CHAPTER 3
SOLID-STATE DIODES AND DIODE CIRCUITS
80
第3章 固态二极管和二极管电路
本章提纲
3.1 pn结二极管
3.2 二极管的i-v特性
3.3 二极管方程：二极管的数学模型
3.4 二极管特性之反偏、零偏和正偏
3.5 二极管的温度系数
3.6 反偏下的二极管
3.7 pn结电容
3.8 肖特基势垒二极管
3.9 二极管的SPICE模型及版图
3.10 二极管电路分析
3.11 多二极管电路
3.12 二极管工作在击穿区域的分析
3.13 半波整流电路
3.14 全波整流电路
3.15 全波桥式整流
3.16 整流器的比较和折中设计
3.17 二极管的动态开关行为
3.18 光电二极管、太阳能电池和发光二极管
本章目标
理解二极管结构及基本版图；
了解pn结的电学特性；
熟悉各种二极管模型，包括数学模型、理想模型及恒定压降模型；
理解二极管的SPICE描述及二极管的模型参数；
熟悉二极管的工作区，包括正向偏置、反向偏置以及反向击穿；
能在电路分析中应用不同模型；
熟悉不同类型二极管，包括齐纳二极管、变容二极管、肖特基势垒二极管、太阳能电池和发光二极管（LED）；
熟悉 pn 结二极管的动态开关行为；
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81
熟悉二极管整流器；
熟悉二极管电路SPICE。
本章导读
本章主要研究了固态二极管相关的原理、特性、模型及电路应用。首先从pn结入手，介绍pn结的电学特
性。pn结二极管是p型半导体和n型半导体的紧密接触形成的。在pn结二极管中，冶金结附近存在较大的浓度
梯度，由此产生较大的电子和空穴扩散电流。在零偏条件下，二极管两端以及空间电荷区中均没有电流存
在。空间电荷区将产生内建电势和内部电场，而内部电场将导致电子和空穴电流的漂移电流，正好与扩散电
流相抵消。
接下来，研究了二极管两端施加电压时的i-v特性以及二极管模型方程。在二极管两端施加电压时，二
极管内部结区域的平衡遭到破坏，产生电流的传导。根据二极管的i-v特性可以通过二极管方程实现精确建
模。根据二极管两端加载电压的方向不同，分为二极管反偏和正偏。反偏条件下二极管电流等于-IS，这个
值非常小。正偏条件下，电流可以很大，二极管的压降为0.6~0.7V。室温下，二极管电压每变化60mV可
以引起10倍的二极管电流变化；室温下，硅二极管的温度系数为-1.8mV/℃。
击穿现象是二极管电路中的常见现象。本章对击穿现象的定义、预防及利用都进行了介绍。如果二极管
两端的反向偏压过大，内部电场会使二极管发生击穿，包括齐纳击穿和雪崩击穿。工作在击穿区的二极管具
有基本固定的压降，必须严格限制二极管上的电流，否则很容易烧毁器件。工作在击穿区的齐纳二极管可用
于稳压器电路设计。电压调整率和负载调整率分别用于表征输出电压随输入电压和输出电流的变化。
二极管电容对于二极管的性能具有很大的影响。本章给出了二极管电容产生的原因以及影响。如果二极
管两端电压发生变化，则在空间电荷区附近存储的电荷量也会随之变化，这意味着二极管模型中应包含电
容。二极管反向偏置时，电容与外加电压的平方根成反比；正向偏置时，电容与工作电流和二极管传输时间
成正比。由于这一电容的存在，使得二极管不能立即截止或导通，在截止时会形成一个电荷存储延迟。
本章重点给出了二极管电路计算和模拟的主要方法，包括迭代法、负载线分析法、恒压降法、SPICE
等。在电路计算中，迭代法主要针对直接使用二极管方程时的情况。负载线分析法、理想二极管模型以及恒
压降法常被用于二极管电路的简化分析。SPICE电路分析程序包含了精确描述理想和非理想二极管特性的内
建模型，能够方便地分析含有二极管的电路特性。
在二极管的应用方面，本章重点讲解了半波、全波以及全波桥式整流电路。整流电路主要将交流电压转
换为直流电压。在电源电路中使用的整流器必须能够经受大的周期峰值电流，以及刚上电时的浪涌电流。在
整流电路中，滤波电容的设计将决定纹波电压和二极管的导通角。由于存在内部电容，二极管的导通和截止
必须经过电容充放电，因而不能瞬间完成。导通时间通常都是很小的，但截止过程则要慢得多，它必须将二
极管中的存储电荷都转移掉，这样就产生一个存储延迟τS。在存储延迟期间，可能出现较大的反向电流。
本章最后研究了pn结发光和检测光的能力，讨论了光电二极管、太阳能电池和发光二极管的基本特性。
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CHAPTER O UTL INE
3.1 The pn Junction Diode
3.2 The i-v Characteristics of the Diode
3.3 The Diode Equation: A Mathematical Model
for the Diode
3.4 Diode Characteristics Under Reverse, Zero, and
Forward Bias
3.5 Diode Temperature Coefficient
3.6 Diodes Under Reverse Bias
3.7 pn Junction Capacitance
3.8 Schottky Barrier Diode
3.9 Diode SPICE Model and Layout
3.10 Diode Circuit Analysis
3.11 Multiple-Diode Circuits
3.12 Analysis of Diodes Operating in the Breakdown
Region
3.13 Half-Wave Rectifier Circuits
3.14 F ul l-W ave Rectifier Circuits
3.15 F ul l-W ave Bridge Rectification
3.16 R ec ti fier Comparison and Design Tradeoffs
3.17 Dynamic Switching Behavior of the Diode
3.18 Photo Diodes, Solar Cells, and Light-Emitting
Diodes
Summary
Key Terms
Reference
Additional Reading
Problems
CHAPTER G OALS
? Understand diode structure and basic layout
? Develop electrostatics of the pn junction
? Explore various diode models including the
mathematical model, the ideal diode model,
and the constant voltage drop model
? Understand the SPICE representation and model
parameters for the diode
? Define regions of operation of the diode, including
forward and reverse bias and reverse breakdown
? Apply the various types of models in circuit analysis
? Explore different types of diodes including Zener,
variable capacitance, and Schottky barrier diodes
as well as solar cells and light emitting diodes (LEDs)
? Discuss the dynamic switching behavior of the
pn junction diode
? Explore diode rectifiers
? Practice simulating diode circuits using SPICE
Photograph of an
assortment of diodes
Fabricated diode
The first electronic circuit element that we explore is the
solid-state pn junction diode. The diode is an extremely
important device in its own right with many important
applications including ac-dc power conversion (rectification),
solar power generation, and high-frequency mixers
for RF communications. In addition, the pn junction diode
is a fundamental building block for other solid-state devices.
In later chapters, we will find that two closely coupled
diodes are used to form the bipolar junction transistor
(BJT), and two diodes form an integral part of the metaloxide-
semiconductor field-effect transistor (MOSFET) and
the junction field-effect transistor (JFET). Gaining an understanding
of diode characteristics is prerequisite to understanding
the behavior of the field-effect and bipolar
transistors that are used to realize both digital logic circuits
and analog amplifiers.
The pn junction diode is formed by fabricating adjoining
regions of p-type and n-type semiconductor material.
Another type of diode, called the Schottky barrier
diode, is formed by a non-ohmic contact between a metal
such as aluminum, palladium, or platinum and an n-type
or p-type semiconductor. Both types of solid-state diodes
are discussed in this chapter. The vacuum diode, which was
used before the advent of semiconductor diodes, still finds
application in very high voltage situations.
82
3.1 The pn Junction Diode 83
The pn junction diode is a nonlinear element, and for
many of us, this will be our first encounter with a nonlinear
device. The diode is a two-terminal circuit element similar
to a resistor, but its i-v characteristic, the relationship
between the current through the element and the voltage
across the element, is not a straight line. This nonlinear
behavior allows electronic circuits to be designed to provide
many useful operations, including rectification, mixing (a
form of multiplication), and wave shaping. Diodes can also
be used to perform elementary logic operations such as the
AND and OR functions.
This chapter begins with a basic discussion of the structure and behavior of the pn junction diode
and its terminal characteristics. Next is an introduction to the concept of modeling, and several different
models for the diode are introduced and used to analyze the behavior of diode circuits. We
begin to develop the intuition needed to make choices between models of various complexities in
order to simplify electronic circuit analysis and design. Diode circuits are then explored, including
the detailed application of the diode in rectifier circuits. The characteristics of Zener diodes, photo
diodes, solar cells, and light-emitting diodes are also discussed.
3.1 THE pn JUNCTION DIODE
The pn junction diode is formed by fabrication of a p-type semiconductor region in intimate contact
with an n-type semiconductor region, as illustrated in Fig. 3.1. The diode is constructed using the
impurity doping process discussed in the last section of Chapter 2.
An actual diode can be formed by starting with an n-type wafer with doping ND and selectively
converting a portion of the wafer to p-type by adding acceptor impurities with NA > ND. The point
at which the material changes from p-type to n-type is called the metallurgical junction. The p-type
region is also referred to as the anode of the diode, and the n-type region is called the cathode of
the diode.
Figure 3.2 gives the circuit symbol for the diode, with the left-hand end corresponding to the
p-type region of the diode and the right-hand side corresponding to the n-type region. We will see
shortly that the “arrow” points in the direction of positive current in the diode.
3.1.1 pn JUNCTION ELECTROSTATICS
Consider a pn junction diode similar to Fig. 3.1 having NA = 1017/cm3 on the p-type side and
ND = 1016/cm3 on the n-type side. The hole and electron concentrations on the two sides of the
junction will be
p-type side: pp = 1017 holes/cm3 np = 103 electrons/cm3
n-type side: pn = 104 holes/cm3 nn = 1016 electrons/cm3 (3.1)
Metallurgical junction
Metal contact
Anode Cathode
p n
pp = NA nn = ND
np = NNAA
ni
2
pn = NNAD
ni
2
Figure 3.1 Basic pn junction diode.
p n
Anode Cathode
Figure 3.2 Diode circuit symbol.
84 Chapter 3 Solid-State Diodes and Diode Circuits
Hole diffusion
Hole current density
–xp xn
–xp xn
x
p(x)
pp
np
pn
nn
x
p n
p, n (log scale)
jp = –qDp
dp
dx
dp
(Note that d x < 0.)
1017/cm3
103/cm3 104/cm3
1016/cm3
(a)
(b)
Electron diffusion
Electron current density
–xp xn
x
jn = qDn
n(x)
dn
dx
dn
(Note that d x > 0.)
(c)
Figure 3.3 (a) Carrier concentrations; (b) hole diffusion current in the space charge region; (c) electron diffusion current in
the space charge region.
As shown in Fig. 3.3(a), a very large concentration of holes exists on the p-type side of the metallurgical
junction, whereas a much smaller hole concentration exists on the n-type side. Likewise,
there is a very large concentration of electrons on the n-type side of the junction and a very low
concentration on the p-type side.
From our knowledge of diffusion from Chapter 2, we know that mobile holes will diffuse from
the region of high concentration on the p-type side toward the region of low concentration on the
n-type side and that mobile electrons will diffuse from the n-type side to the p-type side, as in
Fig. 3.3(b) and (c). If the diffusion processes were to continue unabated, there would eventually be
a uniform concentration of holes and electrons throughout the entire semiconductor region, and the
pn junction would cease to exist. Note that the two diffusion current densities are both directed in
the positive x direction, but this is inconsistent with zero current in the open-circuited terminals of
the diode.
Asecond, competing process must be established to balance the diffusion current. The competing
mechanism is a drift current, as discussed in Chapter 2, and its origin can be understood by focusing
on the region in the vicinity of the metallurgical junction shown in Fig. 3.4. As mobile holes
move out of the p-type material, they leave behind immobile negatively charged acceptor atoms.
Correspondingly, mobile electrons leave behind immobile ionized donor atoms with a localized
positive charge. A space charge region (SCR), depleted of mobile carriers, develops in the region
immediately around the metallurgical junction. This region is also often called the depletion region,
or depletion layer.
From electromagnetics, we know that a region of space charge ρc (C/cm3) will be accompanied
by an electric field E measured in V/cm through Gauss’ law
? · E =
ρc
εs
(3.2)
3.1 The pn Junction Diode 85
– – + +
– – + +
– – + +
p n
Metallurgical
junction
Ionized donor
atom
Ionized acceptor
atom
Space charge region (SCR)
or depletion region
Neutral
region
Neutral
region
–xp xn
Hole drift
Electron drift
Hole diffusion
Electron diffusion
E(x)
Figure 3.4 Space charge region formation near the metallurgical junction.
+qND
?qNA
–xp
xn
x
ρc(ρ x)
(a)
–xp
xn
x
E(x)
–EMAX
(b)
–xp xn
x
φφ(x)
φφn
φφp
φφj
(c)
Figure 3.5 (a) Charge density (C/cm3), (b) electric field (V/cm), and (c) electrostatic potential (V) in the space charge region
of a pn junction.
written assuming a constant semiconductor permittivity εs (F/cm). In one dimension, Eq. (3.2) can
be rearranged to give
E(x) =
1
εs  ρc(x) dx (3.3)
Figure 3.5 illustrates the space charge and electric field in the diode for the case of uniform
(constant) doping on both sides of the junction. As illustrated in Fig. 3.5(a), the value of the space
charge density on the p-type side will be ?qNA and will extend from the metallurgical junction at
x = 0 to ?xp, whereas that on the n-type side will be +qND and will extend from 0 to +xn. The
overall diode must be charge neutral, so
qNAxp = qNDxn (3.4)
The electric field is proportional to the integral of the space charge density and will be zero in the
(charge) neutral regions outside of the depletion region. Using this zero-field boundary condition
yields the triangular electric field distribution in Fig. 3.5(b).
Figure 3.5(c) represents the integral of the electric field and shows that a built-in potential or
junction potential φj exists across the pn junction space charge region according to
φj = ? E(x) dx V (3.5)
86 Chapter 3 Solid-State Diodes and Diode Circuits
φj represents the difference in the internal chemical potentials between the n and p sides of the
diode, and it can be shown [1] to be given by
φj = VT lnNAND
n2
i  (3.6)
where the thermal voltage VT = kT/q was originally defined in Chapter 2.
Equations (3.3) to (3.6) can be used to determine the total width of the depletion region wdo in
terms of the built-in potential:
wdo = (xn + xp) = 2εs
q  1
NA +
1
ND φj m (3.7)
From Eq. (3.7), we see that the doping on the more lightly doped side of the junction will be the
most important in determining the depletion-layer width.
EXAMPLE 3.1 DIODE SPACE CHARGE REGION WIDTH
When diodes are actually fabricated, the doping levels on opposite sides of the pn junction tend
to be quite asymmetric, and the resulting depletion layer tends to extend primarily on one side of
the junction and is referred to as a “one-sided” step junction or one-sided abrupt junction. The pn
junction that we analyze provides an example of the magnitudes of the distances involved in such
a pn junction.
PROBLEM Calculate the built-in potential and depletion-region width for a silicon diode with NA = 1017/cm3 on the p-type side and ND = 1020/cm3 on the n-type side.
SOLUTION Known Information and Given Data: On the p-type side, NA = 1017/cm3; on the n-type side,
ND = 1020/cm3. Theory describing the pn junction is given by Eqs. (3.4) through (3.7).
Unknowns: Built-in potential φj and depletion-region width wdo
Approach: Find the built-in potential using Eq. (3.6); use φj to calculate wdo in Eq. (3.7).
Assumptions: The diode operates at room temperature with VT = 0.025 V. There are only donor
impurities on the n-type side and acceptor impurities on the p-type side of the junction. The doping
levels are constant on each side of the junction.
Analysis: The built-in potential is given by
φj = VT lnNAND
n2
i  = (0.025 V) ln (1017/cm3)(1020/cm3)
(1020/cm6)  = 0.979 V
For silicon, εs =11.7εo, where εo =8.85 × 10?14 F/cm represents the permittivity of free space.
wdo = 2εs
q  1
NA +
1
ND φj
wdo = 2 · 11.7 · (8.85 × 10?14 F/cm)
1.60 × 10?19 C  1
1017/cm3 +
1
1020/cm30.979 V = 0.113 m
3.1 The pn Junction Diode 87
Check of Results: The built-in potential should be less than the bandgap of the material. For silicon
the bandgap is approximately 1.12V(see Table 2.3), so φj appears reasonable. The depletion-layer
width seems quite small, but a double check of the numbers indicates that the calculation is correct.
Discussion: The numbers in this example are fairly typical of a pn junction diode. For the normal
doping levels encountered in solid-state diodes, the built-in potential ranges between 0.5 V and
1.0 V, and the total depletion-layer width wdo can range from a fraction of 1 m in heavily doped
diodes to tens of microns in lightly doped diodes.
EXERCISE: Calculate the built-in potential and depletion-region width for a silicon diode if NA
is increased to 2 × 1018/cm3 on the p-type side and ND = 1020/cm3 on the n-type side.
A NSWERS: 1.05 V; 0.0263 m
3.1.2 INTERNAL DIODE CURRENTS
Remember that the electric field E points in the direction that a positive carrier will move, so electrons
drift toward the positive x direction and holes drift in the negative x direction in Fig. 3.4. Because
the terminal currents must be zero, a dynamic equilibrium is established in the junction region. Hole
diffusion is precisely balanced by hole drift, and electron diffusion is exactly balanced by electron
drift. This balance is stated mathematically in Eq. (3.8), in which the total hole and electron current
densities must each be identically zero:
j T
n = qnμnE + qDn
?n
?x = 0 and j T
p = qpμpE ? qDp
?p
?x = 0 A/cm2 (3.8)
The difference in potential in Fig. 3.5(c) represents a barrier to both hole and electron flow
across the junction. When a voltage is applied to the diode, the potential barrier is modified, and the
delicate balances in Eq. (3.8) are disturbed, resulting in a current in the diode terminals.
EXAMPLE 3.2 DIODE ELECTRIC FIELD AND SPACE-CHARGE REGION EXTENTS
Now we find the value of the electric field in the diode and the size of the individual depletion
layers on either side of the pn junction.
PROBLEM Find xn, xp, and EMAX for the diode in Ex. 3.1.
SOLUTION Known Information and Given Data: On the p-type side, NA = 1017/cm3; on the n-type side,
ND = 1020/cm3. Theory describing the pn junction is given by Eqs. (3.4) through (3.7). From
Ex. 3.1, φj = 0.979 V and wdo = 0.113 m.
Unknowns: xn, xp, and EMAX
Approach: Use Eqs. (3.4) and (3.7) to find xn and xp; use Eq. (3.5) to find EMAX.
Assumptions: Room temperature operation
Analysis: Using Eq. (3.4), we can write
wdo = xn + xp = xn 1 +
ND
NA  and wdo = xn + xp = xp 1 +
NA
ND 
88 Chapter 3 Solid-State Diodes and Diode Circuits
Solving for xn and xp gives
xn =
wdo
1 +
ND
NA  =
0.113 m
1 +
1020/cm3
1017/cm3 = 1.13 × 10?4 m
and
xp =
wdo
1 +
NA
ND  =
0.113 m
1 +
1017/cm3
1020/cm3 = 0.113 m
Equation (3.5) indicates that the built-in potential is equal to the area under the triangle in Fig. 3.5(b).
The height of the triangle is (?EMAX) and the base of the triangle is xn + xp = wdo:
φj =
1
2
EMAXwdo and EMAX =
2φj
wdo =
2(0.979 V)
0.113 m = 173 kV/cm
Check of Results: From Eqs. (3.3) and (3.4), EMAX can also be found from the doping levels and
depletion-layer widths on each side of the junction. The equation in the next exercise can be used
as a check of the answer.
EXERCISE: Using Eq. (3.3) and Fig. 3.5(a) and (b), show that the maximum field is given by
EMAX =
qNAxp
εs =
qNDxn
εs
Use this formula to find EMAX.
A NSWER: 175 kV/cm
EXERCISE: Calculate EMAX, xp, and xn for a silicon diode if NA = 2 × 1018/cm3 on the p-type
side and ND = 1020/cm3 on the n-type side. Use φ j = 1.05 V and wdo=0.0263 m.
A NSWERS: 798 kV/cm; 5.16 × 10?4 m; 0.0258 m
3.2 THE i -v CHARACTERISTICS OF THE DIODE
The diode is the electronic equivalent of a mechanical check valve—it permits current to flow in one
direction in a circuit, but prevents movement of current in the opposite direction. We will find that
this nonlinear behavior has many useful applications in electronic circuit design. To understand this
phenomenon, we explore the relationship between the current in the diode and the voltage applied to
the diode. This information, called the i -v characteristic of the diode, is first presented graphically
and then mathematically in this section and Sec. 3.3.
The current in the diode is determined by the voltage applied across the diode terminals, and the
diode is shown with a voltage applied in Fig. 3.6. Voltage vD represents the voltage applied between
the diode terminals; iD is the current through the diode. The neutral regions of the diode represent a
low resistance to current, and essentially all the external applied voltage is dropped across the space
charge region.
The applied voltage disturbs the balance between the drift and diffusion currents at the junction
specified in the two expressions in Eq. (3.8). A positive applied voltage reduces the potential barrier
for electrons and holes, as in Fig. 3.7, and current easily crosses the junction. A negative voltage
3.2 The i-v Characteristics of the Diode 89
vD
iD
SCR
Metal
contact
p
v ? 0 v ? 0
n
vD
Figure 3.6 Diode with external applied voltage vD.
xn φj φ ? vD
x
vD < 0
φ(x)
–xp
vD = 0
vD > 0
φ
Figure 3.7 Electrostatic junction potential for different
applied voltages.
Diode current (A)
Diode voltage (V)
0.10
0.08
0.02
0.04
0.06
0.00
–0.02
–1.5 –1.0 –0.5 0.0 0.5 1.0 1.5
Figure 3.9
Turn-on
voltage
Figure 3.8 Graph of the i-v characteristics of a pn junction diode.
–0.2 –0.1 0.0 0.1
–2.00  10–15
0
2.00  10–15
4.00  10–15
6.00  10–15
8.00  10–15
1.00  10–14
1.20 10–14
Diode current (A)
Diode voltage (V)
IS
–IS
Figure 3.9 Diode behavior near the origin with IS = 10?15 A and n = 1.
increases the potential barrier, and although the balance in Eq. (3.8) is disturbed, the increased barrier
results in a very small current.
The most important details of the diode i-v characteristic appear in Fig. 3.8. The diode characteristic
is definitely not linear. For voltages less than zero, the diode is essentially nonconducting,
with iD ～=
0. As the voltage increases above zero, the current remains nearly zero until the
voltage vD exceeds approximately 0.5 to 0.7 V. At this point, the diode current increases rapidly,
and the voltage across the diode becomes almost independent of current. The voltage required to
bring the diode into significant conduction is often called either the turn-on or cut-in voltage of
the diode.
Figure 3.9 is an enlargement of the region around the origin in Fig. 3.8. We see that the
i -v characteristic passes through the origin; the current is zero when the applied voltage is zero.
For negative voltages the current is not actually zero but reaches a limiting value labeled as ?IS for
voltages less than ?0.1 V. IS is called the reverse saturation current, or just saturation current,
of the diode.
90 Chapter 3 Solid-State Diodes and Diode Circuits
3.3 THE DIODE EQUATION: A MATHEMATICAL MODEL FOR THE DIODE
When performing both hand and computer analysis of circuits containing diodes, it is very helpful
to have a mathematical representation, or model, for the i -v characteristics depicted in Figs. 3.8
and 3.9. In fact, solid-state device theory has been used to formulate a mathematical expression that
agrees amazingly well with the measured i -v characteristics of the pn junction diode.We study this
extremely important formula called the diode equation in this section.
A positive voltage vD is applied to the diode in Fig. 3.10; in the figure the diode is represented
by its circuit symbol from Fig. 3.2. Although we will not attempt to do so here, Eq. (3.8) can be
solved for the hole and electron concentrations and the terminal current in the diode as a function
of the voltage vD across the diode. The resulting diode equation, given in Eq. (3.9), provides a
mathematical model for the i-v characteristics of the diode:
iD = ISexpqvD
nkT ? 1 = IS exp vD
nVT ? 1 (3.9)
where IS =reverse saturation current of diode (A) T = absolute temperature (K)
vD =voltage applied to diode (V) n = nonideality factor (dimensionless)
q =electronic charge (1.60 × 10?19 C) VT = kT /q = thermal voltage (V)
k =Boltzmann’s constant (1.38 × 10?23 J/K)
The total current through the diode is iD, and the voltage drop across the diode terminals is
vD. Positive directions for the terminal voltage and current are indicated in Fig. 3.10. VT is the
thermal voltage encountered previously in Chapter 2 and will be assumed equal to 0.025 V at room
temperature. IS is the (reverse) saturation current of the diode encountered in Fig. 3.9, and n is a
dimensionless parameter discussed in more detail shortly. The saturation current is typically in the
range
10?18 A≤ IS ≤10?9 A (3.10)
From device physics, it can be shown that the diode saturation current is proportional to n2
i , where
ni is the density of electrons and holes in intrinsic semiconductor material. After reviewing Eq. (2.1)
in Chapter 2, we realize that IS will be strongly dependent on temperature. Detailed discussion of
this temperature dependence is in Sec. 3.5.
Parameter n is termed the nonideality factor. For most silicon diodes, n is in the range 1.0
to 1.1, although it approaches a value of 2 in diodes operating at high current densities. From
this point on, we assume that n = 1 unless otherwise indicated, and the diode equation will be
written as
iD = IS expvD
VT ? 1 (3.11)
It is difficult to distinguish small variations in the value of n from an uncertainty in our knowledge in
iD
vD
vD
Figure 3.10 Diode with applied voltage vD.
3.3 The Diode Equation: A Mathematical Model for the Diode 91
the absolute temperature. This is one reason that we will assume that n = 1 in this text. The problem
can be investigated further by working on the next exercise.
EXERCISE: For n = 1 and T = 300 K, n(KT/q) = 25.9 mV. Verify this calculation. Now, suppose
n = 1.03. What temperature gives the same value for nVT?
A NSWER: 291 K
The mathematical model in Eq. (3.11) provides a highly accurate prediction of the i-v characteristics
of the pn junction diode. The model is useful for understanding the detailed behavior of diodes
and also forms the heart of the diode model in the SPICE circuit simulation program. It provides a
basis for understanding the i-v characteristics of the bipolar transistor in Chapter 5.
The static i -v characteristics of the diode are well-characterized by three parameters: saturation
current IS, temperature via the thermal voltage VT , and nonideality factor n.
iD = IS exp vD
nVT ? 1
DESIGN
NOTE
EXAMPLE 3.3 DIODE VOLTAGE AND CURRENT CALCULATIONS
In this example, we calculate some typical values of diode voltages for several different current
levels and types of diodes.
PROBLEM (a) Find the diode voltage for a silicon diode with IS = 0.1 fA operating at room temperature at
a current of 300 A. What is the diode voltage if IS = 10 fA? What is the diode voltage if the
current increases to 1 mA?
(b) Find the diode voltage for a silicon power diode with IS = 10 nA and n = 2 operating at room
temperature at a current of 10 A.
(c) A silicon diode is operating with a temperature of 50?C and the diode voltage is measured to
be 0.736 V at a current of 2.50 mA. What is the saturation current of the diode?
SOLUTION (a) Known Information and Given Data: The diode currents are given and the saturation current
parameter IS is specified.
Unknowns: Diode voltage at each of the operating currents
Approach: Solve Eq. (3.9) for the diode voltage and evaluate the expression at each operating
current.
Assumptions: At room temperature, we will use VT = 0.025 V = 1/40 V; assume n = 1, since
it is not specified otherwise; assume dc operation: iD = ID and vD = VD.
92 Chapter 3 Solid-State Diodes and Diode Circuits
Analysis: Solving Eq. (3.9) for VD with ID = 0.1 fA yields
VD = nVT ln1 +
ID
IS  = 1(0.025 V) ln1 +
3 × 10?4 A
10?16 A  = 0.718V
For IS = 10 fA:
VD = nVT ln1 +
ID
IS  = 1(0.025V) ln1 +
3 × 10?4 A
10?14 A  = 0.603V
For ID = 1 mA with IS = 0.1 fA:
VD = nVT ln1 +
ID
IS  = 1(0.025V) ln1 +
10?3 A
10?16 A = 0.748V
Check of Results: The diode voltages are all between 0.5 V and 1.0 V and are reasonable (the
diode voltage should not exceed the bandgap for n = 1).
SOLUTION (b) Known Information and Given Data: The diode current is given and the values of the saturation
current parameter IS and n are both specified.
Unknowns: Diode voltage at the operating current
Approach: Solve Eq. (3.9) for the diode voltage and evaluate the resulting expression.
Assumptions: At room temperature, we will use VT = 0.025 V = 1/40 V.
Analysis: The diode voltage will be
VD = nVT ln1 +
ID
IS  = 2(0.025V) ln1 +
10 A
10?8 A = 1.04V
Check of Results: Based on the comment at the end of part (a) and realizing that n =2, voltages
between 1 V and 2 V are reasonable for power diodes operating at high currents.
SOLUTION (c) Known Information and Given Data: The diode current is 2.50 mA and voltage is 0.736 V. The
diode is operating at a temperature of 50?C.
Unknowns: Diode saturation current IS
Approach: Solve Eq. (3.9) for the saturation current and evaluate the resulting expression. The
value of the thermal voltage VT will need to be calculated for T = 50?C.
Assumptions: The value of n is unspecified, so assume n = 1.
Analysis: Converting T = 50?C to Kelvins, T = (273 + 50) K = 323 K, and
VT =
kT
q =
(1.38 × 10?23 J/K)(323 K)
1.60 × 10?19 ?C = 27.9 mV
Solving Eq. (3.9) for IS yields
IS =
ID
exp VD
nVT ? 1
=
2.5mA
exp 0.736V
0.0279V? 1
= 8.74 × 10?15 A = 8.74 fA
Check of Results: The saturation current is within the range of typical values specified in
Eq. (3.10).
3.4 Diode Characteristics Under Reverse, Zero, and Forward Bias 93
EXERCISE: A diode has a reverse saturation current of 40 fA. Calculate i D for diode voltages of
0.55 and 0.7 V. What is the diode voltage if i D = 6 mA?
A NSWERS: 143 A; 57.9 mA; 0.643 V
3.4 DIODE CHARACTERISTICS UNDER REVERSE, ZERO, AND FORWARD BIAS
When a dc voltage or current is applied to an electronic device, we say that we are providing a dc
bias or simply a bias to the device. As we develop our electronics expertise, choosing the bias will
be important to all of the circuits that we analyze and design. We will find that bias determines
device characteristics, power dissipation, voltage and current limitations, and other important circuit
parameters such as impedance levels and voltage gain. For a diode, there are three important bias
conditions. Reverse bias and forward bias correspond to vD < 0 V and vD > 0 V, respectively. The
zero bias condition, with vD = 0 V, represents the boundary between the forward and reverse bias
regions. When the diode is operating with reverse bias, we consider the diode “off” or nonconducting
because the current is very small (iD = ?IS). For forward bias, the diode is usually in a highly
conducting state and is considered “on.”
3.4.1 REVERSE BIAS
For vD < 0, the diode is said to be operating under reverse bias. Only a very small reverse leakage
current, approximately equal to IS, flows through the diode. This current is small enough that we
usually think of the diode as being in the nonconducting or off state when it is reverse-biased. For
example, suppose that a dc voltage V = ?4VT = ?0.1 V is applied to the diode terminals so that
vD = ?0.1 V. Substituting this value into Eq. (3.11) gives
iD = IS expvD
VT ? 1 = IS[ ??→
negligible
exp(?4) ? 1] ≈ ?IS (3.12)
because exp(?4) = 0.018. For a reverse bias greater than 4VT , that is, vD ≤ ?4VT = ?0.1 V, the
exponential term exp(vD/VT ) is much less than 1, and the diode current will be approximately equal
to ?IS, a very small current. The current IS was identified in Fig. 3.9.
EXERCISE: A diode has a reverse saturation current of 5 fA. Calculate i D for diode voltages of
?0.04 V and ?2 V (see Sec. 3.6).
A NSWERS: ?3.99 fA; ?5 fA
The situation depicted in Fig. 3.9 and Eq. (3.12) actually represents an idealized picture of the
diode. In a real diode, the reverse leakage current is several orders of magnitude larger than IS due
to the generation of electron–hole pairs within the depletion region. In addition, iD does not saturate
but increases gradually with reverse bias as the width of the depletion layer increases with reverse
bias. (See Sec. 3.6.1).
3.4.2 ZERO BIAS
Although it may seem to be a trivial result, it is important to remember that the i-v characteristic of
the diode passes through the origin. For zero bias with vD = 0, we find iD = 0. Just as for a resistor,
there must be a voltage across the diode terminals in order for a nonzero current to exist.
94 Chapter 3 Solid-State Diodes and Diode Circuits
10–12
10–13
1.0
10–1
Diode current (A)
Diode voltage (V)
10–2
10–3
10–4
10–5
10–6
10–7
10–8
10–9
10–10
10–11
10–14
10–15
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
Slope ? 1 decade/60 mV
IS = 10–15 A
VT = 0.025 V
Figure 3.11 Diode i-v characteristic on semilog scale.
3.4.3 FORWARD BIAS
For the case vD > 0, the diode is said to be operating under forward bias, and a large current can be
present in the diode. Suppose that a voltage vD ≥ +4VT = +0.1 V is applied to the diode terminals.
The exponential term exp(vD/VT ) is now much greater than 1, and Eq. (3.9) reduces to
iD = IS expvD
VT ?	
negligible
1 ～= IS expvD
VT  (3.13)
The diode current grows exponentially with applied voltage for a forward bias greater than approximately
4VT .
The diode i-v characteristic for forward voltages is redrawn in semilogarithmic form in Fig. 3.11.
The straight line behavior predicted by Eq. (3.13) for voltages vD ≥ 4VT is apparent. A slight
curvature can be observed near the origin, where the ?1 term in Eq. (3.13) is no longer negligible.
The slope of the graph in the exponential region is very important. Only a 60-mV increase in the
forward voltage is required to increase the diode current by a factor of 10. This is the reason for the
almost vertical increase in current noted in Fig. 3.8 for voltages above the turn-on voltage.
EXAMPLE 3.4 DIODE VOLTAGE CHANGE VERSUS CURRENT
The slope of the diode i -v characteristic is an important number for circuit designers to remember.
PROBLEM Use Eq. (3.13) to accurately calculate the voltage change required to increase the diode current by
a factor of 10.
SOLUTION Known Information and Given Data: The current changes by a factor of 10.
Unknowns: The diode voltage change corresponding to a one decade change in current; the
saturation current has not been given.
Approach: Form an expression for the ratio of two diode currents using the diode equation. The
saturation current will cancel out and is not needed.
3.4 Diode Characteristics Under Reverse, Zero, and Forward Bias 95
Assumptions: Room temperature operation with VT = 25.0 mV; assume ID 
 IS.
Analysis: Let
iD1 = IS expvD1
VT  and iD2 = IS expvD2
VT 
Taking the ratio of the two currents and setting it equal to 10 yields
iD2
iD1 = expvD2 ? vD1
VT  = expvD
VT  = 10 and vD = VT ln 10 = 2.3VT
Therefore VD = 2.3VT = 57.5 mV (or approximately 60 mV) at room temperature.
Check of Results: The result is consistent with the logarithmic plot in Fig. 3.11. The diode voltage
changes approximately 60 mV for each decade change in forward current.
EXERCISE: A diode has a saturation current of 2 fA. (a) What is the diode voltage at a diode
current of 40 A (assume VT = 25.0 mV)? Repeat for a diode current of 400 A. What is the
difference in the two diode voltages? (b) Repeat for VT = 25.9 mV.
A NSWERS: 0.593 V, 0.651 V, 57.6 mV; 0.614 V, 0.674 V, 59.6 mV
The diode voltage changes by approximately 60 mV per decade change in diode current. Sixty
mV/decade often plays an important role in our thinking about the design of circuits containing
both diodes and bipolar transistors and is a good number to remember.
DESIGN
NOTE
Figure 3.12 compares the characteristics of three diodes with different values of saturation
current. The saturation current of diode A is 10 times larger than that of diode B, and the saturation
current of diode B is 10 times that of diode C. The spacing between each pair of curves is
Diode current (A)
Diode voltage (V)
0.10
0.08
0.02
0.04
0.06
0.00
?0.02
0.0 0.2 0.4 0.6 0.8 1.0
A
B
~60 mV C
IS
Figure 3.12 Diode characteristics for three different reverse saturation currents: (a) 10?12 A, (b) 10?13 A, and (c) 10?14 A.
96 Chapter 3 Solid-State Diodes and Diode Circuits
approximately 60 mV. If the saturation current of the diode is reduced by a factor of 10, then the
diode voltage must increase by approximately 60 mV to reach the same operating current level.
Figure 3.12 also shows the relatively low sensitivity of the forward diode voltage to changes in the
parameter IS. For a fixed diode current, a change of two orders of magnitude in IS results in a diode
voltage change of only 120 mV.
3.5 DIODE TEMPERATURE COEFFICIENT
Another important number to keep in mind is the temperature coefficient associated with the diode
voltage vD. Solving Eq. (3.11) for the diode voltage under forward bias
vD = VT lniD
IS + 1 =
kT
q
lniD
IS + 1～=
kT
q
lniD
IS  V (3.14)
and taking the derivative with respect to temperature yields
dvD
dT =
k
q
lniD
IS ?
kT
q
1
IS
d IS
dT =
vD
T ? VT
1
IS
d IS
dT =
vD ? VGO ? 3VT
T
V/K (3.15)
where it is assumed that iD 
 IS and IS ∝ n2
i . In the numerator of Eq. (3.15), vD represents the
diode voltage, VGO is the voltage corresponding to the silicon bandgap energy at 0 K (VGO = EG/q),
and VT is the thermal potential. The last two terms result from the temperature dependence of n2
i
as defined by Eq. (2.2). Evaluating the terms in Eq. (3.15) for a silicon diode with vD = 0.65 V,
EG = 1.12 eV, and VT = 0.025 V yields
dvD
dT =
(0.65 ? 1.12 ? 0.075) V
300 K = ?1.82 mV/K (3.16)
The forward voltage of the diode decreases as temperature increases, and the diode exhibits a
temperature coefficient of approximately ?1.8 mV/?C at room temperature.
DESIGN
NOTE
EXERCISE: (a) Verify Eq. (3.15) using the expression for n2
i from Eq. (2.1). (b) A silicon diode is
operating at T = 300 K, with i D = 1 mA, and vD = 0.680 V. Use the result from Eq. (3.16) to
estimate the diode voltage at 275 K and at 350 K.
A NSWERS: 0.726 V; 0.589 V
3.6 DIODES UNDER REVERSE BIAS
We must be aware of several other phenomena that occur in diodes operated under reverse bias. As
depicted in Fig. 3.13, the reverse voltage vR applied across the diode terminals is dropped across the
space charge region and adds directly to the built-in potential of the junction:
vj = φj + vR for vR > 0 (3.17)
The increased voltage results in a larger internal electric field that must be supported by additional
charge in the depletion layer, as defined by Eqs. (3.2) to (3.5). Using Eq. (3.7) with the voltage
3.6 Diodes Under Reverse Bias 97
(x)
E(x) (x)
p
–qNA
(a) Space charge density (b) Electric field (c) Electrostatic potential
+qND
–xp –xp xn –xp xn
x
xn
(j+ vR)
x x
Figure 3.13 The pn junction diode under reverse bias.
from Eq. (3.17), the general expression for the depletion-layer width wd for an applied reverse-bias
voltage vR becomes
wd = (xn + xp) = 2εs
q  1
NA +
1
ND (φj + vR)
or
(3.18)
wd = wdo1 +
vR
φj
where wdo = 2εs
q  1
NA +
1
ND φj
The width of the space charge region increases approximately in proportion to the square root of the
applied voltage.
EXERCISE: The diode in Ex. 3.1 had a zero-bias depletion-layer width of 0.113 mand a built-in
voltage of 0.979 V. What will be the depletion-layer width for a 10-V reverse bias? What is the
new value of EMAX?
A NSWERS: 0.378 m; 581 kV/cm
3.6.1 SATURATION CURRENT IN REAL DIODES
The reverse saturation current actually results from the thermal generation of hole–electron pairs
in the depletion region that surrounds the pn junction and is therefore proportional to the volume
of the depletion region. Since the depletion-layer width increases with reverse bias, as described
by Eq. (3.18), the reverse current does not truly saturate, as depicted in Fig. 3.9 and Eq. (3.9).
Instead, there is gradual increase in reverse current as the magnitude of the reverse bias voltage is
increased.
IS = ISO1 +
vR
φj
(3.19)
Under forward bias, the depletion-layer width changes very little, and IS = ISO for forward
bias.
EXERCISE: A diode has ISO = 10 fA and a built-in voltage of 0.8 V. What is IS for a reverse bias
of 10 V?
A NSWER: 36.7 fA
98 Chapter 3 Solid-State Diodes and Diode Circuits
E L E C T R O N I C S I N A C T I O N
The PTAT Voltage and Electronic Thermometry
The well-defined temperature dependence of the diode voltage discussed in Secs. 3.3 to 3.5
is actually used as the basis for most digital thermometers. We can build a simple electronic
thermometer based on the circuit shown here in which two identical diodes are biased by
current sources I1 and I2.
VCC
D1 VD1 D2
I1 I2
VD2

 VPTAT 
 

Digital thermometer: ? D. Hurst/Alamy RF.
If we calculate the difference between the diode voltages using Eq. (3.14), we discover a voltage
that is directly proportional to absolute temperature (PTAT), referred to as the PTAT voltage
or VPTAT:
VPTAT = VD1 ? VD2 = VT lnID1
IS ? VT lnID2
IS  = VT lnID1
ID2 =
kT
q
lnID1
ID2
The PTAT voltage has a temperature coefficient given by
dVPTAT
dT =
k
q
lnID1
ID2 =
VPTAT
T
By using two diodes, the temperature dependence of IS has been eliminated from the equation.
For example, suppose T = 295 K, ID1 = 250 A, and ID2 = 50 A. Then VPTAT = 40.9 mV
with a temperature coefficient of +0.139 mV/K.
This simple but elegant PTAT voltage circuit forms the heart of most of today’s highly
accurate electronic thermometers as depicted in the block diagram here. The analog PTAT
voltage is amplified and then converted to a digital representation by an A/D converter. The
digital output is scaled and offset to properly represent either the Fahrenheit or Celsius temperature
scales and appears on an alphanumeric display. The scaling and offset shift can also
be done in analog form prior to the A/D conversion operation.
+
–
PTAT
voltage
generator
A/D
converter
Digital
scaling
and offset
Display
Amplification
A n
Block diagram of a digital thermometer.
3.6 Diodes Under Reverse Bias 99
0.1
Diode voltage (V)
Diode current (A)
0.0
–0.1
–0.2
–5 –4 –3 –2 –1 0
5.6 V VZ
TC of VZ
1
–VZ
Slope = 1
RZ
Figure 3.14 i-v characteristic of a diode including the reverse-breakdown region. The inset shows the temperature coefficient
(TC) of VZ .
3.6.2 REVERSE BREAKDOWN
As the reverse voltage increases, the electric field within the device grows, and the diode eventually
enters the breakdown region. The onset of the breakdown process is fairly abrupt, and the current
increases rapidly for any further increase in the applied voltage, as shown in the i-v characteristic of
Fig. 3.14.
The magnitude of the voltage at which breakdown occurs is called the breakdown voltage VZ
of the diode and is typically in the range 2 V ≤ VZ ≤ 2000 V. The value of VZ is determined
primarily by the doping level on the more lightly doped side of the pn junction, but the heavier the
doping, the smaller the breakdown voltage of the diode.
Two separate breakdown mechanisms have been identified: avalanche breakdown and Zener
breakdown. These are discussed in the following two sections.
Avalanche Breakdown
Silicon diodes with breakdown voltages greater than approximately 5.6 V enter breakdown through
a mechanism called avalanche breakdown. As the width of the depletion layer increases under
reverse bias, the electric field increases, as indicated in Fig. 3.13. Free carriers in the depletion region
are accelerated by this electric field, and as the carriers move through the depletion region, they
collide with the fixed atoms. At some point, the electric field and the width of the space charge
region become large enough that some carriers gain energy sufficient to break covalent bonds upon
impact, thereby creating electron–hole pairs. The new carriers created can also accelerate and create
additional electron–hole pairs through this impact-ionization process, as illustrated in Fig. 3.15.
Zener Breakdown
True Zener breakdown occurs only in heavily doped diodes. The high doping results in a very
narrow depletion-region width, and application of a reverse bias causes carriers to tunnel directly
between the conduction and valence bands, again resulting in a rapidly increasing reverse current in
the diode.
Breakdown Voltage Temperature Coefficient
We can differentiate between the two types of breakdown because the breakdown voltages associated
with the two mechanisms exhibit opposite temperature coefficients (TC). In avalanche breakdown,
100 Chapter 3 Solid-State Diodes and Diode Circuits
Ionizing
collision
Figure 3.15 The avalanche breakdown process. (Note
that the positive and negative charge carriers will actually
be moving in opposite directions in the electric field in
the depletion region.)
RZ
VZ
(a) (b)
Figure 3.16 (a) Model
for reverse-breakdown region
of diode. (b) Zener
diode symbol.
VZ increases with temperature; in Zener breakdown, VZ decreases with temperature. For silicon
diodes, a zero temperature coefficient is achieved at approximately 5.6 V. The avalanche breakdown
mechanism dominates in diodes that exhibit breakdown voltages of more than 5.6 V, whereas diodes
with breakdown voltages below 5.6 V enter breakdown via the Zener mechanism.
3.6.3 DIODE MODEL FOR THE BREAKDOWN REGION
In breakdown, the diode can be modeled by a voltage source of value VZ in series with resistor RZ ,
which sets the slope of the i-v characteristic in the breakdown region, as indicated in Fig. 3.14. The
value of RZ is normally small (RZ ≤ 100 ), and the reverse current flowing in the diode must be
limited by the external circuit or the diode will be destroyed.
From the i-v characteristic in Fig. 3.14 and the model in Fig. 3.16, we see that the voltage across
the diode is almost constant, independent of current, in the reverse-breakdown region. Some diodes
are actually designed to be operated in reverse breakdown. These diodes are called Zener diodes1
and have the special circuit symbol given in Fig. 3.16(b). Links to data sheets for a series of Zener
diode can be found on the MCD website.
3.7 pn JUNCTION CAPACITANCE
Forward- and reverse-biased diodes have a capacitance associated with the pn junction. This capacitance
is important under dynamic signal conditions because it prevents the voltage across the diode
from changing instantaneously. The capacitance is referred to as the pn junction capacitance or the
depletion-layer capacitance.
3.7.1 REVERSE BIAS
Under reverse bias, wd increases beyond its zero-bias value, as expressed by Eq. (3.18), and hence
the amount of charge in the depletion region also increases. Since the charge in the diode is changing
with voltage, a capacitance results. Using Eqs. (3.4) and (3.7), the total space charge on the n-side
1 The term Zener diode is typically used to refer to diodes that breakdown by either the Zener or avalanche mechanism.
3.7 pn Junction Capacitance 101
of the diode is given by
Qn = qNDxn A = q  NAND
NA + ND wd A C (3.20)
where A is the cross-sectional area of the diode and wd is described by Eq. (3.18). The capacitance
of the reverse-biased pn junction is given by
Cj =
dQn
dvR =
Cjo A
1 +
vR
φj
where Cjo =
εs
wdo
F/cm2 (3.21)
in which Cjo represents the zero-bias junction capacitance per unit area of the diode.
Figure 3.17 Circuit
symbol for the variable
capacitance diode
(varactor).
Equation (3.21) shows that the capacitance of the diode changes with applied voltage. The
capacitance decreases as the reverse bias increases, exhibiting an inverse square root relationship.
This voltage-controlled capacitance can be very useful in certain electronic circuits. Diodes can be
designed with impurity profiles (called hyper-abrupt profiles) specifically optimized for operation as
voltage-controlled capacitors. As for the case of Zener diodes, a special symbol exists for the variable
capacitance diode, as shown in Fig. 3.17. Remember that this diode is designed to be operated under
reverse bias, but it conducts in the forward direction. Links to data sheets for a series of variable
capacitance diodes can be found on the MCD website.
EXERCISE: What is the value of Cjo for the diode in Ex. 3.1? What is the zero bias value of Cj
if the diode junction area is 100 m × 125 m? What is the capacitance at a reverse bias of
5 V?
A NSWERS: 91.7 nf/cm2; 11.5 pF; 4.64 pF
3.7.2 FORWARD BIAS
When the diode is operating under forward bias, additional charge is stored in the neutral regions
near the edges of the space charge region. The amount of charge QD stored in the diode is proportional
to the diode current:
QD = iDτT C (3.22)
The proportionality constant τT is called the diode transit time and ranges from 10?15 s to more than
10?6 s (1 fs to 1 s) depending on the size and type of diode. Because we know that iD is dependent
on the diode voltage through the diode equation, there is an additional capacitance, the diffusion
capacitance CD, associated with the forward region of operation:
CD =
dQD
dvD =
(iD + IS)τT
VT ～=
iDτT
VT
F (3.23)
in which VT is the thermal voltage. The diffusion capacitance is proportional to current and can
become quite large at high currents.
102 Chapter 3 Solid-State Diodes and Diode Circuits
E L E C T R O N I C S I N A C T I O N
The SPICE Circuit Simulation Program An IEEE Global History Network Milestone2
Courtesy of David Hodges
“SPICE (Simulation Program with Integrated Circuit Emphasis) was created at UC Berkeley
as a class project in 1969–1970. It evolved to become the worldwide standard integrated
circuit simulator. SPICE has been used to train many students in the intricacies of circuit
simulation. SPICE and its descendants have become essential tools employed by virtually all
integrated circuit designers.
SPICEwas the first computer program for simulating the performance of integrated circuits
that was readily available to undergraduate students for study of integrated circuit design.
Hundreds of graduates from UC Berkeley and other universities became the backbone of the
engineering workforce that moved the United States to microelectronics industry leadership
Professor Don Pederson, “Father of
SPICE”, IEEE Medal of Honor Recipient:
Courtesy of David Hodges
in the 1970s. Graduates of Berkeley became leaders of today’s largest
firms delivering design automation capabilities for advanced microelectronics.”
Above is a photograph of the SPICE Commemorative Plaque
that may be found just inside the main entrance to Cory Hall, the Electrical
Engineering Building at the University of California, Berkeley,
CA. Cory Hall is the building where SPICE was developed.
Professor Donald O. Pederson guided the students that developed
the SPICE program, and he was awarded the 1998 IEEE Medal
of Honor “For creation of the SPICE program universally used for the
computer aided design of circuits.” Further information can be found
in the IEEE Solid-State Circuits Magazine, “SPICE Commemorative
Issue,” vol. 3, no. 2, Spring 2011, and on the IEEE Global History
Network website.
2 Quoted from the SPICE Circuit Simulation Program Milestone of the IEEE Global History Network: http://www.ieeeghn
.org/wiki/index.php/Milestones:SPICE Circuit Simulation Program Milestone photograph courtesy of Professor David A.
Hodges, used with permission.
3.9 Diode SPICE Model and Layout 103
Metal
(b)
(a)
Ohmic
contact
Rectifying
contact
Anode n-type n+ Cathode
semiconductor
Figure 3.18 (a) Schottky barrier diode
structure; (b) Schottky diode symbol.
Diode voltage (V)
Diode current (A)
1.00  10–2
8.00  10–3
6.00  10–3
4.00  10–3
2.00  10–3
0
0.0 0.2 0.4 0.6 0.8 1.0
SB
pn
Figure 3.19 Comparison of pn junction ( pn) and Schottky barrier diode
(SB) i -v characteristics.
EXERCISE: A diode has a transit time of 10 ns. What is the diffusion capacitance of the diode
for currents of 10 A, 0.8 mA, and 50 mA at room temperature?
A NSWERS: 4 pF; 320 pF; 20 nF
3.8 SCHOTTKY BARRIER DIODE
In a p+n junction diode, the p-side is a highly doped region (a conductor), and one might wonder if
it could be replaced with a metallic layer. That is in fact the case, and in the Schottky barrier diode,
one of the semiconductor regions of the pn junction diode is replaced by a non-ohmic rectifying
metal contact, as indicated in Fig. 3.18. It is easiest to form a Schottky contact to n-type silicon, and
for this case the metal region becomes the diode anode. An n+ region is added to ensure that the
cathode contact is ohmic. The symbol for the Schottky barrier diode appears in Fig. 3.18(b).
The Schottky diode turns on at a much lower voltage than its pn-junction counterpart, as indicated
in Fig. 3.19. It also has significantly reduced internal charge storage under forward bias.We encounter
an important use of the Schottky diode in bipolar logic circuits in Chapter 9. Schottky diodes also
find important applications in high-power rectifier circuits and fast switching applications.
3.9 DIODE SPICE MODEL AND LAYOUT
The circuit in Fig. 3.20 represents the diode model that is included in SPICE programs. Resistance
RS represents the inevitable series resistance that always accompanies fabrication of, and making
contacts to, a real device structure. The current source represents the ideal exponential behavior of
the diode as described by Eq. 3.12 and SPICE parameters IS, N, and VT . The model equation for
iD also includes a second term, not shown here, that models the effects of carrier generation in the
space charge region in a manner similar to Eq. (3.19).
The capacitor specification includes the depletion-layer capacitance for the reverse-bias region
modeled by SPICE parameters CJO, VJ, and M, as well as the diffusion capacitance associated
104 Chapter 3 Solid-State Diodes and Diode Circuits
CD = TT for vD ≥ 0
iD
NVT
Cj = RAREA for vD ≤ 0
CJO
vD
VJ   1 ?
M
iD = IS exp ?  1
Anode
vD
i'D
i'D
Cathode
+
Anode
+
RS
iD vD C = Cj + CD
Cathode
Figure 3.20 Diode equivalent circuit and simplified versions of the model equations used in SPICE programs.
with the junction under forward bias and defined by N and the transit-time parameter TT. In SPICE,
the “junction grading coefficient” is an adjustable parameter. Using the typical value of M = 0.5
results in Eq. (3.21).
EXERCISE: Find the default values of the seven parameters in Table 3.1 for the SPICE program
that you use in class. Compare to the values in Table 3.1.
TABLE 3.1
SPICE Diode Parameter Equivalences
PARAMETER OUR TEXT SPICE TYPICAL DEFAULT VALUES
Saturation current IS IS 10 fA
Ohmic series resistance RS RS 0 
Ideality factor or emission n N 1
coefficient
Transit time τT TT 0 s
Zero-bias junction capacitance Cjo · A CJO 0 F
for a unit area diode RAREA = 1
Built-in potential φj VJ 1 V
Junction grading coefficient — M 0.5
Relative junction area — RAREA 1
3.9.1 DIODE LAYOUT
Figure 3.21(a) shows the layout of a simple diode fabricated by forming a p-type diffusion in an
n-type silicon wafer, as outlined in Chapter 2. This diode has a long rectangular p-type diffusion
to increase the value of IS, which is proportional to the junction area. Multiple contacts are formed
to the p-type anode, and the p-region is surrounded by a collar of contacts to the n-type region.
Both these sets of contacts are used to minimize the value of the extrinsic series resistance RS of
the diode, as included in the model in Fig. 3.20. Identical contacts are used so that they all tend to
etch open at the same time during the fabrication process. The use of multiple identical contacts
also facilitates calculation of the overall contact resistance. Heavily doped n-type regions are placed
under the n-region contacts to ensure formation of an ohmic contact and prevent formation of a
Schottky barrier diode.
NVT
vD

3.10 Diode Circuit Analysis 105
n-type silicon
n-type silicon
p
pn junction diode Schottky barrier diode
n+
Cathode Anode Cathode Anode
p-type diffusion
C A C C A C
n+
SiO2
(a)
(c)
(b)
Figure 3.21 (a) Layout of a pn junction diode and a Schottky diode (b) pn junction diode photograph (c) Cross-section of
the two diodes (See top view of diode in Chapter 3 opener.)
A conceptual drawing of a metal-semiconductor or Schottky diode also appears in Fig. 3.21(a)
in which the aluminum metallization acts as the anode of the diode and the n-type semiconductor is
the diode cathode. Careful attention to processing details is needed to form a diode rather than just
an ohmic contact.
3.10 DIODE CIRCUIT ANALYSIS
We now begin our analysis of circuits containing diodes and introduce simplified circuit models for
the diode. Figure 3.22 presents a series circuit containing a voltage source, resistor, and diode. Note
that V and R may represent the Th′evenin equivalent of a more complicated two-terminal network.
Also note the notational change in Fig. 3.22. In the circuits that we analyze in the next few sections,
the applied voltage and resulting diode voltage and current will all be dc quantities. (Recall that the
dc components of the total quantities iD and vD are indicated by ID and VD, respectively.)
One common objective of diode circuit analysis is to find the quiescent operating point
(Q-point), or bias point, for the diode. The Q-point consists of the dc current and voltage (ID, VD)
that define the point of operation on the diode’s i-v characteristic. We start the analysis by writing
the loop equation for the circuit of Fig. 3.22:
V = ID R + VD (3.24)
Equation (3.24) represents a constraint placed on the diode operating point by the circuit elements.
The diode i-v characteristic in Fig. 3.8 represents the allowed values of ID and VD as determined by
the solid-state diode itself. Simultaneous solution of these two sets of constraints defines the Q-point.
R
V VD
10 V
ID
10 kΩ
Figure 3.22 Diode circuit containing a voltage source and resistor.
106 Chapter 3 Solid-State Diodes and Diode Circuits
2.00  10?3
1.80  10?3
1.60  10?3
1.40  10?3
1.20  10?3
1.00  10?3
8.00  10?4
6.00  10?4
4.00  10?4
2.00  10?4
0
0 1 2 3 4 5 6
Load line
Q-point
Diode voltage (V)
Diode current (A)
Figure 3.23 Diode i-v characteristic and load line.
We explore several methods for determining the solution to Eq. (3.24), including graphical
analysis and the use of models of varying complexity for the diode. These techniques will include
? Graphical analysis using the load-line technique.
? Analysis with the mathematical model for the diode.
? Simplified analysis with an ideal diode model.
? Simplified analysis using the constant voltage drop model.
3.10.1 LOAD-LINE ANALYSIS
In some cases, the i-v characteristic of the solid-state device may be available only in graphical form,
as in Fig. 3.23. We can then use a graphical approach (load-line analysis) to find the simultaneous
solution of Eq. (3.24) with the graphical characteristic. Equation (3.24) defines the load line for the
diode. The Q-point can be found by plotting the graph of the load line on the i -v characteristic for
the diode. The intersection of the two curves represents the quiescent operating point, or Q-point,
for the diode.
EXAMPLE 3.5 LOAD-LINE ANALYSIS
The graphical load-line approach is an important concept for visualizing the behavior of diode
circuits as well as for estimating the actual Q-point.
PROBLEM Use load-line analysis to find the Q-point for the diode circuit in Fig. 3.22 using the i -v characteristic
in Fig. 3.23.
SOLUTION Known Information and Given Data: The diode i -v characteristic is presented graphically in
Fig. 3.23. Diode circuit is given in Fig. 3.22 with V = 10 V and R = 10 k.
Unknowns: Diode Q-point (ID, VD).
3.10 Diode Circuit Analysis 107
Approach: Write the load-line equation and find two points on the load line that can be plotted
on the graph in Fig. 3.23. The Q-point is at the intersection of the load line with the diode i -v
characteristic.
Assumptions: Diode temperature corresponds to the temperature at which the graph in Fig. 3.23
was measured.
Analysis: Using the values from Fig. 3.22, Eq. (3.24) can be rewritten as
10 = 104 ID + VD (3.25)
Two points are needed to define the line. The simplest choices are
ID = (10 V/10 k) = 1 mA for VD = 0 and VD = 10 V for ID = 0
Unfortunately, the second point is not in the range of the graph presented in Fig. 3.23, but we are
free to choose any point that satisfies Eq. (3.25). Let’s pick VD = 5 V:
ID = (10 ? 5)V/104  = 0.5 mA for VD = 5
These points and the resulting load line are plotted in Fig. 3.23. The Q-point is given by the
intersection of the load line and the diode characteristic:
Q-point = (0.95 mA, 0.6 V)
Check of Results: We can double check our result by substituting the diode voltage found from
the graph into Eq. (3.25) and calculating ID. Using VD = 0.6 V in Eq. (3.25) yields an improved
estimate for the Q-point: (0.94 mA, 0.6 V). [We could also substitute 0.95 mA into Eq. (3.25) and
calculate VD.]
Discussion: Note that the values determined graphically are not quite on the load line since they
do not precisely satisfy the load-line equation. This is a result of the limited precision that we can
obtain by reading the graph.
EXERCISE: Repeat the load-line analysis if V = 5 V and R = 5 k.
A NSWERS: (0.88 mA, 0.6 V)
EXERCISE: Use SPICE to find the Q-point for the circuit in Fig. 3.22. Use the default values of
parameters in your SPICE program.
A NSWERS: (935 A, 0.653 V) for IS = 10 fA and T = 300 K
3.10.2 ANALYSIS USING THE MATHEMATICAL MODEL FOR THE DIODE
We can use our mathematical model for the diode to approach the solution of Eq. (3.25) more directly.
The particular diode characteristic in Fig. 3.23 is represented quite accurately by diode Eq. (3.11),
with IS = 10?13 A, n = 1, and VT = 0.025 V:
ID = IS expVD
VT ? 1 = 10?13[exp(40VD) ? 1] (3.26)
Eliminating ID by substituting Eq. (3.26) into Eq. (3.25) yields
10 = 104 · 10?13[exp(40VD) ? 1] + VD (3.27)
108 Chapter 3 Solid-State Diodes and Diode Circuits
0
?0.0001
ID
iD
VD0
Q-point
VD
VT
vD
rD
1
(a)
iD rD
D +
–
iD +
–
vD vD0
(b)
Figure 3.24 (a) Diode behavior around the Q-point; (b) linear model for the diode at the Q-point.
The expression in Eq. (3.27) is called a transcendental equation and does not have a closed-form
analytical solution, so we settle for a numerical answer to the problem.
One approach to finding a numerical solution to Eq. (3.27) is through simple trial and error.We
can guess a value of VD and see if it satisfies Eq. (3.27). Based on the result, a new guess can be
formulated and Eq. (3.27) evaluated again. The human brain is quite good at finding a sequence of
values that will converge to the desired solution.
On the other hand, it is often preferable to use a computer to find the solution to Eq. (3.27), particularly
if we need to find the answer to several different problems or parameter sets. The computer,
however, requires a much more well-defined iteration strategy than brute force trial and error.
We can develop an iterative solution method for the diode circuit in Fig. 3.22 by creating a linear
model for the diode equation in the vicinity of the diode Q-point as depicted in Fig. 3.24(a). First
we find the slope of the diode characteristic at the operating point:
gD =
?iD
?vD				 Q?Pt =
IS
VT
expVD
VT  =
ID + IS
VT ～=
ID
VT
and rD =
1
gD =
VT
ID
(3.28)
Slope gD is called the diode conductance, and its reciprocal rD is termed the diode resistance. Next
we can use the slope to find the x-axis intercept point VD0:
VD0 = VD ? IDrD = VD ? VT (3.29)
VD0 and rD represent a two-element linear circuit model for the diode as in Fig. 3.24(b), and this
circuit model replaces the diode in the single loop circuit in Fig. 3.25.
Now we can use an iterative process to find the Q-point of the diode in the circuit.
1. Pick a starting guess for ID.
2. Calculate the diode voltage using VD = VT ln1 +
ID
IS .
3. Calculate the values of VD0 and rD.
4. Calculate a new estimate for ID from the circuit in Fig. 3.25(b): ID = V ? VD0
R + rD
.
5. Repeat steps 2 to 4 until convergence is obtained.
3.10 Diode Circuit Analysis 109
(b)
V D V
R
iD
R
(a)
rD
vD vD0
Figure 3.25 (a) Diode circuit; (b) circuit with two-element diode model.
TABLE 3.2
Example of Iterative Analysis
I D (A) VD (V) RD (Ω) VD 0 (V)
1.0000E-03 0.5756 25.80 0.5498
9.4258E-04 0.5742 27.37 0.5484
9.4258E-04 0.5742 27.37 0.5484
Table 3.2 presents the results of performing the above iteration process using a spreadsheet. The
diode current and voltage converge rapidly in only three iterations.
Note that one can achieve answers to an almost arbitrary precision using the numerical approach.
However, in most real circuit situations, we will not have an accurate value for the saturation current of
the diode, and there will be significant tolerances associated with the sources and passive components
in the circuit. For example, the saturation current specification for a given diode type may vary by
factors ranging from 10:1 to as much as 100:1. In addition, resistors commonly have ±5 percent or
±10 percent tolerances, and we do not know the exact operating temperature of the diode (remember
the ?1.8 mV/K temperature coefficient) or the precise value of the parameter n. Hence, it does not
make sense to try to obtain answers with a precision of more than a few significant digits.
An alternative to the use of a spreadsheet is to write a simple program using a high-level language.
The solution to Eq. (3.28) also can be found using the “solver” routines in many calculators, which
use iteration procedures more sophisticated than that just described. MATLAB also provides the
function fzero, which will calculate the zeros of a function as outlined in Ex. 3.6.
EXERCISE: An alternative expression (another transcendental equation) for the basic diode
circuit can be found by eliminating VD in Eq. (3.25) using Eq. (3.14). Show that the result is
10 = 104 ID + 0.025 ln1 +
ID
IS
EXAMPLE 3.6 SOLUTION OF THE DIODE EQUATION USING MATLAB
MATLAB is one example of a computer tool that can be used to find the solution to transcendental
equations.
PROBLEM Use MATLAB to find the solution to Eq. (3.27).
SOLUTION Known Information and Given Data: Diode circuit in Fig. 3.22 with V = 10 V, R = 10 k,
IS = 10?13 A, n = 1, and VT = 0.025 V
Unknowns: Diode voltage VD
110 Chapter 3 Solid-State Diodes and Diode Circuits
Approach: Create a MATLAB “M-File” describing Eq. (3.27). Execute the program to find the
diode voltage.
Assumptions: Room temperature operation with VT = 1/40 V
Analysis: First, create an M-File for the function ‘diode’:
function xd = diode(vd)
xd = 10 ? (10∧(?9)) ? (exp(40 ? vd) ? 1) ? vd;
Then find the solution near 1 V:
fzero(‘diode’,1)
Answer: 0.5742 V
Check of Results: The diode voltage is positive and in the range of 0.5 to 0.8 V, which is expected
for a diode. Substituting this value of voltage into the diode equation yields a current of 0.944 mA.
This answer appears reasonable since we knowthat the diode current cannot exceed 10V/10 k = 1.0 mA, which is the maximum current available from the circuit [i.e., if the diode were replaced
with a short circuit (VD = 0), the current in the circuit would be 1 mA]. See Sec. 3.10.3.
EXERCISE: Use the MATLAB to find the solution to
10 = 104 ID + 0.025 ln1 +
ID
IS for IS = 10?13 A
A NSWER: 942.6 A
EXAMPLE 3.7 EFFECT OF DEVICE TOLERANCES ON DIODE Q-POINTS
Let us now see how sensitive our Q-point results are to the exact value of the diode saturation
current.
PROBLEM Suppose that there is a tolerance on the value of the saturation current such that the value is given by
I nom
S = 10?15 A and 2 × 10?16 A ≤ IS ≤ 5 × 10?15 A
Find the nominal, smallest, and largest values of the diode voltage and current in the circuit in
Fig. 3.22.
SOLUTION Known Information and Given Data: The nominal and worst-case values of saturation current
are given as well as the circuit values in Fig. 3.22.
Unknowns: Nominal and worst-case values for the diode Q-point: (ID, VD)
Approach: Use MATLAB or the solver on our calculator to find the diode voltages and then the
currents for the nominal and worst-case values of IS. Note from Eq. (3.24) that the maximum value
of diode voltage corresponds to minimum current and vice versa.
Assumptions: Room temperature operation with VT = 0.025 V. The voltage and resistance in the
circuit do not have tolerances associated with them.
3.10 Diode Circuit Analysis 111
Analysis: For the nominal case, Eq. (3.28) becomes
f = 10 ? 104(10?15)[exp(40VD) ? 1] ? VD
for which the solver yields
Vnom
D = 0.689 V and I nom
D =
(10 ? 0.689) V
104  = 0.931 mA
For the minimum IS case, Eq. (3.28) is
f = 10 ? 104(2 × 10?16)[exp(40VD) ? 1] ? VD
and the solver yields
Vmax
D = 0.729 V and I min
D =
(10 ? 0.729) V
104  = 0.927 mA
Finally, for the maximum value of IS, Eq. (3.28) becomes
f = 10 ? 104(5 × 10?15)[exp(40VD) ? 1] ? VD
and the solver gives
Vmin
D = 0.649 V and I max
D =
(10 ? 0.649) V
104  = 0.935 mA
Check of Results: The diode voltages are positive and in the range of 0.5 to 0.8 V which is
expected for a diode. The diode currents are all less than the short circuit current available from
the voltage source (10 V/10 k = 1.0 mA).
Discussion: Note that even though the diode saturation current in this circuit changes by a factor
of 5:1 in either direction, the current changes by less than ±0.5%. As long as the driving voltage
in the circuit is much larger than the diode voltage, the current should be relatively insensitive to
changes in the diode voltage or the diode saturation current.
EXERCISE: Find VD and ID if the upper limit on IS is increased to 10?14 A.
A NSWERS: 0.6316 V; 0.9368 A
EXERCISE: Use the Solver function in your calculator to find the solution to
10 = 104 ID + 0.025 ln1 +
ID
IS for IS = 10?13 A and IS = 10?15 A
A NSWERS: 0.9426 mA; 0.9311 mA
3.10.3 THE IDEAL DIODE MODEL
Graphical load-line analysis provides insight into the operation of the diode circuit of Fig. 3.22, and
the mathematical model can be used to provide more accurate solutions to the load-line problem. The
next method discussed provides simplified solutions to the diode circuit of Fig. 3.22 by introducing
simplified diode circuit models of varying complexity.
The diode, as described by its i-v characteristic in Fig. 3.8 or by Eq. (3.11), is obviously a
nonlinear device. However, most, if not all, of the circuit analysis that we have learned in electrical
engineering thus far assumed that the circuits were composed of linear elements. To use this wealth
of analysis techniques, we will use piecewise linear approximations to the diode characteristic.
112 Chapter 3 Solid-State Diodes and Diode Circuits
Ideal diode
characteristic
Ideal diode symbol
iD
vD
iD
vD
A A
C C
Short
circuit
On Off
Open
circuit
C
A
Figure 3.26 (a) Ideal diode i -v characteristics and circuit symbol; (b) circuit models for on and off states of the ideal diode.
The ideal diode model is the simplest model for the diode. The i-v characteristic for the ideal
diode in Fig. 3.26 consists of two straight-line segments. If the diode is conducting a forward or
positive current (forward-biased), then the voltage across the diode is zero. If the diode is reversebiased,
with vD <0, then the current through the diode is zero. These conditions can be stated
mathematically as
vD = 0 for iD > 0 and iD = 0 for vD ≤ 0
The special symbol in Fig. 3.26 is used to represent the ideal diode in circuit diagrams.
We can now think of the diode as having two states. The diode is either conducting in the on
state, or nonconducting and off. For circuit analysis, we use the models in Fig. 3.26(b) for the two
states. If the diode is on, then it is modeled by a “short” circuit, a wire. For the off state, the diode is
modeled by an “open” circuit, no connection.
Analysis Using the Ideal Diode Model
Let us now analyze the circuit of Fig. 3.22 assuming that the diode can be modeled by the ideal
diode of Fig. 3.26(b). The diode has two possible states, and our analysis of diode circuits proceeds
as follows:
1. Select a model for the diode.
2. Identify the anode and cathode of the diode and label the diode voltage vD and current iD.
3. Make an (educated) guess concerning the region of operation of the diode based on the circuit
configuration.
4. Analyze the circuit using the diode model appropriate for the assumption in step 3.
5. Check the results to see if they are consistent with the assumptions.
For this analysis, we select the ideal diode model. The diode in the original circuit is replaced
by the ideal diode, as in Fig. 3.27(b). Next we must guess the state of the diode. Because the voltage
source appears to be trying to force a positive current through the diode, our first guess will be to
10 V
(b)
10 kΩ
Ideal
diode
10 V
(a)
10 kΩ
Figure 3.27 (a) Original diode circuit; (b) circuit modeled by an
ideal diode.
10 V ID VD
10 kΩ
Figure 3.28 Ideal diode replaced with its
model for the on state.
3.10 Diode Circuit Analysis 113
10 V
(b)
10 kΩ
Ideal
diode
10 V
(a)
10 kΩ
Figure 3.29 (a) Circuit with reverse-biased diode; (b) circuit modeled
by ideal diode.
ID = 0 VD
ID
10 V
10 kΩ
Figure 3.30 Ideal diode replaced with its
model for the off region.
assume that the diode is on. The ideal diode of Fig. 3.27(b) is replaced by its piecewise linear model
for the on region in Fig. 3.28, and the diode current is given by
ID =
(10 ? 0) V
10 k = 1.00 mA
The current ID ≥0, which is consistent with the assumption that the diode is on. Based on the ideal
diode model, we find that the diode is forward-biased and operating with a current of 1 mA. The
Q-point is therefore equal to (1 mA, 0 V).
Analysis of a Circuit Containing a Reverse-Biased Diode
A second circuit example in which the diode terminals have been reversed appears in Fig. 3.29; the
ideal diode model is again used to model the diode [Fig. 3.29(b)]. The voltage source now appears
to be trying to force a current backward through the diode. Because the diode cannot conduct in this
direction, we assume the diode is off. The ideal diode of Fig. 3.29(b) is replaced by the open circuit
model for the off region, as in Fig. 3.30.
Writing the loop equation for this case,
10 + VD + 104 ID = 0
Because ID = 0, VD = ?10 V. The calculated diode voltage is negative, which is consistent with the
starting assumption that the diode is off. The analysis shows that the diode in the circuit of Fig. 3.29
is indeed reverse-biased. The Q-point is (0,?10 V).
Although these two problems may seem rather simple, the complexity of diode circuit analysis
increases rapidly as the number of diodes increases. If the circuit has N diodes, then the number
of possible states is 2N. A circuit with 10 diodes has 1024 different possible circuits that could be
analyzed! Only through practice can we develop the intuition needed to avoid analysis of many
incorrect cases. We analyze more complex circuits shortly, but first let’s look at a slightly better
piecewise linear model for the diode.
3.10.4 CONSTANT VOLTAGE DROP MODEL
We know from our earlier discussion that there is a small, nearly constant voltage across the forwardbiased
diode. The ideal diode model ignores the presence of this voltage. However, the piecewise
linear model for the diode can be improved by adding a constant voltage Von in series with the ideal
diode, as shown in Fig. 3.31(b). This is the constant voltage drop (CVD) model. Von offsets the i-v
characteristic of the ideal diode, as indicated in Fig. 3.31(c). The piecewise linear models for the two
states become a voltage source Von for the on state and an open circuit for the off state.We now have
vD = Von for iD > 0 and iD = 0 for vD ≤ Von
We may consider the ideal diode model to be the special case of the constant voltage drop
model for which Von = 0. From the i-v characteristics presented in Fig. 3.8, we see that a reasonable
choice for Von is 0.6 to 0.7 V. We use a voltage of 0.6 V as the turn-on voltage for our diode
circuit analysis.
114 Chapter 3 Solid-State Diodes and Diode Circuits
+
+
–
–
iD iD
Von
(a) (b)
vD
C
A
Von
On Off
Open
circuit
(d) (e)
C
A
+
–
iD
Von
Ideal diode +
source
characteristic
(c)
vD
Figure 3.31 Constant voltage drop model for diode: (a) actual diode; (b) ideal diode plus voltage source Von; (c) composite
i-v characteristic; (d) CVD model for the on state; (e) model for the off state.
10 V
10 kΩ
(a)
10 V
10 kΩ
(b)
0.6 V
Ideal
diode
Constant
voltage drop
model
I
I
10 kΩ
(c)
I = ID
ID
Von = 0.6 V
10 V
Figure 3.32 Diode circuit analysis using constant voltage drop model: (a) original diode circuit; (b) circuit with diode
replaced by the constant voltage drop model; (c) circuit with ideal diode replaced by the piecewise linear model.
Diode Analysis with the Constant Voltage Drop Model
Let us analyze the diode circuit from Fig. 3.22 using the CVD model for the diode. The diode in
Fig. 3.32(a) is replaced by its CVD model in Fig. 3.32(b). The 10-V source once again appears to
be forward biasing the diode, so assume that the ideal diode is on, resulting in the simplified circuit
in Fig. 3.32(c). The diode current is given by
ID =
(10 ? Von) V
10 k =
(10 ? 0.6) V
10 k = 0.940 mA (3.30)
which is slightly smaller than that predicted by the ideal diode model but quite close to the exact
result found earlier in Ex. 3.6.
3.10.5 MODEL COMPARISON AND DISCUSSION
We have analyzed the circuit of Fig. 3.22 using four different approaches; the various results appear
in Table 3.3. All four sets of predicted voltages and currents are quite similar. Even the simple ideal
diode model only overestimates the current by less than 10 percent compared to the mathematical
model.We see that the current is quite insensitive to the actual choice of diode voltage. This is a result
of the exponential dependence of the diode current on voltage as well as the large source voltage
(10 V) in this particular circuit.
TABLE 3.3
Comparison of Diode Circuit Analysis Results
ANALYSIS TECHNIQUE DIODE CURRENT DIODE VOLTAGE
Load-line analysis 0.94 mA 0.6 V
Mathematical model 0.942 mA 0.547 V
Ideal diode model 1.00 mA 0 V
Constant voltage drop model 0.940 mA 0.600 V
3.11 Multiple-Diode Circuits 115
+10 V
R1
R2 R3
D1 D2 D3
A C
10 kΩ
10 kΩ
10 kΩ
– 20 V –10 V
I1
I2 I3
B
Figure 3.33 Example of a circuit containing three diodes.
TABLE 3.4
Possible Diode States
for Circuit in Fig. 3.33
D 1 D 2 D 3
Off Off Off
Off Off On
Off On Off
Off On On
On Off Off
On Off On
On On Off
On On On
Rewriting Eq. (3.30),
ID =
10 ? Von
10 k =
10 V
10 k 1 ?
Von
10  = (1.00 mA)1 ?
Von
10  (3.31)
we see that the value of ID is approximately 1 mA for Von  10 V. Variations in Von have only a
small effect on the result. However, the situation would be significantly different if the source voltage
were only 1 V for example (see Prob. 3.62).
3.11 MULTIPLE-DIODE CIRCUITS
The load-line technique is applicable only to single-diode circuits, and the mathematical model, or
numerical iteration technique, becomes much more complex for circuits with more than one nonlinear
element. In fact, the SPICE electronic circuit simulation program referred to throughout this book
is designed to provide numerical solutions to just such complex problems. However, we also need
to be able to perform hand analysis to predict the operation of multidiode circuits as well as to build
our understanding and intuition for diode circuit operation. In this section we discuss the use of the
simplified diode models for hand analysis of more complicated diode circuits.
As the complexity of diode circuits increases, we must rely on our intuition to eliminate unreasonable
solution choices. Even so, analysis of diode circuits may require several iterations. Intuition
can only be developed over time by working problems, and here we analyze a circuit containing
three diodes.
Figure 3.33 is an example of a circuit with several diodes. In the analysis of this circuit, we will
use the CVD model to improve the accuracy of our hand calculations.
EXAMPLE 3.8 ANALYSIS OF A CIRCUIT CONTAINING THREE DIODES
Now we will attempt to find the solution for a three-diode circuit. Our analysis will employ the
CVD model.
PROBLEM Find the Q-points for the three diodes in Fig. 3.33. Use the constant voltage drop model for the
diodes.
SOLUTION Known Information and Given Data: Circuit topology and element values in Fig. 3.33
116 Chapter 3 Solid-State Diodes and Diode Circuits
+10 V
R1
R2 R3
B
0.6 V 0.6 V 0.6 V
A C
10 kΩ
10 kΩ
10 kΩ
–20 V –20 V –10 V
I1
ID3
ID1 ID2
I2 I3
(a) (b)
D1 D2
– 30 V +
+10 V
20 V –
D3
–10 V
+ – 10 V +
+10 V
–10 V
Figure 3.34 (a) Three diode circuit model with all diodes off; (b) circuit model for circuit of Fig. 3.33 with all diodes on.
Unknowns: (ID1, VD1), (ID2, VD2), (ID3, VD3)
Approach: With three diodes, there are the eight On/Off combinations indicated in Table 3.4. A
common method that we often use to find a starting point is to consider the circuit with all the
diodes in the off state as in Fig 3.34(a). Here we see that the circuit tends to produce large forward
biases across D1, D2, and D3. So our second step will be to assume that all the diodes are on.
Assumptions: Use the constant voltage drop model with Von = 0.6 V.
Analysis: The circuit is redrawn using the CVD diode models in Fig. 3.34(b). Here we skipped
the step of physically drawing the circuit with the ideal diode symbols but instead incorporated
the piecewise linear models directly into the figure. Working from right to left, we see that the
voltages at nodes C, B, and A are given by
VC = ?0.6V VB = ?0.6 + 0.6 = 0V VA = 0 ? 0.6 = ?0.6V
With the node voltages specified, it is easy to find the current through each resistor:
I1 =
10 ? 0
10
V
k = 1mA I2 = ?0.6 ? (?20)
10
V
k = 1.94mA
I3 = ?0.6 ? (?10)
10
V
k = 0.94mA (3.32)
Using Kirchhoff’s current law, we also have
I2 = ID1 I1 = ID1 + ID2 I3 = ID2 + ID3 (3.33)
Combining Eqs. (3.32) and (3.33) yields the three diode currents:
ID1 = 1.94mA > 0 ? ID2 = ?0.94mA < 0 × ID3 = 1.86mA > 0 ? (3.34)
Check of Results: ID1 and ID3 are greater than zero and therefore consistent with the original
assumptions. However, ID2, which is less than zero, represents a contradiction. So we must try
again.
SECOND For our second attempt, let us assume D1 and D3 are on and D2 is off, as in Fig. 3.35(a). We now
ITERATION have
+10 ? 10,000I1 ? 0.6 ? 10,000I2 + 20 = 0 with I1 = ID1 = I2 (3.35)
3.11 Multiple-Diode Circuits 117
+10 V
VD2
R1
R2 R3
B
0.6 V 0.6 V
A C
10 kΩ
10 kΩ
10 kΩ
–20 V –10 V
I1 ID3
ID1
I2 I3
V2 10 V
R1
R2 R3
D1 D2
D3
10 K
10 K
10 K
V1 20 V V3 10 V
(a) (b)
Figure 3.35 (a) Circuit with diodes D1 and D3 on and D2 off; (b) Circuit for SPICE simulation.
which yields
ID1 =
29.4
20
V
k = 1.47mA > 0 ?
Also
ID3 = I3 = ?0.6 ? (?10)
10
V
k = 0.940mA > 0 ?
The voltage across diode D2 is given by
VD2 = 10 ? 10,000I1 ? (?0.6) = 10 ? 14.7 + 0.6 = ?4.10V < 0 ?
Check of Results: ID1, ID3, and VD2 are now all consistent with the circuit assumptions, so the
Q-points for the circuit are
D1: (1.47mA, 0.6V) D2: (0mA,?4.10V) D3: (0.940mA, 0.6V)
Discussion: The Q-point values that we would have obtained using the ideal diode model are (see
Prob. 3.73):
D1: (1.50mA,0V) D2: (0mA,?5.00V) D3: (1.00mA,0V)
The values of ID1 and ID3 differ by less than 6 percent. However, the reverse-bias voltage on D2
differs by 20 percent. This shows the difference that the choice of models can make. The results
from the circuit using the CVD model should be a more accurate estimate of how the circuit
will actually perform than would result from the ideal diode case. Remember, however, that these
calculations are both just approximations based on our models for the actual behavior of the real
diode circuit.
Computer-Aided Analysis: SPICE analysis yields the following Q-points for the circuit in
Fig. 3.35(b): (1.47 mA, 0.665 V), (?4.02 pA, ?4.01 V), (0.935 mA, 0.653 V). Device parameter
and Q-point information are found directly using the SHOW and SHOWMOD commands in
SPICE. Or, voltmeters and ammeters (zero-valued current and voltage sources) can be inserted in
the circuit in some implementations of SPICE. Note that the ?4 pA current in D2 is much larger
than the reverse saturation current of the diode (IS defaults to 10 fA), and results from a more
complete SPICE model in the author’s version of SPICE.
118 Chapter 3 Solid-State Diodes and Diode Circuits
EXERCISE: Find the Q-points for the three diodes in Fig. 3.33 if R1 is changed to 2.5 k.
A NSWERS: (2.13 mA, 0.6 V); (1.13 mA, 0.6 V); (0 mA, ?1.27 V)
EXERCISE: Use SPICE to calculate the Q-points of the diodes in the previous exercise. Use
IS = 1 fA.
A NSWERS: (2.12 mA, 0.734 V); (1.12 mA, 0.718 V); (0 mA, ?1.19 V)
3.12 ANALYSIS OF DIODES OPERATING IN THE BREAKDOWN REGION
Reverse breakdown is actually a highly useful region of operation for the diode. The reverse breakdown
voltage is nearly independent of current and can be used as either a voltage regulator or voltage
reference. Thus, it is important to understand the analysis of diodes operating in reverse breakdown.
Figure 3.36 is a single-loop circuit containing a 20-V source supplying current to a Zener diode
with a reverse breakdown voltage of 5 V. The voltage source has a polarity that will tend to reversebias
the diode. Because the source voltage exceeds the Zener voltage rating of the diode, VZ = 5 V,
we should expect the diode to be operating in its breakdown region.
3.12.1 LOAD-LINE ANALYSIS
The i-v characteristic for this Zener diode is given in Fig. 3.37, and load-line analysis can be used to
find the Q-point for the diode, independent of the region of operation. The normal polarities for ID
and VD are indicated in Fig. 3.36, and the loop equation is
?20 = VD + 5000ID (3.36)
In order to draw the load line, we choose two points on the graph:
VD = 0, ID = ?4 mA and VD = ?5 V, ID = ?3 mA
In this case the load line intersects the diode characteristic at a Q-point in the breakdown region:
(?2.9 mA,?5.2 V).
3.12.2 ANALYSIS WITH THE PIECEWISE LINEAR MODEL
The assumption of reverse breakdown requires that the diode current ID be less than zero or that the
Zener current IZ = ?ID > 0. We will analyze the circuit with the piecewise linear model and test
this condition to see if it is consistent with the reverse-breakdown assumption.
ID
ID
5 kΩ IZ
VD
VZ = 5 V
RZ = 100 Ω
20 V
+
–
Figure 3.36 Circuit containing a Zener diode with VZ = 5 V
and RZ = 100 .
–6 –5 –4 –3 –2 –1
–0.005
1
0.005
2 3 4 5 6
VD (V)
ID (A)
Q-point
Figure 3.37 Load line for Zener diode.
3.12 Analysis of Diodes Operating in the Breakdown Region 119
I Z
ID
20 V
0.1 kΩ
5 V
5 kΩ
Figure 3.38 Circuit with piecewise linear model for
Zener diode. Note that the diode model is valid only in
the breakdown region of the characteristic.
VI
IS
IZ
IL
20 V
VZ = 5 V
RZ = 0
5 kΩ
5 kΩ RL
R
Figure 3.39 Zener diode voltage regulator circuit.
In Fig. 3.38, the Zener diode has been replaced with its piecewise linear model from Fig. 3.16
in Sec. 3.6, with VZ = 5 V and RZ = 100 . Writing the loop equation this time in terms of IZ :
20 ? 5100IZ ? 5 =0 or IZ =
(20 ? 5) V
5100  = 2.94 mA (3.37)
Because IZ is greater than zero (ID < 0), the solution is consistent with our assumption of Zener
breakdown operation.
It is worth noting that diodes have three possible states when the breakdown region is included,
(on, off, and reverse breakdown), further increasing analysis complexity.
3.12.3 VOLTAGE REGULATION
A useful application of the Zener diode is as a voltage regulator, as shown in the circuit of Fig. 3.39.
The function of the Zener diode is to maintain a constant voltage across load resistor RL . As long as
the diode is operating in reverse breakdown, a voltage of approximately VZ will appear across RL .
To ensure that the diode is operating in the Zener breakdown region, we must have IZ > 0.
The circuit of Fig. 3.39 has been redrawn in Fig. 3.40 with the model for the Zener diode, with
RZ = 0. Using nodal analysis, the Zener current is expressed by IZ = II ? IL . The currents II and
IL are equal to
II =
VI ? VZ
R =
(20 ? 5) V
5 k = 3 mA and IL =
VZ
RL =
5 V
5 k = 1 mA (3.38)
resulting in a Zener current IZ = 2 mA. IZ > 0, which is again consistent with our assumptions. If
the calculated value of IZ were less than zero, then the Zener diode no longer controls the voltage
across RL , and the voltage regulator is said to have “dropped out of regulation.”
For proper regulation to take place, the Zener current must be positive:
IZ = II ? IL =
VI ? VZ
R ?
VZ
RL ≥ 0 (3.39)
II
IZ
IL
R
VI 20 V VZ 5 V
5 kΩ
5 kΩ RL > Rmin
Figure 3.40 Circuit with a constant voltage model for the Zener diode.
120 Chapter 3 Solid-State Diodes and Diode Circuits
Solving for RL yields a lower bound on the value of load resistance for which the Zener diode will
continue to act as a voltage regulator.
RL >
R
VS
VZ ? 1 = Rmin (3.40)
EXERCISE: What is the value of Rmin for the Zener voltage regulator circuit in Figs. 3.39 and
3.40? What is the output voltage for RL = 1 k? For RL = 2 k?
A NSWERS: 1.67 k; 3.33 V; 5.00 V
3.12.4 ANALYSIS INCLUDING ZENER RESISTANCE
The voltage regulator circuit in Fig. 3.39 has been redrawn in Fig. 3.41 and now includes a nonzero
Zener resistance RZ . The output voltage is now a function of the current IZ through the Zener diode.
For small values of RZ , however, the change in output voltage will be small.
IZ
20 V
5 V
0.1 kΩ
5 kΩ
VI 5 kΩ
RZ
RL VL
VZ
R
IL
Figure 3.41 Zener diode regulator circuit, including Zener resistance.
EXAMPLE 3.9 DC ANALYSIS OF A ZENER DIODE REGULATOR CIRCUIT
Find the operating point for a Zener-diode-based voltage regulator circuit.
PROBLEM Find the output voltage and Zener diode current for the Zener diode regulator in Figs. 3.39 and
3.41 if RZ = 100  and VZ = 5 V.
SOLUTION Known Information and Given Data: Zener diode regulator circuit as modeled in Fig. 3.41 with
VI = 20 V, R = 5 k, RZ = 0.1 k, and VZ = 5 V
Unknowns: VL , IZ
Approach: The circuit contains a single unknown node voltage VL , and a nodal equation can be
written to find the voltage. Once VL is found, IZ can be determined using Ohm’s law.
Assumptions: Use the piecewise linear model for the diode as drawn in Fig. 3.41.
Analysis: Writing the nodal equation for VL yields
VL ? 20 V
5000  +
VL ? 5 V
100  +
VL
5000  = 0
3.12 Analysis of Diodes Operating in the Breakdown Region 121
Multiplying the equation by 5000  and collecting terms gives
52VL = 270 V and VL = 5.19 V
The Zener diode current is equal to
IZ =
VL ? 5 V
100  =
5.19 V ? 5 V
100  = 1.90 mA > 0
Check of Results: IZ > 0 confirms operation in reverse breakdown.We see that the output voltage
of the regulator is slightly higher than for the case with RZ = 0, and the Zener diode current is
reduced slightly. Both changes are consistent with the addition of RZ to the circuit.
Computer-Aided Analysis: We can use SPICE to simulate the Zener circuit if we specify the
breakdown voltage using SPICE parameters BV, IBV, and RS. BV sets the breakdown voltage,
and IBV represents the current at breakdown. Setting BV=5 V and RS = 100  and letting
IBV default to 1 mA yields VL =5.21 V and IZ = 1.92 mA, which agree well with our hand
calculations. A transfer function analysis from VS to VL gives yields a sensitivity of 21 mV/V and
an output resistance of 108 . The meaning of these numbers is discussed in the next section.
EXERCISE: Find VL, I Z, and the Zener power dissipation in Fig. 3.41 if R = 1 k.
A NSWERS: 6.25 V; 12.5 mA; 78.1 mW
3.12.5 LINE AND LOAD REGULATION
Two important parameters characterizing a voltage regulator circuit are line regulation and load
regulation. Line regulation characterizes how sensitive the output voltage is to input voltage changes
and is expressed as V/V or as a percentage. Load regulation characterizes how sensitive the output
voltage is to changes in the load current withdrawn from the regulator and has the units of Ohms.
Line regulation =
dVL
dVI
and Load regulation =
dVL
d IL
(3.41)
We can find expressions for these quantities from a straightforward analysis of the circuit in Fig. 3.41
similar to that in Ex. 3.9:
VL ? VI
R +
VL ? VZ
RZ + IL = 0 (3.42)
For a fixed load current, we find the line regulation as
Line regulation =
RZ
R + RZ
(3.43)
and for changes in IL as
Load regulation = ?(RZR) (3.44)
The load regulation should be recognized as the Th′evenin equivalent resistance looking back into
the regulator from the load terminals.
EXERCISE: What are the values of the load and line regulation for the circuit in Fig. 3.41?
A NSWERS: 19.6 mV/V; 98.0 . Note that these are close to the SPICE results in Ex. 3.9.
122 Chapter 3 Solid-State Diodes and Diode Circuits
3.13 HALF-WAVE RECTIFIER CIRCUITS
Rectifiers represent an application of diodes that we encounter frequently every day, but they may
not be recognized as such. The basic rectifier circuit converts an ac voltage to a pulsating dc voltage.
An LC or RC filter is then added to eliminate the ac components of the waveform and produce a
nearly constant dc voltage output. Virtually every electronic device that is plugged into the wall
utilizes a rectifier circuit to convert the 120-V, 60-Hz ac power line source to the various dc voltages
required to operate electronic devices such as personal computers, audio systems, radio receivers,
televisions, and the like. All of our battery chargers and “wall-warts” contain rectifiers. As a matter
of fact, the vast majority of electronic circuits are powered by a dc source, usually based on some
form of rectifier.
This section explores half-wave rectifier circuits with capacitor filters that form the basis for
many dc power supplies. Up to this point, we have looked at only steady-state dc circuits in which
the diode remained in one of its three possible states (on, off, or reverse breakdown). Now, however,
the diode state will be changing with time, and a given piecewise linear model for the circuit will be
valid for only a certain time interval.
3.13.1 HALF-WAVE RECTIFIER WITH RESISTOR LOAD
A single diode is used to form the half-wave rectifier circuit in Fig. 3.42. A sinusoidal voltage
source vI = VP sin ωt is connected to the series combination of diode D1 and load resistor R. During
the first half of the cycle, for which vI > 0, the source forces a current through diode D1 in the
forward direction, and D1 will be on. During the second half of the cycle, vI < 0. Because a negative
current cannot exist in the diode (unless it is in breakdown), it turns off. These two states are modeled
in Fig. 3.43 using the ideal diode model.
When the diode is on, voltage source vS is connected directly to the output and vO = vI . When
the diode is off, the current in the resistor is zero, and the output voltage is zero. The input and
output voltage waveforms are shown in Fig. 3.44(b), and the resulting current is called pulsating
D1
iD
vI = VP sin ωω t R vO
Figure 3.42 Half-wave rectifier
circuit.
iD
vI > 0 R vO
D1 on
vI < 0 R vO
D1 off
Figure 3.43 Ideal diode models for the two half-wave rectifier states.
Time (s)
0.00 0.01 0.02
Time (s)
Output voltage (V)
0.00 0.01 0.02
Input voltage (V)
15.00
10.00
5.00
0.00
–5.00
–10.00
–15.00
Diode
off
Diode
off
Diode
on
Diode
on
15.00
10.00
5.00
0.00
–5.00
–10.00
–15.00
Figure 3.44 Sinusoidal input voltage vS and pulsating dc output voltage vO for the half-wave rectifier circuit.
3.13 Half-Wave Rectifier Circuits 123
Von
vI R vO
D1 on
Figure 3.45 CVD model for the
rectifier on state.
Time (s)
Voltage (V)
15
10
5
0
–5
–10
–15
0.000 0.005 0.010 0.015 0.020
Output
voltage
Output
voltage
Input
voltage
Input
voltage
Figure 3.46 Half-wave rectifier output voltage
with VP = 10 V and Von = 0.7 V.
v1 vI = VP sin ωt R vO
+
–
+
–
ω
Figure 3.47 Transformer-driven half-wave rectifier.
v1 vI = VP sin ωt vO
+
–
+
–
ω C
Figure 3.48 Rectifier with capacitor load (peak detector).
direct current. In this circuit, the diode is conducting 50 percent of the time and is off 50 percent of
the time.
In some cases, the forward voltage drop across the diode can be important. Figure 3.45 shows
the circuit model for the on-state using the CVD model. For this case, the output voltage is one
diode-drop smaller than the input voltage during the conduction interval:
vO = (VP sin ωt) ? Von for VP sin ωt ≥ Von (3.45)
The output voltage remains zero during the off-state interval. The input and output waveforms for the
half-wave rectifier, including the effect of Von, are shown in Fig. 3.46 for VP = 10 V and Von = 0.7V.
In many applications, a transformer is used to convert from the 120-V ac, 60-Hz voltage available
from the power line to the desired ac voltage level, as in Fig. 3.47. The transformer can step the
voltage up or down depending on the application; it also enhances safety by providing isolation
from the power line. From circuit theory we know that the output of an ideal transformer can be
represented by an ideal voltage source, and we use this knowledge to simplify the representation of
subsequent rectifier circuit diagrams.
The unfiltered output of the half-wave rectifier in Fig. 3.42 or 3.47 is not suitable for operation
of most electronic circuits because constant power supply voltages are required to establish proper
bias for the electronic devices. A filter capacitor (or more complex circuit) can be added to filter
the output of the circuit in Fig. 3.47 to remove the time-varying components from the waveform.
3.13.2 RECTIFIER FILTER CAPACITOR
To understand operation of the rectifier filter, we first consider operation of the peak-detector circuit
in Fig. 3.48. This circuit is similar to that in Fig. 3.47 except that the resistor is replaced with a
capacitor C that is initially discharged [vO(0) = 0].
Models for the circuit with the diode in the on and off states are in Fig. 3.49, and the input and
output voltage waveforms associated with this circuit are in Fig. 3.50. As the input voltage starts to
124 Chapter 3 Solid-State Diodes and Diode Circuits
vI
iD
Von vO
+
–
C
Diode on
0 ≤ t ≤ T/2
(a)
vI VP – Von
+
–
C
Diode off
t ≥ T/2
(b)
Figure 3.49 Peak-detector circuit models (constant voltage drop model). (a) The diode
is on for 0 ≤ t ≤ T/2. (b) The diode is off for t > T/2.
Voltage (V)
Time
0 T 2T
0
VP – Von
vO
vI
Figure 3.50 Input and output waveforms for the
peak-detector circuit.
rise, the diode turns on and connects the capacitor to the source. The capacitor voltage equals the
input voltage minus the voltage drop across the diode.
At the peak of the input voltagewaveform, the current through the diode tries to reverse direction
because iD = C[d(vI ? Von)/dt] < 0, the diode cuts off, and the capacitor is disconnected from the
rest of the circuit. There is no circuit path to discharge the capacitor, so the voltage on the capacitor
remains constant. Because the amplitude of the input voltage source vI can never exceed VP , the
capacitor remains disconnected from vI for t > T/2. Thus, the capacitor in the circuit in Fig. 3.48
charges up to a voltage one diode-drop below the peak of the input waveform and then remains
constant, thereby producing a dc output voltage
Vdc = VP ? Von (3.46)
3.13.3 HALF-WAVE RECTIFIER WITH RC LOAD
To make use of this output voltage, a load must be connected to the circuit as represented by the
resistor R in Fig. 3.51. Now there is a path available to discharge the capacitor during the time the
diode is not conducting. Models for the conducting and nonconducting time intervals are shown in
Fig. 3.52; the waveforms for the circuit are shown in Fig. 3.53. The capacitor is again assumed to be
vS vI = VP sin ωt vO
+
–
ω C R
+
–
(a) (b)
Figure 3.51 (a) Half-wave rectifier circuit with filter capacitor; (b) a-175,000-F, 15-V filter capacitor. Capacitance tolerance
is ?10 percent, +75 percent.
3.13 Half-Wave Rectifier Circuits 125
iD
Von
vI R vO vI C C R vO
(a) Diode on (b) Diode off
Figure 3.52 Half-wave rectifier circuit models.
Time, t
Voltage (V)
0
0
T
T 2T
Vr
VP – Von
vO
vI
t'
ΔT
Figure 3.53 Input and output voltage waveforms for
the half-wave rectifier circuit (RC 
 T ).
initially discharged and the time constant RC is assumed to be 
 T . During the first quarter cycle,
the diode conducts, and the capacitor is rapidly charged toward the peak value of the input voltage
source. The diode cuts off at the peak of vI , and the capacitor voltage then discharges exponentially
through the resistor R, as governed by the circuit in Fig. 3.52(b). The discharge continues until the
voltage vI ? von exceeds the output voltage vO, which occurs near the peak of the next cycle. The
process is then repeated once every cycle.
3.13.4 RIPPLE VOLTAGE AND CONDUCTION INTERVAL
The output voltage is no longer constant as in the ideal peak-detector circuit but has a ripple voltage
Vr . In addition, the diode only conducts for a short timeT during each cycle. This timeT is called
the conduction interval, and its angular equivalent is the conduction angle θ c where θc = ωT .
The variables T , θc, and Vr are important values related to dc power supply design, and we will
now develop expressions for these parameters.
During the discharge period, the voltage across the capacitor is described by
vo(t) = (VP ? Von) exp?
t
RC  for t = t ?
T
4  ≥ 0 (3.47)
We have referenced the t time axis to t = T/4 to simplify the equation. The ripple voltage Vr is
given by
Vr = (VP ? Von) ? vo(t) = (VP ? Von) 1 ? exp?
T ? T
RC  (3.48)
A small value of Vr is desired in most power supply designs; a small value requires RC to be much
greater than T ? T . Using exp(?x)～=
1 ? x for small x results in an approximate expression for
the ripple voltage:
Vr～=
(VP ? Von)
T
RC 1 ?
T
T  (3.49)
A small ripple voltage also meansT  T , and the final simplified expression for the ripple voltage
becomes
Vr～=
(VP ? Von)
R
T
C = Idc
T
C
(3.50)
126 Chapter 3 Solid-State Diodes and Diode Circuits
where
Idc =
VP ? Von
R
(3.51)
The approximation of the exponential used in Eqs. (3.49) and (3.50) is equivalent to assuming that the
capacitor is being discharged by a constant current so that the discharge waveform is a straight line.
The ripple voltage VR can be considered to be determined by an equivalent dc current discharging
the capacitor C for a time period T (that is, V = (Idc/C) T ).
Approximate expressions can also be obtained for conduction angle θC and conduction interval
T . At time t =
5
4
T ? T , the input voltage just exceeds the output voltage, and the diode is
conducting. Therefore, θ = ωt = 5π/2 ? θC and
Vp sin5
2
π ? θC? Von = (VP ? Von) ? Vr (3.52)
Remembering that sin(5π/2 ? θC) = cos θC, we can simplify the above expression to
cos θC = 1 ?
Vr
VP
(3.53)
For small values of θC, cos θC～=
1?θ2C
/2. Solving for the conduction angle and conduction interval
gives
θC = 2Vr
VP
and T =
θC
ω =
1
ω2Vr
VP
(3.54)
EXAMPLE 3.10 HALF-WAVE RECTIFIER ANALYSIS
Here we see an illustration of numerical results for a half-wave rectifier with a capacitive filter.
PROBLEM Find the value of the dc output voltage, dc output current, ripple voltage, conduction interval, and
conduction angle for a half-wave rectifier driven from a transformer having a secondary voltage of
12.6 Vrms (60 Hz) with R = 15  and C = 25,000 F. Assume the diode on-voltage Von = 1 V.
SOLUTION Known Information and Given Data: Half-wave rectifier circuit with RC load as depicted in
Fig. 3.51; transformer secondary voltage is 12.6 Vrms, operating frequency is 60 Hz, R = 15 ,
and C = 25,000 F.
Unknowns: dc output voltage Vdc, output current Idc, ripple voltage Vr , conduction interval T ,
conduction angle θC
Approach: Given data can be used directly to evaluate Eqs. (3.46), (3.50), (3.51), and (3.54).
Assumptions: Diode on-voltage is 1 V. Remember that the derived results assume the ripple
voltage is much less than the dc output voltage (Vr  Vdc) and the conduction interval is much
less than the period of the ac signal (T  T ).
Analysis: The ideal dc output voltage in the absence of ripple is given by Eq. (3.46):
Vdc = VP ? Von = 
12.6√2 ? 1 V = 16.8 V
3.13 Half-Wave Rectifier Circuits 127
The nominal dc current delivered by the supply is
Idc =
VP ? Von
R =
16.8 V
15  = 1.12 A
The ripple voltage is calculated using Eq. (3.50) with the discharge interval T = 1/60 s:
Vr～=
IdC
T
C = 1.12 A
1
60
s
2.5 × 10?2 F = 0.747 V
The conduction angle is calculated using Eq. (3.54)
θc = ωT = 2Vr
VP = 2 · 0.75
17.8 = 0.290 rad or 16.6?
and the conduction interval is
T =
θc
ω =
θc
2π f =
0.29
120π = 0.769 ms
Check of Results: The ripple voltage represents 4.4 percent of the dc output voltage. Thus the assumption
that the voltage is approximately constant is justified. The conduction time
is 0.769 ms out of a total period T = 16.7 ms, and the assumption that T  T is also satisfied.
Discussion: From this example, we see that even a 1-A power supply requires a significant filter
capacitance C to maintain a low ripple percentage. In this case, C = 0.025F = 25,000 F.
EXERCISE: Find the value of the dc output voltage, dc output current, ripple voltage, conduction
interval, and conduction angle for a half-wave rectifier that is being supplied from a transformer
having a secondary voltage of 6.3 Vrms (60 Hz) with R = 0.5  and C = 500,000 F. Assume
the diode on voltage Von = 1 V.
A NSWERS: 7.91 V; 15.8 A; 0.527 V; 0.912 ms; 19.7?
EXERCISE: What are the values of the dc output voltage and dc output current for a halfwave
rectifier that is being supplied from a transformer having a secondary voltage of 10 Vrms
(60 Hz) and a 2- load resistor? Assume the diode on voltage Von = 1 V. What value of filter
capacitance is required to have a ripple voltage of no more than 0.1 V? What is the conduction
angle?
A NSWERS: 13.1 V; 6.57 A; 1.10 F; 6.82?
3.13.5 DIODE CURRENT
In rectifier circuits, a nonzero current is present in the diode for only a very small fraction of the
period T, yet an almost constant dc current is flowing out of the filter capacitor to the load. The total
charge lost from the capacitor during each cycle must be replenished by the current through the diode
during the short conduction interval T , which leads to high peak diode currents. Figure 3.54 shows
128 Chapter 3 Solid-State Diodes and Diode Circuits
0 s 10 ms 20 ms 30 ms 40 ms
200 A
100 A
0 A
Diode current
Voltage
25 V
0 V
–25 V
vI
Output voltage
Repetitive diode current
Time
Initial surge current
(a) (b)
0 s 10 ms 20 ms 30 ms 40 ms
Time
Figure 3.54 SPICE simulation of the half-wave rectifier circuit: (a) voltage waveforms; (b) diode current.
IP
Idc
ΔT ΔT
T 2T
t
Figure 3.55 Triangular approximation to diode current pulse.
Time
0 T 2T
0
Vdc
vI
Voltage (V)
VP –Von
– VP
~~ 2VP
Figure 3.56 Peak reverse voltage across the diode in
a half-wave rectifier.
the results of SPICE simulation of the diode current. The repetitive current pulse can be modeled
approximately by a triangle of height IP and width T , as in Fig. 3.55.
Equating the charge supplied through the diode during the conduction interval to the charge lost
from the filter capacitor during the complete period yields
Q = IP
T
2 = IdcT or IP = Idc
2T
T
(3.55)
Here we remember that the integral of current over time represents charge Q. Therefore the charge
supplied by the triangular current pulse in Fig. 3.55 is given by the area of the triangle, IPT/2.
For Ex. 3.10, the peak diode current would be
IP = 1.12
2 · 16.7
0.769 = 48.6 A (3.56)
which agrees well with the simulation results in Fig. 3.55. The diode must be built to handle these
high peak currents, which occur over and over. This high peak current is also the reason for the
relatively large choice of Von used in Ex. 3.10. (See Prob. 3.82.)
3.13 Half-Wave Rectifier Circuits 129
EXERCISE: (a) What is the forward voltage of a diode operating at a current of 48.6 A at 300 K
if IS = 10?15 A? (b) At 50 C?
A NSWERS: 0.994 V; 1.07 V
3.13.6 SURGE CURRENT
When the power supply is first turned on, the capacitor is completely discharged, and there will be
an even larger current through the diode, as is visible in Fig. 3.54. During the first quarter cycle, the
current through the diode is given approximately by
id (t) = ic(t)～=
C  d
dt
VP sin ωt = ωCVP cos ωt (3.57)
The peak value of this initial surge current occurs at t =0+ and is given by
ISC = ωCVP = 2π(60 Hz)(0.025 F)(17.8 V) = 168 A
Using the numbers from Ex. 3.10 yields an initial surge current of almost 170 A! This value, again,
agrees well with the simulation results in Fig. 3.54. If the input signal vI does not happen to be
crossing through zero when the power supply is turned on, the situation can be even worse, and
rectifier diodes selected for power supply applications must be capable of withstanding very large
surge currents as well as the large repetitive current pulses required each cycle.
In most practical circuits, the surge current will be large but cannot actually reach the values
predicted by Eq. (3.57) because of series resistances in the circuit that we have neglected. The rectifier
diode itself will have an internal series resistance (review the SPICE model in Sec. 3.9 for example),
and the transformer will have resistances associated with both the primary and secondary windings.
A total series resistance in the secondary of only a few tenths of an ohm will significantly reduce
both the surge current and peak repetitive current in the circuit. In addition, the large time constant
associated with the series resistance and filter capacitance causes the rectifier output to take many
cycles to reach its steady-state voltage. (See SPICE simulation problems at the end of this chapter.)
3.13.7 PEAK-INVERSE-VOLTAGE (PIV) RATING
We must also be concerned about the breakdown voltage rating of the diodes used in rectifier circuits.
This breakdown voltage is called the peak-inverse-voltage (PIV) rating of the rectifier diode. The
worst-case situation for the half-wave rectifier is depicted in Fig. 3.56 in which it is assumed that the
ripple voltage Vr is very small. When the diode is off, as in Fig. 3.52(b), the largest reverse bias across
the diode is equal to Vdc ? vI which occurs when vI reaches its negative peak of ?VP . The diode
must therefore be able to withstand a reverse bias of at least
PIV ≥ Vdc ? vmin
I = VP ? Von ? (?VP ) = 2VP ? Von～=
2VP (3.58)
From Eq. (3.58), we see that diodes used in the half-wave rectifier circuit must have a PIV rating
equal to twice the peak voltage supplied by the source vI . The PIV value corresponds to the minimum
value of Zener breakdown voltage for the rectifier diode. A safety margin of at least 25 to 50 percent
is usually specified for the diode PIV rating in power supply designs.
3.13.8 DIODE POWER DISSIPATION
In high-current power supply applications, the power dissipation in the rectifier diodes can become
significant. The average power dissipation in the diode is defined by
PD =
1
T  T
0
vD(t)iD(t) dt (3.59)
130 Chapter 3 Solid-State Diodes and Diode Circuits
vI = VP v1 sin t vO
+
–
ω C R
+
–
(a)
v1 vI = VP sin t vO
+
–
ω C R
+
–
(b)
Figure 3.57 Half-wave rectifier circuits that develop negative output voltages.
This expression can be simplified by assuming that the voltage across the diode is approximately
constant at vD(t) = Von and by using the triangular approximation to the diode current iD(t) shown
in Fig. 3.55. Eq. (3.59) becomes
PD =
1
T  T
0
VoniD(t) dt =
Von
T  T
T?T
iD(t) dt = Von
IP
2
T
T = Von Idc (3.60)
Using Eq. (3.55) we see that the power dissipation is equivalent to the constant dc output current
multiplied by the on-voltage of the diode. For the half-wave rectifier example, PD = (1 V)(1.1 A) = 1.1W. This rectifier diodewould probably need a heat sink to maintain its temperature at a reasonable
level. Note that the average current through the diode is Idc.
Another source of power dissipation is caused by resistive loss within the diode. Diodes have
a small internal series resistance RS, and the average power dissipation in this resistance can be
calculated using
PD =
1
T  T
0
i2D
(t)RS dt (3.61)
Evaluation of this integral (left for Prob. 3.87) for the triangular current wave form in Fig. 3.55 yields
PD =
1
3
I2P
RS
T
T =
4
3
T
T
I 2
dcRS (3.62)
Using the number from the rectifier example with RS = 0.20  yields PD = 7.3 W! This
is significantly greater than the component of power dissipation caused by the diode on-voltage
calculated using Eq. (3.60). The component of power dissipation described by Eq. (3.62) can be
reduced by minimizing the peak current IP through the use of the minimum required size of filter
capacitor or by using the full-wave rectifier circuits, which are discussed in Sec. 3.14.
3.13.9 HALF-WAVE RECTIFIER WITH NEGATIVE OUTPUT VOLTAGE
The circuit of Fig. 3.51 can also be used to produce a negative output voltage if the top rather than
the bottom of the capacitor is grounded, as depicted in Fig. 3.57(a) or by reversing the direction of
the diode in the original circuit as in Fig. 3.57(b). These two circuits are equivalent. In the circuit in
Fig. 3.57(b), the diode conducts on the negative half cycle of the transformer voltage vI , and the dc
output voltage is Vdc = ?(VP ? Von).
3.13 Half-Wave Rectifier Circuits 131
E L E C T R O N I C S I N A C T I O N
AM Demodulation
Thewaveform for a 100 percent amplitude modulated (AM) signal is shown in the figure below
and described mathematically by vAM = 2 sin ωCt (1 + sin ωMt) V in which ωC is the carrier
frequency ( fC = 50 kHz) and ωM is the modulating frequency ( fM = 5 kHz). The envelope of
4.0 V
2.0 V
0 V
–4.0 V
–2.0 V
0 s 0.1 ms 0.2 ms 0.3 ms 0.4 ms 0.5 ms 0.6 ms 0.7 ms 0.8 ms 0.9 ms 1.0 ms
Time
the AM signal contains the information being transmitted, and the envelope can be recovered
from the signal using a simple half-wave rectifier. In the SPICE circuit below, the signal to be
demodulated is applied as the input signal to the rectifier, and the rectifier, and the R2C1 time
VC
R1 R3
D1
R2
1 K
D1N4148
0
10 K
5 K .02 μF .002 μF
VM
vAM
C1 C2
constant is set to filter out the carrier frequency but follow the signal’s envelope. Additional
filtering is provided by the low-pass filter formed by R3 and C2. SPICE simulation results
appear below along with the results of a Fourier analysis of the demodulated signal. The plots
of vC1 and vC2 represent the voltages across capacitors C1 and C2 respectively.
1.6 V
1.2 V
0.4 V
0.8 V
1.0 ms 1.1 ms 1.2 ms 1.3 ms 1.4 ms 1.5 ms 1.6 ms 1.7 ms 1.8 ms 1.9 ms 2.0 ms
Time
vC2
vC1
SPICE
Results for
Spectral
Content of
vC2 (V)
5 kHz 0.330
10 kHz 0.046
15 kHz 0.006
20 kHz 0.001
–
45 kHz 0.006
50 kHz 0.007
55 kHz 0.004
132 Chapter 3 Solid-State Diodes and Diode Circuits
3.14 FULL-WAVE RECTIFIER CIRCUITS
Full-wave rectifier circuits cut the capacitor discharge time in half and offer the advantage of requiring
only one-half the filter capacitance to achieve a given ripple voltage. The full-wave rectifier circuit
in Fig. 3.58 uses a center-tapped transformer to generate two voltages that have equal amplitudes
but are 180 degrees out of phase.With voltage vI applied to the anode of D1, and ?vI applied to the
anode of D2, the two diodes form a pair of half-wave rectifiers operating on alternate half cycles of
the input waveform. Proper phasing is indicated by the dots on the two halves of the transformer.
For vI > 0, D1 will be functioning as a half-wave rectifier, and D2 will be off, as indicated in
Fig. 3.59. The current exits the upper terminal of the transformer, goes through diode D1, through
the RC load, and returns back into the center tap of the transformer.
For vI < 0, D1 will be off, and D2 will be functioning as a half-wave rectifier as indicated
in Fig. 3.60. During this portion of the cycle, the current path leaves the bottom terminal of the
transformer, goes through D2, down through the RC load, and again returns into the transformer
center tap. The current direction in the load is the same during both halves of the cycle; one-half of
the transformer is utilized during each half cycle.
The load, consisting of the filter capacitor C and load resistor R, now receives two current pulses
per cycle, and the capacitor discharge time is reduced to less than T/2, as indicated in the graph
in Fig. 3.61. An analysis similar to that for the half-wave rectifier yields the same formulas for dc
output voltage, ripple voltage, and T , except that the discharge interval is T/2 rather than T. For
a given capacitor value, the ripple voltage is one-half as large, and the conduction interval and peak
vI
D1
D2
vI
+
+ –
–
C R
vS = VP sin ω t
Figure 3.58 Full-wave rectifier circuit using two diodes
and a center-tapped transformer. This circuit produces a
positive output voltage.
vI
D1
D2 off
vI
+
+ –
–
C R
vS = VP sin ω t
Figure 3.59 Equivalent circuit for vI > 0.
D2
D1 off
vI C R
vS = VP sin ω t
vI
Figure 3.60 Equivalent circuit for vI < 0.
Time
Voltage
0
0 T 2T
vI
–vI
ΔT
T
2
VP – Von
Figure 3.61 Voltage waveforms for the full-wave
rectifier.
3.15 Full-Wave Bridge Rectification 133
vI
vI
+
+ –
vO
+
–
–
D1
D2
C R
Figure 3.62 Full-wave rectifier with negative output
voltage.
current are reduced. The peak-inverse-voltage waveform for each diode is similar to the one shown
in Fig. 3.56 for the half-wave rectifier, with the result that the PIV rating of each diode is the same
as in the half-wave rectifier.
3.14.1 FULL-WAVE RECTIFIER WITH NEGATIVE OUTPUT VOLTAGE
By reversing the polarity of the diodes, as in Fig. 3.62, a full-wave rectifier circuit with a negative
output voltage is realized. Other aspects of the circuit remain the same as the previous full-wave
rectifiers with positive output voltages.
3.15 FULL-WAVE BRIDGE RECTIFICATION
The requirement for a center-tapped transformer in the full-wave rectifier can be eliminated through
the use of two additional diodes in the full-wave bridge rectifier circuit configuration shown in
Fig. 3.63. For vI > 0, D2 and D4 will be on and D1 and D3 will be off, as indicated in Fig. 3.64.
Current exits the top of the transformer, goes through D2 into the RC load, and returns to the
transformer through D4. The full transformer voltage, now minus two diode voltage drops, appears
across the load capacitor yielding a dc output voltage
Vdc = VP ? 2Von (3.63)
The peak voltage at node 1, which represents the maximum reverse voltage appearing across D1, is
equal to (VP ? Von). Similarly, the peak reverse voltage across diode D3 is (VP ?2Von)?(?Von) =
(VP ? Von).
C R
D2
D4
D1
D3
vI = VP sin t
vI
+
–
ω
Figure 3.63 Full-wave bridge rectifier circuit with positive output voltage.
R
C
VP – 2Von
D2
D4
D1 off
D3 off
vI > 0
vI
+
–
3
1
Figure 3.64 Full-wave bridge rectifier circuit for vI > 0.
For vI < 0, D1 and D3 will be on and D2 and D4 will be off, as depicted in Fig. 3.65. Current
leaves the bottom of the transformer, goes through D3 into the RC load, and back through D1 to the
transformer. The full transformer voltage is again being utilized. The peak voltage at node 3 is now
equal to (VP ? Von) and is the maximum reverse voltage appearing across D4. Similarly, the peak
reverse voltage across diode D2 is (VP ? 2Von) ? (?Von) = (VP ? Von).
134 Chapter 3 Solid-State Diodes and Diode Circuits
R
C
VP – 2Von
D2 off
D4 off
D1
D3
vI < 0
vI
+
–
3
1
Figure 3.65 Full-wave bridge rectifier circuit for vI < 0.
R
C
D2
D4
D1
D3
vI = VP sin ωt
vI
+
–
ω
vO < 0
Figure 3.66 Full-wave bridge rectifier circuit with vO < 0.
From the analysis of the two half cycles, we see that each diode must have a PIV rating given by
PIV = VP ? Von～=
VP (3.64)
As with the previous rectifier circuits, a negative output voltage can be generated by reversing the
direction of the diodes, as in the circuit in Fig. 3.66.
3.16 RECTIFIER COMPARISON AND DESIGN TRADEOFFS
Tables 3.5 and 3.6 summarize the characteristics of the half-wave, full-wave, and full-wave bridge
rectifiers introduced in Secs. 3.13 to 3.15. The filter capacitor often represents a significant economic
factor in terms of cost, size, and weight in the design of rectifier circuits. For a given ripple voltage,
the value of the filter capacitor required in the full-wave rectifier is one-half that for the half-wave
rectifier.
The reduction in peak current in the full-wave rectifier can significantly reduce heat dissipation
in the diodes. The addition of the second diode and the use of a center-tapped transformer represent
additional expenses that offset some of the advantage. However, the benefits of full-wave rectification
usually outweigh the minor increase in circuit complexity.
The bridge rectifier eliminates the need for the center-tapped transformer, and the PIV rating of
the diodes is reduced, which can be particularly important in high-voltage circuits. The cost of the
extra diodes is usually negligible, particularly since four-diode bridge rectifiers can be purchased in
single-component form.
TABLE 3.5
Rectifier Equation Summary
HALF-WAVE RECTIFIER FULL-WAVE RECTIFIER FULL-WAVE BRIDGE RECTIFIER
Vdc = VP ? Von Idc =
(VP ? Von)
R
Vdc = VP ? Von Idc =
(VP ? Von)
R
Vdc = VP ? 2Von Idc =
(VP ? 2Von)
R
Vr =
(VP ? Von)
R
T
C = Idc
T
C
Vr =
(VP ? Von)
R
T
2C = Idc
T
2C
Vr =
(VP ? 2Von)
R
T
2C = Idc
T
2C
T =
1
ω2Vr
VP
θc = ωT T =
1
ω2Vr
VP
θc = ωT T =
1
ω2Vr
VP
θc = ωT
IP = Idc
2T
T
PIV = 2VP IP = Idc
T
T
PIV = 2VP IP = Idc
T
T
PIV = VP
3.16 Rectifier Comparison and Design Tradeoffs 135
TABLE 3.6
Comparison of Rectifiers with Capacitive Filters
RECTIFIER HALF-WAVE FULL-WAVE FULL-WAVE BRIDGE
PARAMETER RECTIFIER RECTIFIER RECTIFIER
Filter capacitor C
C
2
C
2
PIV rating 2VP 2VP VP
Peak diode current
(constant Vr )
Highest Reduced Reduced
IP
IP
2
IP
2
Surge current Highest Reduced (∝ C) Reduced (∝ C)
Comments Least complexity Smaller capacitor Smaller capacitor
Requires center-tapped Four diodes
transformer No center tap
Two diodes on transformer
D E S I G N RECTIFIER DESIGN
EXAMPLE 3.11
Now we will use our rectifier theory to design a rectifier circuit that will provide a specified output
voltage and ripple voltage.
PROBLEM Design a rectifier to provide a dc output voltage of 15 V with no more than 1 percent ripple at a
load current of 2 A.
SOLUTION Known Information and Given Data: Vdc = 15 V, Vr < 0.15 V, Idc = 2 A
Unknowns: Circuit topology, transformer voltage, filter capacitor, diode PIV rating, diode repetitive
current rating, diode surge current rating.
Approach: Use given data to evaluate rectifier circuit equations. Let us choose a full-wave bridge
topology that requires a smaller value of filter capacitance, a smaller diode PIV voltage, and no
center tap in the transformer.
Assumptions: Assume diode on-voltage is 1 V. The ripple voltage is much less than the dc output
voltage (Vr  Vdc), and the conduction interval should be much less than the period of the ac
signal (T  T ).
Analysis: The required transformer voltage is
V =
VP √2 =
Vdc + 2Von √2 =
15 + 2
√2
V = 12.0 Vrms
The filter capacitor is found using the ripple voltage, output current, and discharge interval:
C = IdcT/2
Vr  = 2A 1
120
s 1
0.15V = 0.111 F
136 Chapter 3 Solid-State Diodes and Diode Circuits
To find IP , the conduction time is calculated using Eq. (3.54)
T =
1
ω2Vr
VP =
1
120π2(0.15)V
17V = 0.352 ms
and the peak repetitive current is found to be
IP = Idc 2
T T
2  = 2A
(1/60) s
0.352 ms = 94.7 A
The surge current estimate is
Isurge = ωCVP = 120π(0.111)(17) = 711 A
The minimum diode PIV is VP = 17 V. A choice with a safety margin would be PIV>20 V. The
repetitive current rating should be 95 A with a surge current rating of 710 A. Note that both of
these calculations overestimate the magnitude of the currents because we have neglected series
resistance of the transformer and diode. The minimum filter capacitor needs to be 111,000 F.
Assuming a tolerance of ?30 percent, a nominal filter capacitance of 160,000 F would be
required.
Check of Results: The ripple voltage is designed to be 1 percent of the dc output voltage. Thus
the assumption that the voltage is approximately constant is justified. The conduction time is
0.352 ms out of a total period T = 16.7 ms. Thus the assumption that T  T is satisfied.
Computer-Aided Analysis: This design example represents an excellent place where simulation
can be used to explore the magnitude of the diode currents and improve the design so that we
don’t over-specify the rectifier diodes. A SPICE simulation with RS = 0.1 , n = 2, IS = 1 A,
and a transformer series resistance of 0.1  yields a number of unexpected results: IP = 11 A,
Isurge = 70 A, and Vdc = 13 V! The surge current and peak repetitive current are both reduced by
almost an order of magnitude compared to our hand calculations! In addition the output voltage
is lower than expected. If we think further, a peak current of 11 A will cause a peak voltage drop
of 2.2 V across the total series resistance of 0.2 , so it should not be surprising that the output
voltage is 2 V lower than originally expected. The series resistances actually help to reduce the
stress on the diodes. The time constant of the series resistance and the filter capacitor is 0.44 s,
so the circuit takes many cycles to reach the steady-state output voltage.
EXERCISE: Repeat the rectifier design assuming the use of a half-wave rectifier.
A NSWERS: V = 11.3 Vrms; C = 222,000 F; I P = 184 A; I SC = 1340 A
E L E C T R O N I C S I N A C T I O N
Power Cubes and Cell Phone Chargers
We actually encounter the unfiltered transformer driven half-wave rectifier circuit depicted in
Fig. 3.47 frequently in our everyday lives in the form of “power cubes” and battery chargers
for many portable electronic devices. An example is shown in the accompanying figure. The
3.16 Rectifier Comparison and Design Tradeoffs 137
power cube contains only a small transformer and rectifier diode. The transformer is wound
with small wire and has a significant resistance in both the primary and secondary windings.
In the transformer in the photograph, the primary resistance is 600  and the secondary
resistance is 15 , and these resistances actually help provide protection from failure of the
transformer windings. Load resistance R in Fig. 3.51 represents the actual electronic device
that is receiving power from the power cube and may often be a rechargable battery. In some
cases, a filter capacitor may be included as part of the circuit that forms the load for the power
cube.
Part (c) of the figure below shows a much more complex device used for recharging the
batteries in a cell phone. The simplified schematic in part (c) utilizes a full-wave bridge rectifier
with filter capacitor connected directly to the ac line. The rectifier’s high voltage output is
filtered by capacitor C1 and feeds a switching regulator consisting of a switch, the transformer
driving a half-wave rectifier with pi-filter (D5, C2, L, and C3), and a feedback circuit that
controls the output voltage by modulating the duty cycle of the switch. The transformer steps
down the voltage and provides isolation from the high voltage ac line input. Diode D6 and R
clamp the inductor voltage when the switch opens. The feedback signal path is isolated from
the input using an optical isolator. (See Electronics in Action in Chapter 5 for discussion of
an optical isolator.) Note the wide range of input voltages accomodated by the circuit. Thus,
most international voltage standards can be accommodated by one adopter.
(a) (b)
C1
C2 C3
D6
D5 R
Feedback control
circuitry
Optically isolated
switch
Isolation and step-down
transformer
Full-wave bridge rectifier
with capacitor filter
85?265 V
ac input L +
–
VO
(c)
(a) Inside a simple power cube; (b) cell phone charger; (c) simplified schematic for the cell phone charger.
138 Chapter 3 Solid-State Diodes and Diode Circuits
3.17 DYNAMIC SWITCHING BEHAVIOR OF THE DIODE
Up to this point, we have tacitly assumed that diodes can turn on and off instantaneously. However,
an unusual phenomenon characterizes the dynamic switching behavior of the pn junction diode.
SPICE simulation is used to illustrate the switching of the diode in the circuit in Fig. 3.67, in which
diode D1 is being driven from voltage source vI through resistor R1.
The source is zero for t < 0. At t = 0, the source voltage rapidly switches to +1.5 V, forcing a
current into the diode to turn it on. The voltage remains constant until t = 7.5 ns. At this point the
source switches to ?1.5 V in order to turn the diode back off.
The simulation results are presented in Fig. 3.68. Following the voltage source change at t = 0+,
the current increases rapidly, but the internal capacitance of the diode prevents the diode voltage from
changing instantaneously. The current actually overshoots its final value and then decreases as the
diode turns on and the diode voltage increases to approximately 0.7 V. At any given time, the current
flowing into the diode is given by
iD(t) =
v1(t) ? vD(t)
0.75 k
(3.65)
The initial peak of the current occurs when vI reaches 1.5 V and vD is still nearly zero:
iDmax =
1.5 V
0.75 k = 2.0 mA (3.66)
After the diode voltage reaches its final value with Von ≈ 0.7 V, the current stabilizes at a forward
current IF of
IF =
1.5 ? 0.7
0.75 k = 1.1 mA (3.67)
At t = 7.5 ns, the input source rapidly changes polarity to ?1.5 V, and a surprising thing
happens. The diode current also rapidly reverses direction and is much greater than the reverse
saturation current of the diode! The diode does not turn off immediately. In fact, the diode actually
vI
R1 = 0.75 kΩ
1.5 V
–1.5 V
7.5ns 15ns
v1
iD
t
vD D1
Figure 3.67 Circuit used to explore diode-switching behavior.
Voltage
2.0 V
0 V
–2.0 V
Recovery
transient
Turn-on transient
Input voltage
Diode voltage
0 A
–2.0 mA
0 s 2 ns 4 ns 6 ns 8 ns 10 ns 12 ns
2.0 mA
Diode current
–4.0 mA
IR
IF
Time
0 s 2 ns 4 ns 6 ns 8 ns 10 ns 12 ns
Time
Storage time
Figure 3.68 SPICE simulation results for the diode circuit in Fig. 3.67. (The diode transit time is equal to 5 ns.)
3.18 Photo Diodes, Solar Cells, and Light-Emitting Diodes 139
remains forward-biased by the charge stored in the diode, with vD = Von, even though the current
has changed direction! The reverse current IR is equal to
IR = ?1.5 ? 0.7
0.75 k = ?2.9 mA (3.68)
The current remains at ?2.9 mA for a period of time called the diode storage time τS, during which
the internal charge stored in the diode is removed. Once the stored charge has been removed, the
voltage across the diode begins to drop and charges toward the final value of ?1.5 V. The current in
the diode drops rapidly to zero as the diode voltage begins to fall.
The turn-on time and recovery time are determined primarily by the charging and discharging
of the nonlinear depletion-layer capacitance Cj through the resistance RS. The storage time is
determined by the diffusion capacitance and diode transit time defined in Eq. (3.22) and by the
values of the forward and reverse currents IF and IR:
τS = τT ln 1 ?
IF
IR  = 5 ln1 ?
1.1 mA
?2.9 mA ns = 1.6 ns (3.69)
The SPICE simulation results in Fig. 3.68 agree well with this value.
Always remember that solid-state devices do not turn off instantaneously. The unusual storage
time behavior of the diode is an excellent example of the switching delays that occur in pn junction
devices in which carrier flow is dominated by the minority-carrier diffusion process. This behavior
is not present in field-effect transistors, in which current flow is dominated by majority-carrier drift.
3.18 PHOTO DIODES, SOLAR CELLS, AND LIGHT-EMITTING DIODES
Several other important applications of diodes include photo detectors in communication systems,
solar cells for generating electric power, and light-emitting diodes (LEDs). These applications all
rely on the solid-state diode’s ability to detect on produce optical emissions.
3.18.1 PHOTO DIODES AND PHOTODETECTORS
If the depletion region of a pn junction diode is illuminated with light of sufficiently high energy, the
photons can cause electrons to jump the semiconductor bandgap, creating electron–hole pairs. For
photon absorption to occur, the incident photons must have an energy Ep that exceeds the bandgap
of the semiconductor:
Ep = hν =
hc
λ ≥ EG (3.70)
where h =Planck’s constant (6.626 × 10?34 J · s) λ= wavelength of optical illumination
ν =frequency of optical illumination c = velocity of light (3 × 108 m/s)
The i-v characteristic of a diode with and without illumination is shown in Fig. 3.69. The original
diode characteristic is shifted vertically downward by the photon-generated current. Photon absorption
creates an additional current crossing the pn junction that can be modeled by a current source
iPH in parallel with the pn junction diode, as shown in Fig. 3.70.
Based on this model, we see that the incident optical signal can be converted to an electrical
signal voltage using the simple photodetector circuit in Fig. 3.71. The diode is reverse-biased to
enhance the width and electric field in the depletion region. The photon-generated current iPH will
flow through resistor R and produce an output signal voltage given by
vo = iPHR (3.71)
140 Chapter 3 Solid-State Diodes and Diode Circuits
–0.5 1.5
–0.002
Diode current (A)
Diode voltage (V)
–1.5
0.000
0.002
0.004
0.006
0.008
0.010
0.5
No illumination
Illuminated
Figure 3.69 Diode i-v characteristic with and without optical
illumination.
vD
iD
λ ≡ iPH
Figure 3.70 Model for optically illuminated diode.
iPH represents the current generated by absorption of
photons in the vicinity of the pn junction.
+VB
vo
vo
iPH
R
R
+VB
λ
(a) (b)
Figure 3.71 Basic photodetector circuit (a) and
model (b).
iPH
IC
VC
Figure 3.72 pn Diode under
steady-state illumination as a
solar cell.
1
–1.0
Cell current IC (A)
Cell voltage VC (V)
–1
0.0
–0.5
0.5
1.0
1.5
0
Pmax
VOC
ISC
Figure 3.73 Terminal characteristics for a pn
junction solar cell.
In optical fiber communication systems, the amplitude of the incident light is modulated by rapidly
changing digital data, and iPH includes a time-varying signal component. The time-varying signal
voltage at vo is fed to additional electronic circuits to demodulate the signal and recover the original
data that were transmitted down the optical fiber.
3.18.2 POWER GENERATION FROM SOLAR CELLS
In solar cell applications, the optical illumination is constant, and a dc current IPH is generated. The
goal is to extract power from the cell, and the i-v characteristics of solar cells are usually plotted in
terms of the cell current IC and cell voltage VC, as defined in Fig. 3.72.
The i-v characteristic of the pn junction used for solar cell applications is plotted in terms of
these terminal variables in Fig. 3.73. Also indicated on the graph are the short-circuit current ISC ,
3.18 Photo Diodes, Solar Cells, and Light-Emitting Diodes 141
the open-circuit voltage VOC, and the maximum power point Pmax. ISC represents the maximum
current available from the cell, and VOC is the voltage across the open-circuited cell when all the
photo current is flowing into the internal pn junction. For the solar cell to supply power to an external
circuit, the product IC ×VC must be positive, corresponding to the first quadrant of the characteristic.
An attempt is made to operate the cell near the point of maximum output power Pmax.
E L E C T R O N I C S I N A C T I O N
Solar Energy
The photograph below depicts the Long Island Solar Farm installation on the Brookhaven
National Laboratory (BNL) site in the center of Long Island, New York. The installation, consisting
of 164,312 solar panels utilizing crystalline silicon technology, is capable of a generating
a peak power of 32 MW with an estimated annual energy output of 44 million kilowatt-hours,
enough to power an estimated 4500 homes for a year. The project was a collaboration between
the Department of Energy, BP Solar, and the Long Island Power Authority and became operational
near the end of 2011. The Long Island Power Authority purchases 100 percent of
the power generated by the installation that is estimated to offset production of approximately
30,000 metric tons of CO2 per year as well as significant amounts of other pollutants.
Long Island Solar Farm Installation. The main BNL campus is at the upper center of the picture.
Courtesy Brookhaven National Laboratory.
3.18.3 LIGHT-EMITTING DIODES (LEDs)
Light-emitting diodes, or LEDs, rely on the annihilation of electrons and holes through recombination
rather than on the generation of carriers, as in the case of the photo diode. When a hole and
electron recombine, an energy equal to the bandgap of the semiconductor can be released in the
form of a photon. This recombination process is present in the forward-biased pn junction diode. In
silicon, the recombination process actually involves the interaction of photons and lattice vibrations
called phonons, so the optical emission process in silicon is not nearly as efficient as that in the III–V
compound semiconductor GaAs or the ternary materials such as GaIn1?xAsx and GaIn1?xPx . LEDs
in these compound semiconductor materials provide visible illumination, and the color of the output
can be controlled by varying the fraction x of arsenic or phosphorus in the material which changes
this bandgap energy.
142 Chapter 3 Solid-State Diodes and Diode Circuits
S U M M A R Y
In this chapter we investigated the detailed behavior of the solid-state diode.
? A pn junction diode is created when p-type and n-type semiconductor regions are formed in
intimate contact with each other. In the pn diode, large concentration gradients exist in the vicinity
of the metallurgical junction, giving rise to large electron and hole diffusion currents.
? Under zero bias, no current can exist at the diode terminals, and a space charge region forms in the
vicinity of the pn junction. The region of space charge results in both a built-in potential and an
internal electric field, and the electric field produces electron and hole drift currents that exactly
cancel the corresponding components of diffusion current.
? When a voltage is applied to the diode, the balance in the junction region is disturbed, and the
diode conducts a current. The resulting i -v characteristics of the diode are accurately modeled by
the diode equation:
iD = ISexp vD
nVT ? 1
where IS = reverse saturation current of the diode
n = nonideality factor (approximately 1)
VT = kT/q = thermal voltage (0.025 V at room temperature)
? Under reverse bias, the diode current equals ?IS, a very small current.
? For forward bias, however, large currents are possible, and the diode presents an almost constant
voltage drop of 0.6 to 0.7 V.
? At room temperature, an order of magnitude change in diode current requires a change of less than
60 mV in the diode voltage. At room temperature, the silicon diode voltage exhibits a temperature
coefficient of approximately ?1.8 mV/?C.
? One must also be aware of the reverse-breakdown phenomenon that is not included in the diode
equation. If too large a reverse voltage is applied to the diode, the internal electric field becomes
so large that the diode enters the breakdown region, either through Zener breakdown
or avalanche breakdown. In the breakdown region, the diode again represents an almost fixed
voltage drop, and the current must be limited by the external circuit or the diode can easily be
destroyed.
? Diodes called Zener diodes are designed to operate in breakdown and can be used in simple voltage
regulator circuits. Line regulation and load regulation characterize the change in output voltage of
a power supply due to changes in input voltage and output current, respectively.
? As the voltage across the diode changes, the charge stored in the vicinity of the space charge region
of the diode changes, and a complete diode model must include a capacitance. Under reverse bias,
the capacitance varies inversely with the square root of the applied voltage. Under forward bias, the
capacitance is proportional to the operating current and the diode transit time. These capacitances
prevent the diode from turning on and off instantaneously and cause a storage time delay during
turn-off.
? Direct use of the nonlinear diode equation in circuit calculations usually requires iterative numeric
techniques. Several methods for simplifying the analysis of diode circuits were discussed, including
the graphical load-line method and use of the ideal diode and constant voltage drop models.
? SPICE circuit analysis programs include a comprehensive built-in model for the diode that accurately
reproduces both the ideal and nonideal characteristics of the diode and is useful for exploring
the detailed behavior of circuits containing diodes.
Key Terms 143
? Important applications of diodes include half-wave, full-wave, and full-wave bridge rectifier circuits
used to convert from ac to dc voltages in power supplies. Simple power supply circuits use
capacitive filters, and the design of the filter capacitor determines power supply ripple voltage and
diode conduction angle. Diodes used as rectifiers in power supplies must be able to withstand large
peak repetitive currents as well as surge currents when the power supplies are first turned on. The
reverse-breakdown voltage of rectifier diodes is referred to as the peak-inverse-voltage, or PIV,
rating of the diode.
? Real diodes cannot turn on or off instantaneously because the internal capacitances of the diodes
must be charged and discharged. The turn-on time is usually quite short, but diodes that have been
conducting turn off much less abruptly. It takes time to remove stored charge within the diode, and
this time delay is characterized by storage time τs . During the storage time, it is possible for large
reverse currents to occur in the diode.
? Finally, the ability of the pn junction device to generate and detect light was discussed, and the
basic characteristics of photo diodes, solar cells, and light-emitting diodes were presented.
K E Y T E R M S
Anode
Avalanche breakdown
Bias current and voltage
Breakdown region
Breakdown voltage
Built-in potential (or voltage)
Cathode
Center-tapped transformer
Conduction angle
Conduction interval
Constant voltage drop (CVD) model
Cut-in voltage
Depletion layer
Depletion-layer capacitance
Depletion-layer width
Depletion region
Diffusion capacitance
Diode equation
Diode SPICE parameters (IS, RS, N, TT,
CJO, VJ, M)
Filter capacitor
Forward bias
Full-wave bridge rectifier circuit
Full-wave rectifier circuit
Half-wave rectifier circuit
Ideal diode
Ideal diode model
Impact-ionization process
Junction capacitance
Junction potential
Light-emitting diode (LED)
Line regulation
Load line
Load-line analysis
Load regulation
Mathematical model
Metallurgical junction
Nonideality factor (n)
Peak detector
Peak inverse voltage (PIV)
Photodetector circuit
Piecewise linear model
pn junction diode
Q-point
Quiescent operating point
Rectifier circuits
Reverse bias
Reverse breakdown
Reverse saturation current (IS)
Ripple current
Ripple voltage
Saturation current
Schottky barrier diode
Solar cell
Space charge region (SCR)
Storage time
Surge current
Thermal voltage (VT )
Transit time
Turn-on voltage
Voltage regulator
Voltage transfer characteristic (VTC)
Zener breakdown
Zener diode
Zero bias
Zero-bias junction capacitance
144 Chapter 3 Solid-State Diodes and Diode Circuits
R E F E R E N C E
1. G. W. Neudeck, The PN Junction Diode, 2d ed. Pearson Education, Upper Saddle River, NJ:
1989.
A D D I T I O N A L R E A D I N G
PSPICE, ORCAD, now owned by Cadence Design Systems, San Jose, CA.
LTspice available from Linear Technology Corp.
Tina-TI SPICE-based analog simulation program available from Texas Instruments.
T. Quarles, A. R. Newton, D. O. Pederson, and A. Sangiovanni-Vincentelli, SPICE3 Version 3f3
User’s Manual. UC Berkeley: May 1993.
A. S. Sedra, and K. C. Smith. Microelectronic Circuits. 5th ed. Oxford University Press, New York:
2004.
P R O B L E M S
3.1 The pn Junction Diode
3.1. A diode is doped with NA = 1018/cm3 on the
p-type side and ND = 1019/cm3 on the n-type side.
(a) What is the depletion-layer width wdo? (b) What
are the values of xp and xn? (c) What is the value
of the built-in potential of the junction? (d) What is
the value of EMAX? Use Eq. (3.3) and Fig. 3.5.
3.2. A diode is doped with NA =1018/cm3 on the
p-type side and ND =1018/cm3 on the n-type side.
(a) What are the values of pp, pn, np, and nn?
(b) What are the depletion-region width wdo and
built-in voltage?
3.3. Repeat Prob. 3.2 for a diode with NA = 1016/cm3 on
the p-type side and ND = 1020/cm3 on the n-type
side.
3.4. Repeat Prob. 3.2 for a diode with NA = 1018/cm3
on the p-type side and ND =1018/cm3 on the
n-type side.
3.5. A diode has wdo = 1 m and φj = 0.7 V. (a) What
reverse bias is required to double the depletionlayer
width? (b) What is the depletion region width
if a reverse bias of 12 V is applied to the diode?
3.6. A diode has wdo = 0.4 m and φj = 0.85 V.
(a) What reverse bias is required to triple the
depletion-layer width? (b) What is the depletion
region width if a reverse bias of 7 V is applied to
the diode?
3.7. Suppose a drift current density of 5000 A/cm2 exists
in the neutral region on the n-type side of a
diode that has a resistivity of 0.5  · cm. What is
the electric field needed to support this drift current
density?
3.8. Suppose a drift current density of 2000 A/cm2 exists
in the neutral region on the p-type side of a
diode that has a resistivity of 2.5 · cm. What is
the electric field needed to support this drift current
density?
3.9. The maximum velocity of carriers in silicon is
approximately 107 cm/s. (a) What is the maximum
drift current density that can be supported
in a region of p-type silicon with a doping of
5 × 1017/cm3? (b) Repeat for a region of n-type
silicon with a doping of 4 × 1015/cm3?
??3.10. Suppose that NA(x) = No exp(?x/L) in a region
of silicon extending from x = 0 to x = 8m, where
No is a constant. Assume that p(x) = NA(x). Assuming
that jp must be zero in thermal equilibrium,
show that a built-in electric field must exist and find
its value for L = 1 m and No = 1018/cm3.
3.11. What carrier gradient is needed to generate a
diffusion current density of jn = 1500 A/cm2 if
μn = 500 cm2/V · s?
3.12. Use the solver routine in your calculator to find the
solution to Eq. (3.25) for IS = 10?16 A.
3.13. Use a spreadsheet to iteratively find the solution to
Eq. (3.25) for IS = 10?13 A.
3.14. (a) Use MATLAB or MATHCAD to find the solution
to Eq. (3.25) for IS = 10?13 A. (b) Repeat for
IS = 10?15 A.
3.2 –3.4 The i -v Characteristics of the Diode;
the Diode Equation: a Mathematical Model for
the Diode; and Diode Characteristics Under
Reverse, Zero, and Forward Bias
Problems 145
3.15. To what temperature does VT = 0.025 V actually
correspond? What is the value of VT for temperatures
of ?55?C, 0?C, and +85?C?
3.16. (a) Plot a graph of the diode equation similar to
Fig. 3.8 for a diode with IS =10?12 A and n =1.
(b)Repeat for n =2. (c)Repeat(a) for IS =10?15 A.
3.17. A diode has n = 1.06 at T = 320 K. What is the
value of n · VT ? What temperature would give the
same value of n · VT if n = 1.00?
3.18. Plot the diode current for a diode with ISO = 15 fA
and φj = 0.75 V for ?10 V ≤ vD ≤ 0 V using
Eq. (3.19).
?3.19. What are the values of IS and n for the diode in the
graph in Fig. P3.19? Assume VT = 0.0259 V.
10–2
10–3
10–4
10–5
10–6
10–7
10–8
10–9
10–10
10–11
0.0 0.2 0.4
Diode voltage (V)
Diode current (A)
0.6 0.8
Figure P3.19
3.20. A diode has IS = 10?17 A and n = 1.05. (a) What
is the diode voltage if the diode current is 70 A?
(b) What is the diode voltage if the diode current is
5 A? (c) What is the diode current for vD = 0 V?
(d) What is the diode current for vD = ?0.075 V?
(e) What is the diode current for vD = ?5 V?
3.21. A diode has IS = 5 aA and n = 1. (a) What is
the diode voltage if the diode current is 100 A?
(b) What is the diode voltage if the diode current is
10 A? (c) What is the diode current for vD = 0V?
(d) What is the diode current for vD = ?0.06 V?
(e) What is the diode current for vD = ?4 V?
3.22. A diode has IS = 0.2 fA and n = 1. (a) What is
the diode current if the diode voltage is 0.675 V?
(b) What will be the diode voltage if the current
increases by a factor of 3?
3.23. A diode has IS = 10?10 A and n = 2. (a) What
is the diode voltage if the diode current is 40 A?
(b) What is the diode voltage if the diode current is
100 A?
3.24. A diode is operating with iD = 2 mA and vD = 0.82 V. (a) What is IS if n = 1? (b) What is the
diode current for vD = ?5 V?
3.25. A diode is operating with iD = 300 A and
vD = 0.75 V. (a) What is IS if n = 1.07? (b) What
is the diode current for vD = ?3 V?
3.26. The saturation current for diodes with the same part
number may vary widely. Suppose it is known that
10?14 A ≤ IS ≤ 10?12 A. What is the range of forward
voltages that may be exhibited by the diode if
it is biased with iD = 2 mA?
3.27. A diode is biased by a 0.9-V dc source, and its current
is found to be 100 A at T = 35 ?C. (a) At
what temperature will the current double? (b) At
what temperature will the current be 50 A?
??3.28. The i -v characteristic for a diode has been measured
under carefully controlled temperature conditions
(T = 307 K), and the data are in Table P3.28.
TABLE P3.28
Diode i-v Measurements
DIODE VOLTAGE DIODE CURRENT
0.500 6.591 × 10?7
0.550 3.647 × 10?6
0.600 2.158 × 10?5
0.650 1.780 × 10?4
0.675 3.601 × 10?4
0.700 8.963 × 10?4
0.725 2.335 × 10?3
0.750 6.035 × 10?3
0.775 1.316 × 10?2
Use a spreadsheet orMATLAB to find the values of
IS and n that provide the best fit of the diode equation
to the measurements in the least-squares sense.
[That is, find the values of IS and n that minimize
the function M = n
m=1(im
D ? IDm)2, where iD is
the diode equation from Eq. (3.1) and IDm are the
measured data.] For your values of IS and n, what
is the minimum value of M = n
m=1(im
D ? IDm)2?
3.5 Diode Temperature Coefficient
3.29. What is the value of VT for temperatures of ?40?C,
0?C, and +50?C?
3.30. A diode has IS = 10?16 A and n =1. (a) What is
the diode voltage if the diode current is 250 A
146 Chapter 3 Solid-State Diodes and Diode Circuits
at T = 25?C? (b) What is the diode voltage at
T = 85?C? Assume the diode voltage temperature
coefficient is ?2 mV/K at 55?C.
3.31. A diode has IS = 20 fA and n = 1. (a) What is
the diode voltage if the diode current is 100 A
at T = 25?C? (b) What is the diode voltage at
T = 50?C? Assume the diode voltage temperature
coefficient is ?1.8 mV/K at 0?C.
?3.32. The temperature dependence of IS is described
approximately by
IS = CT3 exp?
EG
kT 
What is the diode voltage temperature coefficient
based on this expression and Eq. (3.15) if EG = 1.12 eV, VD = 0.7 V, and T = 315 K?
3.33. The saturation current of a silicon diode is described
by the expression in Prob. 3.32. (a) What temperature
change will cause IS to double? (b) To increase
by 10 times? (c) To decrease by 100 times?
3.6 Diodes under Reverse Bias
3.34. A diode has wdo = 1.5 m and φj = 0.8 V.
(a) What is the depletion layer width for VR = 5V?
(b) For VD = ?10 V?
3.35. A diode has a doping of ND = 1015/cm3 on the
n-type side and NA = 1017/cm3 on the p-type
side. What are the values of wdo and φj ? What
is the value of wd at a reverse bias of 10 V?
At 100 V?
3.36. A diode has a doping of ND = 1020/cm3 on the
n-type side and NA = 1018/cm3 on the p-type side.
What are the values of wdo and φj ? What is the
value of wd at a reverse bias of 5 V? At 25 V?
?3.37. A diode has wdo = 2 m and φj = 0.6 V. If the
diode breaks down when the internal electric field
reaches 300 kV/cm, what is the breakdown voltage
of the diode?
?3.38. Silicon breaks down when the internal electric field
exceeds 300 kV/cm. At what reverse bias do you
expect the diode of Prob. 3.2 to break down?
3.39. What are the breakdown voltage VZ and Zener resistance
RZ of the diode depicted in Fig. P3.39?
??3.40. A diode is fabricated with NA 
 ND. What value
of doping is required on the lightly doped side to
achieve a reverse-breakdown voltage of 750 V if
the semiconductor material breaks down at a field
of 300 kV/cm?
–7 –6 –5 –4 –3 –2 –1 1 2 3 4 5 6 7
0.002
0.001
–0.001
–0.002
vD (V)
iD (A)
Figure P3.39
3.7 pn Junction Capacitance
3.41. What is the zero-bias junction capacitance/cm2 for
a diode with NA = 1018/cm3 on the p-type side
and ND = 1020/cm3 on the n-type side? What is
the diode capacitance with a 3-V reverse bias if the
diode area is 0.05 cm2?
3.42. What is the zero-bias junction capacitance per cm2
for a diode with NA = 1018/cm3 on the p-type side
and ND = 1015/cm3 on the n-type side. What is
the diode capacitance with a 9 V reverse bias if the
diode area is 0.02 cm2?
3.43. Adiode is operating at a current of 250A. (a) What
is the diffusion capacitance if the diode transit time
is 100 ps? (b) How much charge is stored in the
diode? (c) Repeat for iD = 3 mA.
3.44. A diode is operating at a current of 2 A. (a) What is
the diffusion capacitance if the diode transit time is
10 ns? (b) How much charge is stored in the diode?
(c) Repeat for iD = 100 mA.
3.45. A square pn junction diode is 5 mm on a side. The
p-type side has a doping concentration of 1019/cm3
and the n-type side has a doping concentration of
1016/cm3. (a) What is the zero-bias capacitance of
the diode? What is the capacitance at a reverse bias
of 4 V? (b) Repeat for an area of 104 m2.
3.46. A variable capacitance diode with Cjo = 39 pF and
φj = 0.80 V is used to tune a resonant LC circuit
as shown in Fig. P3.46. The impedance of the RFC
(radio frequency choke) can be considered infinite.
What are the resonant frequencies ( fo = 1
2π√LC
)
for VDC = 1 V and VDC = 9 V?
Problems 147
VDC
RFC
C L 10 μH
L
Figure P3.46
3.8 Schottky Barrier Diode
3.47. A Schottky barrier diode is modeled by the diode
equation in Eq. (3.11) with IS = 10?11 A.
(a) What is the diode voltage at a current of 4 mA?
(b) What would be the voltage of a pn junction
diode with IS = 10?14 A operating at the same
current?
3.48. Suppose a Schottky barrier diode can be modeled by
the diode equation in Eq. (3.11) with IS = 10?7 A.
(a) What is the diode voltage at a current of 50 A?
(b) What would be the voltage of a pn junction
diode with IS = 10?15 A and n = 2?
3.9 Diode SPICE Model and Layout
3.49. (a) A diode has IS = 5 × 10?16 A and RS = 10 
and is operating at a current of 1mAat room temperature.
What are the values of VD and VD ? (b) Repeat
for RS = 100 .
3.50. A pn diode has a resistivity of 1 · cm on the ptype
side and 0.02 · cm on the n-type side. What
is the value of RS for this diode if the cross-sectional
area of the diode is 0.01 cm2 and the lengths of the
p- and n-sides of the diode are each 250 m?
?3.51. A diode fabrication process has a specific contact
resistance of 10 · m2. If the contacts are each
1 m × 1 m in size, what are the total contact
resistances associated with the anode and cathode
contacts to the diode in Fig. 3.21(a).
3.10 Diode Circuit Analysis
3.52. (a) Plot the load line and find the Q-point for
the diode circuit in Fig. P3.53 if V = 5 V and
R = 10 k. Use the i-v characteristic in Fig. P3.39.
(b) Repeat for V = ?6Vand R = 3 k. (c) Repeat
for V = ?3 V and R = 3 k.
3.53. (a) Plot the load line and find the Q-point for the
diode circuit in Fig. P3.53 if V = 10 V and R = 5 k. Use the i -v characteristic in Fig. P3.39.
(b) Repeat for V = ?10 V and R = 5 k.
(c) Repeat for V = ?2 V and R = 2 k.
R
V
Figure P3.53
3.54. Simulate the circuit in Prob. 3.53 with SPICE and
compare the results to those in Prob. 3.53. Use
IS = 10?15 A.
3.55. Use the i -v characteristic in Fig. P3.39. (a) Plot the
load line and find the Q-point for the diode circuit
in Fig. P3.53 if V = 6 V and R = 4 k. (b) For
V = ?6 V and R = 3 k. (c) For V = ?3 V
and R = 3 k. (d) For V = 12 V and R = 8 k.
(e) For V = ?25 V and R = 10 k.
Iterative Analysis and the Mathematical Model
3.56. Use direct trial and error to find the solution to the
diode circuit in Fig. 3.22 using Eq. (3.27).
3.57. Repeat the iterative procedure used in the spreadsheet
in Table 3.2 for initial guesses of 1 A, 5 mA,
and 5 A and 0 A. How many iterations are required
for each case? Did any problem arise? If so, what
is the source of the problem?
3.58. A diode has IS = 0.1 fA and is operating at
T = 300 K. (a) What are the values of VDO and
rD if ID = 200 A? (b) If ID = 2.0 mA? (c) If
ID = 20 mA?
3.59. (a) Use the iterative procedure in the spreadsheet in
Table 3.2 to find the diode current and voltage for
the circuit in Fig. 3.22 if V = 2.5 V and R = 3 k.
(b) Repeat for V = 7.5 V and R = 15 k.
3.60. (a) Use the iterative procedure in the spreadsheet in
Table 3.2 to find the diode current and voltage for
the circuit in Fig. 3.22 if V = 1 V and R = 15 k.
(b) Repeat for V = 3 V and R = 6.2 k.
3.61. Use MATLAB or MATHCAD to numerically find
the Q-point for the circuit in Fig. 3.22 using the
equation in the exercise on page 99.
Ideal Diode and Constant Voltage Drop Models
?3.62. Find the Q-point for the circuit in Fig. 3.22 using
the same four methods as in Sec. 3.10 if the voltage
source is 1 V. Compare the answers in a manner
similar to Table 3.3.
3.63. Find the Q-point for the diode in Fig. P3.63 using
(a) the ideal diode model and (b) the constant voltage
drop model with Von = 0.6 V. (c) Discuss the
148 Chapter 3 Solid-State Diodes and Diode Circuits
results. Which answer do you feel is most correct?
(d) Use iterative analysis to find the actual Q-point
if IS = 0.1 fA.
2 kΩ 2 kΩ
2 kΩ
+3 V
3 kΩ I
– V +
Figure P3.63
3.64. Simulate the circuit of Fig. P3.63 and find the
diode Q-point. Compare the results to those in
Prob. 3.63.
3.65. (a) Find the worst-case values of the Q-point current
for the diode in Fig. P3.63 using the ideal
diode model if the resistors all have 10 percent
tolerances. (b) Repeat using the CVD model with
Von = 0.6 V.
3.66. (a) Find I and V in the four circuits in Fig. P3.66
using the ideal diode model. (b) Repeat using the
constant voltage drop model with Von = 0.65 V.
16 kΩ
+5 V
I
V
–5 V
(a)
16 kΩ
+7 V
I
V
–3 V
(c)
–5 V
+5 V
I
16 kΩ
V
(d)
16 kΩ
V
–7 V
+3 V
I
(b)
Figure P3.66
3.67. (a) Find I and V in the four circuits in Fig. P3.66 using
the ideal diode model if the resistor values are
changed to 68 k. (b) Repeat using the constant
voltage drop model with Von = 0.6 V.
3.11 Multiple Diode Circuits
3.68. Find the Q-points for the diodes in the four circuits
in Fig. P3.68 using (a) the ideal diode model and
(b) the constant voltage drop model with Von = 0.7 V.
43 kΩ
22 kΩ
–9 V
+6 V
D2
D1
43 kΩ
22 kΩ
–9 V
+6 V
D2
D1
43 kΩ
22 kΩ
–9 V
+6 V
D2
D1
(c) (d)
(b)
22 kΩ
43 kΩ
–6 V
+9 V
D2
D1
(a)
Figure P3.68
3.69. Find the Q-points for the diodes in the four circuits
in Fig. P3.68 if the values of all the resistors
are changed to 15 k using (a) the ideal diode
model and (b) the constant voltage drop model with
Von = 0.60 V.
3.70. Find the Q-point for the diodes in the circuits in
Fig. P3.70 using the ideal diode model.
3.71. Find the Q-point for the diodes in the circuits in
Fig. P3.70 using the constant voltage drop model
with Von = 0.65 V.
Problems 149
–10 V
0 V
+2 V
8.2 kΩ 12 kΩ –5 V
10 kΩ
(a)
+10 V
+5 V
0 V
8.2 kΩ 12 kΩ –5 V
10 kΩ
(b)
+10 V
+5 V
0 V
3.3 kΩ 6.8 kΩ –5 V
2.4 kΩ
(c)
–5 V
+5 V
0 V
4.7 kΩ
4.7 kΩ 4.7 kΩ +12 V
(d)
Figure P3.70
3.72. Simulate the diode circuits in Fig. P3.70 and compare
your results to those in Prob. 3.70.
3.73. Verify that the values presented in Ex. 3.8 using the
ideal diode model are correct.
3.74. Simulate the circuit in Fig. 3.33 and compare to the
results in Ex. 3.8.
3.12 Analysis of Diodes Operating in the
Breakdown Region
3.75. Drawthe load line for the circuit in Fig. P3.75 on the
characteristics in Fig. P3.39 and find the Q-point.
3.6 kΩ
10 kΩ
24 V RZ = 0
VZ = 6 V
Figure P3.75
3.76. (a) Find the Q-point for the Zener diode in
Fig. P3.75. (b) Repeat if RZ = 100 .
3.77. What is maximum load current IL that can be drawn
from the Zener regulator in Fig. P3.77 if it is to
maintain a regulated output? What is the minimum
value of RL that can be used and still have a regulated
output voltage?
RL
IL
15 kΩ
30 V VZ = 9 V
RZ = 0
Figure P3.77
3.78. What is power dissipation in the Zener diode in
Fig. P3.77 for (a) RL = 2 k (b) RL = 4.7 k
(c) RL = 15 k (d) RL =∞?
3.79. Load resistor RL in Fig. P3.77 is 12 k. What are
the nominal and worst-case values of Zener diode
current and power dissipation if the power supply
voltage, Zener breakdown voltage and resistors all
have 5 percent tolerances?
3.80. What is power dissipation in the Zener diode in
Fig. P3.80 for (a) RL = 100 ? (b) RL =∞?
RL
150 Ω
50 V
VZ = 15 V
RZ = 0
Figure P3.80
150 Chapter 3 Solid-State Diodes and Diode Circuits
3.81. Load resistor RL in Fig. P3.80 is 100 . What are
the nominal and worst-case values of Zener diode
current and power dissipation if the power supply
voltage, Zener breakdown voltage, and resistors all
have 10 percent tolerances?
3.13 Half-Wave Rectifier Circuits
3.82. A power diode has a reverse saturation current of
10?9 A and n = 1.6. What is the forward voltage
drop at the peak current of 48.6 A that was calculated
in the example in Sec. 3.13.5?
3.83. A power diode has a reverse saturation current of
10?8 Aand n = 2. What is the forward voltage drop
at the peak current of 100 A? What is the power
dissipation in the diode in a half-wave rectifier application
operating at 60 Hz if the series resistance
is 0.01  and the conduction time is 1 ms?
?3.84. (a) Use a spreadsheet or MATLAB or write a computer
program to find the numeric solution to the
conduction angle equation for a 60 Hz half-wave
rectifier circuit that uses a filter capacitance of
100,000 F. The circuit is designed to provide 5 V
at 5 A. {That is, solve [(VP ? Von) exp(?t/RC) = VP cos ωt ? Von]. Be careful! There are an infinite
number of solutions to this equation. Be sure your
algorithm finds the desired answer to the problem.} Assume Von = 1 V. (b) Compare to calculations
using Eq. (3.57).
3.85. What is the actual average value (the dc value)
of the rectifier output voltage for the waveform in
Fig. P3.85 if Vr is 10 percent of VP ? Von = 18 V?
0 T 2T
t
VP
_ Von
vO
Vr
Figure P3.85
3.86. Draw the voltage waveforms, similar to those
in Fig. 3.53, for the negative output rectifier in
Fig. 3.57(b).
?3.87. Show that evaluation of Eq. (3.61) will yield the
result in Eq. (3.62).
3.88. The half-wave rectifier in Fig. P3.88 is operating at
a frequency of 60 Hz, and the rms value of the transformer
output voltage vI is 12.6V±10%. What are
the nominal and worst case values of the dc output
voltage VO if the diode voltage drop is 1 V?
C R
D1
vI
+
–
vO
+
–
Figure P3.88
3.89. The half-wave rectifier in Fig. P3.88 is operating
at a frequency of 60 Hz, and the rms value of the
transformer output voltage is 6.3 V. (a) What is the
value of the dc output voltage VO if the diode voltage
drop is 1 V? (b) What is the minimum value
of C required to maintain the ripple voltage to less
than 0.25 V if R = 0.5 ? (c) What is the PIV rating
of the diode in this circuit? (d) What is the surge
current when power is first applied? (e) What is the
amplitude of the repetitive current in the diode?
3.90. Simulate the behavior of the half-wave rectifier in
Fig. P3.88 for vI = 10 sin 120πt, R = 0.025 
and C = 0.5 F. (Use IS = 10?10 A, RS = 0, and
RELTOL = 10?6.) Compare the simulated values
of dc output voltage, ripple voltage, and peak diode
current to hand calculations. Repeat simulation with
RS = 0.02 .
3.91. (a) Repeat Prob. 3.89 for a frequency of 400 Hz.
(b) Repeat Prob. 3.89 for a frequency of 70 kHz.
3.92. A 3.3-V, 30-A dc power supply is to be designed
with a ripple of less than 1.5 percent. Assume that a
half-wave rectifier circuit (60 Hz) with a capacitor
filter is used. (a) What is the size of the filter capacitor
C? (b) What is the PIV rating for the diode?
(c) What is the rms value of the transformer voltage
needed for the rectifier? (d) What is the value of the
peak repetitive diode current in the diode? (e) What
is the surge current at t = 0+?
3.93. A 2500-V, 2-A, dc power supply is to be designed
with a ripple voltage ≤ 0.5 percent. Assume that a
half-wave rectifier circuit (60 Hz) with a capacitor
filter is used. (a) What is the size of the filter capacitor
C? (b) What is the minimum PIV rating for
the diode? (c) What is the rms value of the transformer
voltage needed for the rectifier? (d) What is
the peak value of the repetitive current in the diode?
(e) What is the surge current at t = 0+?
Problems 151
?3.94. Draw the voltage waveforms at nodes vO and v1 for
the “voltage-doubler” circuit in Fig. P3.94 for the
first two cycles of the input sine wave. What is the
steady-state output voltage if VP = 17 V?
C
C
D1
D2
v1
vO
vI = VP sin t
+
–
~ ω
Figure P3.94
3.95. Simulate the voltage-doubler rectifier circuit
in Fig. P3.94 for C =500 F and vI = 1500 sin 2π(60)t with a load resistance of RL = 3000  added between vO and ground. Calculate
the ripple voltage and compare to the simulation.
3.96. Estimate the maximum surge current in a halfwave
rectifier with a transformer having an rms secondary
voltage of 50 V and a secondary resistance
of 0.25 . Assume the filter capacitance is 0.5 F
and a frequency of 60 Hz.
3.14 Full-Wave Rectifier Circuits
3.97. The full-wave rectifier in Fig. P3.97 is operating
at a frequency of 60 Hz, and the rms value of the
transformer output voltage is 18 V. (a) What is the
value of the dc output voltage if the diode voltage
drop is 1 V? (b) What is the minimum value of C
required to maintain the ripple voltage to less than
0.25 V if R = 0.5 ? (c) What is the PIV rating
of the diode in this circuit? (d) What is the surge
current when power is first applied? (e) What is the
amplitude of the repetitive current in the diode?
D1 C R
vI D2
vI
+
+ –
–
Figure P3.97
3.98. Repeat Prob. 3.97 if the rms value of the transformer
output voltage vI is 15 V.
3.99. A 60-Hz full-wave rectifier is built with a transformer
having an rms secondary voltage of 20 V
and filter capacitance C = 150,000 F. What is the
largest current that can be supplied by the rectifier
circuit if the ripple must be less than 0.3 V?
3.100. Simulate the behavior of the full-wave rectifier in
Fig. P3.97 for R = 3  and C = 22,000 F. Assume
that the rms value of vI is 10.0 V and the
frequency is 400 Hz. (Use IS = 10?10 A, RS = 0,
and RELTOL = 10?6.) Compare the simulated values
of dc output voltage, ripple voltage, and peak
diode current to hand calculations. Repeat simulation
with RS = 0.25.
3.101. Repeat Prob. 3.92 for a full-wave rectifier
circuit.
3.102. Repeat Prob. 3.93 for a full-wave rectifier circuit.
?3.103. The full-wave rectifier circuit in Fig. P3.103(a) was
designed to have a maximum ripple of approximately
1 V, but it is not operating properly. The
measured waveforms at the three nodes in the circuit
are shown in Fig. P3.103(b). What is wrong
with the circuit?
C
vI
v2
vI
v1 v3
+
+ –
–
~
R
D1
D2
Figure P3.103(a)
20 V
20 V
10 V
0 V
0 V
–20 V
0 s 10 ms 20 ms 30 ms 40 ms 50 ms
Time
v3
v2
v1
Figure P3.103(b) Waveforms for the circuit in Fig. P3.103(a).
152 Chapter 3 Solid-State Diodes and Diode Circuits
3.104. For the Zener regulated power supply in Fig.
P3.104, the rms value of vI is 15V, the operating frequency
is 60 Hz, R = 100 ,C = 1000F, the onvoltage
of diodes D1 and D2 is 0.75V, and the Zener
voltage of diode D3 is 15V. (a) What type of rectifier
is used in this power supply circuit? (b) What is the
dc voltage at V1? (c) What is the dc output voltage
VO? (d) What is the magnitude of the ripple voltage
at V1? (e) What is the minimum PIV rating for the
rectifier diodes? (f) Draw a new version of the circuit
that will produce an output voltage of ?15 V.
+
–
vI
+
–
vI
+
–
vO
+
–
V1 D3
D1
D2
C
R
Figure P3.104
3.15 Full-Wave Bridge Rectification
3.105. Repeat Prob. 3.97 for a full-wave bridge rectifier
circuit. Draw the circuit.
3.106. Repeat Prob. 3.92 for a full-wave bridge rectifier
circuit. Draw the circuit.
3.107. Repeat Prob. 3.93 for a full-wave bridge rectifier
circuit. Draw the circuit.
?3.108. What are the dc output voltages V1 and V2 for the
rectifier circuit in Fig. P3.108 if vI = 40 sin 377t
and C = 20,000 F?
C
C
vI
vI V1
V2
+
+ –
–
D1
D3
D2
D4
~
Figure P3.108
3.109. Simulate the rectifier circuit in Fig. P3.108 for
C = 100mFand vI = 40 sin 2π(60)t with a 500-
load connected between each output and ground.
3.110. Repeat Prob. 3.97 if the full-wave bridge circuit is
used instead of the rectifier in Fig. P3.97. Draw the
circuit!
3.16 Rectifier Comparison and Design
Tradeoffs
3.111. A 3.3-V, 15-A dc power supply is to be designed
to have a ripple voltage of no more than 10 mV.
Compare the pros and cons of implementating this
power supply with half-wave, full-wave, and fullwave
bridge rectifiers.
3.112. A 200-V, 0.5-A dc power supply is to be designed
with less than a 2 percent ripple voltage. Compare
the pros and cons of implementing this power supply
with half-wave, full-wave, and full-wave bridge
rectifiers.
3.113. A 3000-V, 1-A dc power supply is to be designed
with less than a 4 percent ripple voltage. Compare
the pros and cons of implementing this power supply
with half-wave, full-wave, and full-wave bridge
rectifiers.
3.17 Dynamic Switching Behavior of the Diode
?3.114. (a) Calculate the current at t = 0+ in the circuit
in Fig. P3.114. (b) Calculate IF , IR, and the storage
time expected when the diode is switched off if
τT = 8 ns.
R1 = 1.0 kΩ
v1 vD
v1
t
D1
iD
+
–
5 V
–3 V 10 ns 20 ns
Figure P3.114
3.115. (a) Simulate the switching behavior of the circuit in
Fig. P3.114. (b) Compare the simulation results to
the hand calculations in Prob. 3.114.
?3.116. (a) Calculate the current at t = 0+ in the circuit
in Fig. P3.114 if R1 is changed to 5 . (b) Calculate
IF , IR, and the storage time expected when the
diode is switched off at t = 10 s if τT = 250 ns.
??3.117. The simulation results presented in Fig. 3.68 were
performed with the diode transit time τT = 5 ns.
(a) Repeat the simulation of the diode circuit in
Fig. 3.117(a) with the diode transit time changed
to τT = 50 ns. Does the storage time that you
observe change in proportion to the value of τT in
your simulation? Discuss. (b) Repeat the simulation
with the input voltage changed to the one in
Fig. P3.117(b), in which it is assumed that v1 has
been at 1.5 V for a long time, and compare the
Problems 153
results to those obtained in (a). What is the reason
for the difference between the results in (a) and (b)?
R1 = 0.75 kΩ
v1 vD
v1
t
D1
iD
+
–
1.5 V
7.5 ns 15 ns
–1.5 V
(a)
v1
t
1.5 V
7.5 ns 15 ns
–1.5 V
(b)
Figure 3.117
3.18 Photo Diodes, Solar Cells, and LEDs
?3.118. The output of a diode used as a solar cell is
given by
IC = 1 ? 10?15[exp(40VC) ? 1] amperes
What operating point corresponds to Pmax? What is
Pmax? What are the values of ISC and VOC?
?3.119. Three diodes are connected in series to increase the
output voltage of a solar cell. The individual outputs
of the three diodes are given by
IC1 = 1.05 ? 10?15[exp(40VC1) ? 1] A
IC2 = 1.00 ? 10?15[exp(40VC2) ? 1] A
IC3 = 0.95 ? 10?15[exp(40VC3) ? 1] A
(a) What are the values of ISC and VOC for the series
connected cell? (b) What is the value of Pmax?
3.120. Write an expression for the total photo current
iPH for a diode having dc plus signal current components.
??3.121. The bandgaps of silicon and gallium arsenide
are 1.12 eV and 1.42 eV, respectively. What
are the wavelengths of light that you would expect
to be emitted from these devices based
on direct recombination of holes and electrons?
To what “colors” of light do these wavelengths
correspond?
??3.122. Repeat Prob. 3.121 for Ge, GaN, InP, InAs, BN,
SiC, and CdSe.
154 Chapter 3 Solid-State Diodes and Diode Circuits
术语对照
Anode
Avalanche breakdown
Bias current and voltage
Breakdown region
Breakdown voltage
Built-in potential(or voltage)
Cathode
Center-tapped transformer
Conduction angle
Conduction interval
Constant voltage drop(CVD) mode
Cut-in voltage
Depletion layer
Depletion-layer capacitance
Depletion-layer width
Depletion region
Diffusion capacitance
Diode equation
Diode SPICE parameters(IS, RS,N,TT,CJO, VJ,M）
Filter capacitor
Forward bias
Full-wave bridge rectifier circuit
Full-wave rectifier circuit
Half-wave rectifier circuit
Ideal diode
Ideal diode model
impact-ionization process
Junction capacitance
Junction potential
Light-emitting diode(LED)
Line regulation
Load line
阳极
雪崩击穿
偏置电流和电压
故障区域
击穿电压
内置电位（或电压）
阴极
中心抽头变压器
导通角
传导时间间隔
恒压降（CVD）模型
接通电压
耗尽层
耗尽层电容
耗尽层宽度
耗尽区
扩散电容
二极管方程
二极管SPICE参数
滤波电容器
正向偏压
电桥式整流电路
全波整流电路
半波整流电路
理想二极管
理想二极管模型
碰撞电离过程
结电容
结面电位
发光二极管（LED）
电源调整率
载重线
术语对照155
载重线分析
负载调整率
数学模型
冶金结
非理想性因素（n）
峰值检波器
峰值反向电压（PIV）
光电探测器电路
分段线性模型
pn结二极管
与绝对温度（PTAT）成正比
Q点
静态操作点
整流电路
反向偏压
反向击穿
反向饱和电流（IS）
波纹电流
波纹电压
饱和电流
Load-line analysis
Load regulation
Mathematical model
Metallurgical junction
Nonideality factor(n)
Peak detector
Peak inverse voltage(PIV)
Photodetector circuit
Piecewise linear model
pn junction diode
proportional to absolute temperature(PTAT)
Q-point
Quiescent operating point
Rectifier circuits
Reverse bias
Reverse breakdown
Reverse saturation current(IS)
Ripple current
Ripple voltage
Saturation current
Schottky barrier diode
Solar cell
Space charge region(SCR)
storage time
Surge current
Thermal voltage(VT)
transit time
Turn-on voltage
Voltage regulator
Voltage transfer characteristic(VTC)
Zener breakdown
Zener diode
Zero bias
Zero-bias junction capacitance
肖特基势垒二极管
太阳能电池
空间电荷区域（SCR）
存储时间
冲击电流
热电压（VT）
通过时间
开启电压
稳压器
电压传递特性（VTC）
齐纳击穿
齐纳二极管
零偏差
零偏压结电容