Chapter 5, Problem 7P

5.7 The op amp in Fig. 5.46 has R_i = 100 kΩ, R_o = 100 Ω, A = 100,000. Find the differential voltage v_d and the output voltage v_o.

Figure 5.46

For Prob. 5.7.

To determine

Calculate the differential voltage $v_d$ and the output voltage $v_o$ of the op amp in Figure 5.46.

Answer to Problem 7P

The differential voltage $v_d$ is $\underset{\_}{-100\text{nV}}$ and the output voltage $v_o$ is $\underset{\_}{-10\text{mV}}$.

Explanation of Solution

Given data:

Refer Figure 5.46 in the textbook for the op amp circuit.

The open-loop gain A is 100,000,

The input resistance $R_i$ is $100\text{kΩ}$, and

The output resistance $R_o$ is $100\text{Ω}$.

Calculation:

The equivalent circuit of the the given 741 op amp is drawn and it is shown in Figure 1.

Apply Kirchhoff's current law at node 1 in Figure 1.

$\begin{array}{l}\frac{v_1-v_s}{10\text{kΩ}}+\frac{v_1-0}{100\text{kΩ}}+\frac{v_1-v_o}{100\text{kΩ}}=0\\ \frac{10\left(v_1-v_s\right)+v_1+v_1-v_o}{100\text{kΩ}}=0\\ 10v_s-10v_1=v_1+v_1-v_o\\ 10v_s+v_o=12v_1\end{array}$

Simplify the equation as follows.

$v_1=\frac{10v_s+v_o}{12}$ (1)

From Figure 1, consider the expression for the voltage $v_d$.

$v_d=v_1$

Apply Kirchhoff's current law at node 2 in Figure 1.

$\begin{array}{l}\frac{v_o-\left(-A{v}_d\right)}{100\text{Ω}}+\frac{v_o-v_1}{100\text{kΩ}}=0\\ \frac{{10}^3\left(v_o+A{v}_d\right)+v_o-v_1}{100\text{kΩ}}=0\\ {10}^3\left(v_o+A{v}_d\right)+v_o-v_1=0\end{array}$

Substitute 100,000 for A and $v_1$ for $v_d$.

$\begin{array}{l}{10}^3\left(v_o+100,000v_1\right)+v_o-v_1=0\\ v_1-v_o=1000\left(v_o+100,000v_1\right)\end{array}$

$1001v_o+99,999,999v_1=0$ (1)

Substitute equation (1) in (2).

$\begin{array}{l}1001v_o+99,999,999\left[\frac{\left(10v_s+v_o\right)}{12}\right]=0\\ 83,333,332.5v_s+8,334,334.25v_o=0\end{array}$

Simplify the equation as follows.

$\frac{v_o}{v_s}=-10$

Substitute 1 mV for $v_s$.

$\begin{array}{l}\frac{v_o}{1\text{mV}}=-10\\ v_o=-10\text{mV}\end{array}$

From Figure 1, consider the expression for the output voltage $v_o$.

$v_o=A{v}_d$

Substitute 100,000 for A.

$\begin{array}{l}{v}_o=\left(100,000\right)v_d\\ v_d=\frac{v_o}{100,000}\end{array}$

Substitute $-10\text{mV}$ for $v_o$.

$\begin{array}{c}{v}_d=\frac{-10\text{mV}}{100,000}\\ =-\frac{10\times {10}^{-3}\text{V}}{{10}^5}\\ =-100\times {10}^{-9}\text{V}\\ =-100\text{nV}\end{array}$

Conclusion:

Thus, the differential voltage $v_d$ is $\underset{\_}{-100\text{nV}}$ and the output voltage $v_o$ is $\underset{\_}{-10\text{mV}}$.