Chapter 6, Problem 69P

An op amp integrator with R = 4 MΩ and C = 1 μF has the input waveform shown in Fig. 6.88. Plot the output waveform.

Figure 6.88

For Prob. 6.69.

To determine

Sketch the output voltage waveform of an op-amp integrator.

Explanation of Solution

Given data:

The value of the resistor $\left(R\right)$ is $4\text{M}\Omega$.

The value of the capacitor $\left(C\right)$ is $1\mu \text{F}$.

Refer to Figure 6.88 in the textbook.

Formula used:

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

$v_o=-\frac{1}{RC}{\displaystyle \underset{0}{\overset{t}{\int }}{v}_{i}dt}+v_o\left(0\right)$ (1)

Here,

$R$ is the value of the resistor,

$C$ is the value of the capacitor,

$v_i$ is the value of the input voltage, and

$v_o\left(0\right)$ is the voltage at time $t=0$.

Calculation:

Substitute $4\text{M}\Omega$ for $R$, $1\mu \text{F}$ for $C$ and $0\text{V}$ for $v_o\left(0\right)$ in equation (1) to find $v_o$.

$\begin{array}{c}{v}_o=-\frac{1}{\left(4\text{M}\Omega \right)\left(1\mu \text{F}\right)}{\displaystyle \underset{0}{\overset{t}{\int }}{v}_{i}dt}+0\text{V}\\ =-\frac{1}{\left(4\times {10}^6\right)\left(1\times {10}^{-6}\right)}{\displaystyle \underset{0}{\overset{t}{\int }}{v}_{i}dt}\frac{1}{\Omega \cdot \text{F}}\\ =-\frac{1}{4}{\displaystyle \underset{0}{\overset{t}{\int }}{v}_{i}dt}\frac{1}{\Omega \cdot \text{F}}\end{array}$

$v_o=\left(-0.25\right){\displaystyle \underset{0}{\overset{t}{\int }}{v}_{i}dt}\frac{1}{\Omega \cdot \text{F}}$ (2)

The given input voltage waveform is redrawn as Figure 1.

Refer to Figure 1, the input voltage $v_i$ is expressed as,

$v_i=\{\begin{array}{l}20\text{mV,}\text{0}\text{ms}<t<1\text{ms}\\ 10\text{mV,}\text{1}\text{ms}<t<2\text{ms}\\ -20\text{mV,}\text{2}\text{ms}<t<4\text{ms}\\ -10\text{mV,}4\text{ms}<t<5\text{ms}\\ 20\text{mV,}5\text{ms}<t<6\text{ms}\end{array}$

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

Substitute $20\text{mV}$ for $v_i$ in equation (2) to find $v_o$.

$\begin{array}{c}{v}_o=\left(-0.25\right){\displaystyle \underset{0}{\overset{t}{\int }}\left(20\text{mV}\right)dt}\frac{1}{\Omega \cdot \text{F}}\\ =\left(-0.25\right)\left(20\times {10}^{-3}\right){\displaystyle \underset{0}{\overset{t}{\int }}\left(1\right)dt}\frac{\text{V}}{\Omega \cdot \text{F}}\left\{\because 1\text{m}={10}^{-3}\right\}\\ =-\left(5\times {10}^{-3}\right){\left[t\right]}_0^t\frac{\text{V}\cdot \text{s}}{\Omega \cdot \text{F}}\\ =-\left(5\times {10}^{-3}\right)\left[t-0\right]\frac{\text{V}\cdot \text{F}}{\text{F}}\left\{\because 1\text{F}=\frac{\text{s}}{\text{Ω}}\right\}\end{array}$

Simplify the equation to find $v_o$.

$v_o=-5t\times {10}^{-3}\text{V}$

$v_o=-5t\text{mV}\left\{\because 1\text{m}={10}^{-3}\right\}$ (3)

Substitute $1$ for $t$ in equation (3) to find $v\left(1\right)$.

$\begin{array}{c}v\left(1\right)=-5\left(1\right)\text{mV}\\ =-5\text{mV}\end{array}$

Case (ii): For $\text{1}\text{ms}<t<2\text{ms}$

Equation (2) can be rewritten as following equation to find the output voltage for $\text{1}\text{ms}<t<2\text{ms}$.

$v_o=\left(-0.25\right){\displaystyle \underset{1}{\overset{t}{\int }}{v}_{i}dt}\frac{1}{\Omega \cdot \text{F}}+v\left(1\right)$ (4)

Substitute $10\text{mV}$ for $v_i$ and $-5\text{mV}$ for $v\left(1\right)$ in equation (4) to find $v_o$.

$\begin{array}{c}{v}_o=\left(-0.25\right){\displaystyle \underset{1}{\overset{t}{\int }}\left(10\text{mV}\right)dt}\frac{1}{\Omega \cdot \text{F}}-5\text{mV}\\ =\left(-0.25\right)\left(10\times {10}^{-3}\right){\displaystyle \underset{1}{\overset{t}{\int }}\left(1\right)dt}\frac{\text{V}}{\Omega \cdot \text{F}}-\left(5\times {10}^{-3}\right)\text{V}\left\{\because 1\text{m}={10}^{-3}\right\}\\ =-\left(2.5\times {10}^{-3}\right){\left[t\right]}_1^t\frac{\text{V}\cdot \text{s}}{\Omega \cdot \text{F}}-\left(5\times {10}^{-3}\right)\text{V}\\ =-\left(2.5\times {10}^{-3}\right)\left[t-1\right]\frac{\text{V}\cdot \text{F}}{\text{F}}-\left(5\times {10}^{-3}\right)\text{V}\left\{\because 1\text{F}=\frac{\text{s}}{\text{Ω}}\right\}\end{array}$

Simplify the equation to find $v_o$.

$\begin{array}{c}{v}_o=-\left[\left(2.5t\times {10}^{-3}\right)-\left(2.5\times {10}^{-3}\right)\right]\text{V}-\left(5\times {10}^{-3}\right)\text{V}\\ =-\left(2.5t\times {10}^{-3}\right)\text{V}+\left(2.5\times {10}^{-3}\right)\text{V}-\left(5\times {10}^{-3}\right)\text{V}\\ =\left(-\left(2.5t\right)-\left(2.5\right)\right)\times {10}^{-3}\text{V}\end{array}$

$v_o\text{=}\left(-2.5t-2.5\right)\text{mV}\left\{\because 1\text{m}={10}^{-3}\right\}$ (5)

Substitute $2$ for $t$ in equation (5) to find $v\left(2\right)$.

$\begin{array}{c}v\left(2\right)=\left(-2.5\left(2\right)-2.5\right)\text{mV}\\ =\left(-5-2.5\right)\text{mV}\\ =-7.5\text{mV}\end{array}$

Case (iii): For $\text{2}\text{ms}<t<4\text{ms}$

Equation (2) can be rewritten as following equation to find the output voltage for $\text{2}\text{ms}<t<4\text{ms}$.

$v_o=\left(-0.25\right){\displaystyle \underset{2}{\overset{t}{\int }}{v}_{i}dt}\frac{1}{\Omega \cdot \text{F}}+v\left(2\right)$ (6)

Substitute $-20\text{mV}$ for $v_i$ and $-7.5\text{mV}$ for $v\left(2\right)$ in equation (6) to find $v_o$.

$\begin{array}{c}{v}_o=\left(-0.25\right){\displaystyle \underset{2}{\overset{t}{\int }}\left(-20\text{mV}\right)dt}\frac{1}{\Omega \cdot \text{F}}-7.5\text{mV}\\ =\left(0.25\right)\left(20\times {10}^{-3}\right){\displaystyle \underset{2}{\overset{t}{\int }}\left(1\right)dt}\frac{\text{V}}{\Omega \cdot \text{F}}-\left(7.5\times {10}^{-3}\right)\text{V}\left\{\because 1\text{m}={10}^{-3}\right\}\\ =\left(5\times {10}^{-3}\right){\left[t\right]}_2^t\frac{\text{V}\cdot \text{s}}{\Omega \cdot \text{F}}-\left(7.5\times {10}^{-3}\right)\text{V}\\ =\left(5\times {10}^{-3}\right)\left[t-2\right]\frac{\text{V}\cdot \text{F}}{\text{F}}-\left(7.5\times {10}^{-3}\right)\text{V}\left\{\because 1\text{F}=\frac{\text{s}}{\text{Ω}}\right\}\end{array}$

Simplify the equation to find $v_o$.

$\begin{array}{c}{v}_o=\left[\left(5t\times {10}^{-3}\right)-\left(10\times {10}^{-3}\right)\right]\text{V}-\left(7.5\times {10}^{-3}\right)\text{V}\\ =\left(5t\times {10}^{-3}\right)\text{V}-\left(10\times {10}^{-3}\right)\text{V}-\left(7.5\times {10}^{-3}\right)\text{V}\\ =\left(\left(5t\right)-\left(17.5\right)\right)\times {10}^{-3}\text{V}\end{array}$

$v_o\text{=}\left(5t-17.5\right)\text{mV}\left\{\because 1\text{m}={10}^{-3}\right\}$ (7)

Substitute $4$ for $t$ in equation (7) to find $v\left(4\right)$.

$\begin{array}{c}v\left(4\right)\text{=}\left(5\left(4\right)-17.5\right)\text{mV}\\ \text{=}\left(20-17.5\right)\text{mV}\\ =2.5\text{mV}\end{array}$

Case (iv): For $4\text{ms}<t<5\text{ms}$

Equation (2) can be rewritten as following equation to find the output voltage for $4\text{ms}<t<5\text{ms}$.

$v_o=\left(-0.25\right){\displaystyle \underset{4}{\overset{t}{\int }}{v}_{i}dt}\frac{1}{\Omega \cdot \text{F}}+v\left(4\right)$ (8)

Substitute $-10\text{mV}$ for $v_i$ and $2.5\text{mV}$ for $v\left(4\right)$ in equation (8) to find $v_o$.

$\begin{array}{c}{v}_o=\left(-0.25\right){\displaystyle \underset{4}{\overset{t}{\int }}\left(-10\text{mV}\right)dt}\frac{1}{\Omega \cdot \text{F}}+2.5\text{mV}\\ =\left(0.25\right)\left(10\times {10}^{-3}\right){\displaystyle \underset{4}{\overset{t}{\int }}\left(1\right)dt}\frac{\text{V}}{\Omega \cdot \text{F}}+\left(2.5\times {10}^{-3}\right)\text{V}\left\{\because 1\text{m}={10}^{-3}\right\}\\ =\left(2.5\times {10}^{-3}\right){\left[t\right]}_4^t\frac{\text{V}\cdot \text{s}}{\Omega \cdot \text{F}}+\left(2.5\times {10}^{-3}\right)\text{V}\\ =\left(2.5\times {10}^{-3}\right)\left[t-4\right]\frac{\text{V}\cdot \text{F}}{\text{F}}+\left(2.5\times {10}^{-3}\right)\text{V}\left\{\because 1\text{F}=\frac{\text{s}}{\text{Ω}}\right\}\end{array}$

Simplify the equation to find $v_o$.

$\begin{array}{c}{v}_o=\left[\left(2.5t\times {10}^{-3}\right)-\left(10\times {10}^{-3}\right)\right]\text{V}+\left(2.5\times {10}^{-3}\right)\text{V}\\ =\left(2.5t\times {10}^{-3}\right)\text{V}-\left(10\times {10}^{-3}\right)\text{V}+\left(2.5\times {10}^{-3}\right)\text{V}\\ =\left(\left(2.5t\right)-\left(7.5\right)\right)\times {10}^{-3}\text{V}\end{array}$

$v_o\text{=}\left(2.5t-7.5\right)\text{mV}\left\{\because 1\text{m}={10}^{-3}\right\}$ (9)

Substitute $5$ for $t$ in equation (9) to find $v\left(5\right)$.

$\begin{array}{c}v\left(5\right)=\left(2.5\left(5\right)-7.5\right)\text{mV}\\ =\left(12.5-7.5\right)\text{mV}\\ =5\text{mV}\end{array}$

Case (v): For $5\text{ms}<t<6\text{ms}$

Equation (2) can be rewritten as following equation to find the output voltage for $5\text{ms}<t<6\text{ms}$.

$v_o=\left(-0.25\right){\displaystyle \underset{5}{\overset{t}{\int }}{v}_{i}dt}\frac{1}{\Omega \cdot \text{F}}+v\left(5\right)$ (10)

Substitute $20\text{mV}$ for $v_i$ and $5\text{mV}$ for $v\left(5\right)$ in equation (10) to find $v_o$.

$\begin{array}{c}{v}_o=\left(-0.25\right){\displaystyle \underset{5}{\overset{t}{\int }}\left(20\text{mV}\right)dt}\frac{1}{\Omega \cdot \text{F}}+5\text{mV}\\ =-\left(0.25\right)\left(20\times {10}^{-3}\right){\displaystyle \underset{5}{\overset{t}{\int }}\left(1\right)dt}\frac{\text{V}}{\Omega \cdot \text{F}}+\left(5\times {10}^{-3}\right)\text{V}\left\{\because 1\text{m}={10}^{-3}\right\}\\ =-\left(5\times {10}^{-3}\right){\left[t\right]}_5^t\frac{\text{V}\cdot \text{s}}{\Omega \cdot \text{F}}+\left(5\times {10}^{-3}\right)\text{V}\\ =-\left(5\times {10}^{-3}\right)\left[t-5\right]\frac{\text{V}\cdot \text{F}}{\text{F}}+\left(5\times {10}^{-3}\right)\text{V}\left\{\because 1\text{F}=\frac{\text{s}}{\text{Ω}}\right\}\end{array}$

Simplify the equation to find $v_o$.

$\begin{array}{c}{v}_o=\left[-\left(5t\times {10}^{-3}\right)+\left(25\times {10}^{-3}\right)\right]\text{V}+\left(5\times {10}^{-3}\right)\text{V}\\ =-\left(5t\times {10}^{-3}\right)\text{V+}\left(25\times {10}^{-3}\right)\text{V}+\left(5\times {10}^{-3}\right)\text{V}\\ =\left(-\left(5t\right)+\left(30\right)\right)\times {10}^{-3}\text{V}\end{array}$

$v_o\text{=}\left(-5t+30\right)\text{mV}\left\{\because 1\text{m}={10}^{-3}\right\}$ (11)

Substitute $6$ for $t$ in equation (11) to find $v\left(6\right)$.

$\begin{array}{c}v\left(6\right)=\left(-5\left(6\right)+30\right)\text{mV}\\ =\left(-30+30\right)\text{mV}\\ =0\text{mV}\end{array}$

Therefore, the output voltage function for Figure 1 is expressed as,

$v_o=\{\begin{array}{l}-5t\text{mV,}\text{0}\text{ms}<t<1\text{ms}\\ \left(-2.5t-2.5\right)\text{mV,}\text{1}\text{ms}<t<2\text{ms}\\ \left(5t-17.5\right)\text{mV,}\text{2}\text{ms}<t<4\text{ms}\\ \left(2.5t-7.5\right)\text{mV,}4\text{ms}<t<5\text{ms}\\ \left(-5t+30\right)\text{mV,}5\text{ms}<t<6\text{ms}\end{array}$

The output voltage waveform is drawn as Figure 2.

Conclusion:

Thus, the output voltage waveform of an op-amp integrator is sketched.