Chapter 7, Problem 73P

For the op amp circuit of Fig. 7.138, let R₁ = 10 kΩ, R_f = 30 kΩ, C = 20 μF, and v(0) = 1 V. Find v₀.

To determine

Find the output voltage $v_o$ of the op amp circuit in the given Figure 7.138.

Answer to Problem 73P

The output voltage $v_o$ of the op amp circuit is $-9e^{-5t}u\left(t\right)\text{V}$.

Explanation of Solution

Given data:

Refer to Figure 7.138 in the textbook.

The value of capacitance $C$ is $20\mu \text{F}$.

The source voltage $v_s$ is $4u\left(t\right)\text{V}$.

The value of resistance $R_{\text{1}}$ is $10\text{k}\Omega$.

The value of feedback resistance $R_f$ is $30\text{k}\Omega$.

The initial voltage v(0) or $V_0$ is 1 V.

Formula used:

Write the expression to find the time constant for an RC circuit.

$\tau =R_{\text{Th}}C$ (1)

Here,

$R_{\text{Th}}$ is the Thevenin resistance, and

C is the capacitance of the capacitor.

Write the general expression for the unit step function.

$u(t)=\{\begin{array}{l}0,t<0\\ 1,t>0\end{array}$ (2)

Calculation:

The given Figure 7.138 is redrawn as shown in Figure 1.

The given source voltage is,

$v_s=4u\left(t\right)\text{V}$ (3)

Apply the unit step function in equation (2) to equation (3).

$v_s=\{\begin{array}{l}0\text{V},t<0\\ 4\text{V},t>0\end{array}$

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 source voltage is,

$v_s=v_1=4\text{V}$

In Figure 1, apply Kirchhoff’s current law at node $v_2$.

$\frac{v_1-v_2}{R_1}=C\frac{dv}{dt}$ (4)

In Figure 1, apply Kirchhoff’s current law at node $v_3$.

$C\frac{dv}{dt}=\frac{v_3-v_o}{R_f}$ (5)

From Figure 1, the voltages are,

$v_3=0\text{V}$ and

$v=v_2-v_3$

Substitute $0\text{V}$ for $v_3$ to find the capacitor voltage v.

$\begin{array}{c}v=v_2-0\text{V}\\ =v_2\end{array}$

Substitute $v$ for $v_2$ in equation (4).

$\begin{array}{l}\frac{v_1-v}{R_1}=C\frac{dv}{dt}\\ v_1-v=R_{1}C\frac{dv}{dt}\\ \frac{dv}{dt}=\frac{v_1-v}{R_{1}C}\\ \frac{dv}{dt}=\frac{v_1}{R_{1}C}-\frac{v}{R_{1}C}\end{array}$

Rearrange the equation as follows,

$\frac{dv}{dt}+\frac{v}{R_{1}C}=\frac{v_1}{R_{1}C}$ (6)

The equation is similar to the equation (7.42) in the textbook.

Hence,

$v\left(t\right)=\{\begin{array}{l}{V}_0,t<0\\ v_1+\left(V_0-v_1\right)e^{-\frac{t}{\tau }},t>0\end{array}$ (7)

In Figure 1, the Thevenin resistance $R_{\text{Th}}$ is equal to the resistance $R_1=10\text{k}\Omega$.

Substitute $10\text{k}\Omega$ for $R_{\text{Th}}$ and $20\mu \text{F}$ for C in equation (1) to find the time constant $\tau$.

$\tau =\left(10\text{k}\Omega \right)\left(20\mu \text{F}\right)$

$\tau =\left(10\times {10}^3\Omega \right)\left(20\times {10}^{-6}\text{F}\right)\left\{\because 1\mu ={10}^{-6},1\text{k}={10}^3\right\}$ (8)

Substitute the units $\frac{\text{V}}{\text{A}}$ for $\Omega$ and $\frac{\text{A}\cdot \text{s}}{\text{V}}$ for F in equation (8) to find the time constant $\tau$ in seconds.

$\begin{array}{c}\tau =\left(10\times {10}^3\frac{\text{V}}{\text{A}}\right)\left(20\times {10}^{-6}\frac{\text{A}\cdot \text{s}}{\text{V}}\right)\\ =0.2\text{s}\end{array}$

Substitute 1 V for $V_0$, 4 V for $v_1$, and $0.2\text{s}$ for $\tau$ in equation (7).

$\begin{array}{c}v\left(t\right)=\{\begin{array}{l}1\text{V},t<0\\ 4\text{V}+\left(1\text{V}-4\text{V}\right)e^{-\frac{t}{0.2\text{s}}},t>0\end{array}\\ =\{\begin{array}{l}1\text{V},t<0\\ 4-3e^{-5t}\text{V,}t>0\end{array}\end{array}$

On differentiating the above equation as follows,

$\begin{array}{l}\frac{dv\left(t\right)}{dt}=\{\begin{array}{l}0,t<0\\ 0-3\left(-5\right)e^{-5t}\frac{\text{V}}{\text{s}}\text{,}t>0\end{array}\\ \frac{dv}{dt}=\{\begin{array}{l}0,t<0\\ 15e^{-5t}\frac{\text{V}}{\text{s}}\text{,}t>0\end{array}\left\{\because v\left(t\right)=v\right\}\end{array}$

Substitute $0\text{V}$ for $v_3$ in equation (5).

$\begin{array}{l}C\frac{dv}{dt}=\frac{0\text{V}-v_o}{R_f}\\ C\frac{dv}{dt}=\frac{-v_o}{R_f}\\ v_o=-R_{f}C\frac{dv}{dt}\end{array}$

Substitute $30\text{k}\Omega$ for $R_f$, $20\mu \text{F}$ for C, and $15e^{-5t}\frac{\text{V}}{\text{s}}$ for $\frac{dv}{dt}$ to find the output voltage $v_o$ of op amp for $t>0$ in volts.

$\begin{array}{c}{v}_o=-\left(30\text{k}\Omega \right)\left(20\mu \text{F}\right)\left(15e^{-5t}\frac{\text{V}}{\text{s}}\right)\left\{\because 1\Omega =\frac{1\text{V}}{1\text{A}},1\text{F}=\frac{\text{1}\text{A}\cdot 1\text{s}}{\text{1}\text{V}},1\mu ={10}^{-6},1\text{k}={10}^3\right\}\\ =-\left(30\times {10}^3\frac{\text{V}}{\text{A}}\right)\left(20\times {10}^{-6}\frac{\text{A}\cdot \text{s}}{\text{V}}\right)\left(15e^{-5t}\frac{\text{V}}{\text{s}}\right)\end{array}$

$v_o=-9e^{-5t}\text{V}$ (9)

Apply the unit step function in equation (2) to equation (9).

$\begin{array}{c}{v}_o=\left(-9e^{-5t}\text{V}\right)u\left(t\right)\\ =-9e^{-5t}u\left(t\right)\text{V}\end{array}$

Conclusion:

Thus, the output voltage $v_o$ of the op amp circuit is $-9e^{-5t}u\left(t\right)\text{V}$.