Question

Plot the output voltage of an integrator 1, 3, 5, and 7 s after the start of integration if the input resistor is 2.0 MV, the feedback capacitor is 0.25 μF, and the input voltage is 4.0 mV.

Short answer

The obtained plot of the output voltage versus time is a linear graph in the negative y-direction.

Step-by-step solution

Step 1. Given Information

Given an integrator with input resistor 2.0 MV.

The feedback capacitor is 0.25 μF.

The input voltage is 4.0 mV.

Step 2.  Output voltage 

We know that the output voltage of an integrator circuit is given by;

$v_o=\frac{-1}{R_i{C}_f}{\int }_0^t{v}_{i}dt$

Substituting the known values, we get;

localid="1645775128952" $$\begin{gathered} v_o=\frac{-1}{(2\times {10}^6)\times (0.25\times {10}^{-6})}{\int }_0^t(4\mathrm{mV})dt \\ =-8\mathrm{mV}\times t \\ =\left(8t\right)\mathrm{mV} \end{gathered}$$

Step 3. Plotting the Graph

First, we make a table with the output voltage and the corresponding time.

Time (in seconds)	Output Voltage Vo (in mV)
1	-8
3	-24
5	-40
7	-56

So, the graph of output voltage versus time can be plotted as shown below;