$v_x=v_{\text{in}}\left(\frac{R_1+R_{f_1}}{R_1}\right)$ (3)

Substitute $2\text{V}$ for $v_{\text{in}}$, $10\Omega$ for $R_1$ and $15\text{}\Omega$ for $R_{f_1}$ in equation (3).

$\begin{array}{c}{v}_x=v_{\text{in}}\left(\frac{10\Omega +15\Omega }{10\Omega }\right)\\ =v_{\text{in}}\left(\frac{25\Omega }{10\Omega }\right)\end{array}$

$v_x=2.5v_{\text{in}}$ (4)

Substitute $2\text{V}$ for $v_{\text{in}}$ in the equation (4).

$v_x=2.5\times 2\text{V}$

$v_x=\text{5}\text{V}$ (5)

The expression for the nodal analysis at node voltage $v_c$ of 2^nd op amp is,

$\frac{v_c-v_x}{R_4}+\frac{v_c-{v^{\prime }}_x}{R_{f_2}}=0$ (6)

Here,

$v_c$ is the voltage node across 2^nd op amp,

${v^{\prime }}_x$ is the output voltage of 2^nd op amp and

$R_1$ and $R_{f_2}$ is the resistance of 2^nd op amp

The expression for the virtual ground concept across 2^nd op amp is as follows,

$v_c=v_d$

$v_c=0\text{V}$ (7)

Substitute $0\text{V}$ for $v_c$ in equation (6).

$\begin{array}{l}\frac{0\text{V}-v_x}{R_4}+\frac{0\text{V}-{v^{\prime }}_x}{R_{f_2}}=0\\ -\frac{v_x}{R_4}-\frac{{v^{\prime }}_x}{R_{f_2}}=0\end{array}$

Rearrange for ${v^{\prime }}_x$.

${v^{\prime }}_x=-\left(\frac{R_{f_2}}{R_4}\right)v_x$ (8)

Substitute $5\text{V}$ for $v_x$, $5\text{k}\Omega$ for $R_{f_2}$ and $2\text{k}\Omega$ for $R_4$ in equation (8).

${v^{\prime }}_x=-\left(\frac{5\text{k}\Omega }{2\text{k}\Omega }\right)\times \left(5\text{V}\right)$

${v^{\prime }}_x=-12.5\text{V}$ (9)

The expression for the nodal analysis at node voltage $v_e$ of 3^rd op amp is,

$\frac{v_e-{v^{\prime }}_x}{R_2}+\frac{v_e-v_x\text{'}\text{'}}{R_{f_3}}=0$ (10)

Here,

$v_e$ is the voltage node across 3^rd op amp,

${v^{\prime }}_x$ is the output voltage of 3^rd op amp and

$R_2$ and $R_{f_3}$ is the resistance of 3^rd op amp.

The expression for the virtual ground concept across 3^rd op amp is as follows.

$v_e=v_f$

$v_e=0\text{V}$ (11)

Substitute $0\text{V}$ for $v_c$ from equation (11) in equation (10),

$\begin{array}{l}\frac{0\text{V}-{v^{\prime }}_x}{R_2}+\frac{0\text{V}-v_x\text{'}\text{'}}{R_{f_3}}=0\\ -\frac{{v^{\prime }}_x}{R_2}-\frac{v_x\text{'}\text{'}}{R_{f_3}}=0\end{array}$

Rearrange for $v_x\text{'}\text{'}$,

$v_x\text{'}\text{'}=-\left(\frac{R_{f_3}}{R_2}\right){v^{\prime }}_x$ (12)

Substitute $-12.5\text{V}$ for ${v^{\prime }}_x$, $1.5\text{k}\Omega$ for $R_{f_3}$ and $500\Omega$ for $R_2$ in equation (12),

$\begin{array}{c}{v}_x\text{'}\text{'}=-\left(\frac{1.5\text{k}\Omega }{500\Omega }\right)\times \left(-12.5\text{V}\right)\\ =\left(\frac{1.5\times {10}^3}{500\Omega }\right)\times \left(12.5\text{V}\right)\\ =3\times \left(12.5\text{V}\right)\end{array}$

$v_x\text{'}\text{'}=37.5\text{V}$

The expression for the nodal analysis at node voltage $v_g$ of 4^th op amp is,

$\frac{v_g-v_x\text{'}\text{'}}{R_3}+\frac{v_g-v_{\text{out}}}{R_{f_4}}=0$ (13)

Here,

$v_g$ is the voltage node across 4^th op amp,

$v_{\text{out}}$ is the output voltage of 4^th op amp and

$R_3$ and $R_{f_4}$ is the resistance of 4^th op amp.

The expression for the nodal analysis at node voltage $v_h$ of 4^th op amp is,

$\frac{v_h-v_1}{R_5}+\frac{v_h}{R_6}=0$ (14)

Here,

$v_h$ is the node voltage of 4^th op amp,

$v_1$ is the input voltage of 4^th op amp and

$R_5$ and $R_6$ is the resistance of 4^th op amp.

The expression for the virtual ground concept across 3^rd op amp is as follows,

$v_g=v_h$ (15)

Simplify equation (14) for $v_h$,

$\begin{array}{c}\frac{v_h}{R_5}+\frac{v_h}{R_6}=\frac{v_1}{R_5}\\ v_h\left(\frac{R_5+R_6}{R_5{R}_6}\right)=\frac{v_1}{R_5}\end{array}$

Rearrange for $v_h$,

$v_h=\left(\frac{R_6}{R_5+R_6}\right)v_1$ (16)

Substitute $\left(\frac{R_6}{R_5+R_6}\right)v_1$ for $v_g$ from equation (16) in equation (14),

$\begin{array}{c}\frac{\left(\frac{R_6}{R_5+R_6}\right)v_1-v_x\text{'}\text{'}}{R_3}+\frac{\left(\frac{R_6}{R_5+R_6}\right)v_1-v_{\text{out}}}{R_{f_4}}=0\\ \frac{R_6{v}_1}{R_3\left(R_5+R_6\right)}-\frac{v_x\text{'}\text{'}}{R_3}+\frac{R_6{v}_1}{R_{f_4}\left(R_5+R_6\right)}-\frac{v_{\text{out}}}{R_{f_4}}=0\end{array}$

Rearrange for $v_{out}$,

$\frac{R_6{v}_1}{R_3\left(R_5+R_6\right)}-\frac{v_x\text{'}\text{'}}{R_3}+\frac{R_6{v}_1}{R_{f_4}\left(R_5+R_6\right)}-\frac{v_{\text{out}}}{R_{f_4}}=0$

Rearrange for $v_{out}$,

$\frac{v_{out}}{R_{f_4}}=\frac{R_{f_4}{R}_6{v}_1+R_3{R}_6{v}_1}{R_{f_4}{R}_3\left(R_5+R_6\right)}-\frac{v_x\text{'}\text{'}}{R_3}$

$v_{out}=R_{f_4}\left(\frac{\left(R_3+R_{f_4}\right)R_6{v}_1}{R_{f_4}{R}_3\left(R_5+R_6\right)}-\frac{v_x\text{'}\text{'}}{R_3}\right)$ (17)

Substitute $37.5\text{V}$ for $v_x\text{'}\text{'}$, $5\text{k}\Omega$ for $R_5$, $5\text{k}\Omega$ for $R_6$, $5\text{k}\Omega$ for $R_{f_4}$, $1\text{V}$ for $v_1$ and $5\text{k}\Omega$ for $R_3$ in equation (17),

$\begin{array}{c}{v}_{out}=5\text{k}\Omega \left(\frac{\left(5\text{k}\Omega +5\text{k}\Omega \right)\times 5\text{k}\Omega \times 1\text{V}}{\left(5\text{k}\Omega \times 5\text{k}\Omega \right)\times \left(5\text{k}\Omega +5\text{k}\Omega \right)}-\frac{37.5\text{V}}{5\text{k}\Omega }\right)\\ =5\text{k}\Omega \left(\frac{\left(10\text{k}\Omega \right)\times 5\text{k}\Omega }{\left(5\text{k}\Omega \times 5\text{k}\Omega \right)\times \left(10\text{k}\Omega \right)}-\frac{37.5\text{V}}{5\text{k}\Omega }\right)\end{array}$

Solve for $v_{out}$,

$\begin{array}{c}{v}_{out}=5\text{k}\Omega \left(\frac{1}{5\text{k}\Omega }-\frac{37.5\text{V}}{5\text{k}\Omega }\right)\\ =\left(1-37.5\text{V}\right)\\ =-36.5\text{V}\end{array}$

Conclusion:

Thus, the output voltage of combined circuit is $-36.5\text{V}$.

(b)

To determine

Find the output voltage of the circuit.

Answer to Problem 24E

The output voltage of combined circuit is $18.75\text{V}$.

Explanation of Solution

Given Data:

Value of resistance $R_4$ is $2\text{k}\Omega$ and

Value of input voltage of 1^st op amp $v_{\text{in}}$ is $1\text{V}$ and

Value of input voltage of 4th op amp $v_1$ is $0\text{V}$.

Calculation:

Refer to the Figure 1,

Substitute $1\text{V}$ for $v_{\text{in}}$ in equation (4),

$\begin{array}{c}{v}_x=2.5\times 1\text{V}\\ =\text{2}\text{.5}\text{V}\end{array}$

Substitute $2.5\text{V}$ for $v_x$, $5\text{k}\Omega$ for $R_{f_2}$ and $2\text{k}\Omega$ for $R_4$ in equation (8),

$\begin{array}{c}{v}_x\text{'}=-\left(\frac{5\text{k}\Omega }{2\text{k}\Omega }\right)\times \left(2.5\text{V}\right)\\ =-6.25\text{V}\end{array}$

Substitute $-6.25\text{V}$ for $v_x\text{'}$, $1.5\text{k}\Omega$ for $R_{f_3}$ and $500\Omega$ for $R_2$ in equation (12),

$\begin{array}{c}{v}_x\text{'}\text{'}=-\left(\frac{1.5\text{k}\Omega }{500\Omega }\right)\times \left(-6.25\text{V}\right)\\ =\left(\frac{1.5\times {10}^3}{500\Omega }\right)\times \left(6.25\text{V}\right)\\ =3\times \left(6.25\text{V}\right)\\ =18.75\text{V}\end{array}$

Substitute $18.75\text{V}$ for $v_x\text{'}\text{'}$, $5\text{k}\Omega$ for $R_5$, $5\text{k}\Omega$ for $R_6$, $5\text{k}\Omega$ for $R_{f_4}$, $0\text{V}$ for $v_1$ and $5\text{k}\Omega$ for $R_3$ in equation (17),

$\begin{array}{c}{v}_{out}=5\text{k}\Omega \left(\frac{\left(5\text{k}\Omega +5\text{k}\Omega \right)\times 5\text{k}\Omega \times 0\text{V}}{\left(5\text{k}\Omega \times 5\text{k}\Omega \right)\times \left(5\text{k}\Omega +5\text{k}\Omega \right)}-\frac{18.75\text{V}}{5\text{k}\Omega }\right)\\ =5\text{k}\Omega \left(0-\frac{18.75\text{V}}{5\text{k}\Omega }\right)\end{array}$

Solve for $v_{out}$,

$\begin{array}{c}{v}_{out}=5\text{k}\Omega \left(-\frac{18.75\text{V}}{5\text{k}\Omega }\right)\\ =-18.75\text{V}\end{array}$

Conclusion:

Thus, the output voltage of combined circuit is $18.75\text{V}$.

(c)

To determine

Find the output voltage of the circuit.

Answer to Problem 24E

The output voltage of combined circuit is $-19.75\text{V}$.

Explanation of Solution

Given Data:

Value of resistance $R_4$ is $2\text{k}\Omega$,

Value of input voltage of 1^st op amp $v_{\text{in}}$ is $1\text{V}$ and

Value of input voltage of 4th op amp $v_1$ is $-1\text{V}$.

Calculation:

Refer to the Figure 1,

Substitute $1\text{V}$ for $v_{\text{in}}$ in equation (4),

$\begin{array}{c}{v}_x=2.5\times 1\text{V}\\ \text{=2}\text{.5}\text{V}\end{array}$

Substitute $2.5\text{V}$ for $v_x$, $5\text{k}\Omega$ for $R_{f_2}$ and $2\text{k}\Omega$ for $R_4$ in equation (8),

$\begin{array}{c}{v}_x\text{'}=-\left(\frac{5\text{k}\Omega }{2\text{k}\Omega }\right)\times \left(2.5\text{V}\right)\\ =-6.25\text{V}\end{array}$

Substitute $-6.25\text{V}$ for $v_x\text{'}$, $1.5\text{k}\Omega$ for $R_{f_3}$ and $500\Omega$ for $R_2$ in equation (12),

$\begin{array}{c}{v}_x\text{'}\text{'}=-\left(\frac{1.5\text{k}\Omega }{500\Omega }\right)\times \left(-6.25\text{V}\right)\\ =\left(\frac{1.5\times {10}^3}{500\Omega }\right)\times \left(6.25\text{V}\right)\\ =3\times \left(6.25\text{V}\right)\end{array}$

$v_x\text{'}\text{'}=18.75\text{V}$

Substitute $18.75\text{V}$ for $v_x\text{'}\text{'}$, $5\text{k}\Omega$ for $R_5$, $5\text{k}\Omega$ for $R_6$, $5\text{k}\Omega$ for $R_{f_4}$, $-1\text{V}$ for $v_1$ and $5\text{k}\Omega$ for $R_3$ in equation (17),

$\begin{array}{c}{v}_{out}=5\text{k}\Omega \left(\frac{\left(5\text{k}\Omega +5\text{k}\Omega \right)\times 5\text{k}\Omega \times \left(-1\text{V}\right)}{\left(5\text{k}\Omega \times 5\text{k}\Omega \right)\times \left(5\text{k}\Omega +5\text{k}\Omega \right)}-\frac{18.75\text{V}}{5\text{k}\Omega }\right)\\ =5\text{k}\Omega \left(\frac{-\left(10\text{k}\Omega \right)\times 5\text{k}\Omega }{\left(5\text{k}\Omega \times 5\text{k}\Omega \right)\times \left(10\text{k}\Omega \right)}-\frac{37.5\text{V}}{5\text{k}\Omega }\right)\\ =5\text{k}\Omega \left(\frac{-1}{5\text{k}\Omega }-\frac{18.75\text{V}}{5\text{k}\Omega }\right)\end{array}$

Solve for $v_{out}$,

$\begin{array}{c}{v}_{out}=5\text{k}\Omega \left(\frac{-1}{5\text{k}\Omega }-\frac{18.75\text{V}}{5\text{k}\Omega }\right)\\ =-1-18.75\text{V}\\ =-19.75\text{V}\end{array}$

Conclusion:

Thus, the output voltage of combined circuit is $-19.75\text{V}$.