6.6 Applications

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6.6 Applications

Circuit elements such as resistors and capacitors are commercially available in either discrete form or integrated-circuit (IC) form. Unlike capacitors and resistors, inductors with appreciable inductance are dif ficult to produce on IC substrates. Therefore, inductors (coils) usually

Figure 6.33 For Example 6.12.

come in discrete form and tend to be more b ulky and e xpensive. For this reason, inductors are not as versatile as capacitors and resistors, and they are more limited in applications. However, there are several applications in which inductors have no practical substitute. They are routinely used in relays, delays, sensing devices, pick-up heads, telephone circuits, radio and TV receivers, power supplies, electric motors, microphones, and loudspeakers, to mention a few.

Capacitors and inductors possess the following three special properties that make them very useful in electric circuits:

• 1. The capacity to store ener gy makes them useful as temporary v oltage or current sources. Thus, they can be used for generating a large amount of current or voltage for a short period of time.
• 2. Capacitors oppose an y abrupt change in v oltage, while inductors oppose any abrupt change in current. This property makes inductors useful for spark or arc suppression and for converting pulsating dc voltage into relatively smooth dc voltage.
• 3. Capacitors and inductors are frequenc y sensiti ve. This property makes them useful for frequency discrimination.

The first two properties are put to use in dc circuits, while the third one is taken advantage of in ac circuits. We will see how useful these properties are in later chapters. F or now, consider three applications involving capacitors and op amps: integrator, differentiator, and analog computer.

6.6.1 Integrator

Important op amp circuits that use energy-storage elements include integrators and dif ferentiators. These op amp circuits often in volve resistors and capacitors; inductors (coils) tend to be more b ulky and expensive.

The op amp integrator is used in numerous applications, especially in analog computers, to be discussed in Section 6.6.3.

An integrator is an op amp circuit whose output is proportional to the integral of the input signal.

If the feedback resistor Rf in the f amiliar in verting amplifier of Fig. 6.35(a) is replaced by a capacitor , we obtain an ideal inte grator, as shown in Fig. 6.35(b). It is interesting that we can obtain a mathematical representation of integration this way. At node a in Fig. 6.35(b),

$$i_R = i_C \tag{6.32}$$

But

$$i_R = \frac{v_i}{R}, \qquad i_C = -C \frac{dv_o}{dt}$$

Substituting these in Eq. (6.32), we obtain

$$\frac{v_i}{R} = -C \frac{dv_o}{dt}$$

(6.33a)

$$dv_o = -\frac{1}{RC}v_i dt$$

(6.33b)

Replacing the feedback resistor in the inverting amplifier in (a) produces an integrator in (b).

Integrating both sides gives

$$v_o(t) - v_o(0) = -\frac{1}{RC} \int_0^t v_i(\tau) \, d\tau \tag{6.34}$$

To ensure that vo(0) = 0, it is always necessary to discharge the integrator’s capacitor prior to the application of a signal. Assuming vo(0) = 0,

$$v_o = -\frac{1}{RC} \int_0^t v_i(\tau) \, d\tau \tag{6.35}$$

which shows that the circuit in Fig. 6.35(b) pro vides an output v oltage proportional to the integral of the input. In practice, the op amp integrator requires a feedback resistor to reduce dc gain and prevent saturation. Care must be tak en that the op amp operates within the linear range so that it does not saturate.

If v^{1 =} 10 cos 2 ^t mV and v^{2 =} 0.5t mV, find vo in the op amp circuit in Example 6.13 Fig. 6.36. Assume that the voltage across the capacitor is initially zero.

Solution:

This is a summing integrator, and

$$v_o = -\frac{1}{R_1C} \int v_1 dt - \frac{1}{R_2C} \int v_2 dt$$

=

$$-\frac{1}{3 \times 10^6 \times 2 \times 10^{-6}} \int_0^t 10 \cos(2\tau) d\tau$$ $$-\frac{1}{100 \times 10^3 \times 2 \times 10^{-6}} \int_0^t 0.5\tau d\tau$$

=

$$-\frac{1}{6} \frac{10}{2} \sin 2t - \frac{1}{0.2} \frac{0.5t^2}{2} = -0.833 \sin 2t - 1.25t$$

The integrator in Fig. 6.35(b) has ^{R =} 100 kΩ, C = 20 μF. Determine the Practice Problem 6.13 output voltage when a dc voltage of 2.5 mV is applied at t = 0. Assume that the op amp is initially nulled.

Answer: −1.25t m V.

6.6.2 Differentiator

A differentiator is an op amp circuit whose output is proportional to the rate of change of the input signal.

In Fig. 6.35(a), if the input resistor is replaced by a capacitor , the resulting circuit is a dif ferentiator, shown in Fig. 6.37. Applying KCL at node a,

$$i_R = i_C \tag{6.36}$$

2 mV But

$$i_R = -\frac{v_o}{R}, \qquad i_C = C \frac{dv_i}{dt}$$

Substituting these in Eq. (6.36) yields

$$v_o = -RC \frac{dv_i}{dt} \tag{6.37}$$

showing that the output is the deri vative of the input. Differentiator circuits are electronically unstable because an y electrical noise within the circuit is exaggerated by the differentiator. For this reason, the differentiator circuit in Fig. 6.37 is not as useful and popular as the integrator. It is seldom used in practice.

Example 6.14 Sketch the output voltage for the circuit in Fig. 6.38(a), given the input voltage in Fig. 6.38(b). Take v^o = 0 at t = 0.

Solution:

This is a differentiator with

$$RC = 5 \times 10^3 \times 0.2 \times 10^{-6} = 10^{-3} \text{ s}$$

For 0 < t < 4 ms, we can express the input voltage in Fig. 6.38(b) as

^{vi =} { ^2000t 8 − 2000t 0 < t < 2 ms 2 < t < 4 ms

This is repeated for 4 < t < 8 ms. Using Eq. (6.37), the output is ob tained as

$$v_o = -RC \frac{dv_i}{dt} = \begin{cases} -2\text{ V} & 0 < t < 2\text{ ms} \\ 2\text{ V} & 2 < t < 4\text{ ms} \end{cases}$$

Thus, the output is as sketched in Fig. 6.39.

vo (V)

8642 2 0 –2 t (ms)

Figure 6.39 Output of the circuit in Fig. 6.38(a).

Practice Problem 6.14 The differentiator in Fig. 6.37 has ^{R =} 100 kΩ and C = 0.1 μF. Given that vⁱ = 5t V, determine the output vo.

Answer: −50 mV.

Figure 6.37 An op amp differentiator.

For Example 6.14.

6.6.3 Analog Computer

Op amps were initially de veloped for electronic analog computers. Analog computers can be programmed to solv e mathematical models of mechanical or electrical systems. These models are usually expressed in terms of differential equations.

To solv e simple dif ferential equations using the analog computer requires cascading three types of op amp circuits: inte grator circuits, summing amplifiers, and inverting/noninverting amplifiers for negative/ positive scaling. The best way to illustrate how an analog computer solves a differential equation is with an example.

Suppose we desire the solution x(t) of the equation

$$a\frac{d^2x}{dt^2} + b\frac{dx}{dt} + cx = f(t), \qquad t > 0$$

\n(6.38)

where a, b, and c are constants, and f(t) is an arbitrary forcing func tion. The solution is obtained by first solving the highest-order derivative term. Solving for d² x∕dt² yields

$$\frac{d^2x}{dt^2} = \frac{f(t)}{a} - \frac{b}{a}\frac{dx}{dt} - \frac{c}{a}x\tag{6.39}$$

To obtain dx∕dt, the d² x∕dt² term is inte grated and inverted. Finally, to obtain x, the dx∕dt term is integrated and inverted. The forcing function is injected at the proper point. Thus, the analog computer for solving Eq. (6.38) is implemented by connecting the necessary summers, inverters, and inte grators. A plotter or oscilloscope may be used to vie w the output x, or dx∕dt, or d² x∕dt² , depending on where it is connected in the system.

Although the above example is on a second-order differential equation, any differential equation can be simulated by an analog computer comprising integrators, inverters, and inverting summers. But care must be exercised in selecting the v alues of the resistors and capacitors, to ensure that the op amps do not saturate during the solution time interval.

The analog computers with vacuum tubes were built in the 1950s and 1960s. Recently their use has declined. They ha ve been superseded by modern digital computers. Ho wever, we still study analog computers for two reasons. First, the a vailability of integrated op amps has made it pos sible to build analog computers easily and cheaply. Second, understanding analog computers helps with the appreciation of the digital computers.

Design an analog computer circuit to solve the differential equation: Example 6.15

$$\frac{d^2v_o}{dt^2} + 2\frac{dv_o}{dt} + v_o = 10\sin 4t, \qquad t > 0$$

subject to vo(0) = −4, v^o ′ (0) = 1, where the prime refers to the time derivative.

Solution:

1. Define. We have a clearly defined problem and expected solution. I might remind the student that many times the problem is not so well defined and this portion of the problem-solving process could

require much more effort. If this is so, then you should always keep in mind that time spent here will result in much less effort later and most likely save you a lot of frustration in the process.

• 2. Present. Clearly, using the devices developed in Section 6.6.3 will allow us to create the desired analog computer circuit. We will need the integrator circuits (possibly combined with a summing capability) and one or more inverter circuits.
• 3. Alternative. The approach for solving this problem is straightforward. We will need to pick the correct values of resistances and capacitors to allow us to realize the equation we are representing. The final output of the circuit will give the desired result.
• 4. Attempt. There are an infinite number of possibilities for picking the resistors and capacitors, many of which will result in correct solutions. Extreme values of resistors and capacitors will result in incorrect outputs. For example, low values of resistors will overload the electronics. Picking values of resistors that are too large will cause the op amps to stop functioning as ideal devices. The limits can be determined from the characteristics of the real op amp.

We first solve for the second derivative as

$$\frac{d^2v_o}{dt^2} = 10\sin 4t - 2\frac{dv_o}{dt} - v_o$$

(6.15.1)

Solving this requires some mathematical operations, includ ing summing, scaling, and integration. Integrating both sides of Eq. (6.15.1) gives

$$\frac{dv_o}{dt} = -\int_0^t \left(-10\sin(4\tau) + 2\frac{dv_o(\tau)}{d\tau} + v_o(\tau)\right) d\tau + v'_o(0) \quad (6.15.2)$$

where v ^o′ (0) = 1. We implement Eq. (6.15.2) using the summing integrator shown in Fig. 6.40(a). The values of the resistors and capacitors have been chosen so that RC = 1 for the term

$$-\frac{1}{RC}\int_0^t v_o(\tau)\,d\tau$$

Other terms in the summing integrator of Eq. (6.15.2) are implemented accordingly. The initial condition dv ^o(0) ∕dt = 1 is imple mented by connecting a 1-V battery with a switch across the capacitor as shown in Fig. 6.40(a).

The next step is to obtain v ^o by integrating dv o ∕dt and inverting the result,

$$v_o = -\int_0^t \frac{dv_o(\tau)}{d\tau} d\tau + v(0)$$

(6.15.3)

This is implemented with the circuit in Fig. 6.40(b) with the battery giving the initial condition of −4 V. We now combine the two circuits in Fig. 6.40(a) and (b) to obtain the complete circuit shown in Fig. 6.40(c). When the input signal 10 sin 4 t is applied, we open the switches at t = 0 to obtain the output waveform v ^o, which may be viewed on an oscilloscope.

For Example 6.15.

5. Evaluate. The answer looks correct, but is it? If an actual solution for vo is desired, then a good check would be to first find the solution by realizing the circuit in PSpice. This result could then be compared with a solution using the differential solution capability of MATLAB.

Since all we need to do is check the circuit and confirm that it represents the equation, we have an easier technique to use. We just go through the circuit and see if it generates the desired equation.

However, we still have choices to make. We could go through the circuit from left to right but that would involve differentiating the result to obtain the original equation. An easier approach would be to go from right to left. This is the approach we will use to check the answer.

Starting with the output, vo, we see that the right-hand op amp is nothing more than an inverter with a gain of one. This means that the output of the middle circuit is −vo. The following represents the action of the middle circuit.

$$-v_o = -\left(\int_0^t \frac{dv_o}{d\tau} d\tau + v_o(0)\right) = -\left(v_o \Big|_0^t + v_o(0)\right)$$

= -\left(v_o(t) - v_o(0) + v_o(0)\right)

where vo(0) = −4 V is the initial voltage across the capacitor.

We check the circuit on the left the same way.

$$\frac{dv_o}{dt} = -\left(\int_0^t -\frac{d^2v_o}{dt^2}d\tau - v'_o(0)\right) = -\left(-\frac{dv_o}{dt} + v'_o(0) - v'_o(0)\right)$$

Now, all we need to verify is that the input to the first op amp is −d² vo∕dt² .

Looking at the input we see that it is equal to

$$-10\sin(4t) + v_o + \frac{1/10^{-6}}{0.5\,\text{M}\Omega} \frac{dv_o}{dt} = -10\sin(4t) + v_o + 2\frac{dv_o}{dt}$$

which does produce −d² vo∕dt² from the original equation.

6. Satisfactory? The solution we have obtained is satisfactory. We can now present this work as a solution to the problem.

Practice Problem 6.15 Design an analog computer circuit to solve the differential equation:

$$\frac{d^2v_o}{dt^2} + 3\frac{dv_o}{dt} + 2v_o = 4\cos 10t, \qquad t > 0$$

subject to vo(0) = 2, v^o ′ (0) = 0.

Answer: See Fig. 6.41, where RC = 1 s.