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Electrical Engineering QUESTION #2181
Question 1
An op-amp integrator has $R = 10\,\text{k}\Omega$ and $C = 0.1\,\mu\text{F}$. The input is a $\pm1\,\text{V}$ square wave at $1\,\text{kHz}$. The peak-to-peak output voltage is:
  • $0.5\,\text{V}$✔️
  • $1.0\,\text{V}$
  • $2.0\,\text{V}$
  • $0.25\,\text{V}$
Correct Answer Explanation
$RC = 10^4 \times 10^{-7} = 1\,\text{ms}$. Each half-period $T/2 = 0.5\,\text{ms}$. Output ramp amplitude $\Delta V = \frac{V_{in}\cdot(T/2)}{RC} = \frac{1\times0.5\times10^{-3}}{10^{-3}} = 0.5\,\text{V}$. Since the output ramps up then down symmetrically, peak-to-peak $= 0.5\,\text{V}$.