Integrators and differentiators are circuits that simulate the mathematical operations of integration and differentiation. It is not necessary for you to understand these operations now to be able to learn how integrators and differentiators work.
The Integrator
An integrator computes the total area underneath the curve of a given waveform. This is basically a summing process. Figure 25.1 shows a basic circuit of an integrator. As you can see this circuit is an inverting amplifier with a feedback branch through a capacitor C. In terms of the mathematical operation of integration^{1}, if we consider the integrator in terms of its inputoutput behavior, when an input signal, v_{i}(t), is applied to the input terminal the device will generate at the output terminal the integral respect to time of the input waveform multiplied by a constant. In equation form
Figure 25.1: A basic integrator using an opamp

As you can see the constant that multiplies the integral is 1/RC.
Figure 25.2 shows the output produced when several input functions are applied at the input terminal of an integrator. As you can see a constant voltage applied to the input of an integrator generates a voltage with a constant negative slope (a ramp), a square wave produces a triangular wave, and a sine functions generates a negative cosine function.
Figure 25.2: basic integrator responses

In this experiment we will concentrate on input functions which are constant during a fixed period of time (the step function and the square wave). One of these functions – the step function – is shown in Fig. 25.1 together with the output waveform generated if the step function is applied to the input of the integrator shown in the figure. At time t = 0 a constant voltage V is applied to the input of the integrator. At the output terminal the integrator produces a negative going ramp as is shown in part (b) of the figure. This ramp has a slope equal to 1/RC and a rate of change given by
  (25.1)

where is the change of the output voltage, and is the change in the time to accomplish .
The following example illustrates this concept.Example:
In Figure 25.1 the opamp saturation voltages are ±12V, the resistance isR = 10kΩ, and C = 0.01mF. If at t = 0 we apply a voltage of V = 7.5V, determine: (a) The value of the output voltage at t = 100µs, and (b) The time to reach saturation.
Solution:
(a) First, let’s determine the rate of change of the output voltage using Eq. 25.1
The change in the output voltage is given by
If we assume that at t = 0 the output voltage is V_{o} = 0V, then after 100 µs the output voltage is
  (25.3)

(b) The time to reach saturation can be found using Eq. 25.2.
Therefore, if the input voltage is kept at 7.5 V for 160 µs or more the output voltage remains at its negative saturation value (12 V) until the input is changed.
^{1} If you do not understand this terminology yet, do not worry at this moment.
Practical Integrator
The integrator of Figure 25.1 is the basic circuit. This circuit has at least the following shortcomings:
1. The input bias current and the offset voltage^{2} at the input of the integrator will be integrated just like any other input signal. The output voltage, in this condition, will not reflect the true purpose of the circuit, which is to integrate a desired input signal.
2. For low frequency signals this circuit is very unstable. As you can see, if the input signal has a low frequency the capacitor looks like an opencircuit that disconnects the feedback path from the circuit. In this situation the circuit behaves like an opamp in openloop. Thus, the output voltage will be in saturation for any input signal.
The solution to these shortcomings is to add two new elements to the basic circuit: a resistor in the feedback path and a resistor in the noninverting input. These changes are shown in Figure 25.3.
Figure 25.3: A practical integrator

The reasons for these changes are explained as follows:
1. A resistor R_{f} is added in the feedback path to avoid instabilities at low frequencies (item 2 above). Also, if properly selected, this resistor will help discharge the integrating capacitor when offset voltage is present at the input (item 1 above). A very large feedback capacitor is used to accomplish the discharge of the offset voltage. A typical design ruleofthumb is to choose
  (25.4)

where R_{1} is the input resistance.
2. To minimize the output offset due to the input bias currents a resistor, R_{2} is added to the noninverting input, as is shown in Fig 25.3. A ruleofthumb for choosing this resistor is to have
  (25.5)


^{2} See Experiment No. 22.
The Differentiator
A differentiator is a circuit that calculates the instantaneous slope of the line at every point on a waveform. The output of the differentiator is always proportional to the rate of change of the input voltage. Figure 25.4 shows a basic circuit for a differentiator. This circuit is an inverting amplifier but instead of a resistor a capacitor is used as the input element of the system. When a signal, v_{i}(t), is applied to the input terminal the output will be the derivative^{3} with respect to time of the input signal multiplied by a constant factor. In equation form
Figure 25.4: A basic differentiator using an opamp

As you can see the constant that multiplies the derivative is –RC.
Figure 25.5 shows the output produced when several input functions are applied to the input terminal of a differentiator. Notice that the functions are exactly opposite to the integrator actions shown in Fig. 25.3. When a triangular wave is applied to the input the output will be a negative square wave; if the input is a triangular wave the output produces a negative triangular signal; and when the input is a sine wave the output is a negative cosine function.
In this experiment we will concentrate on ramp input functions. If a ramp of certain slope is applied to the input terminal of the differentiator, a constant voltage is produced at the output^{4} for as long as the input is unchanged.
Figure 25.5: basic differentiator responses

The value of the voltage at the output is given by the following equation:
  (25.8)

or,
  (25.9)

where slope is the slope of the ramp , and R and C are the circuit elements. The above equation is applied every time there is a constant slope in the input signal.
To illustrate this concept we present in part (b) of Fig.25.4 a triangular input waveform being applied to the differentiator. The corresponding output voltage is as indicated. For the first ramp (from t = 0 to t = t_{1}) the slope of the input voltage is V/t_{1}, where V is the input voltage reached at t = t_{1}. For this time interval the output voltage is (V / t_{1}) RC as indicated. For the second ramp (from t = t_{1} to t = 2t_{1}) the output voltage is given by (V / t_{1})RC. Thus, a triangular wave input produces a square wave output.
The following example shows how to use the formulas.
Example:
Sketch the output waveform of the following differentiator when the triangular wave shown is applied to the input.
Figure 25.6: Example 2

Solution:
Because the input is a triangular wave, the output voltage is a square wave as shown in the figure. The maximum and minimum values are given by Eq. 25.9
The sketch of the output is shown in Fig. 25.6

^{3} Again the student should not be concerned about this high mathematics term. I include it here just for completeness of my presentation.
Practical Differentiator
The differentiator of Fig. 25.4 is an ideal circuit. It is not, however, stable and it is very susceptible to high frequency noise. To improve the circuit and make it suitable for practical applications, a resistor is added in series with the input capacitor. This resistor acts to reduce the highfrequency gain and improves stability of the circuit.
A more general circuit is shown below (Fig. 25.7) where a feedback capacitor, C_{f}, is connected in parallel with the feedback resistor, and there is a resistor in the noninverting input.
Figure 25.7: practical circuit

In this experiment, however, we will use the circuit shown for our calculations.
By adding the capacitor in the input terminal the differentiator behaves like a lowpass filter with a critical frequency given by
  (25.10)

The output voltage of the practical differentiator is given by
  (25.11)

In other words, Eq. 25.11 tells us that if the frequency of the input signal (f_{i}) is smaller than the critical frequency of the circuit given by Eq. 25.10, the circuit behaves like a normal differentiator, whereas if the frequency of the input signal is bigger than the critical frequency, the circuit approaches an inverting amplifier with a voltage gain of R_{f} / R_{1}.
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