Chapter 6, Problem 1E

For the op amp circuit shown in Fig. 6.39, calculate v_out if (a) R₁ = R₂ = 100 Ω, and v_in = 5 V; (b) R₂ = 200R₁ and v_in = 1 V; (c) R₁ = 4.7 kΩ, R₂ = 47 kΩ, and v_in = 20 sin 5t V.

FIGURE 6.39

To determine

Find the value of $v_{\text{out}}$.

Answer to Problem 1E

The value of $v_{\text{out}}$ is $-5\text{ V}$.

Explanation of Solution

Formula used:

Refer to the FIGURE 6.39 in the Textbook.

The expression for the relation of $v_{\text{out}}$ and $v_{\text{in}}$ for inverting amplifier is as follows.

$v_{\text{out}}=\left(\frac{-R_2}{R_1}\right)v_{\text{in}}$ (1)

Here,

$R_1$ and $R_2$ are the resistances in the circuit.

$v_{\text{out}}$ is the output voltage.

$v_{\text{in}}$ is the input voltage.

Calculation:

Refer to the FIGURE 6.39 in the Textbook.

Substitute $100\Omega$ for $R_1$, $100\Omega$ for $R_2$ and $5\text{V}$ for $v_{\text{in}}$ in equation (1).

$\begin{array}{c}{v}_{\text{out}}=\left(\frac{-100\Omega }{100\Omega }\right)\left(5\text{ V}\right)\\ =-5\text{ V}\end{array}$

Conclusion:

Thus, the value of $v_{\text{out}}$ is $-5\text{ V}$.

To determine

Find the value of $v_{\text{out}}$.

Answer to Problem 1E

The value of $v_{\text{out}}$ is $-200\text{ V}$.

Explanation of Solution

Calculation:

Refer to the FIGURE 6.39 in the Textbook.

Substitute $200R_1$ for $R_2$ and $1\text{V}$ for $v_{\text{in}}$ in equation (1).

$\begin{array}{c}{v}_{\text{out}}=\left(\frac{-200R_1}{R_1}\right)\left(\text{1 V}\right)\\ =-200\text{ V}\end{array}$

Conclusion:

Thus, the value of $v_{\text{out}}$ is $-200\text{ V}$.

To determine

Find the value of $v_{\text{out}}$.

Answer to Problem 1E

The value of $v_{\text{out}}$ is $-200\left(\sin \text{ 5}t\right)\text{V}$.

Explanation of Solution

Calculation:

Refer to the FIGURE 6.39 in the TEXTBOOK.

Substitute $4.7\text{ k}\Omega$ for $R_1$, $47\text{ k}\Omega$ for $R_2$ and $20\left(\sin \text{ 5}t\right)\text{V}$ for $v_{\text{in}}$ in equation (1).

$\begin{array}{c}{v}_{\text{out}}=\left(\frac{-47\text{ k}\Omega }{4.7\text{ k}\Omega }\right)\left(20\left(\sin \text{ 5}t\right)\text{V}\right)\\ =-200\left(\sin \text{ 5}t\right)\text{V}\end{array}$

Conclusion:

Thus, the value of $v_{\text{out}}$ is $-200\left(\sin \text{ 5}t\right)\text{V}$.