Electronics · Ch 5 — Operational Amplifiers

Op-Amp as Differentiator and Integrator

5.6

Op-Amp as Differentiator and Integrator

This section covers two non-linear applications of the op-amp: the differentiator, whose output follows the derivative of the input, and the integrator, whose output follows the integral of the input.

Op-amp as a differentiator

A differentiator gives an output proportional to the rate of change (derivative) of its input voltage. In figure 5.6.1 the input $V_i$ reaches the inverting terminal (node A) through a capacitor $C$ ; the non-inverting terminal is grounded and $R_f$ is the feedback resistor.

For the ideal op-amp $V_B = V_A = 0$ and $i_b = 0$ , so KCL at node A gives $i_i = i_f$ . Writing the capacitor's charging current as the rate of change of charge, $\dfrac{dQ}{dt} = \dfrac{V_A - V_O}{R_f}$ , substituting $Q = C(V_i - V_A)$ , and putting $V_A = 0$ :

C\dfrac{d(V_i)}{dt} = -\dfrac{V_O}{R_f} \quad\Rightarrow\quad V_O = -R_f C\,\dfrac{d(V_i)}{dt}

So the output is proportional to the derivative of the input. For proper working the time constant must satisfy $RC < 10T$ , where $T$ is the period of the input waveform. Differentiators are used in wave-shaping circuits, as high-pass filters, and similar applications. Figure 5.6.2 shows the response: a sine input gives an inverted cosine output, a square input produces sharp spikes at its edges, and a triangular input yields a square wave.

Op-amp as an integrator

An integrator gives an output proportional to the integral of its input. In figure 5.6.3 the input $V_i$ reaches the inverting terminal through a resistor $R_i$ , the non-inverting terminal is grounded, and a capacitor $C$ forms the feedback element. …

Figure 1Circuit of an op-amp differentiator: input Vi fed to the inverting terminal through a capacitor C with a feedback resistor Rf.

Fig. 1 — Circuit of an op-amp differentiator: input Vi fed to the inverting terminal through a capacitor C with a feedback resistor Rf.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

Figure 5.6.1 shows the op-amp differentiator. The input Vi reaches the inverting terminal (node A) through a series capacitor C (input current i_i); the feedback resistor Rf carries i_f from the output back to node A, and the non-inverting terminal (node B) is grounded. Supply pins +Vcc and -Vee, output Vo. The capacitor at the input and the resistor in feedb …

Formula 2Output voltage of the op-amp differentiator

$V_O = -R_f C\,\dfrac{d(V_i)}{dt}$ — the output is proportional to the derivative of the input. Proper functioning requires the time constant $RC < 10T$ , where $T$ is th …

Figure 3Input and output waveforms of a differentiator: sine gives inverted cosine, square gives spikes, triangular gives a square wave.

Fig. 3 — Input and output waveforms of a differentiator: sine gives inverted cosine, square gives spikes, triangular gives a square wave.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

Figure 5.6.2 shows three input-output waveform pairs for the differentiator. A sine input produces an inverted cosine output; a square input produces a train of alternating positive and negative spikes at its edges; a triangular input produces a square wave. Each pair is plotted as v_i …

Figure 4Circuit of an op-amp integrator: input Vi fed to the inverting terminal through resistor Ri with a feedback capacitor C.

Fig. 4 — Circuit of an op-amp integrator: input Vi fed to the inverting terminal through resistor Ri with a feedback capacitor C.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

Figure 5.6.3 shows the op-amp integrator. The input Vi reaches the inverting terminal (node A) through a series resistor Ri (input current i_i); a feedback capacitor C carries i_f from the output back to node A, and the non-inverting terminal (node B) is grounded. Supply pins +Vcc and -Vee, output Vo. The resistor at the input and the capacitor …

Formula 5Output voltage of the op-amp integrator

$V_O = -\dfrac{1}{R_i C}\displaystyle\int V_i\,dt$ — the output is proportional to the integral of the input. Proper functioning requires the time constant $RC > 10T$ , where $T$ is …

Figure 6Input and output waveforms of an integrator: sine gives a cosine, square gives a triangular wave.

Fig. 6 — Input and output waveforms of an integrator: sine gives a cosine, square gives a triangular wave.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

Figure 5.6.4 shows two input-output waveform pairs for the integrator. A sine input produces a cosine output; a square input produces a triangular wave. Each pair is plotted as v_in (top) against …

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