The opamps in the circuit shown are ideal, but have saturation voltages of $\pm 10$ V.

Assume that the initial inductor current is 0 A. The input voltage ($V_i$) is a triangular signal with peak voltages of $\pm 2$ V and time period of 8 $\mu$s. Which one of the following statements is true?

Question

Prepp Team · Accepted Answer

To identify the correct statement, we analyze the two operational amplifier stages separately: the first as a Schmitt trigger and the second as an integrator.

1. Analysis of OPA1 (Schmitt Trigger)

OPA1 is configured as a non-inverting Schmitt trigger with positive feedback. The voltage at the non-inverting terminal ($V_+$) is determined by the voltage divider between $V_i$ and $V_{o1}$:

$$V_+ = \frac{V_i \cdot 100\text{k}\Omega + V_{o1} \cdot 10\text{k}\Omega}{100\text{k}\Omega + 10\text{k}\Omega}$$

The op-amp switches its output when $V_+$ crosses the voltage at the inverting terminal (which is $0\text{ V}$). Setting $V_+ = 0$:

$$100 V_i + 10 V_{o1} = 0 \implies 10 V_i = -V_{o1} \implies V_i = -\frac{V_{o1}}{10}$$

Given the saturation voltages $V_{sat} = \pm 10\text{ V}$:

If $V_{o1} = -10\text{ V}$, the switching threshold is $V_i = +1\text{ V}$.
If $V_{o1} = +10\text{ V}$, the switching threshold is $V_i = -1\text{ V}$.

The input $V_i$ is a triangular wave with a peak of $\pm 2\text{ V}$ and a period of $8\text{ }\mu\text{s}$. The slope of the ramp is:

$$\text{Slope} = \frac{\Delta V}{\Delta t} = \frac{2\text{ V}}{2\text{ }\mu\text{s}} = 1\text{ V/}\mu\text{s}$$

If $V_i = 0$ at $t = 0$, it reaches the threshold of $1\text{ V}$ at $t = 1\text{ }\mu\text{s}$. Thus, the switching of $V_{o1}$ is delayed by $1\text{ }\mu\text{s}$ relative to the zero-crossing of $V_i$.

2. Analysis of OPA2 (Integrator)

OPA2 is an inverting integrator using an inductor ($L = 1\text{ mH}$) and a resistor ($R_f = 1\text{ k}\Omega$). Due to the virtual ground at the inverting input:

$$V_{o1} = L \frac{di_L}{dt} \implies i_L = \frac{1}{L} \int V_{o1} dt$$

The output voltage is given by $V_{o2} = -i_L \cdot R_f$, so:

$$V_{o2} = -\frac{R_f}{L} \int V_{o1} dt$$

Since $V_{o1}$ is a square wave, its integral is a triangular wave. We check if the output exceeds the saturation limits ($\pm 10\text{ V}$):

$$\Delta V_{o2} = \left| -\frac{1000}{10^{-3}} \int_{0}^{4\mu\text{s}} 10 \, dt \right| = 10^6 \times (10 \times 4 \times 10^{-6}) = 40\text{ V}$$

The theoretical peak-to-peak swing is $40\text{ V}$, but the op-amp saturates at $\pm 10\text{ V}$ (total range of $20\text{ V}$). Therefore, the triangle wave will be clipped at the saturation levels, resulting in a trapezoidal waveform.