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Unformatted text preview: r i r'w: imiroclpr'llom 486 Chapter 10 Diodes var) m 7 m (b) Source voltage versus time (a) Circuit diagram Figure 10.24 Halfwave rectifier with resistive load. ,..__.__J 10.6 RECTIFIER CIRCUITS (c) Load voltage versus time Now that we have introduced the diode and some methods for analysis of diode
circuits, we consider some additional practical circuits. First, we consider several
types of rectiﬁers, which convert ac power into dc power. These rectiﬁers form the
basis for electronic power supplies and batterycharging circuits. Typically, a power
supply takes power from a raw source, which is often the (SO—Hz ac power line, and
delivers steady dc voltages to a load such as computer circuits or television circuits.
Other applications for rectiﬁers are in signal processing, such as demodulation of a radio signal. (Demodulation is the process of retrieving the message, such as a voice
or video signal.) Another application is precision conversion of an ac voltage to d0 in an electronic voltmeter. HalfWave Rectifier Circuits A halfwave rectiﬁer with a sinusoidal source and resistive load RL is shown in
Figure 10.24. When the source voltage 115(1) is positive, the diode is in the forward' bias region. If an ideal diode is assumed, the source voltage appears acros
load. For a typical reai diode, the output voltage is less than the source voltage by
an amount equal to the drop across the diode, which is approximately 0.7 V s the for silicon diodes at room temperature. When the source voltage is negative. the diOdE
is reverse biased and no current ﬂows through the load. Even for typical real diodes
only a very small reverse current ﬂows. Thus, only the positive halfecycles 0f the source voltage appear across the load. Battery—Charging Circuit. We can use a haltwave rectiﬁer to charge a hatter}i 3: .2
. .. v [F
shown in Figure 10.25. Current flows whenever the instantaneous ac source Vomit“ is higher than the battery voltage. As shown in the ﬁgure, it is necessary [ oad resistance to the CtrCUtt to limit the magnitude of the current. When the aC 50”"; voitage is less than the battery voltage. the diode is reverse biased and the Cu Item is zero. Hence, the current ﬂows only in the direction that charges the batter)“ a? J Section 10.6 Rectifier Circuits 487 Current—limiting resistor V sintwrl {m _ VB m Figure 10.25 Halfwave rectifier used to Charge a battery. Half—Wave Rectifier with Smoothing Capacitor. Often, we want to convert
an ac voltage into a nearly constant dc voltage to be used as a power suppiy for
electronic circuits. One approach to smoothing the rectiﬁer output voltage is to
place a large capacitance across the output terminals of the rectiﬁer. The circuit and
waveforms of current and voltage are shown in Figure 10.26. When the ac source
reaches a positive peak, the capacitor is charged to the peak voltage (assuming
an ideal diode). When the source voltage drops below the voltage stored on the m Diode oft" Ideal _9 _> Load a (b) Voltage waveforms (a) Circuit diagram (cl Current waveforms Flgure 10.26 Halfwave rectifier with smoothing capacitor. 488 Chapter 10 Diodes capacitor, the diode is reverse biased and no current ﬂows through the diode. The
capacitor continues to supply current to the load, slowly discharging until the next
positive peak of the ac input. As shown in the ﬁgure, current flows through the diode
in pulses that recharge the capacitor. Because of the charge and discharge cycle, the load voltage contains a small
ac component called ripple. Usually, it is desirable to minimize the amplitude of
the ripple, so we choose the largest capacitance vaiue that is practical. In this
case, the capacitor discharges for nearly the entire cycle, and the charge removed
from the capacitor during one discharge cycle is Q ’2“ [LT (108) where IL is the average load current and T is the period of the ac voltage. Since the
charge removed from the capacitor is the product of the change in voltage and the capacitance, we can also write
Q : VrC where Vr is the peaktopeak ripple voltage and C is the capacitance. Equating the
righthand sides of Equations 10.8 and 10.9 allows us to solve for C: (10.9) C = [LT (1010) Vr In practice, Equation 10.10 is approximate because the load current varies and
because the capacitor does not discharge for a complete cycle. However, it gives
a good starting value for calculating the capacitance required in the design 0f
powersupply circuits. The average voltage supplied to the load it a smoothing capacitor is used is ap'
proximately midway between the minimum and maximum voltages. Thus, referring
to Figure 1026, the average load voltage is H? (10.10 Vr
V 1/}11 7 A;
L 2 Peak Inverse Voltage An important aspect of rectiﬁer circuits is the peak inverse voltage (PIV) across the
diodes. Of course, the breakdown speciﬁcation of the diodes should be greater In
magnitude than the PIV. For example, in the halfkwave circuit with a resistive 103d=
shown in Figure 10.24, the PIV is Vm. The addition of a smoothing capacitor in parallel with the load increases the
PIV to (approximately) 2V},2 . Referring to Figure 10.26, for the negative peak Ofthe
ac input, we see that the reverse bias of the diode is the sum of the source voliﬁg‘3
and the voltage stored on the capacitor. FullWave Rectifier Circuits . . . i . C
Several fullwave rectiﬁer c1rcu1ts are in common use. One approach uses 1W0 Bis
. . . .  1
sources and two diodes, as shown in Figure 10.27(a). One feature of this diagram
the ground symbol. Usually in electronic circuits, many components are connects Section 10.6 Rectifier Circuits 489 Pitt) Diode A Diode B
on On v
Ri /— Ground symbol it unlit”) 11: Ideal (a) Circuit diagram (13} Figure 1027 Full—wave rectifier. to a common point known as ground. Often, the chassis containing the circuit is the
electrical ground. Therefore, in Figure 10.27(a), the lower end of RL and the point
between the voltage sources are connected together. When the upper source applies a positive voltage to the leﬂ~hand end of diode
A, the lower source applies a negative voltage to the lefthand end of diode B, and
Vice versa. We say that the sources are out of phase. Thus, the circuit consists of
two halfewave rectiﬁers with outofphasc source voltages and a common load. The
diodes conduct on alternate halfcycles. Usually, the two outof—phase ac voltages are provided by a transformer. (Trans
formers arc discussed in Chapter 15.) Besides providing the outofphase ac voltages.
the transformer also allows the designer to adjust V,,, by selection of the turns ratio.
This is important, because the ac voltage available is often not of a suitable amplitude
for direct rectiﬁcation—usually either a higher or lower dc voltage is required. A second type of fullewave rectiﬁer uses the diode bridge shown in Figure 1028.
When the ac voltage is positive, current ﬂows through diode A, then through the
load, and returns through diode B, as shown in the ﬁgure. For the opposite polarity,
current ﬂows through diodes C and D. Notice that in either case, current ﬂows in
the same direction through the load. Usually, neither of the ac source terminals is connected to ground. This is
necessary if one side of the load is to be connected to ground, as shown in the ﬁgure.
(If both the ac source and the load have a common ground connection, part of the
circuit is shorted.) If we wish to smooth the voltage across the load, a capacitor can be placed in
parallel with the load, similar to the halfewave circuit discussed earlier. In the full,
Wave circuits, the capacitor discharges for only a halfcycle before being recharged.
Hence, the capacitance required is only half as much in the full—wave circuit as for
the halfwave circuit. Therefore, we modify Equation 10.10 to obtain rLr
C =
2V. for the fullﬁwave rectiﬁer with a capacitive ﬁlter. (10.12) 144 490 Chapter 10 Diodes A clipper circuit “clips off
part of the input waveform
to produce the output
waveform. Current path for
positive half—cycle + d!) Vm sin(mt) Figure 10.28 Diodebridge fullwave rectifier. Exercise 10.11 Consider the battery~charging circuit of Figure 10.25 with V,” :
20 V. R z 10 S2, and VB 2 14 V. a. Find the peak current assuming an ideal diode.
b. Find the percentage of each cycie for which the diode is in the on state. Answer a. [peak = 600 mA; b. the diode is on for 25.3 percent of each Cycle. L Exercise 10.12 A powersupply circuit is needed to deliver 0.1 A and 15 V
(average) to a load. The ac source has a frequency of 60 Hz. Assume that the circuit
of Figure 10.26 is to be used. The peak—to—peak ripple voltage is to be 0.4 V. Instead
of assuming an ideal diode, allow 0.7 V for forward diode dr0p. Find the peak ac
voltage Vm needed and the approximate value of the smoothing capacitor. (Hint
To achieve an average load voltage of 15 V with a ripple of 0.4 V, design for a peak
load voltage of 15.2 V.) Answer Vm = 15.9, C = 4166 uF. 9
Exercise 10.13 Repeat Exercise 10.12 using the circuit of Figure 10.28 with the
smoothing capacitor in parallel with the load R L. Answer Vm : 16.6. C = 2083 ,uF. D 10.7 WAVESHAPING CIRCUITS A wide variety of waveshaping circuits are used in electronic systems. These circlllts
are used to transform one waveform into another. Numerous examples of wave'
shaping circuits can be found in transmitters and receivers for televiSion or rad”
In this section, we discuss a few examples of wave—shaping circuits that 0311 be
constructed with diodes. Clipper Circuits Diodes can be used to form clipper circuits, in which a portion of an input Sigma;
waveform is “clipped” off, For example. the circuit of Figure 10.29 clips Off am]
part of the input waveform above 6 V or less than —9 V. (We are assuming “ﬂed
diodes.) When the input voltage is between 79 and +6 V, both diodes are Offanl
no current ﬂows. Then, there is no drop across R and the output voltage UK, is eqﬁa J. 496 Chapter 10 Diodes Figure 10.35 Answer for Exer
cise 10.17. C
R
vin + U0
5.4V ; 15 V
_I _5 Figure 1036 Answer for Exercise 10.18. +0 Answer a. For villa) = 0, we have 715 : —5 V; b. see Figure 10.34(b); c. see
Figure 10.34(c). D Exercise 10.17 Design a circuit that clamps the negative peaks of an ac sig
nal to +6 V. You can use batteries, resistors, and capacitors of any value de'
sired in addition to Zener or conventional diodes. Allow 0.6 V for the forward
drop. Answer A solution is shown in Figure 10.35. Other solutions are possible. 5 Exercise 10.18 Repeat Exercise 10.17 for a circuit that clamps the positive peaks
to +6 V. Answer A solution is shown in Figure 1036. Other solutions are possible. ‘3 10.8 LINEAR SMALL—SIGNAL EQUIVALENT CIRCUITS We will encounter many examples of electronic circuits in which dc supply Voltﬂg‘gfs
are used to bias a nonlinear device at an operating point, and a small ac sigﬂai ‘5
injected into the circuit. We often split the analysis of such circuits into two Pa?“
First, we analyze the dc circuit to ﬁnd the operating point. In this analysis Of blag
conditions, we must deal with the nonlinear aspects of the device. In the 53.5011
part of the analysis, we consider the small ac signal. Since virtually any nonlultear
characteristic is approximately linear (straight) if we consider a sufﬁciently 531?
portion, we can ﬁnd a linear smallsignal equivalent circuit for the nonlinear dam“
to use in the ac analysis. Often, the main concern in the design of such circuits is what happens [0 the, ac
signal. The dc supply voltages simply bias the device at a suitable operating Poln‘ Section 10.8 Linear Small—Signal Equivalent Circuits For example, in a portable radio, the main interest is the signal being received,
demodulated, ampliﬁed, and delivered to the speaker. The dc currents supplied by
the battery are required for the devices to perform their intended function on the
ac Signals. However, most of our design time is spent in consideration of the small
as Signals to be processed. The small—signal linear equivalent circuit is an important analysis approach
that applies to many types of electronic circuits. In this section, we demonstrate
the principles with a simple diode circuit. in Chapters 12 and 13, we use similar
techniques for transistor ampliﬁer circuits. Now, we show that in the case of a diode, the small—signal equivalent circuit con~
Sists simply of a resistance. Consider the diode characteristic shown in Figure 10.37.
Assume that the dc supply voltage results in operation at the quiescent point, or
Q point, indicated on the characteristic. Then, a small ac signal injected into the
circuit swings the instantaneous point of operation slightly above and below the Q
point. For a sufﬁciently small ac signal, the characteristic is straight. Thus, we can write .
(it D Ai E — Av
D (dvD)Q D where ND is the small change in diode current from the Q point current caused by
the ac signal, Avg is the change in the diode voltage from the (2—point value, and
{dip/dUD)Q is the slope of the diode characteristic evaluated at the Q point. Notice
that the slope has the units of inverse resistance. Hence, we deﬁne the dynamic resistance of the diode as (10.14) 71
d. rd = ] (10.15) d'U'D Q and Equation 10.14 becomes
A
Air) ’2“ 1"” (10.16)
rd Figure 10.37 Diode characteristic, illus
trating the Q point. 497 The smallesignal equivalent
circuit for a diode is a
resistance. 498 Chapter 10 Diodes We ﬁnd it convenient to drop the A notation and denote changes of current
and voltage from the Qpoint values as vd and id. (Notice that lowercase subscripts
are used for the small changes in current and voltage.) Therefore, for these small ac signals, we write id x 3% (10.17) rd As shown by Equation 10.15, we can ﬁnd the equivalent resistance of the diode
for the small ac signal as the reciprocal of the slope of the characteristic curve.
The current of a junction diode is given by the Shockley equation (Equation 10.1),
repeated here for convenience: in = I; [exp(%/D;) — 1] The slope of the characteristic can be found by differentiating the Shockley equation, resulting in
dip 1 DD
— = 15— exp 7
dub “VT "VT Substituting the voltage at the Q point, we have dip 1 VDQ
—H = IS—exp
dvo Q IIVT nVy For forwardbias conditions with VDQ at least several times as large as VT, the 1
inside the brackets of the Shockley equation is negligible. Thus, we can write V
IDQ "5 IS exp(ﬂ) 11 VI (10.18) (10.19) (10.21))
Substituting this into Equation 10.19, we have (Eta _ IDQ dUD Q W nVT Taking the reciprocal and substituting into Equation 10.15, we have the dynamic
smallsignal resistance of the diode at the Q point: (10.20 "VT (10.22) r : M
a 109 To summarize, for signals that cause small changes from the Q point. W?can
treat the diode simply as a linear resistance. The value of the resistance iS Swen
by Equation 10.22 (provided that the diode is forward biased). As the Q‘Polné
current IDQ increases, the resistance becomes smaller. Thus, an ac voltage of ﬁxe
amplitude produces an ac current that has higher amplitude as the Q point move higher. This is illustrated in Figure 10.38. Section “18 Linear Smallisignal Equivalent Circuits '499 Equaleamplitude voltage signals
VD Figure 10.38 M the Q point moves higher, a fixedramplitude
ac voltage produces an at current of iarger amplitude. Notation for Currents and Voltages in Electronic Circuits Perhaps we shouid review the notation we have used for the diode currents and
voltages, because we use similar notation throughout this book: I v0 and iD represent the total instantaneous diode voltage and current. At times,
we may wish to emphasize the timevarying nature of these quantities, and then
we use van“) and mm. l VDQ and IDQ represent the dc diode current and voltage at the quiescent point. I ad and id represent the (smali) ac signals. If we wish to emphasize their
timevarying nature, we use 110(1) and 1'40). This notation is illustrated for the waveform shown in Figure 10.39. Exercise 10.19 Compute the dynamic resistance of a junction diode having 11 = 1
at a temperature of 300 K for IDQ : a. 0.1 mA; I). 1 mA; c. 10 mA. Answer a. 260 S2; b. 26 S2; c. 2.6 Q. n Voltage—Controlled Attenuator NOW, we consider an example of linearequivalentcircuit analysis for the relatively
Simple, but useful circuit shown in Figure 1040. The function of this circuit is to
produce an output signal yea) that is a variable fraction of the ac input signal 12mm.
It is similar to the resistive voltage divider (see Section 2.3), except that in this case, 500 Chapter 10 Diodes Figure 10.39 illustration of diode currents. Coupling VC “PWQ‘C RC 1 C
l R T 2
ﬂ( W l( 3 Din“) UN“) R1, s_ i— .0 Figure 10.40 Variable attenuator using a diode as a
controlled resistance. we want the division ratio to depend on another voltage VC called the control signal
We refer to the process of reduction of the amplitude of a signal as attenuation
Thus, the circuit to be studied is called a voltagecontrolled attenuator. The degree of attenuation depends on the value of the dc control voltage Vci
Notice that the ac signal to be attenuated is connected to the circuit by a coup capacitor; The output voltage is connected to the load RL by a second coupling
capacitor. Recall that the impedance of a capacitance is given by 1
ZC : 7.
ij ling tance Values
1‘ However.
ting P01m
S can 5
1d affecl
rabiel in which a) is the angular frequency of the ac signal. We select the capaci
large enough so that they are effectively short circuits for the ac signa
the coupling capacitors are open circuits for dc. Thus, the quiescent opera
(Q point) of the diode is unaffected by the signal source or the loads Thi
important for a circuit that must work for various sources and loads that con the Q point. Furthermore, the coupling capacitors prevent (sometimes undﬁsi
dc currents from flowing in the source or the load. 1'66 ing
ing 165 Section 10.8 Linear SmallSignal Equivalent Circuits 501 Because of the coupling capacitors, we only need to consider VF. RC. and the
diode t0 pﬁltform the bias analysis t0 ﬁnd the Q point. Hence, the dc circuit is shown
in Figure 10.41. We can use any of the techniques discussed earlier in this chapter to
ﬁnd the Q point. Once it is known, the onint value of the diode current IDQ can
be substituted into Equation 10.22 to determine the dynamic resistance of the diode. NOW, we turn our attention to the ac signal. The dc control source should be
COnsidered as a short circuit for ac signals. The signal source causes an ac current to
ﬂow through the Vc source. Hovvever, VC is a dc voltage source, and by deﬁnition,
the voltage across it is constant. Since the dc voltage source has an ac component of
current, but no ac voltage, the dc voltage source is equivalent to a short circuit for
ac signals. This is an important concept that we will use many times in drawing ac
equjvalent Circuits. ‘ The equivalent circuit for ac signals is shown in Figure 10.42. The control source
and the capacitors have been replaced by short circuits, and the diode has been
replaced by its dynamic resistance. This circuit is a voltage divider and can be
analyzed by ordinary linear~circuit analysis. The parallel combination of RC, R1,,
and rd is denoted as RP, given by 1
RF : m (10.23)
l/RC +1/RL + 1m,
Then, the voltage gain of the circuit is
A, : i 2 RP (10.24)
vm R + RP (Of course, AU is less than unity.) Exercise 10.20 Suppose that the circuit of Figure 10.40 has R : 100 92, RC : 2 k9,
and RI, ; 2 k9. The diode has n : 1 and is at a temperature of 300 K. For purposes
of Qpoint analysis, assume a constant diode voltage of 0.6 V. Find the Qpoint
value of the diode current and AU for VC : a. 1.6 V; b. 10.6 V. Answer a. IDQ : 0.5 mA and AU 2 0.331; b. IDQ : 5 mA and Av = 0.0492. n An application for voltagercontrolled attenuators occurs in tape recorders. A
problem frequently encountered in recording a conversation is that some persons
speak quietly, while others speak loudly. Furthermore, some may be far from the
microphone, while others are close. If an ampliﬁer with ﬁxed gain is used between
the microphone and the tape head, either the weak signals are small compared with Dina) Figure 10.42 Smallsignal ac equivalent circuit
for Figure 1040. Figure 1041 Dc circuit
equivalent to Figure 10.40
for onint anaiysis. Dc sources and coupling
capacitors are replaced by
short circuits in small—signal
ac equivalent circuits. Diodes
are replaced with their
dynamic resistances. 502 Chapter 10 Diodes Control voltage \ i Microphone I I
1‘ Voltage— I
controlled Ampliﬁer Rcigigmg
attenuator Figure 10.43 The voltagecontrolled attenuator is useful in maintaining a suitable
' signal amplitude at the recording head. the noise level or the strong signals drive the recording head nonlinear so that severe
distortion occurs. A solution is to use a voltageAcontrolled attenuator in a system such as the one
shown in Figure 10.43. The attenuator is placed between the microphone and a
highgain ampliﬁer. When the signal being recorded is weak, the control voltage is
small and very little attenuation occurs. On the other hand, when the signal is strong,
the control voltage is large so that the signal is attenuated, preventing distortion. The control voltage is generated by rectifying the output of the ampliﬁer. The
rectiﬁed signal is ﬁltered by a longtime—constant RC ﬁlter so that the attenuation
responds to the average signal amplitude rather than adjusting too rapidly. With
proper design, this system can provide an acceptable signal at the recording head
for a wide range of input signal amplitudes. 'i 1. A pn junction diode is a two—terminal device that 6. The ideal—diode model is a short circuit (on) if Cur’
conducts Current easily in one direction (from an rent flows in the forward direction and an 0P?“
ode to cathode}, but not in the opposite direction. circuit (off) if voltage is applied in the revel'Se d“
The volt—ampere characteristic has three regions: rection.
forward bias, reverse bias, and reverse breakdown. a state
f 2 The Shockle e I. l t d 1 f 7: In the method ofassumed states,we assume. d
i ' , _ qua} 10“ re a 65 current an V0 [age for each diode (on or off), analyze the circuit, a“
i m a P’I'luncuon dmde' check to see if the assumed states are conSiéiem
3. Nonlinear circuits,such as those containinga diode, with the current directions and voltage polaritleé'
can be analyzed by using the load—line technique. This process is repeated until a valid set Of States 15
4. Zencr diodes are intended to be operated in the found“
reversebreakdown region as constant—voltage ref— 8. In a piecewiseghncar model for a nonlinear devicﬁ:
Cfences' the volteampere characteristic is approximated 5 Voltage regulators are circuits that produce a straightline segments. On each segment, the d:
a nearly constant output voltage while operating vice is modeled as a voltage source in Series “nth
mu“ 3 variable source. resistance.
“ ...
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This note was uploaded on 02/06/2012 for the course EECS 70B taught by Professor Henrylee during the Spring '08 term at UC Irvine.
 Spring '08
 HENRYLEE

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