Question:
In the circuit shown, \(V(t) = 2\sin(2000\pi t)\) Volts, where \(t\) is in seconds. The source \(V(t)\) drives the non-inverting (+) input of an opamp directly, the inverting (-) input is tied to ground (0 V), and the opamp is powered from +15 V and -15 V rails with no feedback network between its output and either input. Take the opamp to be ideal. Which of the following options is correct?
In the circuit shown, \(V(t) = 2\sin(2000\pi t)\) Volts, where \(t\) is in seconds. The source \(V(t)\) drives the non-inverting (+) input of an opamp directly, the inverting (-) input is tied to ground (0 V), and the opamp is powered from +15 V and -15 V rails with no feedback network between its output and either input. Take the opamp to be ideal. Which of the following options is correct?
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With no feedback, an ideal opamp comparator output only ever sits at one of its two supply rails, comparing \(V(t)\) against the grounded inverting input tells you when it flips.
Updated On: Jul 28, 2026
- \(V_{out}(t)\) is square wave with peak-to-peak voltage = 30 V and time period is 1 ms.
- \(V_{out}(t)\) is a sine wave with peak-to-peak voltage = 4 V and time period is 1 ms.
- \(V_{out}(t)\) is sine wave with peak-to-peak voltage = 30 V and time period of 1 ms.
- \(V_{out}(t)\) is square wave with peak-to-peak voltage = 4 V and time period is 1 ms.
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept:
The opamp here has no feedback path from its output back to either input, so it is not being used as an amplifier, it is being used as an open-loop comparator. An ideal opamp with no feedback has effectively infinite gain, so the tiniest difference between its two inputs drives the output all the way to one power rail or the other.
Step 2: Key Formula or Approach:
Comparator rule: \(V_{out} = +V_{sat}\) when \(V_+ > V_-\), and \(V_{out} = -V_{sat}\) when \(V_+ < V_-\). Here \(V_+ = V(t)\) and \(V_- = 0\) (grounded), and the rails are \(\pm15\) V, so \(V_{sat}=15\) V.
Step 3: Detailed Explanation:
Since \(V_- = 0\), the comparison is simply whether \(V(t)\) is positive or negative.
When \(V(t) = 2\sin(2000\pi t) > 0\), the output snaps to \(+15\) V.
When \(V(t) < 0\), the output snaps to \(-15\) V.
This makes \(V_{out}(t)\) a square wave (not a scaled copy of the sine wave) that flips sign exactly when \(V(t)\) crosses zero.
Peak-to-peak amplitude: \[ V_{pp} = 15 - (-15) = 30\ \text{V} \]
Time period: \(V(t)\) has angular frequency \(\omega = 2000\pi\) rad/s, and the output switches at the same zero crossings as \(V(t)\), so it repeats with the same period as \(V(t)\): \[ T = \frac{2\pi}{\omega} = \frac{2\pi}{2000\pi} = \frac{1}{1000}\ \text{s} = 1\ \text{ms} \]
Step 4: Why the other options are wrong.
Options (B) and (C) both say the output is a sine wave. An open-loop opamp with no feedback cannot produce a sine wave, its output can only sit at one of the two rails, so it must be a square wave.
Option (D) has the right waveform shape but the wrong amplitude, 4 V is nowhere close to the rail-to-rail swing of 30 V that an ideal saturated opamp actually produces.
Final Answer:
The comparator output is a 30 V peak-to-peak square wave with a 1 ms period. \[ \boxed{V_{pp} = 30\ \text{V},\ T = 1\ \text{ms}} \]
The opamp here has no feedback path from its output back to either input, so it is not being used as an amplifier, it is being used as an open-loop comparator. An ideal opamp with no feedback has effectively infinite gain, so the tiniest difference between its two inputs drives the output all the way to one power rail or the other.
Step 2: Key Formula or Approach:
Comparator rule: \(V_{out} = +V_{sat}\) when \(V_+ > V_-\), and \(V_{out} = -V_{sat}\) when \(V_+ < V_-\). Here \(V_+ = V(t)\) and \(V_- = 0\) (grounded), and the rails are \(\pm15\) V, so \(V_{sat}=15\) V.
Step 3: Detailed Explanation:
Since \(V_- = 0\), the comparison is simply whether \(V(t)\) is positive or negative.
When \(V(t) = 2\sin(2000\pi t) > 0\), the output snaps to \(+15\) V.
When \(V(t) < 0\), the output snaps to \(-15\) V.
This makes \(V_{out}(t)\) a square wave (not a scaled copy of the sine wave) that flips sign exactly when \(V(t)\) crosses zero.
Peak-to-peak amplitude: \[ V_{pp} = 15 - (-15) = 30\ \text{V} \]
Time period: \(V(t)\) has angular frequency \(\omega = 2000\pi\) rad/s, and the output switches at the same zero crossings as \(V(t)\), so it repeats with the same period as \(V(t)\): \[ T = \frac{2\pi}{\omega} = \frac{2\pi}{2000\pi} = \frac{1}{1000}\ \text{s} = 1\ \text{ms} \]
Step 4: Why the other options are wrong.
Options (B) and (C) both say the output is a sine wave. An open-loop opamp with no feedback cannot produce a sine wave, its output can only sit at one of the two rails, so it must be a square wave.
Option (D) has the right waveform shape but the wrong amplitude, 4 V is nowhere close to the rail-to-rail swing of 30 V that an ideal saturated opamp actually produces.
Final Answer:
The comparator output is a 30 V peak-to-peak square wave with a 1 ms period. \[ \boxed{V_{pp} = 30\ \text{V},\ T = 1\ \text{ms}} \]
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