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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 reversebias 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)
—eL
Anode O‘Mho Cathode
+ v _
D Reverse I ‘ I
breakdown Iﬁ‘ Reversebras 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
reversebias voltage is applied
to the diode, operation enters
the reversebreakdown 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 oneway valve in a fluidflow system, as illustrated in Figure 10nd), If a sufﬁciently large reverse—bias voltage is applied to the diode, operation
enters the reversebreakdown 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 ntype 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 ptype 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 reversebias 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 forwardbias 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 forwardbias region, smallsignal 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 forwardbias 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 reversebias 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 LoadLine 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 Loadline 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 : 50uti0n 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 operatingpoint 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
Val1d 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 LoadLine 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 SV change in the supply voltage
results in only a 0.5V 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 VoltageRegulator 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
twoterminal 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 loadline 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 Zenerdiode 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 opencircuit 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 loadline 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 IDEALDIODE 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 idealdiode 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 voleampere 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 IdealDiode 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 idealdiode
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 idealdiode 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 idealdiode modei. See Example 10.5. IdealDiode 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 idealdiode 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 constantvoltage 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 (107) 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 straightline 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 straightline 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 PiecewiseLinear Model for a Zener Diode Find circuit models for the Zener—diode volt—ampere characteristic shown in Fig
ure10.19. Use the straightline 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 1082 resistance in series with a 0.6V 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 PiecewiseLinear 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 3V 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 PiecewiseLinear Diode Equivalent Circuit Figure 10.23 shows a simple piecewiselinear 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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 Spring '08
 HENRYLEE

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