Diode+circuit+notes+1+from+Humbley

Diode+circuit+notes+1+from+Humbley - Section 10.1 10.1...

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Unformatted text preview: Section 10.1 10.1 BASiC DIODE CONCEPTS The diode is a basic but very important device that has two terminals, the anode and the cathode. The circuit symbol for a diode is shown in Figure 10.1(a), and a typical VDtttfiampere characteristic is shown in Figure 10103). As shown in Figure 10.1(a), the voltage U D across the diode is referenced positive at the anode and negative at the cathode. Similarly, the diode current in is referenced positive from anode to cathode. Notice in the characteristic that if the voltage at) applied to the diode is positive, retatively large amounts of current flow for small voltages. This condition is called Wward bias. Thus, current flows easily through the diode in the direction of the arrowhead of the circuit symbol. On the other hand, for moderate negative values of Up, the current to is very Small in magnitude. This is called the reverse-bias region, as shown on the diode Characteristic. In many applications, the ability of the diode to conduct current easily in one direction, but not in the reverse direction, is very useful, For example, in an automobile, diodes allow current from the alternator to charge the battery When the engine is running. However, when the engine stops, the diodes prevent the Basic Diode Concepts 469 Diodes readily conduct Current from anode to cathode (in the direction of the arrow), but do not readily allow current to flow in the opposite direction. I’D it) —e-L Anode O‘Mho Cathode + v _ D Reverse I ‘ I breakdown Ifi‘ Reverse-bras Forward—bias region Teglon region (a) Circuit symbol (b) Volt—ampere charac Free electrons eristic Cathode Electric field barrier (c) Simplified physical structure FiQure 10.1 Semiconductor diode. M (d) Fluideflow analogy: flapper valve 470 Chapter 10 Diodes If a sufficiently large reverse-bias voltage is applied to the diode, operation enters the reverse-breakdown region of the characteristic, and currents of large magnitude flow. battery from discharging through the alternator. In these applications, the diode is analogous to a one-way valve in a fluid-flow system, as illustrated in Figure 10nd), If a sufficiently large reverse—bias voltage is applied to the diode, operation enters the reverse-breakdown region of the characteristic, and currents of large magnitude flow. Provided that the power dissipated in the diode does not raise its temperature too high, operation in reverse breakdown is not destructive to the device. In fact, we will see that diodes are sometimes deliberately operated in the reverseebreakdown region Brief Sketch of Diode Physics We concentrate our discussion on the external behavior of diodes and some of their circuit applications. However, at this point, we give a thumbnail sketch of the internal physics of the diode. The diodes that we consider consist of a junction between two types of semi— conducting material (usually, silicon with carefully selected impurities). On one side of the junction, the impurities create n-type material, in which large numbers of electrons move freely. On the other side of the junction, different impurities are employed to create (in effect) positively charged particles known as holes. Semiconductor material in which holes predominate is called p-type material. Most diodes consist of a junction between n—type material and p—type material, as show in Figure 10.1(c). . Even with no external applied voltage, an electricnfield barrier appears naturally at the pn junction. This barrier holds the free electrons on the n—side and the holes on the p—side of the junction. If an external voltage is applied with positive polarity on the aside, the barrier is enhanced and the charge carriers cannot cross the junction. Thus, virtually no current flows. On the other hand, if a voltage is appll'6d with positive polarity on the p—side, the barrier is reduced and large currents cross the junction. Thus, the diode conducts very little current for one polarity and larg‘5 current for the other polarity of applied voltage. The anode corresponds to the petype material and the cathode is the n—side. Small—Signal Diodes Various materials and structures are used to fabricate diodes. For now, we COHfine our discussion to smallesignal silicon diodes, which are the most common type found in low— and medium—power electronic circuits. The characteristic curve of a typical small—signal silicon diode operated at 3 temperature of 300 K is shown in Figure 10.2. Notice that the voltage and Curfent scales for the forward—bias region are different than for the reverse-bias T3310” This is necessary for displaying details of the characteristic, because the Curran magnitudes are much smaller in the reverseebias region than in the forwafd'tjlz15 region. Furthermore, the forward-bias voltage magnitudes are much less than ill-“Ca breakdown voltages. Section iOJ It) mA W 400V Figure 10.2 Volt—ampere characteristic for a typical small—signal silicon diode at a temperature of 300 K. Notice the change of scale for negative current and voltage. In the forward-bias region, small-signal silicon diodes conduct very little current (much less than 1 mA) until a forward voltage of about 0.6 V is applied (assuming that the diode is at a temperature of about 300 K). Then, current increases very rapidly as the voltage is increased. We say that the forward-bias characteristic displays a knee in the forward bias characteristic at about 0.6 V. (The exact value of the knee voltage depends on the device. its temperature, and the current magnitude. Typical values are 0.6 or 0.7 V.) As temperature is increased, the knee voltage decreases by about 2 mV/K. (Because of the linear change in voltage with temperature, diodes are useful as temperature sensors. The diode is operated at a fixed current, and the voltage across the diode depends on its temperature. Electronic thermometers used by physicians contain a diode sensor, amplifiers, and other electronic circuits that drive the liquid—crystal temperature display.) In the reverse-bias region, a typical current is about 1 nA for small~signal silicon diodes at room temperature. As temperature increases, reverse current increases in magnitude. A rule of thumb is that the reverse current doubles for each 10—K increase in temperature. When reverse breakdown is reached, current increases in magnitude very rapidly. The voltage for which this occurs is called the breakdown voltage. For example, the breakdown voltage of the diode characteristic shown in Figure 10.2 is approximately 7100 V. Breakdown~voltage magnitudes range from several volts to sexieral hundred volts. Some applications call for diodes that operate in the forward- bias and nonconducting reverse—bias regions without entering the breakdown region. Diodes intended for these applications have a specification for the minimum magnitude of the breakdown voltage. Basic Diode Concepts 471 472 Chapter l0 Diodes Figure 10.3 Zener—diode symbol. Shockley Equation Under certain simplifying assumptions, theoretical considerations result in the following relationship between current and voltage for a junction diode: - a £13.. a. tD—Isl:exp(nVT) 1] This is known as the Shockley equation. 1,, the saturation current, has a value on the order of 10—14 A for small—signal junction diodes at 300 K. (1, depends on temperature, doubling for each SuK increase in temperature for silicon devices.) The parameter a, known as the emission coefficient, takes values between 1 and 2, depending on details of the device structure. The voltage VT is given by [J Vrzw 9' and is called the thermal voltage. The temperature of the junction in kelvin is represented by T. Furthermore, k : 1.38 x 10‘23 J/K is Boltzmann’s constant, and q : 1.60 x 10’“) C is the magnitude of the electrical charge of an electron. At a temperature of 300 K, we have VT 2 0.026 V. If we solve the Shockley equation for the diode voltage, we find that tip : 11V?“ Jr 3 (70.1) (10.2) (10.3) For small—signal junction diodes operated at forward currents between 0.01 #A and 10 mA, the Shockley equation with it taken as unity is usually very accurate Because the derivation of the Shockley equation ignores several phenomena, the equation is not accurate for smaller or larger currents. For example, under reverse bias, the Shockley equation predicts in E 715, but we usually find that the reverse current is much larger in magnitude than 1,. (although still small). Furthermore, thB Shockley equation does not account for reverse breakdovvn. With forward bias of at least several tenths of 3 volt, the exponential in Ills Shockley equation is much larger than unity; with good accuracy, we have _ ' Up 2 Le — lD ‘5 Xp<nVr) This approximate form of the equation is often easier to use. . Occasion ally, we are able to derive useful analytical results for electronic cirCUlES by use of the Shockley equation, but much simpler models for diodes are usuall.V more useful. (10.4) Zener Diodes Diodes that are intended to operate in the breakdown region are called Z3“er diodes. Zener diodes are useful in applications for which a constant voltage In breakdown is desirable. Therefore, manufacturers try to optimize Zener diodes for a nearly vertical characteristic in the breakdown region. The modified diode Symbol shown in Figure 10.3 is used for Zener diodes. Zener diodes are available Wlt breakdown voltages that are specified to a tolerance of :l:5 %. Section 10.2 LoadALine Anaiysis oi Diode Circuits 473 Exercise 10.1 At a temperature of 300 K. a certain junction diode has Q; = A for U” : 0.6 V, Assume that n is unity and use VT : 0.026 V. Find the Bilgof the saturation current [A . Then. compute the diode current at 111') : 0.65 V and at 0.70 V. Answer 15 : 9.50 x 10 5 A. in : 0.684 mA, in : 4.68 mA. g Exercise 30.2 Consider a diode under forward bias so that the approximate form ofthe Shockley equation (Equation 10.4) applies. Assume that VT : 0.026 V and {- R n 2 1.3. By what increment must no increase to double the current? h. To increase 3 —> “A . the Current by a factor of 10'? + 1, — Answer 3. Av}; = 18 mV; b. Aug) : 59.9 mV. 3 10,2 LOAD—LINE ANALYSIS OF DIODE CIRCUITS In Section 10.1, we learned that the volt—ampere characteristics of diodes are nonlinear. We will see shortly that other electronic devices are also nonlinear, On the other hand, resistors have linear volt—ampere characteristics, as shown in Figure 10.4. Because of this nonlinearity, many of the techniques that we have studied for linear circuits in Chapters 1 through 6 do not apply to circuits involving diodes. In fact, much of the study of electronics is concerned with techniques for Figure “14 In contrast to analysis of circuits containing nonlinear elements. diodes, resistors have “near Graphical methods provide one approach to analysis of nonlinear circuits. For volt—ampere characteristics. example, consider the circuit shown in Figure 30.5. By application of Kirchhoff’s voltage law, we can write the equation R Vss = Rio + Up (10.5) We assume that the values of V55 and R are known and that we wish to find in and 1);). Thus, Equation 10.5 has two unknowns, and another equation (or its equivalent) is needed before a solution can be found. This is available in graphical form in Figure 10.6, which shows the volt—ampere characteristic of the diode. Figure 105 Circuit for load- We can obtain a solution by plotting Equation 10.5 on the same set of axes line analysis. used for the diode characteristic. Since Equation 10.5 is linear, it plots as a straight line, which can be drawn if two points satisfying the equation are located. A simple method is to assume that i0 : 0, and then Equation 10.5 yields v0 : V55. This pair of values is shown as point A in Figure 10.6. A second point results if we assume that v3 = 0, for which the equation yields iD : Vgs/R. The pair of values is shown as point B in Figure 10.6. Then, connecting points A and B results in a plot called the load line. The operating point is the intersection of the load line and the diode characteristic. This point represents the simultaneous solution of Equation 10.5 and the diode characteristic. Example 10.1 Load-Line Analysis If the circuit of Figure 10.5 has V55 2 2 V, R z 1 k9, and a diode with the Characteristic shOWn in Figure 10.7, find the diode voltage and current at the Operating point. 474 Chapter 10 Diodes Figure 10.6 Load-line analysis of the circuit of Figure 10.5. Solution First, we locate the ends of the load line. Substituting vD : 0 and the values given for V55 and R into Equation 10.5 yields in = 2 mA. These values plot as point B in Figure 10.7. Substitution of i0 = O and circuit values results in Up 3 2 V. These values plot as point A in the figure. Constructing the load line results in an operating point of VDQ E 0.7 V and [pg '2” 1.3 mA, as shown in the figure. I iD (mA) / Point 8 2.0 Diode characteristic 1.3 Load line for Exam le 10.2 1.0 K p k Point D (an: 2 v. 50 = 0.3 mA) for Example 1011 Pomt A l / yo (V) 0.7 1.0 2.0 | | | : Load line I | Figure 10.7 Loaduline analysis for Examples 101 and 10.2. Section 10.3 ZenereDiode VoltageeRegulator Circuits Example 10.2 Load—Line Analysis — Repeat Example V53 2 V and R : 50|uti0n If we let an : O and substitute values into Equation 10.5. we find that in L. 1 mA. This is plotted as point C in Figure 10.7. If we proceed as before by assuming that in = 0, we find that up : 10 V. This is a perfectly valid point on the load line, but it plots at a point far off the page. Of Course, we can use any other point satisfying Equation 10.5 to locate the load line. since we already have point C on the in axis, a good point to use would be on the rightehand edge of Figure 10.7. Thus, we assume that DD = 2 V and substitute Vglues into Equation 10.5, resulting in 2'3 : 0.8 mA. These values plot as point D. Then! we can draw the load line and find that the operating-point values are VDQ E V and [pg ; 0.93 I Exercise 10.3 Find the operating point for the circuit of Figure 10.5 if the diode characteristic is shown in Figure 10.8 and a. V55 : 2 V and R = 100 S2; 1). V55 : 15 Val-1d R=1kQ;C. V53 2 Answer a. VDQ g 1.1 V, IDQ r“: 9.0 mA; b. VDQ E 1.2 V, IDQ E 13.8 mA; '2. VDQ E V, [DQ g D 10.3 ZENER—DIODE VOLTAGE—REGULATOR CIRCUITS Sometimes, a circuit that produces constant output voltage while operating from a variable supply voltage is needed. Such circuits are called voltage regulators. For example, if we wanted to operate computer Circuits from the battery in an automobile, a voltage regulator would be needed. Automobile battery voltage typically varies between about 10 and 14 V (depending on the state of the battery in (mm 20 10 l : vi) (V) 0 0.5 1.0 1.5 2.0 2.5 3.0 Figure 10.8 Diode characteristic for Exercise 10.3. 475 When an intercept of the load line falls off the page, we select a point at the edge of the page. A voitage regulator circuit provides a nearly constant voltage to a load from a variable source. 476 Chapter 10 Diodes Figure 10.9 A simple regulator cir- K cuit that provides a nearly constant output voltage v0 from a variable supply voltage. Variable supply and whether or not the engine is running). Many computer circuits require a nearly constant voltage of 5 V. Thus, a regulator is needed that operates from the 10 to 14 V supply and produces a nearly constant S—V output. In this section, we use the load—line technique that we introduced in Section 10.2 to analyze a simple regulator circuit. The regulator circuit is shown in Figure 10.9. (For proper operation, it is necessary for the minimum value of the variable source voltage to be somewhat larger than the desired output voltage.) The Zencr diode has a breakdown voltage equal to the desired output voltage. The resistor R limits the diode current to a safe value so that the Zener diode does not overheat. Assuming that the characteristic for the diode is available, we can construct a load line to analyze the operation of the circuit. As before, we use Kirchhoff S voltage law to write an equation relating v5 and in. (In this circuit, the diode operates in the breakdown region with negative values for 119 and i9.) For the circuit of Figure 109, we obtain Vss + Rin + up = 0 (10.6) Once again, this is the equation of a Straight line, so location of any two points is sufficient to construct the load line. The intersection of the load line with the diode characteristic yields the operating point. Example 10.3 Load-Line Analysis of a Zener—Diode Voltage Regulator The voltage—regulator circuit of Figure 10.9 has R = ‘1 kg and uses 3 cher diOdS havmg the Characteristic Shown in Figure 10.10. Find the output voltage lOF Vss : 15 V. Repeat for V55 = 20 V. Solution The load lines for both values of V55 are shown in Figure 1010, The output voltages are determined from the points where the load lines intersecI the diode characteristic. The output voltages are found to be no 2 10.0 V for Vs: ‘ 15 V and v0 : 10.5 V for V3; 2 20 V. Thus, a S-V change in the supply voltage results in only a 0.5-V change in the regulated output voltage. Actual Zener diodes are capable of much better performance than this. slope of the characteristic has been accentuated in Figure 10.10 for clarity—eawill cher diodes have a more nearly vertical slope in breakdown. The al I Section 10.3 Zener—Diode Voltage-Regulator Circuits [D(U1A) Up (V) 1.. Load line for V55 2 15 v J Load line for j V55 = 20 V 710 —15 Zenerediode characteristic with accentuated slope 720 Figure 10.10 See Example 103. Slope of the Load Line Notice that the two load lines shown in Figure 10.10 are parallel. Inspection of Equation 10.5 or Equation 10.6 shows that the slope of the load line is 71 / R. Thus, a change of the supply voltage changes the position, but not the slope of the load line. Load—Line Analysis of Complex Circuits Any circuit that contains resistors, voltage sources, current sources, and a single two-terminal nonlinear element can be analyzed by the load—line technique. First, the Thévenin equivalent is found for the linear portion of the circuit as illustrated in Figure 10.11. Then, a load line is constructed to find the operating point on the characteristic of the nonlinear device. Once the operating point of the nonlinear element is known, voltages and currents can be determined in the original circuit. Nonlinear element RT ' I \ IIv VT :— O C her—J Linear circuit containing Thévenin voltage sources, current equivalent sources, and resistors circuit (a) Original circuit (b) Simplified circuit Figure 10.11 Analysis of a circuit containing a single non- linear element can be accomplished by load-line analysis of a simplified circuit. 477 Load lines for different source voltages (but the same resistance) are parallel. 478 Chapter 10 Diodes Example 10.4 Analysis of a Zener—Diode Regulator with a Load Consider the Zener-diode regulator circuit shown in Figure 10.12(a). The diode characteristic is shown in Figure 10.13. Find the load voltage UL and source current nirws=24V,R=12konmdaL:6ko. Solution First, consider the circuit as redrawn in Figure 10.12(b), in which we have grouped the linear elements together on the leftrhand side of the diode. Next, we find the Thévenin equivalent for the linear portion of the circuit. The Thévenin voltage is the open-circuit voltage (i.e., the voltage across [6,5 with the diode replaced by an open circuit), which is given by RL R + RI. The Thévenin resistance can be found by zeroing the voitage source and looking back into the circuit from the diode terminals. This is accomplished by reducing V55 to zero so that the voltage source becomes a short circuit. Then, we have R and RL in parallel, so the Thévenin resistance is i RRL _R+RL The resulting equivalent circuit is shown in Figure 10.12(c). Now, we can use Kirchhoff’s voltage law to write the load-line equation from the equivalent circuit as =20V VT : V35 RT :1k9 VT+RriD+UD:0 Using the values found for VT and RT, we can construct the load line shown in Figure 10.13 and locate the operating point. This yields UL : —vD : 10.0 V. 1"s (b) Circuit of (a) redrawn (c) Circuit with linear portion replaced by Thévenin equivalent Figure 10.12 See Exampie 10.4. Section lO.3 ZenereDiode Voltage~Regulator Circuits 479 5D (mA) J30 718 A16 —|4 ei2 —l€] —8 76 —4 72 (V) 110 Figure 10.13 Zener—diode characteristic for Example 10.4 and Ex— ercise 10.4. Once UL is known, we can find the voltages and currents in the original circuit. For example, using the output voltage value of 10.0 V in the original circuit of Figure 10.l2(a), we find that [5 2 (V3; — UL)/R : 11.67 mA. I Exercise 10.4 Find the voltage across the load in Example 104 if 3. RL 2 1.2 k9; i). R; : 400 {2. Answer a. 1),; ’5 9.4 V; 13. UL E 6.0 V. (Notice that this regulator is not perfect because the load voltage varies as the load current changes.) in Exercise 10.5 Consider the circuit of Figure 10.14(a). Assume that the breakdovm characteristic is vertical, as shown in Figure 10.14(b). Find the output voltage 229 for 3.111 : 0; b. ti = 20 mA; c. ti = 100 mA. [Hint Applying Kirchhoff’s voltage law to the circuit, we have 15 = 7 i9) — DD Construct a different load line for each value of iL.] k Zero current in reverse~ Zero voitage in forward Vertical bias region bias region k____ Load (a) Circuit diagram (b) Ideal Zener—diode characteristic Figure 10.14 See Exercise 10.5. ‘- I..- 480 Chapter 10 Diodes The ideal diode acts as a short circuit for forward currents and as an open circuit with reverse voltage applied. Answer a. v0 2 10.0 V; b. no : 10.0 V; c. v,7 : 5.0 V. (Notice that the regulator is not effective for large load currents.) D 1 0.4 IDEAL-DIODE MODEL Graphical load—line analysis is useful for some circuits, such as the voltage regulator studied in Section 10.3. However, it is too cumbersome for more complex circuits. Instead, we often use simpler models to approximate diode behavior. One model for a diode is the ideal diode, which is a perfect conductor with zero voltage drop in the forward direction. In the reverse direction, the ideal diode is an Open circuit. We use the ideal-diode assumption if our judgment tells us that the forward diode voltage drop and reverse current are negligible, or if we want a basic understanding of a circuit rather than an exact analysis. The vole-ampere characteristic for the ideal diode is shown in Figure 10.15.1f i D is positive, on is zero, and we say that the diode is in the on state. On the other hand, if an is negative, in is zero, and we sayr that the diode is in the off state. Assumed States for Analysis of Ideal-Diode Circuits In analysis of a circuit containing ideal diodes, we may not know in advance which diodes are on and which are off. Thus, we are forced to make a considered guess. Then, we analyze the circuit to find the currents in the diodes assumed to be on and the voltages across the diodes assumed to be off. If ED is positive for the diodes as- sumed to be on and if DD is negative for the diodes assumed to be off, our assumptions are correct, and we have solved the circuit. (We are assuming that ip is referenced positive in the forward direction and that Up is referenced positive at the anode-) Otherwise, we must make another assumption about the diodes and try again. After a little practice, our first guess is usually correct, at least for simple circuits. A step—by—step procedure for analyzing circuits that contain ideal diodes is 101 1. Assume a state for each diode, either on (i.e., a short circuit) or off (i.e., an Open circuit). For n diodes there are 2” possible combinations of diode states. Figure 10.15 ideal-diode volt—ampere characteristic. Section 10.4 2' Analyze the circuit to determine the current through the diodes assumed to be on and the voltage across the diodes assumed to be off. 3_ Check to see if the result is consistent with the assumed state for each diode. Current must flow in the forward direction for diodes assumed to be on. Furthermore, the voitage across the diodes assumed to be off must be positive at the cathode (ire, reverse bias). 4. If the results are consistent with the assumed states, the analysis is finished. Otherwise, return to step 1 and choose a different combination of diode states. Example 10.5 Analysis by Assumed Diode States Use the ideal-diode model to analyze the circuit shown in Figure 10.16(a). Start by assuming that [)1 is off and [)2 is on. Solution With Dr off and D2 on, the equivalent circuit is shown in Figure 10.l6(b). Solving results in in; : 0.5 mA. Since the current in D; is positive, our assumption that D; is on seems to be correct. However, continuing the solution of the circuit of Figure 10.l6(b), we find that um : +7 V. This is not consistent with the assumption that DI is off. Therefore, we must try another assumption. This time, we assame that Dr is on and D2 is off. The equivalent circuit for these assumptions is shown in Figure 10.16(c). We can soive this circuit to find that im : 1 mA and em = 73 V. These values are consistent with the assumptions about the diodes (D; on and D2 off) and, therefore, are correct. I 10V (b) Equivalent circuit assuming Dl off and D2 on (since vm 2+7 V, this assumption is not correct) (a) Circuit diagram (C) Equivalent circuit assuming D; on and D2 off (this is the correct assumption since im turns out to be a positive value and Um turns out negative) Figure 10.16 Analysis of a diode circuit, using the ideal-diode modei. See Example 10.5. Ideal-Diode Model 481 482 Chapter 10 Diodes In general, we cannot decide on the state of a particular diode until we have found a combination of states that works for all of the diodes in the circuit. (a) (b) (C) Figure 10.17 Circuits for Exercise 10.8. Notice in Example 10.5 that even though current flows in the forward direction of D2 for our first guess about diode states (01 off and D2 on), the correct solution is that D2 is off. Thus, in general, we cannot decide on the state of a particular diode until we have found a combination of states that works for all the diodes in the circuit. For a circuit containing n diodes, there are 2” possible states. Thus, an exhaustive search eventually yields the solution for each circuit. Exercise 10.6 Show that the condition D1 off and D2 off is not valid for the circuit of Figure 10.16(a). D Exercise 10.7 Show that the condition D1 on and D; on is not vaiid for the circuit of Figure 10.16(a). D Exercise 10.8 Find the diode states for the circuits shown in Figure 10.17. Assume ideal diodes. Answer a. D1 is on; b. D; is off; c. D3 is off and D4 is on. D 10.5 PIECEWISE—LINEAR DIODE MODELS Sometimes, we want a more accurate model than the ideal-diode assumption, but do not want to resort to nonlinear equations or graphical techniques. Then, We can use piecewise—linear models for the diodes. First1 we approximate the actual volt—ampere characteristic by straight~1ine segments. Then, we model each sectiOI1 of the diode characteristic with a resistance in series with a constant-voltage source Different resistance and voltage values are used in the various sections of the characteristic. Consider the resistance 1?a in series with a voltage source Va shown in Fig” ure 10.18(a). We can write the following equation, relating the voltage and cum?nt of the series combination: 7} = Raf + va (10-7) The current i is plotted versus a in Figure 10.18(b). Notice that the intercept on the voltage axis is at v = ‘Va and that the slope of the line is l/Ra. Given a straight-line volt—ampere characteristic, we can work backward t0 find the corresponding series voltage and resistance. Thus, after a nonlinear volteamper‘e characteristic has been approximated by several straight-line segments, :3 CirCmt model consisting of a voltage source and series resistance can be found for each segment. Section 10.5 Piecewise—Linear Diode Models 483 (I Ru [1 1L Va —" Figure 10.18 Circuit and 1— — volt»ampere characteristic for piecewiseqinear model; (a) Circuit diagram (b) Voltkampere characteristic Example 10.6 Piecewise-Linear Model for a Zener Diode Find circuit models for the Zener—diode volt—ampere characteristic shown in Fig- ure10.19. Use the straight-line segments shown. Solution For line segment A of Figure 10.19, the intercept on the voltage axis is 0.6V and the reciprocal of the slope is 10 S2. Hence, the circuit model for the diode on this segment is a 10-82 resistance in series with a 0.6-V source, as shown in the figure. Line Segment B has zero current, and therefore, the equivalent circuit for segment 8 is an open circuit, as illustrated in the figure. Finally, line segment C has an intercept of —6 V and a reciprocal slope of 12 S2. resulting in the equivalent circuit shown. Thus, this diode can be approximated by one of these linear circuits, depending on where the operating point is located. I Example 10.7 Analysis Using a Piecewise-Linear Model Use the circuit models found in Example 10.6 to solve for the current in the circuit of Figure 10.20(a). Solution Since the 3-V source has a polarity that results in forward bias of the diode, we assume that the operating point is on line segment A of Figure 10.19. Consequently, the equivalent circuit for the diode is the one for segment A. Using this equivalent circuit, we have the circuit of Figure 10.20(b). Solving, we find that it) = 80 mA. I Exercise 10.9 Use the appropriate circuit model from Figure 10.19 to solve for v0 iflthe circuit of Figure 10.21 if :1. RL : 10 k9 and b. RL : 1 k9. (Hint: Be sure that Your answers are consistent with your choice of equivalent circuit for the diodeithe Various equivalent circuits are valid only for specific ranges of diode voltage and Current. The answer must fall into the valid range for the equivalent circuit used.) Answer a. v0 = 6.017 V; b. v0 : 3.333 V. D 434 Chapter 10 Diodes IGQ segment C + 0.6 V 100 mA — — _ - — — — — I Line : segment A I k ' | | | Diode = open Circuit : i —7.2 7 6.0 \ // t 1 ,’ I r I + a ,’ j 0.6 1.6 __ 6 V : Line 7 u I segment B _ I 12 Q | Line 1 | i i I Diode of Figure 10.19 (a) Circuit diagram (b) Circuit with diode modeled by the equivalent circuit for the forward—bias region Figure 10.20 Circuit for Example 10.7. Figure 1021 Exercise 10.9. Circuit for Diode of Figure |().19 1:3 (V) Section i0.5 Piecewise—Linear Diode Models i (mA) 12.5 ‘ — 50.0 — — — — — ‘ _ m i i ’ Segment B a f; 5 0 7 7 7 fl _ _ _ _ _ Segment A: O—W—LO ' Segrent \ 400 Q a + _ b 00/) Segment B: ill—0 100 Q 15 V L) a b a _ + j, fl segmentc: O—VW—iiEi—O u 800 (2 55 V Figure (a) Voltiampere characteristic (b) Equivalent circuits 10.22 Hypothetical nonlinear device for Exercise 10.10. Exercise 10.10 Find a circuit model for each line segment shown in Figure 1022(3). Draw the circuit models identifying terminals :1 and b for each equivalent circuit. Answer See to terminals a and b. D Figure 10.2203). Notice the polarity of the voltage sources with respect Simple Piecewise-Linear Diode Equivalent Circuit Figure 10.23 shows a simple piecewise-linear equivalent circuit for diodes that is often sufficiently accurate. It is an open circuit in the reverse~bias regioa and a constant voltage drop in the forward direction. This model is equivalent to a battery in series with an ideal diode. : open circuit Figure 10.23 Simple piecewiseAiinear equivalent for the diode. 485 ...
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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+1+from+Humbley - Section 10.1 10.1...

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