If the input voltage waveform \(V_{IN}\) is a ramp function (as shown in the \(V_{IN} - t\) plot below), then the output wave form (\(V_{OUT}\)) for the given circuit diagram having an ideal operational amplifier (Op-Amp) is
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Carefully identify the Op-Amp circuit configuration.
- **Integrator**: Resistor in input, Capacitor in feedback. Output is the negative integral of the input.
- **Differentiator**: Capacitor in input, Resistor in feedback. Output is the negative derivative of the input.
Memorizing the output for standard inputs (step, ramp, sine) for both circuits is very helpful. For an integrator: input step \(\to\) output ramp; input ramp \(\to\) output parabola.
Step 1: Understanding the Concept:
The given circuit is an inverting differentiator. An operational amplifier (Op-Amp) with a capacitor in the input path and a resistor in the feedback path acts as a differentiator. The output voltage is proportional to the negative time derivative of the input voltage. We are given a ramp function as input and need to find the corresponding output waveform. Step 2: Key Formula or Approach:
For an ideal Op-Amp in the inverting configuration, the virtual ground principle applies, meaning the voltage at the inverting input (\(V_-\)) is equal to the voltage at the non-inverting input (\(V_+\)). Here, \(V_+ = 0\) (grounded), so \(V_- = 0\).
The current flowing through the capacitor is \(I_C = C_F \frac{d(V_{IN} - V_-)}{dt}\). Since \(V_-=0\), \(I_C = C_F \frac{dV_{IN}}{dt}\).
The current flowing through the feedback resistor is \(I_R = \frac{V_- - V_{OUT}}{R_{IN}}\). Since \(V_-=0\), \(I_R = -\frac{V_{OUT}}{R_{IN}}\).
For an ideal Op-Amp, the input current is zero, so \(I_C = I_R\).
Therefore, \(C_F \frac{dV_{IN}}{dt} = -\frac{V_{OUT}}{R_{IN}}\).
This gives the input-output relationship for a differentiator circuit:
\[ V_{OUT}(t) = -R_{IN} C_F \frac{dV_{IN}(t)}{dt} \]
\textit{Note: The component labels in the diagram are swapped. The input element should be the capacitor and the feedback element should be the resistor for a standard differentiator. The given circuit is an integrator. Let's solve for the given circuit.}
The given circuit is an inverting integrator. The input is through a resistor \(R_{IN}\) and the feedback element is a capacitor \(C_F\).
For an integrator, the output voltage is:
\[ V_{OUT}(t) = -\frac{1}{R_{IN}C_F} \int_0^t V_{IN}(\tau) d\tau + V_{OUT}(0) \]
Step 3: Detailed Explanation:
1. Analyze the input signal: The input \(V_{IN}\) is a ramp function. This means \(V_{IN}(t) = kt\) for some positive constant \(k\). The graph shows a straight line with a positive slope starting from the origin.
2. Apply the integrator formula: We need to integrate the input signal. Assume the initial voltage across the capacitor (and hence the initial output voltage) is zero, i.e., \(V_{OUT}(0) = 0\).
\[ V_{OUT}(t) = -\frac{1}{R_{IN}C_F} \int_0^t (k\tau) d\tau \]
\[ V_{OUT}(t) = -\frac{k}{R_{IN}C_F} \left[ \frac{\tau^2}{2} \right]_0^t \]
\[ V_{OUT}(t) = -\frac{k}{2R_{IN}C_F} t^2 \]
3. Analyze the output waveform: The output voltage is \(V_{OUT}(t) = - \alpha t^2\), where \(\alpha = \frac{k}{2R_{IN}C_F}\) is a positive constant.
This equation describes a parabola that opens downwards and starts from the origin (0,0).
Let's check the given options:
(A) Shows a function that increases faster than a line (like \(t^2\)), but it is positive. This is incorrect due to the negative sign.
(B) Shows a constant negative output. This would be the output if the input were a constant positive voltage. (Incorrect)
(C) Shows a negative step function. This is incorrect.
(D) Shows a parabola opening downwards, starting from the origin. This matches our derived result \(V_{OUT}(t) = -\alpha t^2\).
Step 4: Final Answer:
The circuit is an inverting integrator. The integral of a ramp function (\(kt\)) is a parabolic function (\(-\alpha t^2\)). The plot that correctly represents a downward-opening parabola starting from the origin is (D).
