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A sine wave of 1vpeak at 1000Hz is applied to a differentiator with the following specification: RF =1kΩ and C1=0.33µF, find the output waveform?

I had been asked this question in final exam.

My question comes from Differentiator topic in section Operational Amplifier Applications of Linear Integrated Circuits

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**Differentiator Circuit Output Analysis**

**Given Parameters:**

- Input Signal: Sine wave, 1V peak at 1000 Hz

- Feedback Resistor (RF) = 1 kΩ

- Input Capacitor (C1) = 0.33 μF

- Circuit Type: Op-amp differentiator

**Differentiator Function:**

A differentiator circuit produces an output proportional to the rate of change (derivative) of the input signal with respect to time.

**Transfer Function:**

Vout = -RF × C1 × (dVin/dt)

**Output Voltage Calculation:**

For a sinusoidal input: Vin = A × sin(2πft)

Where A = 1V (peak amplitude), f = 1000 Hz

The derivative of sine is cosine:

dVin/dt = A × 2πf × cos(2πft) = 1 × 2π × 1000 × cos(2πft) = 6283.19 × cos(2πft)

**Peak Output Voltage:**

Vout(peak) = -RF × C1 × dVin/dt(peak)

Vout(peak) = -1000 × 0.33 × 10^-6 × 6283.19

Vout(peak) = -330 × 10^-6 × 6283.19

Vout(peak) = -2.073V

Vout(peak) ≈ 2.07V (ignoring polarity)

**Output Waveform Characteristics:**

1. **Waveform Shape**: The output will be a COSINE waveform (90° phase shift from input sine wave)

2. **Peak Amplitude**: Approximately 2.07V

3. **Frequency**: Same as input frequency (1000 Hz)

4. **Phase Relationship**: The output leads the input by 90° (cosine is 90° ahead of sine)

5. **Polarity**: Inverted due to the negative sign in the transfer function

**Detailed Waveform Description:**

- When the input sine wave has maximum positive slope (crossing zero going upward), the output is at maximum negative peak (-2.07V)

- When the input is at positive peak, the output crosses zero

- When the input has maximum negative slope (crossing zero going downward), the output is at maximum positive peak (+2.07V)

- When the input is at negative peak, the output crosses zero again

**Time Domain Expression:**

Vout(t) = -2.07 × cos(2π × 1000 × t) V

Or in terms of sine:

Vout(t) = -2.07 × sin(2π × 1000 × t + 90°) V

**Practical Observations:**

- The output is a pure cosine waveform with 2.07V peak amplitude

- The frequency remains 1000 Hz (same as input)

- The output oscillates between -2.07V and +2.07V

- The derivative relationship ensures that rapid changes in input produce large output voltages

- This circuit is ideal for detecting the rate of change of signals in applications like zero-crossing detection and high-frequency emphasis filters

**Circuit Behavior Summary:**

The differentiator converts the input 1V peak sine wave at 1000 Hz into a 2.07V peak cosine waveform, demonstrating the classic property of differentiators to perform time differentiation and produce a 90° phase lead in the output signal.

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