Lab 7: Op-Amp Differentiator and Integrator
Using op-amp circuits to differentiate and integrate signals is common in electronic circuits. The two circuits below show "ideal" op-amp integrators and differentiators. However, these "ideal" circuits are never used in practice for reasons discussed below. In this lab, we will build real differentiators and integrators using op-amps, resistors and capacitors.
Section 7-3 of Thomas and Rosa shows that, for the ideal differentiator,
and that for the ideal integrator,
Use sinusoidal steady state analysis to show this result; i.e., that vo(jw) = -jwRCvs(jw) for the ideal differentiator and vo(jw) = -vs(jw)/jwRC for the ideal integrator. A. Differentiator The ideal differentiator is inherently unstable. Every electronic system has some noise at high frequencies. An ideal differentiator would amplify this small noise. (e.g., if is differentiated, the output would be . Even if A = 1microV when f = 10 MHz, vout would have an amplitude of 63V!) To circumvent this problem, it is traditional to include a series resistor at the input, and a parallel capacitor across the feedback resistor, converting the differentiator to an integrator at high frequencies. Plot gain vs. frequency for the circuit shown in figure 1. For what frequencies does the circuit act as a differentiator? What is the response of the circuit to high frequencies? Construct this real op-amp differentiator. Drive it with a 1 kHz sine wave, a 1 kHz square wave, and a 1 kHz triangle wave. For each input signal, plot the input and output waveforms. Are the output waveforms and their amplitudes what you would expect? (i.e., does the circuit differentiate the signal? Is the amplitude correct?) B. Integrator Op-amps allow you to make nearly perfect integrators. Figure 2 shows a practical integrator. Plot gain vs. frequency for this circuit. For what frequencies does this circuit act as an integrator? How does the circuit respond to low frequencies? (The circuit incorporates a large resistor in parallel with the feedback capacitor. This is necessary because real op-amps have a small current flowing at their input terminals called the "bias current". This current is typically a few nanoamps, and is negligible in circuits where the current is in the microamp to milliamp range. However, if you integrate a nanoamp current, it won't take long until it charges up the 0.1microF capacitor! The resistor provides a path for the bias current to flow through. The effect of the resistor on the response is negligible at all but the lowest frequencies.) Build the circuit. Drive the input with a 500 Hz square wave of 2 V p-p amplitude. Plot the input and output waveforms. Has the input been integrated? Does the amplitude of the output agree with what it should be from the circuit values? (i.e.the calculated output waveform.) Repeat, using a sine wave and a triangle wave. © Copyright 1996 New Mexico Institute of Mining and Technology
A. Differentiator
The ideal differentiator is inherently unstable. Every electronic system has some noise at high frequencies. An ideal differentiator would amplify this small noise. (e.g., if is differentiated, the output would be . Even if A = 1microV when f = 10 MHz, vout would have an amplitude of 63V!) To circumvent this problem, it is traditional to include a series resistor at the input, and a parallel capacitor across the feedback resistor, converting the differentiator to an integrator at high frequencies.
Plot gain vs. frequency for the circuit shown in figure 1. For what frequencies does the circuit act as a differentiator? What is the response of the circuit to high frequencies?
Construct this real op-amp differentiator. Drive it with a 1 kHz sine wave, a 1 kHz square wave, and a 1 kHz triangle wave. For each input signal, plot the input and output waveforms. Are the output waveforms and their amplitudes what you would expect? (i.e., does the circuit differentiate the signal? Is the amplitude correct?)
B. Integrator
Op-amps allow you to make nearly perfect integrators. Figure 2 shows a practical integrator. Plot gain vs. frequency for this circuit. For what frequencies does this circuit act as an integrator? How does the circuit respond to low frequencies? (The circuit incorporates a large resistor in parallel with the feedback capacitor. This is necessary because real op-amps have a small current flowing at their input terminals called the "bias current". This current is typically a few nanoamps, and is negligible in circuits where the current is in the microamp to milliamp range. However, if you integrate a nanoamp current, it won't take long until it charges up the 0.1microF capacitor! The resistor provides a path for the bias current to flow through. The effect of the resistor on the response is negligible at all but the lowest frequencies.)
Build the circuit. Drive the input with a 500 Hz square wave of 2 V p-p amplitude. Plot the input and output waveforms. Has the input been integrated? Does the amplitude of the output agree with what it should be from the circuit values? (i.e.the calculated output waveform.) Repeat, using a sine wave and a triangle wave.
