Chapter 6, Problem 23E

Obtain an expression for v_out as labeled in the circuit of Fig. 6.50 if v₁ equals (a) 0 V; (b) 1 V; (c) −5 V; (d) 2 sin 100t V.

FIGURE 6.50

To determine

Find the output voltage $v_{\text{out}}$ in the circuit.

Answer to Problem 23E

The output voltage $v_{\text{out}}$ in the circuit is $4.5\text{V}$.

Explanation of Solution

Given data:

The input voltage $v_1$ across 2^nd op amp is $0\text{V}$.

Calculation:

The redrawn circuit is shown in Figure 1 as follows.

Refer to the Figure 1.

The expression of output voltage for 1^st inverting op amp is,

$v_2=\left(\frac{-R_{f_1}}{R^{\text{'}}}\right)\times v$ (1)

Here,

$v_2$ is the voltage output of 1^st op amp,

$R_{f_1}$ is the feedback resistance of 1^stop amp,

$R^{\text{'}}$ is the input resistance of 1^st op amp and

$v$ is the voltage supply across 1^st op amp.

The expression by the nodal analysis at node $v_a$ for 2^nd op amp is,

$\left(\frac{v_a-v_2}{R_1}\right)+\left(\frac{v_a-v_{\text{out}}}{R_{f_2}}\right)=0$ (2)

Here,

$R_1$ and $R_{f_2}$ are the resistances in the circuit.

$v_2$ is the input voltage for 2^nd op amp

$v_a$ is the node voltage.

The expression by the nodal analysis at node $v_b$

$\frac{v_1-v_b}{R_2}=\frac{v_b}{R_3}$ (3)

Here,

$R_2$ and $R_3$ are the resistances in the circuit.

$v_1$ is the input voltage.

$v_b$ is the node voltage.

The expression for the virtual ground concept is as follows,

$v_a=v_b$ (4)

Refer to the redrawn Figure 1.

Simplify equation (2) as follows.

$v_a\left(\frac{1}{R_1}+\frac{1}{R_{f_2}}\right)-v_2\left(\frac{1}{R_1}\right)=\frac{v_{\text{out}}}{R_{f_2}}$ (5)

Rearrange equation (3) for $v_b$.

$v_b=\frac{R_3{v}_1}{R_2+R_3}$ (6)

Substitute $\frac{R_3{v}_1}{R_2+R_3}$ for $v_a$ in equation (5).

$\left(\frac{R_3{v}_1}{R_2+R_3}\right)\left(\frac{1}{R_1}+\frac{1}{R_{f_2}}\right)-v_2\left(\frac{1}{R_1}\right)=\frac{v_{\text{out}}}{R_{f_2}}$

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

$\frac{v_{\text{out}}}{R_{f_2}}=\left(\frac{R_3}{R_2+R_3}\right)\left(\frac{1}{R_{f_2}}+\frac{1}{R_1}\right)v_1-\left(\frac{1}{R_1}\right)v_2$

$v_{\text{out}}=\left(\frac{R_3}{R_2+R_3}\right)\left(1+\frac{R_{f_2}}{R_1}\right)v_1-\left(\frac{R_{f_2}}{R_1}\right)v_2$ (7)

Substitute for $v_2$, $1.5\text{V}$ for $v$, $5\Omega$ for $R_{f_1}$ and $10\Omega$ for $R^{\text{'}}$ in the equation (1),

$\begin{array}{c}{v}_2=\left(\frac{-1.5\text{k}\Omega }{500\Omega }\right)\times 1.5\text{V}\\ \text{=}\left(\frac{-1.5\times {10}^3\Omega }{500\Omega }\right)\times 1.5\text{V}\left\{\because 1\text{k}\Omega =\text{1}\times {\text{10}}^3\Omega \right\}\\ =-4.5\text{V}\end{array}$

Substitute $-4.5\text{V}$ for $v_2$, $0\text{V}$ for $v_1$, $5\text{k}\Omega$ for $R_1$, $5\text{k}\Omega$ for $R_2$, $5\text{k}\Omega$ for $R_3$ and $5\text{k}\Omega$ for $R_{f_2}$ in the equation (7).

$\begin{array}{c}{v}_{\text{out}}=\left(\frac{5\text{k}\Omega }{5\text{}\text{k}\Omega +5\text{k}\Omega }\right)\left(1+\frac{5\text{k}\Omega }{5\text{k}\Omega }\right)\times \left(0\right)-\left(\frac{5\text{k}\Omega }{5\text{k}\Omega }\right)\times \left(-4.5\text{V}\right)\\ =-1\times \left(-4.5\text{V}\right)\\ =4.5\text{V}\end{array}$

Conclusion:

Thus, the output voltage $v_{\text{out}}$ in the circuit is $4.5\text{V}$.

To determine

Find the output voltage $v_{\text{out}}$ in the circuit.

Answer to Problem 23E

The output voltage $v_{\text{out}}$ in the circuit is $5.5\text{V}$.

Explanation of Solution

Given data:

Value of input voltage $v_1$ across 2^nd op amp is $1\text{V}$.

Calculation:

Refer to the redrawn Figure 1.

Substitute $-4.5\text{V}$ for $v_2$, $1\text{V}$ for $v_1$, $5\text{k}\Omega$ for $R_1$, $5\text{k}\Omega$ for $R_2$, $5\text{k}\Omega$ for $R_3$ and $5\text{k}\Omega$ for $R_{f_2}$ in equation (7).

$\begin{array}{c}{v}_{\text{out}}=\left(\frac{5\text{k}\Omega }{5\text{}\text{k}\Omega +5\text{k}\Omega }\right)\left(1+\frac{5\text{k}\Omega }{5\text{k}\Omega }\right)\left(1\text{V}\right)-\left(\frac{5\text{k}\Omega }{5\text{k}\Omega }\right)\left(-4.5\text{V}\right)\left\{\because 1\text{ k}\Omega =\text{1}\times {\text{10}}^3\Omega \right\}\\ =\left(\frac{5\times {10}^3\Omega }{5\times {10}^3\Omega +5\times {10}^3\Omega }\right)\left(1+\frac{5\times {10}^3\Omega }{5\times {10}^3\Omega }\right)\left(1\text{V}\right)-\left(\frac{5\times {10}^3\Omega }{5\times {10}^3\Omega }\right)\left(-4.5\text{V}\right)\\ =\left(\frac{1}{2}\right)\left(1+1\right)\left(1\text{V}\right)-\left(1\right)\left(-4.5\text{V}\right)\end{array}$

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

$\begin{array}{c}{v}_{\text{out}}=1\text{V}+\text{4}\text{.5}\text{V}\\ \text{=5}\text{.5}\text{V}\end{array}$

Conclusion:

Thus, the output voltage $v_{\text{out}}$ in the circuit is $5.5\text{V}$.

To determine

Find the output voltage $v_{\text{out}}$ in the circuit.

Answer to Problem 23E

The output voltage $v_{\text{out}}$ in the circuit is $-0.5\text{V}$.

Explanation of Solution

Given data:

Value of input voltage $v_1$ across 2^nd op amp is $-5\text{V}$.

Calculation:

Refer to the redrawn Figure 1.

Substitute $-4.5\text{V}$ for $v_2$, $-5\text{V}$ for $v_1$, $5\text{k}\Omega$ for $R_1$, $5\text{k}\Omega$ for $R_2$, $5\text{k}\Omega$ for $R_3$ and $5\text{k}\Omega$ for $R_{f_2}$ in equation (7).

$\begin{array}{c}{v}_{\text{out}}=\left(\frac{5\text{k}\Omega }{5\text{}\text{k}\Omega +5\text{k}\Omega }\right)\left(1+\frac{5\text{k}\Omega }{5\text{k}\Omega }\right)\times \left(-5\text{V}\right)-\left(\frac{5\text{k}\Omega }{5\text{k}\Omega }\right)\times \left(-4.5\text{V}\right)\left\{\because 1\text{ k}\Omega \text{=1}\times {\text{10}}^3\Omega \right\}\\ =\left(\frac{5\times {10}^3\Omega }{5\times {10}^3\Omega +5\times {10}^3\Omega }\right)\left(1+\frac{5\times {10}^3\Omega }{5\times {10}^3\Omega }\right)\times \left(-5\text{V}\right)-\left(\frac{5\times {10}^3\Omega }{5\times {10}^3\Omega }\right)\times \left(-4.5\text{V}\right)\\ =\left(\frac{1}{2}\right)\left(1+1\right)\times \left(-5\text{V}\right)-\left(1\right)\times \left(-4.5\text{V}\right)\end{array}$

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

$\begin{array}{c}{v}_{\text{out}}=-5\text{V}+\text{4}\text{.5}\text{V}\\ \text{=}-0.\text{5}\text{V}\end{array}$

Conclusion:

Thus, the output voltage $v_{ou}{}_t$ in the circuit is $-0.5\text{V}$.

To determine

Find the output voltage $v_{\text{out}}$ in the circuit.

Answer to Problem 23E

The output voltage $v_{\text{out}}$ in the circuit is $2\sin \left(\text{100}t\right)\text{V}+4.5\text{V}$.

Explanation of Solution

Given data:

Value of input voltage $v_1$ across 2^nd op amp is $2\text{sin100}t\text{V}$.

Calculation:

Refer to the redrawn Figure 1.

Substitute $-4.5\text{V}$ for $v_2$, $2\text{sin100}t\text{V}$ for $v_1$, $5\text{k}\Omega$ for $R_1$, $5\text{k}\Omega$ for $R_2$, $5\text{k}\Omega$ for $R_3$ and $5\text{k}\Omega$ for $R_{f_2}$ in equation (7).

$\begin{array}{c}{v}_{\text{out}}=\left(\frac{5\text{k}\Omega }{5\text{}\text{k}\Omega +5\text{k}\Omega }\right)\left(1+\frac{5\text{k}\Omega }{5\text{k}\Omega }\right)\left(2\text{sin100}t\text{V}\right)-\left(\frac{5\text{k}\Omega }{5\text{k}\Omega }\right)\left(-4.5\text{V}\right)\left\{\because 1\text{ k}\Omega \text{=1}\times {\text{10}}^3\Omega \right\}\\ =\left(\frac{5\times {10}^3\Omega }{5\times {10}^3\Omega +5\times {10}^3\Omega }\right)\left(1+\frac{5\times {10}^3\Omega }{5\times {10}^3\Omega }\right)\left(2\text{sin100}t\text{V}\right)-\left(\frac{5\times {10}^3\Omega }{5\times {10}^3\Omega }\right)\left(-4.5\text{V}\right)\\ =\left(\frac{1}{2}\right)\left(1+1\right)\left(2\text{sin100}t\text{V}\right)-\left(1\right)\left(-4.5\text{V}\right)\end{array}$

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

$v_{\text{out}}=2\sin \left(\text{100}t\right)\text{V}+4.5\text{V}$

Conclusion:

Thus, the output voltage $v_{\text{out}}$ in the circuit is $2\sin \left(\text{100}t\right)\text{V}+4.5\text{V}$.