7.9 †Applications

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7.9 †Applications

The various devices in which RC and RL circuits find applications include filtering in dc power supplies, smoothing circuits in digital communications, differentiators, integrators, delay circuits, and relay circuits. Some of these applications take advantage of the short or long time constants of the RC or RL circuits. We will consider four simple applications here. The first two are RC circuits, the last two are RL circuits.

7.9.1 Delay Circuits

An RC circuit can be used to pro vide various time delays. Figure 7.73 shows such a circuit. It basically consists of an RC circuit with the ca pacitor connected in parallel with a neon lamp. The voltage source can provide enough voltage to fire the lamp. When the switch is closed, the capacitor voltage increases gradually toward 110 V at a rate determined by the circuit’s time constant, (R1 + R2)C. The lamp will act as an open

An RC delay circuit.

circuit and not emit light until the voltage across it exceeds a particular level, say 70 V. When the voltage level is reached, the lamp fires (goes on), and the capacitor dischar ges through it. Due to the lo w resistance of the lamp when on, the capacitor voltage drops fast and the lamp turns off. The lamp acts again as an open circuit and the capacitor recharges. By adjusting R2, we can introduce either short or long time delays into the circuit and mak e the lamp fire, recharge, and fire repeatedly every time constant τ = (R1 + R2)C, because it takes a time period τ to get the capacitor voltage high enough to fire or low enough to turn off.

The warning blinkers commonly found on road construction sites are one example of the usefulness of such an RC delay circuit.

Example 7.19 Consider the circuit in Fig. 7.73, and assume that R1 = 1.5 MΩ, 0 ^&lt; R2 < 2.5 MΩ. (a) Calculate the extreme limits of the time constant of the circuit. (b) How long does it take for the lamp to glow for the first time after the switch is closed? Let R2 assume its largest value.

Solution:

(a) The smallest value for R2 is 0 Ω, and the corresponding time constant for the circuit is

$$\tau = (R_1 + R_2)C = (1.5 \times 10^6 + 0) \times 0.1 \times 10^{-6} = 0.15 \text{ s}$$

The largest value for R2 is 2.5 MΩ, and the corresponding time constant for the circuit is

$$\tau = (R_1 + R_2)C = (1.5 + 2.5) \times 10^6 \times 0.1 \times 10^{-6} = 0.4 \text{ s}$$

Thus, by proper circuit design, the time constant can be adjusted to introduce a proper time delay in the circuit.

(b) Assuming that the capacitor is initially uncharged, vC(0) = 0, while vC(∞) = 110. But

$$v_C(t) = v_C(\infty) + [v_C(0) - v_C(\infty)]e^{-t/\tau} = 110[1 - e^{-t/\tau}]$$

where τ = 0.4 s, as calculated in part (a). The lamp glows when vC = 70 V. If vC(t) = 70 V at t = t0, then

$$70 = 110[1 - e^{-t_0/\tau}] \qquad \Rightarrow \qquad \frac{7}{11} = 1 - e^{-t_0/\tau}$$

or

$$e^{-t_0/\tau} = \frac{4}{11} \qquad \Rightarrow \qquad e^{t_0/\tau} = \frac{11}{4}$$

Taking the natural logarithm of both sides gives

$$t_0 = \tau \ln \frac{11}{4} = 0.4 \ln 2.75 = 0.4046 \text{ s}$$

A more general formula for finding t0 is

$$t_0 = \tau \ln \frac{-v(\infty)}{v(t_0) - v(\infty)}$$

The lamp will fire repeatedly every t0 seconds if and only if v(t0) < v(∞).

The RC circuit in Fig. 7.74 is designed to operate an alarm which acti - Practice Problem 7.19 vates when the current through it exceeds 90 μA. If 0 ≤ R ≤ 6 kΩ, find the range of the time delay that the variable resistor can create.

Answer: Between 34.47 and 89.26 ms.

7.9.2 Photoflash Unit

An electronic flash unit pro vides a common e xample of an RC circuit. This application exploits the ability of the capacitor to oppose any abrupt change in voltage. Figure 7.75 shows a simplified circuit. It consists essentially of a high-voltage dc supply, a current-limiting large resistor R1, and a capacitor C in parallel with the flashlamp of low resistance R2. When the switch is in position 1, the capacitor charges slowly due to the large time constant ( τ1 = R1C). As shown in Fig. 7.76(a), the capacitor voltage rises gradually from zero to Vs, while its current decreases gradually from I1 = Vs∕R1 to zero. The charging time is approximately five times the time constant,

$$t_{\text{charge}} = 5R_1C \tag{7.65}$$

With the switch in position 2, the capacitor v oltage is dischar ged. The low resistance R2 of the photolamp permits a high discharge current with peak I2 = Vs∕R2 in a short duration, as depicted in Fig. 7.76(b). Discharging takes place in approximately five times the time constant,

$$t_{\text{discharge}} = 5R_2C \tag{7.66}$$

(a) Capacitor voltage showing slow charge and fast discharge, (b) capacitor current showing low charging current I1 = Vs∕R1 and high discharge current I2 = Vs∕R2.

Thus, the simple RC circuit of Fig. 7.75 provides a short-duration, highcurrent pulse. Such a circuit also finds applications in electric spot welding and the radar transmitter tube.

Figure 7.75

Circuit for a flash unit providing slow charge in position 1 and fast discharge in position 2.

Example 7.20 An electronic flashgun has a current-limiting 6-kΩ resistor and 2000-μ^F electrolytic capacitor char ged to 240 V. If the lamp resistance is 12 Ω, find: (a) the peak charging current, (b) the time required for the capaci tor to fully charge, (c) the peak discharging current, (d) the total energy stored in the capacitor, and the average power dissipated by the lamp.

Solution:

(a) The peak charging current is

$$I_1 = \frac{V_s}{R_1} = \frac{240}{6 \times 10^3} = 40 \text{ mA}$$

(b) From Eq. (7.65),

$$t_{\text{charge}} = 5R_1C = 5 \times 6 \times 10^3 \times 2000 \times 10^{-6} = 60 \text{ s} = 1 \text{ minute}$$

(c) The peak discharging current is

$$I_2 = \frac{V_s}{R_2} = \frac{240}{12} = 20 \text{ A}$$

(d) The energy stored is

$$W = \frac{1}{2}CV_s^2 = \frac{1}{2} \times 2000 \times 10^{-6} \times 240^2 = 57.6 \text{ J}$$

The energy stored in the capacitor is dissipated across the lamp during the discharging period. From Eq. (7.66),

$$t_{\text{discharge}} = 5R_2C = 5 \times 12 \times 2000 \times 10^{-6} = 0.12 \text{ s}$$

Thus, the average power dissipated is

$$p = \frac{W}{t_{\text{discharge}}} = \frac{57.6}{0.12} = 480 \text{ watts}$$

The flash unit of a camera has a 2-mF capacitor charged to 40 V. Practice Problem 7.20

• (a) How much charge is on the capacitor?
• (b) What is the energy stored in the capacitor?
• (c) If the flash fires in 0.8 ms, what is the average current through the flashtube?
• (d) How much power is delivered to the flashtube? After a picture has been taken, the capacitor needs to be recharged by a power unit that supplies a maximum of 5 mA. How much time does it take to charge the capacitor?

Answer: (a) 80 mC, (b) 1.6 J, (c) 100 A, (d) 2 kW, 16 s.

7.9.3 Relay Circuits

A magnetically controlled switch is called a relay. A relay is essentially an electromagnetic de vice used to open or close a switch that controls another circuit. Figure 7.77(a) sho ws a typical relay circuit. The coil

circuit is an RL circuit like that in Fig. 7.77(b), where R and L are the resistance and inductance of the coil. When switch S1 in Fig. 7.77(a) is closed, the coil circuit is energized. The coil current gradually increases and produces a magnetic field. Eventually the magnetic field is sufficiently strong to pull the mo vable contact in the other circuit and close switch S2. At this point, the relay is said to be pulled in. The time interval td between the closure of switches S1 and S2 is called the relay delay time.

Relays were used in the earliest digital circuits and are still used for switching high-power circuits.

The coil of a certain relay is operated by a 12-V battery. If the coil has a Example 7.21 resistance of 150 Ω and an inductance of 30 mH and the current needed to pull in is 50 mA, calculate the relay delay time.

Solution:

The current through the coil is given by

$$i(t) = i(\infty) + [i(0) - i(\infty)]e^{-t/\tau}$$

where

$$i(0) = 0,$$

$i(\infty) = \frac{12}{150} = 80 \text{ mA}$

$$\tau = \frac{L}{R} = \frac{30 \times 10^{-3}}{150} = 0.2 \text{ ms}$$

Thus,

$$i(t) = 80[1 - e^{-t/\tau}] \text{ mA}$$

If i(td) = 50 mA, then

50 = 80[1 − e⁻td∕^τ ] ⇒ __5 8 = 1 − e⁻td∕^τ

or

$$e^{-t_d/\tau} = \frac{3}{8} \qquad \Rightarrow \qquad e^{t_d/\tau} = \frac{8}{3}$$

By taking the natural logarithm of both sides, we get

$$t_d = \tau \ln \frac{8}{3} = 0.2 \ln \frac{8}{3} \text{ ms} = 0.1962 \text{ ms}$$

Alternatively, we may find td using

$$t_d = \tau \ln \frac{i(0) - i(\infty)}{i(t_d) - i(\infty)}$$

Practice Problem 7.21

A relay has a resistance of 200 Ω and an inductance of 500 mH. The relay contacts close when the current through the coil reaches 175 μA. What time elapses between the application of 110 V to the coil and contact closure?

Answer: 957.5 μs.

7.9.4 Automobile Ignition Circuit

The ability of inductors to oppose rapid change in current mak es them useful for arc or spark generation. An automobile ignition system tak es advantage of this feature.

The g asoline engine of an automobile requires that the fuel-air mixture in each c ylinder be ignited at proper times. This is achie ved by means of a spark plug (Fig. 7.78), which essentially consists of a pair of electrodes separated by an air g ap. By creating a lar ge voltage (thousands of v olts) between the electrodes, a spark is formed across the air g ap, thereby igniting the fuel. But ho w can such a lar ge voltage be obtained from the car battery, which supplies only 12 V? This is achieved by means of an inductor (the spark coil) L. Since the v oltage across the inductor is v = L di∕dt, we can make di∕dt large by creating a large change in current in a very short time. When the ignition switch in Fig. 7.78 is closed, the current through the inductor increases gradu ally and reaches the final value of i = Vs∕R, where Vs = 12 V. Again, the time taken for the inductor to char ge is five times the time constant of the circuit (τ = L∕R),

$$t_{\text{charge}} = 5 \frac{L}{R} \tag{7.67}$$

Since at steady state, i is constant, di∕dt = 0, and the inductor v oltage v = 0. When the switch suddenly opens, a lar ge voltage is de veloped across the inductor (due to the rapidly collapsing field) causing a spark or arc in the air g ap. The spark continues until the ener gy stored in the inductor is dissipated in the spark dischar ge. In laboratories, when one is working with inductive circuits, this same ef fect causes a v ery nasty shock, and one must exercise caution.

A solenoid with resistance 4 Ω and inductance 6 mH is used in an auto- Example 7.22 mobile ignition circuit similar to that in Fig. 7.78. If the battery supplies 12 V, determine: the final current through the solenoid when the switch is closed, the energy stored in the coil, and the voltage across the air gap, assuming that the switch takes 1 μs to open.

Solution:

The final current through the coil is

I = __ Vs R ⁼___ ¹² 4 = 3 A

The energy stored in the coil is

$$W = \frac{1}{2}L\ l^2 = \frac{1}{2} \times 6 \times 10^{-3} \times 3^2 = 27 \text{ mJ}$$

The voltage across the gap is

$$V = L\frac{\Delta I}{\Delta t} = 6 \times 10^{-3} \times \frac{3}{1 \times 10^{-6}} = 18 \text{ kV}$$

The spark coil of an automobile ignition system has a 20-mH induc - Practice Problem 7.22 tance and a 5-Ω resistance. With a supply voltage of 12 V, calculate: the time needed for the coil to fully charge, the energy stored in the coil, and the voltage developed at the spark gap if the switch opens in 2 μs.

Answer: 20 ms, 57.6 mJ, and 24 kV.