Draw the circuit of an OP-AMP integrator and explain its working. Indicate input and output waveforms.

An operational amplifier (op-amp) integrator is a circuit configuration that performs the mathematical operation of integration, which is essentially a summation over time. The output voltage of an integrator is proportional to the time integral of the input voltage.

Circuit Diagram

Here is a basic circuit diagram of an op-amp integrator:

      V_in
       |
       /
       \ R
       /
       |
       +-----|+\
       |     |  >---- V_out
       |     |-/
       |
       |
      === C
       |
      ---
       -

Components of the Circuit

• V_in: Input voltage
• R: Resistor
• C: Capacitor
• V_out: Output voltage
• Op-amp: Operational Amplifier

Working of the OP-AMP Integrator

The op-amp integrator consists of an op-amp in an inverting configuration with a resistor R in the input path and a capacitor C in the feedback path. The input voltage is applied across the resistor R, and the output is taken across the capacitor C.

Step by Step Explanation

1. Virtual Ground Concept: The inverting input of the op-amp is at virtual ground because the non-inverting input is grounded and the op-amp has a high gain, which forces the inverting input to be at a voltage close to zero.

2. Input Current: The input voltage V_in across the resistor R generates a current I given by Ohm's law: [ I = \frac{V_{in}}{R} ]

3. Charging the Capacitor: This current I flows through the capacitor C, causing it to charge or discharge depending on the polarity of the input voltage. Since the inverting input is at virtual ground, the current through the resistor is constant for a constant input voltage.

4. Integration: The voltage across the capacitor V_c is related to the charge Q on the capacitor and its capacitance C by the relation: [ V_c = \frac{Q}{C} ] Since the current is the rate of change of charge, we can write: [ I = \frac{dQ}{dt} ] Substituting Q from the previous equation, we get: [ I = C \frac{dV_c}{dt} ] Rearranging the terms, we get the output voltage as the integral of the input voltage: [ V_{out} = -\frac{1}{RC} \int V_{in} dt ] The negative sign indicates that the output is 180 degrees out of phase with the input when the op-amp is in the inverting configuration.

5. Steady-State Response: For a constant input voltage, the output voltage will ramp linearly with time, reflecting the integration operation.

Input and Output Waveforms

Let's consider a square wave input to see how the integrator responds:

Input (V_in):   ----| |----| |----| |----
                 ___   ___   ___   ___
Output (V_out):     \_/   \_/   \_/

• When the input is high, the output voltage decreases linearly (negative slope).
• When the input is low, the output voltage increases linearly (positive slope).

The slope of the output waveform is determined by the time constant RC and the magnitude of the input voltage. The higher the input voltage or the larger the time constant, the steeper the slope of the output waveform.

Important Points to Note

Point	Explanation
Linearity	The integrator works well within the linear range of the op-amp. Saturation occurs if the output voltage exceeds the supply voltage of the op-amp.
DC Stability	Without a resistor in parallel with the capacitor, the integrator may drift due to input bias currents or DC offsets.
Frequency Response	The integrator acts as a low-pass filter, with a frequency response that decreases at 20 dB/decade.

Example

If we apply a sinusoidal input to the integrator, the output will be a cosine wave (since the integral of sine is negative cosine), and it will be shifted by 90 degrees with respect to the input signal. If V_in = V_m sin(ωt), then V_out = -(1/ωRC) V_m cos(ωt).

In summary, the op-amp integrator circuit integrates the input voltage over time, and the output voltage is a representation of this integral. The circuit is widely used in analog computers, signal processing, and control systems for performing mathematical operations.