Diodes
1
3.1 THE IDEAL DIODE
3.1.1Current-Voltage Characteristic
i–v characteristic
diode circuit symbol
equivalent circuit in the reverse direction
equivalent circuit in the forward direction
Figure 3.1 The ideal diode: (a) diode circuit symbol; (b) i–v characteristic; (c) equivalent
circuit in the reverse direction; (d) equivalent circuit in the forward direction.
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If a negative voltage (relative to the reference direction indicated in
Fig. 3.1(a)) is applied to the diode, no current flows and the diode
behaves as an open circuit. Diodes operated in this mode are said to
be reverse biased, or operated in the reverse direction.
An ideal diode has zero current when operated in the reverse
direction and is said to be cut off, or simply off.
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On the other hand, if a positive current (relative to the reference
direction indicated in Fig.3.1(a)) is applied to the ideal diode, zero
voltage drop appears across the diode. In other words, the ideal diode
behaves as a short circuit in the forward direction (Fig.3.1(d)); it
passes any current with zero voltage drop.
A forward-biased diode is said to be turned on, or simply on.
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It should be noted that the external circuit must be designed to limit
the forward current through a conducting diode, and the reverse
voltage across a cutoff diode, to predetermined the values.
Figure 3.2 The two modes of operation of ideal diodes and the use of an external circuit to limit the
forward current (a) and the reverse voltage (b).
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The positive terminal of the diode is called the anode and the
negative terminal is called the cathode.
The i-v characteristic of the ideal diode is highly nonlinear;
although it consists of two straight-line segments, they are at 90o to
one another. A nonlinear curve that consists of straight-line
segments is said to be piecewise linear.
If a device having a piecewise-linear characteristic is used in a
particular application in such a way that the signal across its
terminals swings along only one of the linear segments, then the
device can be considered a linear circuit element as far as that
particular circuit application is concerned. On the other hand, if
signals swing past one or more of the break points in the
characteristic, linear analysis is no longer possible.
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3.1.2 A Simple Application: The Rectifier
A fundamental application of the diode. One that makes use of its
severely nonlinear i-v curve, is the rectifier.
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Figure 3.3 (a) Rectifier circuit. (b) Input waveform. (c) Equivalent circuit when vI 0. (d)
Equivalent circuit when vI  0. (e) Output waveform.
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EXERCISE 3.1
For the circuit in Fig. 3.3 (a), sketch the transfer characteristic vO
versus vI.
Figure E3.1
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EXERCISE 3.2
For the circuit in Fig. 3.3 (a), sketch the waveform of vD.
vD+vO= vI  vD=vI - vO
Figure E3.2
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EXAMPLE 3.1
Figure 3.4(a) shows a circuit for charging a 12-V battery. If vs is a
sinusoid with 24-V peak amplitude, find the fraction of each cycle
during which the diode conducts. Also, find the peak value of the
diode current and the maximum reverse-bias voltage that appears
across the diode.
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The diode conducts when vs exceeds 12 V, as shown in Fig. 3.4(b).
The conduction angle is 2, where is given by
24COSθ12  60o
Thus, the conduction angle is 120o, or one-third of a cycle.
Start of a cycle
End of a cycle
24COSθ12
Figure 3.4 Circuit and waveforms for Example 3.1.
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The peak value of the diode current is given by
24 12
Id 
0.12A
100
The maximum reverse voltage across the diode occurs when vs is at
its negative peak and is equal to 24 +12 =36 V.
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3.1.3 Another Application: Diode Logic Gates
Diodes together with resistors can be used to implement digital logic
functions.
Y A B C
Y A 
B
C
Figure 3.5 Diode logic gates: (a) OR gate; (b) AND gate (in a positive-logic system).
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EXAMPLE 3.2
Assume the diodes to be ideal, find the values of I and V in the
circuits of Fig. 3.6
Figure 3.6 Circuits for Example 3.2.
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Assume that both the diodes are conducting.
I D2
At node B
10 0

1mA
10
0 (10)
I 1 
5
Results in I = 1mA. Thus D1 is conducting as originally assumed,
and the final result is I = 1mA and V = 0V.
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For the circuit in Fig. 3.6(b). If we assume that both diodes are
conducting, then VB=0 and V=0. The current in D2 is obtained from
I D2
10 0

2mA
5
At node B
0 (10)
I 2 
10
Which yields I = -1 mA. Since this is not possible, our original
assumption is not correct. We start again, assuming that D1 is
off and D2 is on. The current ID2 is given by
I D2
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10 (10)

1.33mA
15
17
And the voltage at node B is
VB 10 10 1.33 3.3V
Thus D1 is reverse biased as assumed, and the final results is I=0
and V=3.3V.
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3.2 TERMINAL CHARACTERISTICS OF JUNCTION DIODES
Figure 3.7 The i–
v characteristic of a silicon junction diode.
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The characteristic curve
consists of three distinct
regions:
1. The forward-bias,
region, determined by
v > 0
2. The reverse-bias,
region, determined by
v < 0
3. The breakdown region,
determined by
v < -VZK
Figure 3.8 The diode i–v relationship with some scales expanded and others compressed in order to
reveal details.
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3.2.1 The Forward-Bias Region
The forward-bias -or simply forward- region of operation is entered
when the terminal voltage v is positive.
In the forward region the i-v relationship is closely approximated by
i I S (e v / nVT 1)
Where IS , usually called as the saturation current, is a constant for a
given diode at a given temperature. (A formula for IS in terms of the
diode’
s physical parameters and temperature will be given in section
3.7)
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Another name for IS, and one that will occasionally use, is the scale
current. The name arises from the fact that IS is directly proportional
to the cross-sectional area of the diode. Thus doubling of the junction
area results in a diode with double the value of IS and, as the diode
equation indicates, double the value of current I for a given forward
voltage v.
For “
small-signal”diodes, which are small-size diodes intended for
low-power applications, IS is on the order of 10-15 A. The value of IS
is, however, a very strong function of temperature. As a rule of thumb,
IS doubles in value for every 5oC rise in temperature.
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The voltage VT is a constant called the thermal voltage and is given
by
kT
VT 
q
where
k = Boltzmann’
s constant = 1.3810-23 joule/kelvin
T = the absolute temperature in kelvin = 273 + temperature
in oC
q = the magnitude of electron charge = 1.60 10-19 columb
At room temperature (20oC) the value of VT is 25.2 mV. In rapid
approximate circuit analysis we shall use VT 25mV at room
temperature.
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In the diode equation the constant n has a value between 1 and 2,
depending on the material and the physical structure of the diode.
Diode made using the standard integrated-circuit fabrication process
exhibit n = 1 when operated under normal conditions.
Diodes available as discrete two-terminal components generally
exhibits n = 2.
In general, we shall assume n = 1 unless otherwise specified.
i I S (e v / nVT 1)
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For appreciable current i in the forward direction, specifically for
i>>IS we can have
i I S (e v / nVT 1) I S e v / nVT
This relationship can be expressed alternately in the logarithmic
form
i
v nVT ln
IS
nature (base e) logarithmic
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Let us consider the forward i-v relationship and evaluate the current
I1 corresponding to a diode voltage V1:
I1 I S eV1 / nVT
Similarly, if the voltage is V2, the diode current I2 well be
I 2 I S eV2 / nVT
These two equations can be combined to produce
I2
e (V2 V1 ) / nVT
I1
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which can be rewritten as
I2
V2 V1 nVT ln
I1
or, in terms of base-10 logarithms,
I2
V2 V1 2.3nVT log
I1
which means that, for a decade change in current, the diode
voltage drop changes by 2.3nVT per decade of current. (which is
approximately 60 mV for n=1 and 120 mV for n=2)
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Usually, the diode i-v relationship is most conveniently plotted on
semilog paper. Using the vertical, linear axis for v and the
horizontal, log axis for i, one obtains a straight line with a slope of
2.3nVT per decade of current.
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A glance at the i-v characteristics in the forward region reveals that
the current is negligibly small for v smaller than about 0.5V. This
value is usually referred to as the cut-in voltage.
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It should be emphasized however, that this apparent threshold in the
characteristic is simply a consequence of the exponential relationship.
Another consequence of this relationship is the rapid increase of i.
Thus, for a “
fully conducting”diode, the voltage drop lies in a
narrow range, approximately 0.6 to 0.8V. This gives rise to a simple
“
model”for the diode where it is assumed that a conducting diode
has approximately a 0.7-V drop across it.
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Diodes with different current ratings (i.e., different areas and
correspondingly different IS) will exhibit the 0.7-V drop at different
currents. For instance, a small-signal diode may be considered to
have a 0.7-V drop at i=1 mA, while a higher-power diode may have
a drop 0.7-V drop at i=1 A.
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EXAMPLE 3.3
A silicon diode said to be a 1-mA device displays a forward
voltage of 0.7V at a current of 1mA. Evaluate the junction scaling
constant IS in the event that n is either 1 or 2. What scaling
constants would apply for a 1-A diode of the same manufacture
that conducts 1A at 0.7V?
Since
i I S e v / nVT
then
I S ie v / nVT
For the 1-mA diode
If n=1: IS=10-3e-700/25=6.910-16A, or about 10-15A
If n=2: IS=10-3e-700/50=8.310-10A, or about 10-9A
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The diode conducting 1A at 0.7V corresponding to one-thousand 1mA diodes in parallel with a total junction area 1000 times greater.
Thus IS is also 1000 times greater, being 1 pA and 1 A.
From this example it should be apparent that the value of n used can
be quite important.
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Since both IS and VT are functions of temperature, the forward i-v
characteristic varies with temperature, as illustrated in Fig. 3.9. At a
given constant diode current the voltage drop across the diode decreases
by approximately 2 mV for every 1oC increase in temperature.
Figure 3.9 Illustrating the temperature dependence of the diode forward characteristic. At a
constant current, the voltage drop decreases by approximately 2 mV for every 1
C increase in
temperature.
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3.2.2 The Reverse-Bias Region
The reverse-bias region of operation is entered when the diode voltage v
is made negative. If v is negative and a few times larger than VT (25mV)
in magnitude, the exponential term becomes negligibly small compared
to unity, and the diode current become
i I S
That is, the current in the reverse direction is constant and equal to IS.
This constancy is the reason behind the term saturation current.
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Real diodes exhibit reverse currents that, though quite small, are
much larger than IS. For instance, a small-signal diode whose IS is on
the order of 10-14 A to 10-15 A could show a reverse current on the
order of 1 nA.
The reverse current also increases somewhat with the increase in
magnitude of reverse voltage.
A large part of the reverse current is due to leakage effects. These
leakage currents are proportional to the junction area, just as IS is.
Their dependence on temperature, however, is different from that of
IS. Thus, whereas IS doubles for every 5oC rise in temperature, the
corresponding rule of thumb for the temperature dependence of the
reverse current is that it doubles for every 10oC rise in temperature.
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3.2.3 The Breakdown Region
The breakdown region is entered when the magnitude of the reverse
voltage exceeds a threshold value that is specific to the particular diode,
called the breakdown voltage. This is the voltage at the “
knee”of the i-v
curve in Fig.3.8 and is denoted VZK.
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In the breakdown region, the reverse current increases rapidly, with the
associated increase in voltage drop being very small. Diode breakdown is
normally not destructive provided that the power dissipated in the diode
is limited by external circuitry to a “
safe”level.
This safe value is normally specified on the device data sheets. It
therefore is necessary to limit the reverse current in the breakdown
region to a value consistent with the permissible power dissipation.
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3.3 MODELING THE DIODE FORWARD CHARACTERISTIC
Fig. 3.10 shows the circuit employing forward-conducting diodes. We
wish to analyze this circuit to determine the diode voltage VD and current
ID.
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3.3.1 The Exponential Model
The most accurate description of the diode operation in the forward
region is provided by the exponential model. Unfortunately, however, its
severely nonlinear nature makes this model the most difficult to use.
Assume that VDD > 0.5V.
Also
I D I S eVD / nVT
VDD VD
ID 
R
Two alternative ways for obtaining the
solution are graphical analysis and
iterative analysis.
Figure 3.10 A simple circuit used to illustrate the analysis of circuits in which the diode is
forward conducting.
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3.3.2 Graphical Analysis Using the Exponential
I D I S eVD / nVT
VDD VD
ID 
R
Figure 3.11 Graphical analysis of the circuit in Fig. 3.10 using the exponential diode model.
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3.3.3 Iterative Analysis Using the Exponential Model
EXAMPLE 3.4
Determine the current ID and the diode voltage VD for the circuit in
Fig. 3.10 with VDD=5V and R=1k. Assume that the diode has a
current of 1 mA at a voltage of 0.7V and that its voltage drop
changes by 0.1V for every decade changing in current.
Assume that VD=0.7V and
VDD VD 5 0.7
ID 

4.3mA
R
1
We then use the diode equation to obtain a better estimate for VD.
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I2
V2 V1 2.3nVT log
I1
For our case 2.3nVT=0.1. Thus
I2
V2 V1 0.1log
I1
Substituting V1=0.7V, I1=1mA, and I2=4.3mA results in V2=0.763V.
Thus the results of the first iteration are ID=4.3mA and VD=0.763V.
The second iteration proceeds in a similar manner.
5 0.763
ID 
4.237mA
1
4.237 

V2 0.763 0.1log 
0.762V

 4 .3 
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Thus the second iteration yields ID=4.237 mA and VD=0.762V.
Since these values are not much different from the values obtained
after the first iteration, no further iterations are necessary, and the
solution is ID=4.237 mA and VD=0.762V.
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3.3.4 The Need for Rapid Analysis
To speed up the analysis process, we must find simpler models for the
diode forward characteristic.
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3.3.5 The Piecewise-Linear Model
iD=0, vDVD0
iD=(vD-VD0)/rD, vDVD0
For the present case,
VD0=0.65V, rD=20 
Figure 3.12 Approximating the diode forward characteristic with two straight lines: the
piecewise-linear model.
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The piecewise linear model described by the previous equations can
be represented by the equivalent circuit.
Battery-plus-resistance model
Figure 3.13 Piecewise-linear model of the diode forward characteristic and its equivalent
circuit representation.
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EXAMPLE 3.5
Repeat the problem in Example 3.4 utilizing the piecewise-linear
model whose parameters are giving in Fig. 3.12 (VD0=0.65V, rD=20).
Note that the characteristics depicted in this figure are those of the
diode described in Example 3.4 (1mA at 0.7V and 0.1v/decade)
Replacing the diode with the equivalent
circuit model, we can have
V VD 0 5 0.65
I D  DD

4.26mA
R rD
1 0.02
Where from 3.12, VD0=0.65V and rD=20 
Figure 3.14 The circuit of Fig. 3.10 with the diode replaced with its piecewise-linear model of Fig.
3.13.
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Then we can have the diode voltage VD be calculated as
VD VD 0 I D rD 0.65 4.26 0.02 0.735V
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3.3.6 The Constant-Voltage-Drop Model
0.8V at 10 mA
0.6V at 0.1 mA
VD=0.7V
Figure 3.15 Development of the constant-voltage-drop model of the diode forward
characteristics. A vertical straight line (B) is used to approximate the fast-rising exponential.
Observe that this simple model predicts VD to within 0.1 V over the current range of 0.1 mA to
10 mA.
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By using this model and assume that VD=0.7V, we can have
VDD VD 5 0.7
ID 

4.3mA
R
1
Figure 3.16 The constant-voltage-drop model of the diode forward characteristics and its equivalentcircuit representation.
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3.3.7 The Ideal-Diode Model
In applications that involve voltages much greater than the diode
voltage drop (0.6-0.8V), we may neglect the diode voltage drop
altogether while calculating the diode current. The result is idealdiode model.
By using this model and assume that VD=0, we can have
VDD VD 5 0
ID 

5mA
R
1
Which for a very quick analysis would not be bad a gross
estimate. However, with almost no additional work, the 0.7-V
drop model yields much more realistic results.
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3.3.8 The Small-Signal Model
There are applications in which a diode is biased to operate at a point
on the forward i-v characteristic and small ac signal is superimposed
on the dc quantities. For this situation, we first have to determine the
dc operating point (VD and ID) of the diode using one of the models
discussed above. Most frequently, the 0.7-V-drop model is utilized.
Then, for a small-signal operation around the dc bias point, the diode
is best modeled by a resistance equal to the inverse of the slope of
the tangent to the exponential i-v characteristic at the bias point.
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DC + small signal
Figure 3.17 Development of the diode small-signal model. Note that the numerical values shown
are for a diode with n = 2.
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In the absence of the signal vd(t) the diode voltage is equal to VD, and
correspondingly, the diode will conduct a dc current ID given by
I D I S eVD / nVT
When the signal vd(t) is applied, the total instantaneous diode
voltage vD(t) will be given by
vD (t ) VD vd (t )
Correspondingly, the total instantaneous diode current iD(t) will be
iD I S e vD / nVT
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iD I S e
(VD v D ) / nVT
iD (t ) I S e
VD / nVT
Recall that
I D I S eVD / nVT
therefore
iD I D e
e
vd / nVT
vd / nVT
Now if the amplitude of the signal vd(t) is kept sufficiently small
such that
vd
1
nVT
Then we may expand the exponential of Eq. (3.12) in a series and
truncate the series after the first two terms to obtain the
approximate expression
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 vd
iD I D 
1

 nVT




This is the small-signal approximation. It is valid for signals whose
amplitudes are smaller than about 10 mV for the case n =2 and 5 mV
for n =1.
From the above equation, we have
ID
iD I D 
vd
nVT
Thus, superimposed on the dc current ID, we have a signal current
component directly proportional to the signal voltage vd. That is
iD I D id
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ID
id 
vd
nVT
57
The quantity relating the signal current id to the signal voltage vd has
the dimensions of conductance, mho (1/), and is called the diode
small-signal conductance. The inverse of this parameter is the diode
small-signal resistance, or incremental resistance, rd.
nVT
rd 
ID
Note that the value of rd is inversely proportional to the bias current
ID.
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It is easy to see that using the small-signal approximation is
equivalent to assuming that the signal amplitude is sufficiently small
such that the excursion along the i-v curve is limited to a short
almost-linear segment. The slope of this segment, which is equal to
the slope of the tangent to the i-v curve at the operating point Q, is
equal to the small-signal conductance.
The slope of the i-v curve at i = ID is equal to ID/nVT, which is 1/rd,
that is
rd  1
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

iD

v D i I
D
D
59
From the preceding we conclude that superimposed on the quantities
VD and ID that define the dc bias point, or quiescent point, of the
diode will be the small-signal quantities vd(t) and id(t), which are
related by the diode small-signal resistance rd evaluated at the bias
point.
Thus the small-signal analysis can be performed separately from the
dc bias analysis, a great convenience that results from the
linearization of the diode characteristics inherent in the small-signal
approximation.
Specifically, after the dc analysis is performed, the small-signal
equivalent circuit is obtained by eliminating all dc sources and
replacing the diode by its small-signal resistance.
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EXAMPLE 3.6
Consider the circuit shown in Fig. 3.18(a) for the case in which R=10
k. The power supply V+ has a dc value of 10 V on which is
superimposed a 60-Hz sinusoid of 1-V peak amplitude. Calculate both
the dc voltage of the diode and the amplitude of the sine-wave signal
appearing across it. Assume the diode to have a 0.7-V drop at 1-mA
current and n=2.
Figure 3.18 (a) Circuit for Example 3.6. (b) Circuit for calculating the dc operating point. (c) Smallsignal equivalent circuit.
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Consider dc quantities first.
VDD VD 10 0.7
ID 

0.93mA
R
10
Since this value is very close to 1 mA, the diode voltage will be
very close to the assumed value of 0.7 V. At this operating point,
the diode increment resistance rd is
nV
2 25
rd  T 
53.8
ID
0.93
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The signal voltage across the diode can be found from the smallsignal equivalent circuit in Fig. 3.18(c). Here vs denotes the 60 –Hz
1 –V peak sinusoidal component of V+, and vd is the corresponding
signal across the diode. Using the voltage-divider rule provides the
peak amplitude of vd as follows:
 rd
0.0538
vd (peak ) VS
1
5.35mV
R rd
10 0.0538
Finally we note that since this value is quite small, our use of smallsignal model of the diode is justified.
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3.3.9 Use of the Diode Forward Drop in Voltage Regulation
A further application of the diode small-signal model is found in a
popular diode application, namely the use of diodes to create a
regulated voltage. A voltage regulator is a circuit whose purpose is to
provide a constant dc voltage between its output terminals.
The output voltage is required to remain as constant as possible in
spite of (a) changes in the load current drawn from the regulator
output terminal and (b) changes in the dc power-supply voltage that
feeds the regulator circuit.
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Since the forward voltage drop of the diode remains almost constant
at approximately 0.7 V while the current through it varies by
relatively large amounts, a forward-biased diode can make a simple
voltage regulator. For instance, we have seen in Example 3.6 that
while the 10 –V dc supply voltage had a ripple of 2 V peak-to-peak
(a 10% variation), the corresponding ripple in the diode voltage
was only about 5.4 mV (a 0.8% variation).
Regulated voltages greater than 0.7V can be obtained by connecting
a number of diodes in series. For example, the use of three forwardbiased diodes in series provides a voltage of about 2 V. One such
circuit is investigated in the following example.
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EXAMPLE 3.7
Consider the circuit shown in Fig. 3.19. A string of three diodes in used
to provide a constant voltage of about 2.1 V. We want to calculate the
percentage change in this regulated voltage caused by (a)a 10%
change in the power-supply voltage and (b) connection of a 1-kload
resistance. Assume n = 2.
With no load, the nominal value of the
current in the diode string is given by
10 3 0.7
ID 
7.9mA
1
Figure 3.19 Circuit for Example 3.7.
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Thus each diode will have an increment resistance of
nVT 2 25
rd 

6.3
I
7 .9
The three diodes in series will have a total incremental resistance of
r 3rd 18.9
This resistance, along with the resistance R, forms a voltage divider
whose ratio can be used to calculate the change in output voltage
due to a 10% (i.e., 1-V) change in supply voltage. Thus the
peak-to-peak change in output voltage will be
vO 2
r
0.0189
2
37.1mV
r R
0.0189 1
That is, corresponding to the 1-V change in supply voltage, the
output voltage will change by 18.5mV or 0.9%. Since this
implies a change of about 6.2 mV per diode, our use of the
small-signal model is justified.
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67
When a load resistance of 1-kis connected across the diode string, it
draws a current of approximately 2.1 mA. Resulting in a decrease in
voltage across the diode string given by
vO 2.1r 2.118.9 39.7 mV
Since this implies that the voltage across each diode decreases by
about 13.2 mV, our use of the small-signal model is not entirely
justified. Nevertheless, a detailed calculation of the voltage change
using the exponential model results in vO = -35.5mV, which is not
too different from the approximate value obtained using the
incremental model.
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68
3.3.10 Summary
As a summary of this important section on diode modeling, please
refer to Table 3.1(p.165~p.166).
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69
Exponential
model
IS is on the order of 10-12 A to 10-15 A depending on junction area.
VT is about 25 mV, n = 1 to 2.
Physical based and remarkable accurate model. Useful when
accurate analysis is needed.
Table 3.1 Modeling the Diode Forward Characteristic
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70
Piecewiselinear model
1
vD VD 0 
For vDVD0; iD=0. For vDVD0; iD  
rD
Not frequently used!
Table 3.1 Modeling the Diode Forward Characteristic
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Constantvoltage-drop
model
For iD>0, vD=0.7V.
Easy to use and very popular for the quick hand analysis that is
essential in circuit design.
Table 3.1 Modeling the Diode Forward Characteristic
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72
Ideal diode
For iD>0, vD=0V.
Good for determining which
diodes are conducting and
which are cutoff in a multiple
diode circuit.
Good for obtaining very
approximate values for diode
currents, especially when the
circuit voltages are much
greater than VD.
Table 3.1 (Continued)
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Small signal model
For small signals superimposed on VD
and ID:
id=vd/rd
rd=nVT/ID
For n =1, vd is limited to 5 mV, for n=2,
10 mV.
Table 3.1 (Continued)
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3.4 OPERATION IN THE REVERSE BREAKDOWN REGIONZENER DIODES
The very steep i-v curve that the diode exhibits in the breakdown
region and the almost-constant voltage drop suggest that diode
operating in the breakdown region can be used in the design of
voltage regulators.
Special diodes are manufactured to operate specifically in the
breakdown region. Such diodes are called breakdown diodes or,
more commonly, zener diodes.
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75
3.4.1 Specifying and Modeling the Zener Diode
rZ is the
incremental
reistance of
the zener
diode at
operating
point Q, it is
also known
as the
dynamic
resistance.
For I > IZK, the i-v
characteristic is almost
a straight line.
NFUEE Weihsing Liu
Knee current
The manufacturer usually
specify the voltage across the
zener diode VZ at a test
current IZT.
As the current through the zener deviates from
IZT, the voltage across it will change.
76
Typically, rZ is in the range of a few ohms to a few tens of ohms.
Obviously, the lower the value of rZ is, the more constant the zener
voltage remains as its current varies and thus the more ideal its
performance becomes in the design of voltage regulators.
A general design guideline, one should avoid operating the zener in
the low-current region.
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77
Zener diodes are fabricated with voltages VZ in the range of a few
volts to a few hundred volts. In addition to specifying VZ, rZ, and IZK,
the manufacturer also specifies the maximum power that the device
can safely dissipate. Thus a 0.5 W, 6.8V zener diode can operate
safely at currents up to a maximum of about 70 mA.
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78
The almost linear i-v characteristic of the zener diode suggests that
the device can be modeled as indicated in Fig. 3.22 (p. 76). Here
VZ0 denotes the point at which the straight line of slope 1/rZ
intersects the voltage axis. Although VZ0 is shown to be slightly
different from the knee voltage VZK, in practice their values are
almost equal. The equivalent circuit model of Fig. 3.22 can be
analytically described by
VZ VZ 0 rZ I Z
And it applies for IZ > IZK and,
obviously, VZ > VZ0.
Figure 3.22 Model for the zener diode.
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79
3.4.2 Use of the Zener as a Shunt Regulator
We now illustrate, by way of an example, the use of zener diodes in
the design of shunt regulators, so named because the regulator
circuit appears in parallel (shunt) with the load.
EXAMPLE 3.8
The 6.8-V zener diode in the circuit of Fig. 3.23(a) is specified to have
VZ=6.8 V at IZ =5 mA, rZ =20 , and IZK =0.2 mA. The supply voltage
V+ is nominally 10 V but can vary by 1V.
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(a) Find VO with no load and with V+ at its nominal value.
(b) Find the change in VO resulting from the 1-V change in V+. Note
that (VO/ V+ ), usually expressed in mV/V, is known as line
regulation.
(c) Find the change in VO resulting from connecting a load resistance RL
that draws a current IL = 1 mA, and hence find the load regulation (VO/
IL) in mV/mA.
(d) Find the change in VO when RL = 2 k .
(e) Find the change in VO when RL = 0.5 k .
(f) What is the minimum value of RL for which the diode still operates in
the breakdown region?
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81
First we must determine the value of the parameter VZO of the zener
diode model. Substituting VZ=6.8 V at IZ =5 mA, rZ =20 in Eq.
3.20 yields VZ0 = 6.7V. (refer to Fig. 3.23(b))
VZ VZ 0 rZ I Z
Eq. 3.20
(a) With no load connected, the current through the zener is given
by
V VZ 0
10 6.7
I Z I 

6.35mA
R rZ
0.5 0.02
thus
VO VZ 0 rZ I Z 6.7 6.35 0.02 6.83V
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82
Figure 3.23 (a) Circuit for Example 3.8. (b) The circuit with the zener diode
replaced with its equivalent circuit model.
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83
(b) For a 1-V change in V+, the change in output voltage can be found
from
20
 rZ
VO V
1
38.5mV
R rZ
500 20
(c) When a load resistance RL that draws a load current IL = 1mA is
connected, the zener current will decrease by 1 mA. The corresponding
change in zener voltage can be found from
VO rZ 
I Z 20 
(1) 20mV
Thus the load regulation is
V
Load_regulation  O 20mV/mA
I L
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84
(d) When a load resistance of 2 k is connected, the load current will be
approximately 6.8V/ 2 k = 3.4 mA. Thus the change in zener current
will be IZ= -3.4mA, and the corresponding change in zener voltage
(output voltage) will thus be
VO rZ 
I Z 20 
(3.4) 68mV
This calculation, however, is approximate, because it neglects the
change in the current I. A more accurate estimate of VO can be
obtained by analyzing the circuit in Fig. 3.23(b). The result of such an
analysis is VO = -70mV.
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85
(e) An RL of 0.5 k would draw a load current of 6.8/0.5=13.6 mA.
This in not possible, because the current I supplied through R is only
6.4 mA (for V+ = 10V). Therefore, the zener must be cut off. If this is
indeed the case, then VO is determined by the voltage divider formed
by RL and R,
RL
0.5
VO V
10 
5V
R RL
0.5 0.5

Since this voltage is lower than the breakdown voltage of the zener,
the diode is indeed no longer operating in the breakdown region.
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86
(f) For the zener to be at the edge of the breakdown region,
IZ=IZK=0.2mA and VZVZK6.7V. At this point the lowest current
supplied through R is (9-6.7)/0.5=4.6 mA, and thus the load current is
4.6-0.2=4.4mA. The corresponding value of RL is
6. 7
RL  1.5K
4.4
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87
3.4.3 Temperature Effect
The dependence of the zener voltage VZ on temperature is specified in
terms of its temperature coefficient TC, or temco as it is commonly
known, which is usually expressed in mV/oC.
The value of TC depends on the zener voltage, and for a given diode
the TC varies with the operating current. Zener diodes whose VZ are
lower than about 5 V exhibits a negative TC. On the other hand,
zeners with higher voltages exhibit a positive TC.
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88
The TC of a zener diode with a VZ of about 5 V can be made zero by
operating the diode at a specified current. Another commonly used
technique for obtaining a reference voltage with low temperature
coefficient is to connect a zener diode with a positive temperature
coefficient of about 2 mV/oC in series with a forward-conducting
diode. Since the forward-conducting diode has a voltage drop of
0.7V and a TC of about –2 mV/oC, the series combination will
provide a voltage of (VZ+0.7V) with a TC of about zero.
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89
3.5 RECTIFIER CIRCUITS
Figure 3.24 Block diagram of a dc power supply.
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90
3.5.1 The Half-Wave Rectifier
The half-wave rectifier utilizes alternate half-cycles of the input
sinusoid. Fig. 3.25(a) shows the circuit of half-wave rectifier.
From Fig.3.25(b), we can have
vO = 0,
vS<VD0
R
R
vO 
vS VD 0
,
R rD
R rD
vSVD0
The transfer characteristic represented by these equations is
sketched in Fig. 3.25(c). In many applications, rD << R and the
second equation can be simplified to
vO vS VD 0
0.7 or 0.8 V
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91
The output voltage obtained when the input vS is
a sinusoid.
Figure 3.25 (a) Half-wave rectifier. (b) Equivalent circuit of the half-wave rectifier with the
diode replaced with its battery-plus-resistance model. (c) Transfer characteristic of the rectifier
circuit. (d) Input and output waveforms, assuming that rD ! R.
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92
In selecting diodes for rectifier design, two important parameters must
be specified: the current-handling capability required of the diode,
determined by the largest current the diode is expected to conduct,
and the peak inverse voltage (PIV) that the diode must be able to
withstand without breakdown, determined by the largest reverse
voltage that is expected to appear across the diode.
In the rectifier circuit of Fig. 3.25(a), we observe that when vS is
negative the diode will be cut off and vO will be zero. If follows that
the PIV is equal to the peak of vS.
PIV = VS
Note: To select a diode that has a reverse breakdown voltage at least
50% greater than the expected PIV.
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93
3.5.2 The Full-Wave Rectifier
The full-wave rectifier utilizes both halves of the input sinusoid. To
provide a unipolar output, it inverts the negative halves of the sine
wave. One possible implementation is shown in Fig. 3.26(a). Here
the transformer secondary winding is center-tapped to provide two
equal voltages vS across the two halves of the secondary winding
with the polarities indicated.
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94
Figure 3.26 Full-wave rectifier utilizing a transformer with a center-tapped secondary winding: (a)
circuit; (b) transfer characteristic assuming a constant-voltage-drop model for the diodes; (c) input and
output waveforms.
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95
In the positive half cycle, D1 is on and D2 is off.
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96
In the negative half cycle, D1 is off and D2 is on.
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97
To find the PIV, during the positive half-cycle D1 is conducting and
D2 is off. The reverse voltage across D2 will be [vO-(- vS)]=(vO+vS)
D1 is on
vO
-vS
D2 is off
When vS at its positive peak value VS, the corresponding vO is (VS –
VD), therefore the PIV = (VS –VD)+ VS = 2VS –VD which is about
twice that for the case of the half-wave rectifier.
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98
3.5.3 The Bridge Rectifier
Does not need a center-tapped transformer, but needs four diodes.
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99
During the positive half cycle, D1, D2 are on and D3, D4 are off.
Note: vO is lower than vS by two diode drops (compared to one drop
in the circuit previously discussed)
vO = vS –2VD
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100
During the negative half cycle, D3, D4 are on and are D1, D2 off.
Note: During both half-cycles, current flows through R in the same
direction.
vO = vS –2VD
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101
vO = vS –2VD
Figure 3.27 The bridge rectifier: (a) circuit; (b) input and output waveforms.
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102
To determine the PIV, let us consider the positive half cycle. The
reverse voltage across D3 can be determined from the loop formed
by D3, R,and D2 as
vD 3 (reverse) vO vD 2 (forward)
Thus the maximum value of vD3 occurs at the peak of vO and is
given by
About half the
PIV = (VS-2VD)+VD= VS-VD
value for the fullwave rectifier
with a centertapped
transformer.
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103
Note that, only about half as many turns are required for the
secondary winding of the transformer used in the bridge rectifier
than that used in the center-tapped one.
Another way of looking at this point can be obtained by observing
that each half of the secondary winding of the center-tapped
transformer is utilized for only half the time.
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104
3.5.4 The Rectifier with a Filter Capacitor –The Peak Rectifier
The output of the rectifier circuits discussed before is unsuitable
as a dc supply for electronic circuits. A simple way to reduce the
variation of the output voltage is to place a capacitor, which is
called filter capacitor, across the load resistor.
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105
Assume that the diode is ideal and vI is sinusoid with a peak value VP.
As vI goes positive, the diode conducts and the capacitor is charged so
that vO = vI. This situation continues until vI reaches its peak value VP.
Beyond the peak, as vI decreases the diode becomes reverse biased
and the output voltage remains constant at the value VP.
Figure 3.28 (a) A simple circuit used to illustrate the effect of a filter capacitor. (b) Input and output
waveforms assuming an ideal diode. Note that the circuit provides a dc voltage equal to the peak of the
input sine wave. The circuit is therefore known as a peak rectifier or a peak detector.
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106
In fact, theoretically speaking, the capacitor will retain its charge and
hence its voltage indefinitely, because there is no way for the
capacitor to discharge. Thus the circuit provides a dc voltage output
equal to the peak of the input sine wave.
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107
Now consider load resistance R is connected across the capacitor C. As
before, for a sinusoidal input, the capacitor charges to the peak of the
input VP. Then the diode cuts off, and the capacitor discharges through
the load resistance R. The capacitor discharge will continue for almost
the entire cycle, until the time at which vI exceeds the capacitor voltage.
Then the diode turns on again and charges the capacitor up to the peak
of vI, and the process repeats itself.
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108
Note that to keep the
output voltage from
decreasing too much
during capacitor
discharge, C must be
selected large enough
so the time constant
CR is much greater
than the discharge
interval.
Assume CR >>T
iD iC iL
C
dv I
i L
dt
v
iL  O
R
Figure 3.29 Voltage and current waveforms in the peak rectifier circuit with CR @ T. The diode is assumed ideal.
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109
Note that:
1. The diode conducts for a brief interval, t, near the peak of the input
sinusoid and supplies the capacitor with charge equal to that lost
during the much longer discharge interval. The latter is
approximately equal to the period T.
2. Assuming an ideal diode, the diode conduction begins at time t1, at
which the input vI equals the exponentially decaying output vO.
Conduction stops at t2 shortly after the peak of vI; the exact value of
t2 can be determine by setting iD = 0 in the equation
iD iC iL
dv I
C
i L
dt
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110
3. During the diode-off interval, the capacitor C discharges through R,
and thus vO decays exponentially with a time constant CR. The
discharge interval begins just past the peak of vI. At the end of the
discharge interval, which lasts for almost the entire period T, vO =
Vp –Vr, where Vr is the peak-to-peak ripple voltage. When CR >> T,
the value of Vr is small.
4. When Vr is small, vO is almost constant and equal to the peak value
of vI. Thus the dc output voltage is approximately equal to Vp.
Similarly, the current iL is almost constant, and its dc component IL is
given by
Vp
IL 
R
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111
If desired, a more accurate expression for the output dc voltage can be
obtained by taking the average of the extreme value of vO,
1
VO V p  Vr
2
With these observations in hand, we now derive expressions for Vr
and for the average and peak values of the diode current.
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112
During the diode-off interval, vO can be expressed as
vO V p e t / CR
At the end of the discharge interval we have
V p Vr V p e T / CR
Now, since CR >> T, we can use the approximation
e T / CR 1 T
CR
to obtain
T
Vr V p
CR
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113
We observe that to keep Vr small we must select a capacitance C so
that CR >> T. the ripple voltage Vr can be expressed in terms of the
frequency f = 1/T as
Vp
T
Vr V p

CR fCR
Vp
Using I L  we can have Vr as
R
IL
Vr 
fC
The above equation is valid provided that the capacitor discharges
by means of a constant current IL = Vp/R which requires that
Vr<<Vp.
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114
Assuming that diode conduction ceases almost at the peak of vI, we
can determine the conduction interval t from
V p cos( w)
t V p Vr
where w = 2f = 2/T is the angular frequency of vI. Since (wt) is
a small angle, we can employ the approximation cos(wt) 1-1/2
(wt)2 , to obtain
wt 
2Vr
Vp
Note that when Vr << Vp, the conduction angle wt will be small,
as assumed.
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115
To determine the average diode current during conduction, iDav, we
equate the charge that the diode supplies to the capacitor,
Qsupplied iCav t
iCav iDav I L
to the charge that the capacitor loses during the discharge interval
Qlost CVr
to obtain
iDav I L (1 
where
wt 
NFUEE Weihsing Liu
2Vr
Vr
)
Vp
IL 
Vr 
R
fCR
Vp
Vp
2V p
have been used.
116
Vp
Vp
iCav t CVr C 

fCR fR
Vp
 iCav t 
fR
Vp
 ft (iDav I L ) 
R
Vp
1 2Vr


iDav I L 

V
p
2
R
Vp 
2
IL 
2

iDav I L 

2
V
2
V
r
r
R
Vp
Vp
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117
 iDav
IL 
2
2V p
I L 
I L I L
Vr
2Vr
Vp
 iDav I L (1 
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2V p
Vr
)
118
Observe that when Vr << Vp, the average diode current during
conduction is much greater than the dc load current. This is not
surprising, since the diode conducts for a very short interval and
must replenish (補充) the charge lost by the capacitor during the
much longer interval in which it is discharged by IL.
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119
The peak value of the diode current, iDmax, can be determined by
evaluating the expression in
dvI
iD C
iL
dt
at the onset of diode conduction - that is, at t = t1 = - t (where t =
0 is at the peak). Assuming that iL is almost constant at the value
given by
Vp
IL 
R
we obtain
iD max I L (1 2
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2V p
Vr
)
120
Example 3.9
Consider a peak rectifier fed by a 60-Hz sinusoid having a peak Vp =
100 V. Let the load resistance R=10k . Find the value of capacitance
C that will result in a peak-to-peak ripple of 2 V. Also, calculate the
fraction of the cycle during which the diode is conducting and the
average and peak values of the diode current.
From
Vr 
Vp
fCR
we obtain the value of C as
Vp
100
C

83.3μF
3
Vr fR 2 60 10 10
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121
The conduction angle wt is found from wt 
wt  2 2
100
2Vr
Vp
as
0.2rad
Thus the diode conducts for (0.2/2)100=3.18% of the cycle. The
average diode currents is obtained from
iDav I L (1 
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2V p
324mA
2

100
) 10
1




Vr
2


122
The peak diode current is found by
iD max I L (1 2
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2V p
) 10
1 2 2 100 
638mA


Vr
2

123
The circuits shown in Fig. 3.26(a) and 3.27(a) can be converted to
peak rectifier by including a capacitor across the load resistor. As in
the half-wave case, the output dc voltage will be almost equal to the
peak value of the input sine wave. The ripple frequency, however,
will be twice that of the input.
Figure 3.30 Waveforms in the full-wave peak rectifier.
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124
The peak-to-peak ripple voltage, for this case, can be derived using a
procedure identical to that above but with the discharge period T
replaced by T/2, resulting in
Vp
Vr 
2 fCR
While the diode conduction interval, t ,will still be given by
wt 
2Vr
Vp
The peak currents in each of the diodes will be given by
iDav I L (1  V p 2Vr )
iD max I L (1 2 V p 2Vr )
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125
Comparing these expressions with the corresponding ones for the
half-wave case, we note that for the same values of Vp, f, R, and Vr
(and thus the same IL), we need a capacitor half the size of that
required in the half-wave rectifier.
Also, the current in each diode in the full-wave rectifier is
approximately half that which flows in the diode of the half-wave
circuit .
The analysis above assumed ideal diodes. The accuracy of the results
can be improved by taking the diode voltage drop into account. This
can be easily done by replacing the peak voltage Vp to which the
capacitor charges with (Vp –VD) for the half-wave circuit and the
full-wave circuit using a center-tapped transformer and with (Vp –
2VD) for the bridge rectifier case.
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126
We conclude this section by noting that peak-rectifier circuits find
application in signal-processing system where it is required to detect
the peak of an input signal. In such a case, the circuit is referred to as
a peak detector.
A particular application of the peak detector is in the design of a
demodulator for amplitude-modulated (AM) signals.
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3.5.5 Precision Half-Wave Rectifier –The Super Diode
The rectifier circuits studied thus far suffer from having one or two
diode drops in the signal paths. Thus these circuits work well only
when the signal to be rectified is much larger than the voltage drop of
a conducting diode (0.7v or so).
There are other applications, however, where the signal to be
rectified is small (i.e., on the order of 100 mV or so) and thus clearly
insufficient to turn on a diode.
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We first assume that D is off, and the vI is positive, the OPA operates
in the open-loop mode. Then D is on, the OPA operates in the closedloop mode (negative feedback), therefore vO = vI.
Only a few mV will turn on the diode!
v O = v I.
Figure 3.31 The “
superdiode”precision half-wave rectifier and its almost-ideal transfer
characteristic. Note that when vI > 0 and the diode conducts, the op amp supplies the load current,
and the source is conveniently buffered, an added advantage. Not shown are the op-amp power
supplies.
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3.6 LIMITING AND CLAMPING CIRCUITS
3.6.1 Limiter Circuits
For inputs in a certain range
L-/K vI L+/K, the limiter
acts as a linear circuit,
providing an output
proportional to the input vO =
KvI. Although in general K
can be greater than 1, the
circuits discussed in this
section have K 1 and
known as passive limiters.
Figure 3.32 General transfer characteristic for a limiter circuit.
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If vI exceeds the upper threshold L+/K, the output voltage is limited
or clamped to the upper limiting level L+. On the other hand, if vI is
reduced below the lower limiting threshold L-/K, the output voltage
vO is limited to the lower limiting level L-.
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The general transfer characteristic shown below describes a double
limiter –that is, a limiter that works on both the positive and negative
peaks of an input waveform. Single limiters, of course, exist.
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Finally, note that if an input waveform such as that shown in Fig. 3.33
is fed to a double limiter, its two peaks will be clipped off. Limiters
therefore are sometimes referred to clippers.
Figure 3.33 Applying a sine wave to a limiter can result in clipping off its two peaks.
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The limiter whose characteristics are depicted in Fig. 3.32 describes a
hard limiter. Soft limiting is characterized by smoother transitions
between the linear region and the saturation regions and a slope
greater than zero in the saturation regions. Depending on the
application, either hard or soft limiting may be preferred.
Figure 3.34 Soft limiting.
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Limiters find application in a variety of signal-processing systems.
One of their simplest applications is in limiting the voltage between
the two input terminals of an op amp to a value lower than the
breakdown voltage of the transistors that make up the input stage of
the op-amp circuit.
To protect the input stage of the op
amp!
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Figure 3.35 A variety of basic limiting circuits.
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Figure 3.35 A variety of basic limiting circuits.
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Figure 3.35 A variety of basic limiting circuits.
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Assuming the diodes to be ideal, describe the transfer characteristic
of the circuit.
For vI - 5V, D1 on and D2 off
vO [vI (5)]
10k
1
5  vI 2.5
10k 10 K
2
D1
D2
voltage divider
For vI +5V, D2 on and D1 off
vO (vI 5)
10k
1
5  vI 2.5
10k 10 K
2
For -5V vI 5V, D1 and D2
are both off, vO = vI.
Figure E3.27
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3.6.2 The Clamped Capacitor or DC Restorer
If in the basic peak-rectifier circuit the output is taken across the
diode rather than across the capacitor, than a dc restorer is given.
1. Assume that at t =0, there is no
charge stored in the capacitor,
therefore vC = 0.
2. If vI < 0, D is on, vO = -vD0 and
the capacitor is charged to the
most negative peak value of the
input vI, and the polarity is
indicated as shown in the figure.
3. If vI > 0, D is off and vO = vC + vI
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It follows that the output waveform will be identical to that of the
input, except that it is shifted upward by vC volts.
vC = 6V
Figure 3.36 The clamped capacitor or dc restorer with a square-wave input and no load.
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Note, the output waveform will have a finite average value or dc
component. This component is entirely unrelated to the average value
of the input waveform.
As an application, consider a pulse signal being transmitted through a
capacitively coupled or ac-coupled system. The capacitive coupling
will cause the pulse train to lose whatever dc component it originally
had.
Feeding the resulting pulse waveform to a clamping circuit provides it
with a well-determined dc component, a process known as dc
restoration. This is why the circuit is also called a dc restorer.
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When a load resistance R is connected across the diode in a
clamping circuit, as shown below, the situation changes significantly.
D is off during t0 –t1, C
discharges through R (= RC)
Figure 3.37 The clamped capacitor with a load resistance R.
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The diode is not ideal, therefore besides the
VD0, there is some voltage drop across the
equivalent resistance rD.
At t1, the input decreases by Va volts, and the output attempts to
follow. This causes the diode to conduct heavily and to quickly
charge the capacitor.
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At the end of the interval t1 to t2, the output voltage would
normally be a few tenths of a volt negative (i.e. –0.5V). Then, as
the input rises by Va volts (at t2), the output follows, and the
cycle repeats itself.
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In the steady state the charge lost by the capacitor during the interval t0
to t1 is recovered during the interval t1 to t2.
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3.6.3 The Voltage Doubler
During the positive halfcycle, D1 is on and D2 is
off, C1 will be charged to
Vp.
During the negative halfcycle, D2 is on and D1 is
off, C2 will be charged to
-2Vp.
Figure 3.38 Voltage doubler: (a) circuit; (b) waveform of the voltage across D1.
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3.7 PHYSICAL OPERATION OF DIODES
Having studied the terminal characteristics and circuit applications of
junction diodes, we will now briefly consider the physical processes
that give rise to the observed terminal characteristics.
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3.7.1 Basic Semiconductor Concepts
The pn junction
The semiconductor diodes is basically a pn junction, as indicated, the
pn junction consists of p-type semiconductor material brought into
close contact with n-type semiconductor material.
In actual practice, both the p and n regions are part of the same
silicon crystal; that is, the pn junction is formed within a single
silicon crystal by creating regions of different “
dopings”(p and n
regions).
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External wire connections to the p and n regions (i.e., diode
terminals) are made through metal (aluminum) contacts.
Figure 3.39 Simplified physical structure of the junction diode. (Actual geometries are given
in Appendix A.)
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Intrinsic Silicon
A crystal of pure or intrinsic silicon has a regular lattice structure
where the atoms are held in their positions by bonds, called covalent
bonds, formed by the four valence electrons associated with each
silicon atom.
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Each atom shares each of
its four valence electrons
with a neighboring atom,
with each pair of electrons
forming a covalent bond.
Figure 3.40 Two-dimensional representation of the silicon crystal. The circles represent the inner core of silicon atoms,
with +4 indicating its positive charge of +4q, which is neutralized by the charge of the four valence electrons. Observe
how the covalent bonds are formed by sharing of the valence electrons. At 0 K, all bonds are intact and no free electrons
are available for current conduction.
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At sufficiently low temperatures, all covalent bonds are intact and
no(or very few) free electrons are available to conduct electric current.
However, at room temperature, some of the bonds are broken by
thermal ionization and some electrons are freed.
完整無損的
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When a covalent
bond is broken, an
electron leaves its
parent atom; thus a
positive charge,
equal to the
magnitude of the
electron charge, is
left with the parent
atom.
Figure 3.41 At room temperature, some of the covalent bonds are broken by thermal ionization. Each
broken bond gives rise to a free electron and a hole, both of which become available for current
conduction.
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An electron from a neighboring atom may be attracted to this positive
charge, leaving its parent atom. This action fills up the “
hole”that
exists in the ionized atom but creates a new hole in the other atom.
This process may repeats itself, with the result that we effectively
have a positively charged carrier, or hole, moving through the silicon
crystal structure and being available to conduct electric current.
The charge of a hole is equal in magnitude to the charge of an
electron.
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Thermal ionization results in free electrons and holes in equal numbers
and hence equal concentrations. These free electrons and holes move
randomly through the silicon crystal structure, and in the process some
electrons may fill some of the holes.
This process, called recombination, results in the disappearance of free
electrons and holes.
The recombination rate is proportional to the number of free electrons
and holes, which, in turn, is determined by the ionization rate.
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The ionization rate is a strong function of temperature. In thermal
equilibrium, the recombination rate is equal to the ionization or
thermal-generation rate, and one can calculate the concentration of free
electrons n, which is equal to the concentration of holes p,
n = p = ni
Where ni denotes the concentration of free electrons or holes in
intrinsic silicon at a given temperature.
At an absolute temperature T, the intrinsic concentration ni can be
found from
EG
2
3
ni BT e kT (1/cm3)
B is a material-dependent parameter = 5.41031 for silicon. EG is
energy gap, which is 1.12 eV for silicon and k is Boltzmann’
s
constant = 8.62 10-5 eV/K.
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EG represents the minimum energy required to break a covalent
bond and thus generate an electron-hole pair. For room temperature,
ni 1.51015 carriers/cm3..
Note that, the silicon crystal has about .51022 atom/cm3. Thus, at
room temperature, only one of every billion atoms is ionized!
Finally, it should be mentioned that the reason that silicon is called
a semiconductor is that its conductivity, which is determined by the
number of charge carriers available to conduct electric current, is
between that of conductors and that of insulators
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Diffusion and Drift
There are two mechanisms by which holes and electrons move
through a silicon crystal –diffusion and drift.
Diffusion is associated with random motion due to thermal agitation.
In a piece of silicon with uniform concentration of free electrons
and holes, this random motion does not result in a net flow of
charge (current). On the other hand, if by some mechanism the
concentration of say, free electrons is made higher in one part of the
piece of silicon than in another, then electrons will diffuse from the
region of high concentration to the region of low concentration.
This diffusion process gives rise to a net flow of charge, or diffusion
current.
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Figure 3.42 A bar of intrinsic silicon (a) in which the hole concentration profile shown in (b) has been created along the x-axis
by some unspecified mechanism.
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