Chapter 7, Problem 75P

In the circuit of Fig. 7.140, find v_o and i_o, given that v_s = 10[1 – e^–t]u(t) V.

To determine

Find the output voltage $v_o$ and the current $i_o$ of the op amp circuit in the given Figure 7.140.

Answer to Problem 75P

The output voltage $v_o$ is $\left[20-10e^{-t}\right]u\left(t\right)\text{V}$ and the current $i_o$ is $100\mu \text{A}$ of the op amp circuit.

Explanation of Solution

Given data:

Refer to Figure 7.140 in the textbook.

The value of capacitance $\left(C\right)$ is $10\mu \text{F}$.

The source voltage $v_s$ is $10\left[1-e^{-t}\right]u\left(t\right)\text{V}$.

Formula used:

Write the general expression for the unit step function.

$u(t)=\{\begin{array}{l}0,t<0\\ 1,t>0\end{array}$ (1)

Calculation:

The given Figure 7.140 is redrawn as shown in Figure 1.

The given source voltage is,

$v_s=10\left[1-e^{-t}\right]u\left(t\right)\text{V}$ (2)

Apply the unit step function in equation (1) to equation (2).

$v_s=\{\begin{array}{l}0\text{V},t<0\\ 10\left[1-e^{-t}\right]\text{V},t>0\end{array}$

Consider the voltage at the non-inverting terminal as $v_a$ and the voltage at the inverting terminal as $v_b$ in Figure 1.

For $t<0$:

Since the source voltage $v_s=0\text{V}$ for all $t<0$, all the initial voltages are equal to zero.

For $t>0$:

The voltage,

$v_a=v_b=v_s=10\left[1-e^{-t}\right]\text{V}$ (3)

On differentiating the equation (3).

$\begin{array}{c}\frac{d{v}_b}{dt}=10\left[0-\left(-1\right)e^{-t}\right]\frac{\text{V}}{\text{s}}\\ =10e^{-t}\frac{\text{V}}{\text{s}}\end{array}$

In Figure 1, apply Kirchhoff’s current law at node $v_b$.

$\left(10\mu \text{F}\right)\frac{d{v}_b}{dt}+\frac{v_b}{100\text{k}\Omega }+\frac{v_b-v_o}{100\text{k}\Omega }=0$ (4)

Substitute $10e^{-t}\frac{\text{V}}{\text{s}}$ for $\frac{d{v}_b}{dt}$ and $10\left[1-e^{-t}\right]\text{V}$ for $v_b$ in equation (4) to find the output voltage $v_o$ of the op amp in volts.

$\begin{array}{l}\left(10\mu \text{F}\right)\left(10e^{-t}\frac{\text{V}}{\text{s}}\right)+\frac{10\left[1-e^{-t}\right]\text{V}}{100\text{k}\Omega }+\frac{10\left[1-e^{-t}\right]\text{V}-v_o}{100\text{k}\Omega }=0\\ \left(10\times {10}^{-6}\text{F}\right)\left(10e^{-t}\frac{\text{V}}{\text{s}}\right)+\frac{10\left[1-e^{-t}\right]\text{V}}{100\times {10}^3\Omega }+\frac{10\left[1-e^{-t}\right]\text{V}-v_o}{100\times {10}^3\Omega }=0\left\{\because 1\text{k}={10}^3,1\mu ={10}^{-6}\right\}\\ \left(10\times {10}^{-6}\frac{\text{A}\cdot \text{s}}{\text{V}}\right)\left(10e^{-t}\frac{\text{V}}{\text{s}}\right)+\frac{\left[1-e^{-t}\right]\text{V}}{10\times {10}^3\Omega }+\frac{10\left[1-e^{-t}\right]\text{V}-v_o}{100\times {10}^3\Omega }=0\left\{\because 1\text{F}=\frac{\text{A}\cdot \text{s}}{\text{V}}\right\}\\ \left({10}^{-4}{e}^{-t}\text{A}\right)+{10}^{-4}\left[1-e^{-t}\right]\text{A}+{10}^{-4}\left[1-e^{-t}\right]\text{A}-\frac{v_o}{{10}^5\Omega }=0\left\{\because 1\text{A}=\frac{\text{1}\text{V}}{\text{1}\Omega }\right\}\end{array}$

Rearrange the equation as follows,

$\begin{array}{l}\frac{v_o}{{10}^5\Omega }=\left({10}^{-4}{e}^{-t}\text{A}\right)+{10}^{-4}\left[1-e^{-t}\right]\text{A}+{10}^{-4}\left[1-e^{-t}\right]\text{A}\\ \frac{v_o}{{10}^5\Omega }=\left({10}^{-4}{e}^{-t}\text{A}\right)+2\left({10}^{-4}\left[1-e^{-t}\right]\text{A}\right)\\ v_o=\left({10}^5\Omega \right)\left({10}^{-4}{e}^{-t}\text{A}\right)+\left({10}^5\Omega \right)2\left({10}^{-4}\left[1-e^{-t}\right]\text{A}\right)\\ v_o=10e^{-t}\text{V}+20\left[1-e^{-t}\right]\text{V}\end{array}$

Simplify the equation as follows,

$\begin{array}{c}{v}_o=10e^{-t}\text{V}+20\text{V}-20e^{-t}\text{V}\\ =20\text{V}-10e^{-t}\text{V}\end{array}$

$v_o=\left[20-10e^{-t}\right]\text{V}$ (5)

Apply the unit step function in equation (1) to equation (5).

$\begin{array}{c}{v}_o=\left(\left[20-10e^{-t}\right]\text{V}\right)u\left(t\right)\\ =\left[20-10e^{-t}\right]u\left(t\right)\text{V}\end{array}$

The output current $i_o$ of the op amp circuit is calculated as follows,

$i_o=\frac{v_o-v_s}{100\text{k}\Omega }$

Substitute $\left[20-10e^{-t}\right]\text{V}$ for $v_o$ and $10\left[1-e^{-t}\right]\text{V}$ for $v_s$ to find the current $i_o$ in amperes.

$\begin{array}{c}{i}_o=\frac{\left[20-10e^{-t}\right]\text{V}-10\left[1-e^{-t}\right]\text{V}}{100\text{k}\Omega }\\ =\frac{20-10e^{-t}-10+10e^{-t}}{100\times {10}^3}\frac{\text{V}}{\Omega }\left\{\because 1\text{k}={10}^3\right\}\\ =\frac{10}{100\times {10}^3}\text{A}\left\{\because 1\text{A}=\frac{\text{1}\text{V}}{1\Omega }\right\}\\ ={10}^{-4}\text{A}\end{array}$

Convert the unit A to $\mu \text{A}$.

$\begin{array}{c}{i}_o={10}^{-4}\text{A}\times {10}^2\times {10}^{-2}\text{A}\\ =100\times {10}^{-6}\text{A}\\ =100\mu \text{A}\end{array}$

Conclusion:

Thus, the output voltage $v_o$ is $\left[20-10e^{-t}\right]u\left(t\right)\text{V}$ and the current $i_o$ is $100\mu \text{A}$ of the op amp circuit.