Chapter 11

Capacitive Transients,

Pulse, and Waveshaping

11.1 Introduction

Capacitor charging [Fig. 11-1 & 2], page 418

• assume capacitor is uncharged when switch is closed

• current jumps to E/R amps, then decays to zero

• voltage is zero and then gradually climbs to E volts

Steady state conditions

• when capacitor voltage and current reach their final values and stop changing, the circuit is said to be in steady state

• capacitor looks like an open circuit in steady state

• capacitor voltage cannot change instantaneously

• an uncharged capacitor looks like a short circuit when switch closes

Capacitor discharging [Fig. 11-5 & 6], page 419

EXAMPLE 11-1, page 420

For Figure 11-1, E = 40 V, R = 10 , and the capacitor is initially uncharged. The switch is moved to the charge position and the capacitor allowed to charge. Then the switch is moved to the discharge position and the capacitor allowed to discharge. Sketch the voltages and currents and determine the values at switching and steady state.

Solution

The current and voltage curves are shown in Figure 11-7. Initially, i = 0 A since the switch is open. Immediately after it is moved to the charge position, the current jumps to E/R = 40 V/10 = 4 A; then it decays to zero. At the same time, vc starts at 0 V and climbs to 40 V. When the switch is moved to the discharge position, the capacitor looks momentarily like a 40-V source and the current jumps to -40 V/10 = -4 A; then it decays to zero. At the same time, vc also decays to zero.

Refer to Figure 11-7, page 420

11.2 Capacitor Charging Equations

[Fig. 11-8], page 422

E = VR + VC

but VR = R ic and ic = C dvc / dt

E = R C (dvc / dt) + vc

vc = E [1 - exp(-t/RC)]

VR = E exp (-t/RC)

ic = (E/R) exp(-t/RC)

EXAMPLE 11-2, page 423

Suppose E = 100 V, R = 10 k , and C = 10 F:

a. Determine the expression for vc.

b. Determine the expression for ic.

c. Compute the capacitor voltage at t = 150 ms.

d. Compute the capacitor current at t = 150 ms.

e. Locate the computed points on the curves.

Solution

a. RC = (10 x 103 ) (10 x 10-6F) = 0.1 s. From Equation 11-3,

vc = E(1 - e-t/RC) = 100(1 - e-t/0.1) = 100(1 - e-10t) V.

b. From Equation 11-5, ic = (E/R)e-t/RC =(100V/10 k)e-10t

= 10e-10t mA.

c. At t = 0.15 s, vc = 100(1 - e-10t) = 100(1 - e-10(0.15)) = 100(1 - e-1.

5) = 100(1 - 0.223) = 77.7 V.

d. ic = 10e-10t mA = 10e-10(0.15) mA = 10e-1.5 mA = 2.23 mA.

e. The corresponding points are shown in Figure 11-10. [page 423]

Example 11-3, page 424

For the circuit of Figure 11-11, E = 60 V, R = 2 k, and C = 25 F. The switch is closed at t = 0 s, opened 40 ms later and left open. Determine equations for capacitor voltage and current and plot.

Solution

RC = (2 k)(25 F) = 50 ms. From t = 0 s to 40 ms, the following equations hold:

vc = E(1 -e-t/RC) = 60(1 -e-t/50ms) V

ic = (E/R)e-t/RC = 30e-t/50 ms mA

Voltage starts at 0 V and rises exponentially. At t = 40 ms, the switch is opened, interrupting charging. At this instant, vc = 60(1 -e-(40/50)) = 60(1 -e-0.8) = 33.0 V. Since the switch is left open, the voltage remains constant at 33 V thereafter as indicated in Figure 11-12. (The dotted curve shows how the voltage would have kept rising if the switch had remained closed.)

Now consider current. The current starts at 30 mA and decays to ic = 30e-(40/50) mA = 13.5 mA at t = 40 ms. At this point, the switch had remained closed.)

Now consider current. The current starts at 30 mA and decays to ic = 30e-(40/50) mA = 13.5 mA = 13.5 mA at t = 40 ms. At this point, the switch is opened, and the current drops instantly to zero. (The dotted line shows how the current would have decayed if the switch had not been opened.)

Time Constant

= R C (unit: seconds)

Duration of transient

Figures 11-13 and 11.14, pages 425-426

• exp(-t/ ) decreases to zero at infinite time• over 99% of transition takes place after 5• for practical purposes, transients can be ocnsidered to last for only five time constants.

Example 11-4, page 426

For the circuit of Figure 11-11, how long will it take for the capacitor to charge if R = 2 k and C = 10 F?

Solution = RC = (2 k)(10 F) = 20 ms.

Therefore, the capacitor charges in 5 = 100 ms.

Example 11-5, page 426

The transient in a circuit with C = 40 F lasts 0.5 s. What is R?

Solution 5 = 0.5 s. Thus, = 0.1 s and R = / C = 0.1 s / (40 x 10-6 F) = 2.5 k.

Fig. 11-15 [page 426]

It shows how voltage and current are affected by the time constant.

The larger R and C, the larger the time constant and hence the longer it takes for the capacitor to charge.

11.3 Capacitor with an initial voltage (p. 427)

If a capacitor has not been discharged and thus still has voltage, V0, on it

VC = E - ( E - V0) exp (-t/ )

iC = [( E - V0) / R] exp (-t/ )

Example 11-6, [page 427]

11.4 Capacitor Discharging Equations (p. 429)

0 = VR + VC

but iR = iC = VR / R

iC = - (V0/R) exp(-t/RC)

Discharging transients last 5 time constant.

11.5 More Complex Circuits [page 430]• reduced to simple form by using series and parallel combinations, source conversions,

Thevenin‘s theorm, etc.

Example 11-10 [page 431]

The capacitor of Figure 11-21 is initially uncharged. Close the switch at t = 0 s.

a. Determine the expression for vc.

b. Determine the expression for ic.

c. Determine capacitor current and voltage at t = 5ms.

Example 11-11 [page 433]

The capacitor of Figure 11-24(a) is uncharged. The switch is moved to position 1 for 10 ms, then to position 2, where it remains.

a. Determine vc during charge.

b. Determine ic during charge.

c. Determine vc during discharge.

d. Determine ic during discharge.

e. Sketch the charge and discharge waveforms.

Example 11-12 [page 434]

The capacitor of Figure 11-26 is uncharged. The switch is moved to position 1 for 5 ms, then to position 2 and left there.

a. Determine vc while the switch is in position 1.

b. Determine ic while the switch is in position 1.

c. Compute vc and ic at t = 5 ms.

d. Determine vc while the switch is in position 2.

e. Determine ic while the switch is in position 2.

f. Sketch the voltage and current waveforms.

Example 11-13 [page 435]

In Figure 11-28(a), the capacitor is initially uncharged. The switch is moved to the charge position, then to the discharge posiiton, yielding the current shown in (b). The capacitor discharges in 1.75 ms. Determine the following:

a. E b. R1 c. C

Solution

a. Since the capacitor charges fully, it has a value of E volts when switched to discharge. The discharge current spike is

E/(10+25) = -3A

Thus, E = 105 V.

b. The charging current spike has a value of E/(10+ R1) = 7A

Since E = 105V, this yields R1 = 5 .

c. 5 d = 1.75 ms. Therefore d = 350 s. But d = (R2 + R3)C.

Thus, C = 350 s/35 = 10 F.

Example 11-14 [page 436]

The circuit of Figure 11-29(a) has reached steady state. Determine the capacitor voltages.

11.6 An RC Timing Application (p. 438)

• Applications of RC circuits:• delays for alarm• motor control• timing application• Alarm - thresold detector: when input to this detector exceeds a preset value, alarm i

s turned on. [page 438]

Figure 11-32 - Creating a delay with an RC circuit. [page 438]

Refer to example 11-15, page 439

11.7 Pulse Response of RC Circuits [page 440]

Applications: devices and systems utilize pulse or rectangular waveforms e.g. computers

Pulse: voltage or current that changes from one level to the other and back again.

Pulse train: repetitive stream of pulses.

Square wave: waveform’s high time equals its low time.

Period: length of each cycle of a pulse train.

Pulse repetition rate (PRR) or frequency (PRF): number of pulses/second.

Duty cycle: the width of tp relative to its period. [(tp/T) • 100%]

Duty cycle of square wave = 50%

Rise time ( tr ) and fall time ( tf ): measured between 10% and 90% points.

Pulse width: measured at 50% points.

The effect of pulse width

• the width of pulse relative to a circuit‘s time constant determines how it is affected by an RC circuit. [Fig. 11-36, page 441]

Pulse width >> 5

• when the pulse width and time between pulses are very long compared with the circuit time constant, the capacitor charges and discharges fully. [Fig. 11-37, page 442]

• increase rise and fall times of the output.

Pulse width = 5

• capacitor fully charges and discharges during each pulse.

• Differentiator: VR is approximation to the derivative of Vin. [Fig. 11-38, page 443]

Pulse width << 5

• capacitor does not have time to charge significantly before it is switched to discharge, and vice versa. [Fig. 11-39]

• Vc is nearly triangular in shape and have an average = V/2.

• Vc reaches stable state at 5 time constant. [Fig. 11-40]

• Integrator: Vc is approximately integral of Vin.

Capacitive Loading• capacitance occurs whenever conductors are separated by insulating material.• Stray capacitance exists between wires in cables, and traces on PCB.• Stray capacitance causes problems at hifh-speed circuit and is often so small that it can be

neglected at low frequency.• [Fig. 11-41] if the rise and fall times become excessivelylong, the signal reaching the load

may be so degraded that the system malfunction.• Capacitive loading can create serious problems in computers where signal rise and fall ti

mes of a few ns or less may be required.