Topic: Analog Circuits
Which of the following represents the correct output $V_o$ of circuit shown in Figure? Assume $RC=1$
(Circuit is an op-amp differentiator with $C$ at the input and $R$ in feedback. $V_i$ is a triangular pulse: goes linearly from 0 to 1 between t=0 and t=1, falls linearly from 1 to 0 between t=1 and t=2, and is 0 for t>2.)
- A. $V_o$ is -1 for t=0 to 1, and +1 for t=1 to 2
- B. $V_o$ is -0.1 for t=0 to 1, and +0.1 for t=1 to 2
- C. $V_o$ is +1 for t=0 to 1, and -1 for t=1 to 2
- D. $V_o$ is +0.1 for t=0 to 1, and -0.1 for t=1 to 2
Answer: A. $V_o$ is -1 for t=0 to 1, and +1 for t=1 to 2
Explanation: The given circuit is an inverting differentiator. The output voltage is $V_o = -RC \frac{dV_i}{dt}$. Since $RC=1$, $V_o = -\frac{dV_i}{dt}$.
From $t=0$ to $t=1$: $V_i$ increases from 0 to 1, so the slope $\frac{dV_i}{dt} = 1$ V/s. Therefore, $V_o = -1$ V.
From $t=1$ to $t=2$: $V_i$ decreases from 1 to 0, so the slope $\frac{dV_i}{dt} = -1$ V/s. Therefore, $V_o = -(-1) = 1$ V.
This matches the square wave plotted in option (a).