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Study of Op-Amp Circuits: Integrator, Differentiator, Inverting Amplifier, and Instrumentation Amplifier

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Aim

To study and implement the following circuits using an OP-AMP:
  1. Integrator
  2. Differentiator
  3. Inverting Amplifier
  4. Instrumentation Amplifier

Apparatus & Software

ComponentQuantity
Dual Power Supply (±15V)1
DC Power Source1
Function Generator (0–1 MHz)1
Oscilloscope1
Breadboard1
Probes and Connecting WiresAs required
OP071
IC 741C1
Resistor (15 kΩ)1
Resistor (470 kΩ)1
Capacitor (0.015 µF)1
Resistor (10 kΩ)4
Resistor (4.7 kΩ)3

Theory

1. Integrator: An op-amp integrator produces an output proportional to the time integral of the input voltage. The input is applied through a resistor to the inverting terminal, while a capacitor is placed in the feedback path. The output is expressed as:
vo(t)=−1RC∫vin(t) dtv_o(t) = -\frac{1}{RC} \int v_{in}(t)\, dt
The constant of integration depends on the initial output voltage at t = 0. The negative sign shows that the output is inverted with respect to the input. In practice, a resistor is often connected in parallel with the capacitor to prevent output drift and improve stability for long-duration signals.
2. Differentiator: An op-amp differentiator generates an output proportional to how quickly the input changes. The input comes through a capacitor, and there is a resistor in the feedback path. The output is:
vo(t)=−RCdvin(t)dtv_o(t) = -RC \frac{dv_{in}(t)}{dt}
For a sinusoidal input, the output is 90° ahead of the input. A cosine input results in a sine output, while a triangular input produces a square output. To reduce high-frequency noise, practical circuits include a small resistor in series with the capacitor, making the differentiator act like a high-pass filter. Differentiators are often used for edge detection and waveform shaping.
3. Inverting Amplifier: In this circuit, the input goes through R1 to the inverting terminal, and a resistor Rf provides feedback. The non-inverting input is grounded. The voltage gain is:
Av=−RfR1A_v = -\frac{R_f}{R_1}
The output is flipped 180° compared to the input. The gain depends solely on the resistor ratio, and the inverting input node functions as a virtual ground. This setup is common in signal processing because it offers reliable and predictable gain.
4. Instrumentation Amplifier: The instrumentation amplifier is a differential amplifier with high accuracy and common-mode rejection. It usually has three op-amps: two serve as input buffers and the third functions as a difference amplifier. The overall gain is:
Av=(1+2RRg)RfR1A_v = \left(1 + \frac{2R}{R_g}\right) \frac{R_f}{R_1}
The buffered input stage offers high input impedance. The final difference amplifier ensures precise subtraction. This design allows for adjustable gain and effectively rejects unwanted common-mode signals. Instrumentation amplifiers are used in medical instruments, strain gauge sensors, and data acquisition systems.

Pre-Lab / Circuit Diagram

Practical Circuit of Integrator

Practical Circuit of Integrator (R = 15 kΩ, Rf = 470 kΩ, C = 0.015 µF, Op-Amp OP07, ±12V supply).

Practical Circuit of Differentiator

Practical Circuit of Differentiator (capacitor at input, resistor in feedback path).

Circuit of Inverting Amplifier

Circuit of Inverting Amplifier (R1 input resistor, Rf feedback resistor, non-inverting terminal grounded).

Circuit of Instrumentation Amplifier

Circuit of Instrumentation Amplifier (three op-amps: two input buffers and one difference amplifier stage).

Procedure

Part 1 — Integrator:
  1. Assemble the integrator circuit on the breadboard using R = 15 kΩ, Rf = 470 kΩ, and C = 0.015 µF with OP07 powered by ±12V.
  2. Apply a square wave input from the function generator and observe the output on the oscilloscope.
  3. Apply a sinusoidal input and observe the output waveform.
  4. Record and compare input and output waveforms at different frequencies.
Part 2 — Differentiator:
  1. Assemble the differentiator circuit with the capacitor at the input and resistor in the feedback.
  2. Apply a square wave input and observe the output on the oscilloscope.
  3. Apply a sinusoidal input and observe the output waveform.
  4. Record and compare input and output waveforms at different frequencies.
Part 3 — Inverting Amplifier:
  1. Assemble the inverting amplifier circuit with appropriate R1 and Rf values.
  2. Apply a square wave input at 1 kHz and observe the output waveform.
  3. Increase the frequency to 10 kHz and observe the effect of the op-amp slew rate on the output.
  4. Apply a sinusoidal input and verify the phase inversion and gain.
Part 4 — Instrumentation Amplifier:
  1. Assemble the three op-amp instrumentation amplifier circuit using IC 741C and resistors.
  2. Apply an input voltage of 2V Vpp and observe the amplified output on the oscilloscope.
  3. Calculate the measured gain and verify the amplification behavior based on the resistor values.

Simulation / Execution

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Observations

1. Integrator
Integrator output for Square wave input

Figure 1(a): Integrator output for Square wave input — output is triangular, consistent with the integration property.

Integrator output for Sine wave input

Figure 2(b): Integrator output for Sine wave input — output is a cosine-like waveform shifted by 90°.

  • For a square wave input, the output is triangular, consistent with the integration property.
  • For a sinusoidal input, the output is a cosine-like waveform shifted by 90°.
  • At low frequencies, integration is more precise and the expected waveforms are achieved. At higher frequencies, distortion occurs because the capacitor reactance decreases, weakening the ideal integration effect.
2. Differentiator
Differentiator output for Square wave input

Figure 3(a): Differentiator output for Square wave input — output shows sharp spikes at transitions.

Differentiator output for Sine wave input

Figure 4(b): Differentiator output for Sine wave input — output is a cosine-like waveform shifted by 90°.

  • For a square wave input, the output ideally produces sharp spikes at signal transitions.
  • For a sinusoidal input, the output is a cosine-like waveform shifted by 90°.
  • In practice, the output differs from the ideal case due to non-ideal op-amp behavior. The finite open-loop gain decreasing with frequency, and the need for stabilization resistors, result in deviations especially at higher frequencies.
3. Inverting Amplifier
Inverting amplifier output for Square wave input at 1 kHz

Figure (a): Inverting amplifier output for Square wave input at 1 kHz — clean inverted square wave output observed.

Inverting amplifier output for Square wave input at 10 kHz

Figure (b): Inverting amplifier output for Square wave input at 10 kHz — output takes triangular shape due to op-amp slew rate limitation.

Inverting amplifier output for Sine wave input

Figure (c): Inverting amplifier output for Sine wave input — clean inverted sine wave with no significant distortion.

  • At 1 kHz, the square wave input produces a clean inverted square wave output, as expected.
  • At 10 kHz, the output takes the shape of an inverted triangle due to the op-amp slew rate limitation, where the output voltage cannot change fast enough to track the sharp edges.
  • For the sine input, the output is a clean inverted sine wave since the frequency is within the op-amp bandwidth.
4. Instrumentation Amplifier
Output showing amplified signal

Figure: Output showing amplified signal — Vout = 28.8V Vpp.

The power supply is set to ±15V. With an input voltage of 2V Vpp, the output voltage is measured as 28.8V Vpp, resulting in a measured gain of 14.4.

Calculations

Integrator output expression:
vo(t)=−1RC∫vin(t) dtv_o(t) = -\frac{1}{RC} \int v_{in}(t)\, dt
Differentiator output expression:
vo(t)=−RCdvin(t)dtv_o(t) = -RC \frac{dv_{in}(t)}{dt}
Inverting Amplifier gain:
Av=−RfR1A_v = -\frac{R_f}{R_1}
Instrumentation Amplifier measured gain:
Avmeasured=VoutVin=28.8 Vpp2 Vpp=14.4A_{v_{\text{measured}}} = \frac{V_{out}}{V_{in}} = \frac{28.8\,\text{V}_{pp}}{2\,\text{V}_{pp}} = 14.4

Results & Analysis

CircuitInputExpected OutputObserved Output
IntegratorSquare waveTriangular waveTriangular wave (confirmed)
IntegratorSine waveCosine wave (90° shift)Cosine-like wave (confirmed)
DifferentiatorSquare waveSpike pulses at transitionsSpike-like output (confirmed)
DifferentiatorSine waveCosine wave (90° ahead)Cosine-like wave (confirmed)
Inverting AmplifierSquare wave @ 1 kHzInverted square waveClean inverted square wave
Inverting AmplifierSquare wave @ 10 kHzInverted square waveTriangular (slew rate limited)
Inverting AmplifierSine waveInverted sine waveClean inverted sine wave
Instrumentation Amplifier2V VppAmplified output28.8V Vpp (gain = 14.4)
  • All four op-amp circuits functioned as expected with minor practical deviations.
  • The integrator and differentiator outputs matched theoretical waveform transformations at low frequencies.
  • The inverting amplifier demonstrated slew rate limitation at 10 kHz, producing a triangular output instead of a square wave.
  • The instrumentation amplifier achieved a measured gain of 14.4 based on the resistor configuration.

Conclusion

In this experiment, we successfully completed all four parts and confirmed the operation of different amplifier circuits. The results indicated that the practical output closely matched theoretical expectations, with minor deviations arising from component tolerances and non-ideal op-amp behavior such as slew rate limitations. The experiment effectively demonstrated how different feedback configurations with an op-amp lead to distinct signal processing applications such as integration, differentiation, and reliable voltage amplification.

Post-Lab / Viva Voce

  1. Q: In a practical op-amp integrator, why is a resistor connected in parallel with the feedback capacitor, and what effect does it have on the ideal integration behaviour?

    A: Without the parallel resistor, any small DC offset at the op-amp input gets integrated continuously, causing the output to ramp up until it saturates — this is called integrator wind-up. The parallel resistor Rf provides a DC feedback path that clamps the output and stabilises the circuit. However, it introduces a low-frequency pole at f = 1/(2πRfC), so the circuit behaves as a true integrator only for frequencies well above this pole. Below the pole frequency, it behaves like an inverting amplifier with gain −Rf/R1.
  2. Q: For a triangular wave input to a differentiator, what is the ideal output, and why might the practical output look different?

    A: A triangular wave has a constant positive slope for the first half-period and a constant negative slope for the second half-period. Since the differentiator output is proportional to dVin/dt, the ideal output is a square wave — constant positive voltage during the rising ramp and constant negative voltage during the falling ramp, with instantaneous transitions. In practice, the transitions are not instantaneous because the op-amp has a finite slew rate and the practical differentiator includes a small input resistor (in series with the capacitor) which limits high-frequency gain and smooths the edges.
  3. Q: Why does an inverting amplifier present a lower input impedance compared to a non-inverting amplifier built with the same resistor values?

    A: In the inverting amplifier, the input signal is applied to the inverting terminal through R1, and the inverting terminal is held at virtual ground by the feedback. Therefore, the input impedance seen by the source is simply R1 — typically a few kilohms. In contrast, the non-inverting amplifier applies the input directly to the non-inverting terminal, which sees the full input impedance of the op-amp itself (typically in the megaohm to gigaohm range). This makes the inverting topology less suitable when the signal source has high output impedance.
  4. Q: What is virtual ground in an inverting op-amp configuration, and under what conditions does it break down?

    A: Virtual ground is the condition where the inverting input terminal is maintained at 0 V (same as the non-inverting terminal which is grounded), even though it is not physically connected to ground. It arises because the op-amp has very high open-loop gain: any tiny voltage difference between the terminals produces a large output, which through the feedback resistor drives the inverting terminal back to near zero. Virtual ground breaks down when: (1) the output saturates (hits the supply rail), so the feedback can no longer correct the input; (2) the input frequency exceeds the op-amp's gain-bandwidth product, so the open-loop gain is no longer large enough to maintain the condition.
  5. Q: In the instrumentation amplifier, why is the gain controlled by a single external resistor Rg rather than adjusting multiple resistors as in a simple difference amplifier?

    A: In a simple difference amplifier, gain is set by the ratio of resistors across both input and feedback paths. Changing the gain requires changing multiple resistors simultaneously and precisely — any mismatch degrades common-mode rejection ratio (CMRR). In the three-op-amp instrumentation amplifier, the two input buffer op-amps share a single resistor Rg between their inverting terminals. The gain of the input stage depends solely on the ratio 2R/Rg, so adjusting just one resistor changes gain symmetrically for both input channels, preserving CMRR. This makes gain adjustment simple, precise, and non-interactive.
  6. Q: If the output of the differentiator appears noisy even with a clean sinusoidal input, what is the cause and how is it mitigated in a practical design?

    A: A pure differentiator has a transfer function H(s) = −RCs, meaning its gain increases linearly with frequency. High-frequency noise present in any real signal is therefore amplified far more than the signal itself, producing a noisy output. This is mitigated by adding a small resistor Rin in series with the input capacitor, which limits the high-frequency gain to −Rf/Rin (a finite ceiling). This turns the circuit into a band-limited differentiator that differentiates accurately up to a frequency of f = 1/(2πRinC) and attenuates noise above it.

References & Resources (Not Applicable)

This section is not required for this experiment.