fundamentals of electric circuits -...

234 Chapter 6 Capacitors and Inductors

AVW-

(b)Figure 6.35Replacing the feedback resistor in the inverting amplifier in (a) produces an integrator in ibi.

come in discrete form and tend to be more bulky and expensive. 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 energy makes them useful as temporary voltage 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 any abrupt change in voltage, 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 frequency sensitive. 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. For now, consider three applications involving capacitors and op amps: integrator, differentiator, and analog computer.

6.6.1 IntesratorImportant op amp circuits that use energy-storage elements include integrators and differentiators. These op amp circuits often involve resistors and capacitors; inductors (coils) tend to be more bulk) 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 o p am p circuit w hose ou tpu t is proportional to trie integral of the input signal.

If the feedback resistor Rf in the familiar inverting amplifier of Fig. 6.35(a) is replaced by a capacitor, we obtain an ideal integrator, 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. 6.35ibi.

i/t = if <6.321

But

v, dv„

Substituting these in Eq. (6.32). we obtain

v, dv„= ~ C — (6.33a)

R di

J i', di 16.33b In(

6.6 Applications 235

Integrating both sides gives

v j t ) v„(0) = ------RC

Vj(t) dt (6.34)

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

(6.35)

which shows that the circuit in Fig. 6.35(b) provides an output voltage 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 taken that the op amp operates within the linear range so that it does not saturate.

Example 6.132 MF

If y, = 10 cos 21 mV and v 2 = 0.51 mV. find v„ in the op amp circuit in Fig. 6.36. Assume that the voltage across the capacitor is initially zero.

Solution:This is a summing integrator, and

v, dt

1

R^C

3 X 106 X 2 X u r 6

__________ 1________100 X 103 X 2 X 10"

u-. dt

10 cos 21 dt

0.51 dt

1 10= -------- sin It

6 2 0.2 21 0.5/-

~>f---------------= -0 .833 sin 2/ - 1.25/- mV

3 Mt2

ioo kn

Figure 6.36For Example 6.13.

-O

The integrator in Fig. 6.35(b) has R = 100 kli. C = 20/uF. Determine Practice Problem 6.1 3 the output voltage when a dc voltage of 10 mV is applied at / = 0.Assume that the op amp is initially nulled.

Answer: -5 /m V .

6.6.2 Differentiator

A differentiator is an o p am p circuit w hose ou tput 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 differentiator, shown in Fig. 6.37. Applying KCL at node a.

in — >c (6.36)


But

Figure 6.37An op amp differentiator.

* s

r—---\ MA------

h " a>i

/ +vc

VoR

ic = Cdv,dt

Substituting these in Eq. (6.36) yields

(6.371

showing that the output is the derivative of the input. Differentiator circuits are electronically unstable because any 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 6.145 ki2

(a)

(b)


Sketch the output voltage for the circuit in Fig. 6.38(a), given the input voltage in Fig. 6.38(b). Take va = 0 at t = 0.

Solution:This is a differentiator with

RC = 5 x 103 x 0.2 x I0 "6 = 10~3 s

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

f 2000/ 0 < K 2 ms

■ " ' I 8 - 2000/ 2 < / < 4 ms

This is repeated for 4 < / < 8 ms. Using Eq. (6.37), the output is obtained as

dv,vn = - R C —1

dt- 2 V 0 < / < 2 ms

2 V 2 < / < 4 ms

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

v„ (VH

8 /(ms)

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

Practice Problem 6.14 The differentiator in Fig. 6.37 has R = 100 k!2 and C = 0 .1 /iF. Gi\enthat v, = 3/ V. determine the output v„.

Answer: -30 mV

6.6 Applications 237

6.6.3 Analog ComputerOp amps were initially developed for electronic analog computers. Analog computers can be programmed to solve mathematical models of mechanical or electrical systems. These models are usually expressed in terms of differential equations.

To solve simple differential equations using the analog computer requires cascading three types of op amp circuits: integrator 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

d 2x dx „a —y + b ----- 1- cx = f{t), t > 0 (6.38)

d t ' dt

where a, b, and c are constants, and /( /) is an arbitrary forcing function. The solution is obtained by first solving the highest-order derivative term. Solving for d :x / d t 2 yields

d ' x f(t) b dx c—T = '-------------T------- -V (6.39)d t ' a a dt a

To obtain dx/dt , the d 2x/dt~ term is integrated 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 integrators. A plotter or oscilloscope may be used to view the output x, or dx/dt, or d ' x /d t ' . 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 values 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 have been superseded by modem digital computers. However, we still study analog computers for two reasons. First, the availability of integrated op amps has made it possible 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:

+ 2— + v„ = 10 sin At, t > 0 dt dt

subject to uo(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

Example 6.15


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. A ttem pt. 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

drv„ dv„— f = 10sin4r - 2— - v„ (6.15.1)d r dt

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

dv„dt

-1 0 sin 4/ + 2 - ^ + v„ )dt + v'o(0) (6.15.2)

where v ' jO) = 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

RCv„ dt

o

Other terms in the summing integrator of Eq. (6.15.2» are implemented accordingly. The initial condition dv ,,(Oi dr = 1 is implemented by connecting a 1-V batten, with a switch across the capacitor as shown in Fig. 6.40(a).

The next step is to obtain v„ by integrating dv dt and inverting the result.

dv„dt

dt + i'(0) <6.15.31

This is implemented with the circuit in Fig. 6.40<bi with the batter} giving the initial condition of —4 V. We now combine the two circuits in Fig. 6.40(a) and ibi to obtain the complete circuit shown in Fig. 6.40(c). When the input signal 10sin4f is applied, we open the switches at t = 0 to obtain the output waveform v ^ which mav be viewed on an oscilloscope.

66 Applications 239


5. Evaluate. The answer looks correct, but is it? If an actual solution for v„ 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. v,„ 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 — va. The following represents the action of the middle circuit.

' ' dv., J \dt + L’„(0) =~v„ =

dt= ~(l\,(t) - L’„(0) + u„(0))

+ uo(0)0

where u„(0) = —4 V is the initial voltage across the capacitor. We check the circuit on the left the same way.

dv„ ( I"' d~v„ \ ( dv„ \


Now all we need to verify is that the input to the first op amp is- d 2va/ d t 2.

Looking at the input we see that it is equal to

1/10-6 dv„ dv„—10 sin(4/) + v„ + — = — 10 sin(4r) + v„ + 2 —

0.5 M il dt dt

which does produce - d2v„/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:

d- ^ L + 3— + 2v„ = 4 cos 10/, t > 0 dt dt

subject to y„(0) = 2. u„(0) = 0.

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

Figure 6.41For Practice Prob. 6.15.

6.7 Summary1 The current through a capacitor is directly proportional to the time

rate ot change ot the voltage across it.

The current through a capacitor is zero unless the voltage is changing. Thus, a capacitor acts like an open circuit to a dc source

Review Questions 241

2. The voltage across a capacitor is directly proportional to the time integral of the current through it.

I f ' I f 'v = c I ' dt = c ' dt + v(,o)Io

The voltage across a capacitor cannot change instantly.3. Capacitors in series and in parallel are combined in the same way

as conductances.4. The voltage across an inductor is directly proportional to the time

rate of change of the current through it.

div = L—

dt

The voltage across the inductor is zero unless the current is changing. Thus, an inductor acts like a short circuit to a dc source.

5. The current through an inductor is directly proportional to the time integral of the voltage across it.

I f ' J I f ' i = - v dt = — | v d t + i(t0)

— 3C

The current through an inductor cannot change instantly.6. Inductors in series and in parallel are combined in the same way

resistors in series and in parallel are combined.7. At any given time t. the energy stored in a capacitor is \ Cv~, while

the energy stored in an inductor is \Lc.8. Three application circuits, the integrator, the differentiator, and the

analog computer, can be realized using resistors, capacitors, and op amps.

Review Questions

6.1 What charge is on a 5-F capacitor when it is connected across a 120-V source?

(a) 600 C (b) 300 C(c) 24 C (d) 12 C

6.2 Capacitance is measured in:

(a) coulombs (b) joules(c) henrys (d) farads

6.3 When the total charge in a capacitor is doubled, the energy stored:

(a) remains the same (b) is halved(c) is doubled (d) is quadrupled

6.4 Can the voltage w aveform in Fig. 6.42 be associated with a real capacitor ?

(a) Yes (b) No

Figure 6.42For Review Question 6.4.

6.5 The total capacitance of tw o 40-mF series-connected capacitors in parallel w ith a 4-mF capacitor is:

(a) 3.8 mF |b) 5 mF <c)24mF(d)-UmF (e)84mF


6.6 In Fig. 6.43, if i = cos 4t and v = sin At, the element is:

(a) a resistor (b) a capacitor (c) an inductor

“ I '

■ ©

JElement


6.7 A 5-H inductor changes its current by 3 A in 0.2 s. The voltage produced at the terminals of the inductor is:

(a) 75 V

(c) 3 V

(b) 8.888 V

(d) 1.2 V

6.8 If the current through a 10-mH inductor increases from zero to 2 A, how much energy is stored in the inductor?

(a) 40 mJ

(c) 10 mJ

(b) 20 mJ

(d) 5 mJ

6.9 Inductors in parallel can be combined just like resistors in parallel.

(a) Tnie (b) False

6.10 For the circuit in Fig. 6.44. the voltage divider formula is:

(c) vt =L-,

Li + L-)

- v , (b) v,

(d) v,

L, + L^

L2L,

v,

L, ’

-nrw -

"2 3 L->


Answers: 6.1a, 6.2d, 6.3d, 6.4b, 6.5c, 6.6b. 6.7a. 6.Hb, 6.9a. 6.10d.

I ____________________________________________Problems

S e c t io n 6 .2 C a p a c ito rs

6.1 If the voltage across a 5-F capacitor is 2te~y' V. find the current and the power.

6.2 A 20-/xF capacitor has energy w(t) = 10 cos2 377/ J. Determine the current through the capacitor.

6 .3 Design a problem to help other students better e € d understand how capacitors work.

6.4 A current of 6 sin At A flows through a 2-F capacitor. Find the voltage v(t) across the capacitor given that u(0) = 1 V.

6.5 The voltage across a A-^lF capacitor is show n in Fig. 6.45. Find the current waveform.

For Prob. 6.5.

6.6 The voltage waveform in Fig. 6.46 is applied across a 30-/J.F capacitor. Draw the current waveform through it.

Figure 6.46For Prob. 6.6.

6.7 At t = 0. the voltage across a 50-mF capacitor is 10 V Calculate the voltage across the capacitor for t > 0 when current At mA flows through it.

6 .8 A 4-mF capacitor has the terminal voltage

50 V. t < 0A e ' 1"0' + Be "*" V. t > 0

It the capacitor has an initial current o f 2 A. find

la) the constants .A and B. lb) the energy stored in the capacitor at i = 0.

(O the capacitor current for t > 0.

Problems 243

6.9 The current through a 0.5-F capacitor is 6( 1 — e ') A. Determine the voltage and power at t = 2 s. Assume i>(0) = 0.

6.10 The voltage across a 2-mF capacitor is shown inFig. 6.47. Determine the current through the capacitor.

6.15 Two capacitors (20 ^iF and 30 /xF) are connected to a 100-V source. Find the energy stored in each capacitor if they are connected in:

(a) parallel (b) series

6.16 The equivalent capacitance at terminals a-b in the circuit of Fig. 6.50 is 30 /nF. Calculate the value of C.


b o

80 /nF

6.11 A 4-mF capacitor has the current waveform shown in Figure 6.50 Fig. 6.48. Assuming that i>(0) = 10 V. sketch the For Prob. 6.16. voltage waveform v(t).

i(i) (mA)

15 10 5

0 -5

-10


6.12 A voltage of 6 f

6.17 Determine the equivalent capacitance for each of the circuits of Fig. 6.51.

12 F

8 / (s)

(a)

6 F

V appears across a parallel

II

5 F = = 4 F = = 2 F

combination o f a 100-mF capacitor and a 12-11 resistor. Calculate the power absorbed by the parallel combination.

6.13 Find the voltage across the capacitors in the circuit of Fig. 6.49 under dc conditions.

(b) 3 F 6 F

io a 50 Si

IIII II

IIII

II4 F = F 3 F

------A V A -— j-----A W --------

+ | 20 Li +30 Si I c , J = >'l th z

( t ) 60 V

(c)


6.18 Find Ceq in the circuit o f Fig. 6.52 if all capacitors are 4 /nF.


Section 6.3 Series and Parallel Capacitors

6.14 Series-connected 20-pF and 60-pF capacitors are placed in parallel w ith series-connected 30-pF and 70-pF capacitors. Determine the equivalent capacitance.

Figure 6 .5 2For Prob. 6.1 S.

fundamentals of electric circuits -...

Documents