Chapter 6, Problem 74P

The triangular waveform in Fig. 6.91(a) is applied to the input of the op amp differentiator in Fig. 6.91(b). Plot the output.

Figure 6.91

For Prob. 6.74.

To determine

Sketch the output voltage waveform of an op amp differentiator.

Explanation of Solution

Given data:

Refer to Figure 6.91 in the textbook.

Formula used:

Write the expression to calculate the straight line equation for two points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$.

$\left(y-y_1\right)=\frac{y_2-y_1}{x_2-x_1}\left(x-x_1\right)$ (1)

Refer to Figure 6.91(a) in the textbook.

From the given voltage graph, substitute $t$ for $x$ and $v_i\left(t\right)$ for $y$ in equation (1).

$\left(v_i\left(t\right)-y_1\right)=\frac{y_2-y_1}{x_2-x_1}\left(t-x_1\right)$ (2)

Write the expression to calculate the output voltage of an op amp differentiator.

$v_o\left(t\right)=-RC\frac{d{v}_i\left(t\right)}{dt}$ (3)

Here,

$R$ is the value of the resistor,

$C$ is the value of the capacitor, and

$v_i$ is the input voltage.

Calculation:

The given circuit is redrawn as Figure 1.

Substitute $20\text{k}\Omega$ for $R$ and $0.01\mu \text{F}$ for $C$ in equation (3) to find $v_o\left(t\right)$.

$\begin{array}{c}{v}_o\left(t\right)=-\left(20\text{k}\Omega \right)\left(0.01\mu \text{F}\right)\frac{d{v}_i\left(t\right)}{dt}\\ =-\left(20\times {10}^3\right)\left(0.01\times {10}^{-6}\right)\frac{d{v}_i\left(t\right)}{dt}\Omega \cdot \text{F}\left\{1\text{k}={\text{10}}^3,1\mu ={10}^{-6}\right\}\end{array}$

$v_o\left(t\right)=-\left(2\times {10}^{-4}\right)\frac{d{v}_i\left(t\right)}{dt}\Omega \cdot \text{F}$ (4)

The given input voltage waveform is redrawn as Figure 2.

Refer to Figure 2, split up the time period as three divisions such as $0\text{ms}<t<1\text{ms}$, $1\text{ms}<t<3\text{ms}$ and $3\text{ms}<t<4\text{ms}$ to find the respective voltage value.

Case (i): $0\text{ms}<t<1\text{ms}$

The two points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ are $\left(0\text{ms},\text{0}\text{V}\right)$ and $\left(1\text{ms},\text{10}\text{V}\right)$.

Substitute $0\text{ms}$ for $x_1$, $\text{0}\text{V}$ for $y_1$, $1\text{ms}$ for $x_2$ and $\text{10}\text{V}$ for $y_2$ in equation (2) to find $v_i\left(t\right)$.

$\begin{array}{c}\left(v_i\left(t\right)-\text{0}\text{V}\right)=\frac{\text{10}\text{V}-\text{0}\text{V}}{1\text{ms}-0\text{ms}}\left(t-0\text{ms}\right)\\ =\frac{\text{10}\text{V}}{1\text{ms}}\left(t\right)\\ =\frac{\text{10}}{1\times {10}^{-3}}\left(t\right)\text{V}\left\{\because 1\text{m}={\text{10}}^{-3}\right\}\\ =10000t\text{V}\end{array}$

Simplify the equation to find $v_i\left(t\right)$.

$\begin{array}{c}{v}_i\left(t\right)=10000t\text{V}+\text{0}\text{V}\\ =10000t\text{V}\end{array}$

Case (ii): $1\text{ms}<t<3\text{ms}$

The two points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ are $\left(1\text{ms},\text{10}\text{V}\right)$ and $\left(3\text{ms},-\text{10}\text{V}\right)$.

Substitute $1\text{ms}$ for $x_1$, $\text{10}\text{V}$ for $y_1$, $3\text{ms}$ for $x_2$ and $-1\text{0}\text{V}$ for $y_2$ in equation (2) to find $v_i\left(t\right)$.

$\begin{array}{c}\left(v_i\left(t\right)-\text{10}\text{V}\right)=\frac{-1\text{0}\text{V}-\text{10}\text{V}}{3\text{ms}-1\text{ms}}\left(t-1\text{ms}\right)\\ =\frac{-2\text{0}\text{V}}{2\text{ms}}\left(t-1\text{ms}\right)\\ =\frac{-20}{\left(2\times {10}^{-3}\right)}\left(t-\left(1\times {10}^{-3}\right)\right)\text{V}\left\{\because 1\text{m}={\text{10}}^{-3}\right\}\\ =-10000t\text{V}+0.02\text{V}\end{array}$

Simplify the equation to find $v_i\left(t\right)$.

$\begin{array}{c}{v}_i\left(t\right)=-10000t\text{V}+0.02\text{V}+10\text{V}\\ =-10000t\text{V}+10.02\text{V}\\ =\left(-10000t+10.02\right)\text{V}\end{array}$

Case (iii): $3\text{ms}<t<4\text{ms}$

The two points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ are $\left(3\text{ms},-\text{10}\text{V}\right)$ and $\left(4\text{ms},\text{0}\text{V}\right)$.

Substitute $3\text{ms}$ for $x_1$, $-\text{10}\text{V}$ for $y_1$, $4\text{ms}$ for $x_2$ and $\text{0}\text{V}$ for $y_2$ in equation (2) to find $v_i\left(t\right)$.

$\begin{array}{c}\left(v_i\left(t\right)-\left(-1\text{0}\text{V}\right)\right)=\frac{\text{0}\text{V}-\left(-1\text{0}\text{V}\right)}{4\text{ms}-3\text{ms}}\left(t-3\text{ms}\right)\\ \left(v_i\left(t\right)+10\text{V}\right)=\frac{\text{10}\text{V}}{1\text{ms}}\left(t-3\text{ms}\right)\\ =\frac{\text{10}}{1\times {10}^{-3}}\left(t-\left(3\times {10}^{-3}\right)\right)\text{V}\left\{\because 1\text{m}={\text{10}}^{-3}\right\}\\ =10000t\text{V}-0.03\text{V}\end{array}$

Simplify the equation to find $v_i\left(t\right)$.

$\begin{array}{c}{v}_i\left(t\right)=10000t\text{V}-0.03\text{V}-1\text{0}\text{V}\\ =10000t\text{V}-10.03\text{V}\\ =\left(10000t-10.03\right)\text{V}\end{array}$

Therefore, the input voltage function $v_i\left(t\right)$ of the signal in Figure 2 is,

$v_i\left(t\right)=\{\begin{array}{l}10000t\text{V},0\text{ms}<t<1\text{ms}\\ \left(-10000t+10.02\right)\text{V},1\text{ms}<t<3\text{ms}\\ \left(10000t-10.03\right)\text{V},3\text{ms}<t<4\text{ms}\end{array}$

For $0\text{ms}<t<1\text{ms}$:

Substitute $10000t\text{V}$ for $v_i\left(t\right)$ in equation (4) to find $v_o\left(t\right)$.

$\begin{array}{c}{v}_o\left(t\right)=-\left(2\times {10}^{-4}\right)\frac{d\left(10000t\text{V}\right)}{dt}\Omega \cdot \text{F}\\ =-\left(2\times {10}^{-4}\right)\left(10000\right)\frac{d}{dt}\left(t\right)\Omega \cdot \text{F}\cdot \text{V}\\ =-2\left(1\right)\frac{\Omega \cdot \text{F}\cdot \text{V}}{\text{s}}\\ =-2\frac{\text{F}\cdot \text{V}}{\left(\frac{\text{s}}{\Omega }\right)}\end{array}$

Simplify the equation to find $v_o\left(t\right)$.

$\begin{array}{c}{v}_o\left(t\right)=-2\frac{\text{F}\cdot \text{V}}{\text{F}}\left\{\because 1\text{F}=\frac{\text{1s}}{1\Omega }\right\}\\ =-2\text{V}\end{array}$

For $1\text{ms}<t<3\text{ms}$:

Substitute $\left(-10000t+10.02\right)\text{V}$ for $v_i\left(t\right)$ in equation (4) to find $v_o\left(t\right)$.

$\begin{array}{c}{v}_o\left(t\right)=-\left(2\times {10}^{-4}\right)\frac{d\left(\left(-10000t+10.02\right)\text{V}\right)}{dt}\Omega \cdot \text{F}\\ =-\left(2\times {10}^{-4}\right)\frac{d}{dt}\left(-10000t+10.02\right)\Omega \cdot \text{F}\cdot \text{V}\\ =-\left(2\times {10}^{-4}\right)\left(-10000\left(1\right)+0\right)\frac{\Omega \cdot \text{F}\cdot \text{V}}{\text{s}}\\ =-\left(2\times {10}^{-4}\right)\left(-10000\right)\frac{\text{F}\cdot \text{V}}{\left(\frac{\text{s}}{\Omega }\right)}\end{array}$

Simplify the equation to find $v_o\left(t\right)$.

$\begin{array}{c}{v}_o\left(t\right)=2\frac{\text{F}\cdot \text{V}}{\text{F}}\left\{\because 1\text{F}=\frac{\text{1s}}{1\Omega }\right\}\\ =2\text{V}\end{array}$

For $3\text{ms}<t<4\text{ms}$:

Substitute $\left(10000t-10.03\right)\text{V}$ for $v_i\left(t\right)$ in equation (4) to find $v_o\left(t\right)$.

$\begin{array}{c}{v}_o\left(t\right)=-\left(2\times {10}^{-4}\right)\frac{d\left(\left(10000t-10.03\right)\text{V}\right)}{dt}\Omega \cdot \text{F}\\ =-\left(2\times {10}^{-4}\right)\frac{d}{dt}\left(10000t-10.03\right)\Omega \cdot \text{F}\cdot \text{V}\\ =-\left(2\times {10}^{-4}\right)\left(10000\left(1\right)-0\right)\frac{\Omega \cdot \text{F}\cdot \text{V}}{\text{s}}\\ =-\left(2\times {10}^{-4}\right)\left(10000\right)\frac{\text{F}\cdot \text{V}}{\left(\frac{\text{s}}{\Omega }\right)}\end{array}$

Simplify the equation to find $v_o\left(t\right)$.

$\begin{array}{c}{v}_o\left(t\right)=-2\frac{\text{F}\cdot \text{V}}{\text{F}}\left\{\because 1\text{F}=\frac{\text{1s}}{1\Omega }\right\}\\ =-2\text{V}\end{array}$

Therefore, the output voltage function $v_o\left(t\right)$ is expressed as,

$v_o\left(t\right)=\{\begin{array}{l}-2\text{V},0\text{ms}<t<1\text{ms}\\ 2\text{V},1\text{ms}<t<3\text{ms}\\ -2\text{V},3\text{ms}<t<4\text{ms}\end{array}$

The output voltage waveform is drawn as Figure 3.

Conclusion:

Thus, the output voltage waveform of an op amp differentiator $v_o\left(t\right)$ is sketched.