Diode+circuit+notes+2+fromHumbley

Diode+circuit+notes+2+fromHumbley - r i r'w: imiroclpr'llom...

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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 Half-wave 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 battery-charging 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. Half-Wave Rectifier Circuits A half-wave 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 Half-wave 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 Half-wave 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 (10-8) 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 peak-to-peak ripple voltage and C is the capacitance. Equating the right-hand 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 power-supply 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. Full-Wave Rectifier Circuits . . . i . C Several full-wave 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 left-hand 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 out-of-phasc source voltages and a common load. The diodes conduct on alternate half-cycles. Usually, the two out-of—phase ac voltages are provided by a transformer. (Trans- formers arc discussed in Chapter 15.) Besides providing the out-of-phase 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 half-cycle before being recharged. Hence, the capacitance required is only half as much in the full—wave circuit as for the half-wave 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 Diode-bridge full-wave 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 power-supply 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 WAVE-SHAPING CIRCUITS A wide variety of wave-shaping 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 small-signal 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 Q-point 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 forward-bias 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 small-signal 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 time-varying 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 time-varying 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 linear-equivalent-circuit 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 voltage-controlled 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 Small-Signal 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 Q-point analysis, assume a constant diode voltage of 0.6 V. Find the Q-point 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 Small-signal 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 voltage-controlled 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 high-gain 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“ reverse-breakdown 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 straight-line 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.

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Diode+circuit+notes+2+fromHumbley - r i r'w: imiroclpr'llom...

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