Transcript Electrical Engineering

Electrical Engineering
Mao-Hsu Yen
[email protected]
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Introduction
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Expected Context
• Chapter 8 Operational Amplifiers
• Chapter 9 Semiconductors and Diodes
• Chapter 11 Field effect Transistors:
Operation, Circuit, Models, and
Applications
• Chapter 12 Digital Logic Circuit
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Chapter 9
Semiconductors
and Diodes
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Context
• 9.1 Electrical Conduction in Semiconductor
Devices
• 9.2 The pn Jnuction and Semiconductor
Diode
• 9.3 Circuit Models or the Semiconductor
Diode
• 9.4 Rectifier Circuits
• 9.5 DC Power Supplies, Zener Diodes, and
Voltage Regulation
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Electrical Conduction in
Semiconductor Devices
• Elemental or intrinsic semiconductors are
materials consisting of elements from group IV of
the periodic table and having electrical properties
falling somewhere between those of conducting
and of insulating materials.
• A conducting material is characterized by a large
number of conduction band electrons, which have
a very weak bond with the basic structure of the
material.
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• In a semiconductor one needs to
consider the lattice structure of the
material.
• Free electrons enable current flow in
the semiconductor.
• The number of charge carriers depends
on the amount of thermal energy
present in the structure. Many
semiconductor properties are a function
of temperature.
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Current flow in a semiconductor
Free electrons and “holes” in the lattice
structure
• An additional phenomenon, called
recombination, reduces the number of
charge carriers in a semiconductor.
• The number of free electrons available
for a given material is called the
intrinsic concentration ni. For
example, at room temperature, silicon
has:
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• To control the number of charge
carriers in a semiconductor, the process
of doping is usually employed.
• Doping consists of adding impurities to
the crystalline structure of the
semiconductor. If the dopant is an
element from the fifth column of the
periodic table, the end result is that
wherever an impurity is present, an
additional free electron is available for
conducting.
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• The elements providing the impurities
are called donors in the case of group
V elements, since they “donate” an
additional free electron to the lattice
structure.
• An equivalent situation arises when
group III elements are used to dope
silicon.
• An additional hole is created by the
doping element which is called an
acceptor, since it accepts a free electron
from the structure and generates a hole
in doing so.
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• Semiconductors doped with donor elements
conduct current predominantly by means of
free electrons and are therefore called n-type
semiconductors.
• When an acceptor element is used as the
dopant, holes constitute the most common
carrier, the resulting semiconductor is said to
be a p-type semiconductor.
Doped semiconductor
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• If n is the total number of free electrons
and p that of holes, then in an n-type
doped semiconductor, we have:
• Free electrons are the majority
carriers in an n-type material, while
holes are the minority carriers. In a ptype material, the majority and
minority carriers are reversed.
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The pn Junction and The
Semiconductor Diode
• When a section of p-type material and a
section of n-type material are brought in
contact to form a pn junction, a number of
interesting properties arise. The pn junction
forms the basis of the semiconductor diode,
a widely used circuit element.
A pn junction
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Drift and diffusion currents in a pn
junction
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Forward-and reverse-biased pn
junctions
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Equations
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Semiconductor diode i-v
characteristic
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Semiconductor diode circuit
symbol
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The i-v characteristic of the
semiconductor diode
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Large-signal on/off diode
model
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Ideal Diode Model
Circuit of Figure 9.12,
containing ideal diode
Circuit of Figure 9.13,
assuming that the ideal diode
conducts
Circuit of Figure 9.14,
assuming that the ideal
diode does not conduct
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FOCUS ON METHODOLOGY
DETERMING THE CONDUCTION
STATE OF AN IDEAL DIODE
1. Assume a diode conduction state (on or off).
2. Substitute ideal circuit model into circuit (short circuit if
“on,” open circuit if “off”).
3. Solve for diode current and voltage, using linear circuit
analysis techniques.
4. If the solution is consistent with the assumption, then the
initial assumption was correct; if not, the diode conduction
state is opposite to that initially assumed. For example, if
the diode conduction state is opposite to that initially
assumed. For example, if the diode has been assumed to
be “off” but the diode voltage computed after replacing
the diode with an open circuit is a forward bias, then it
must be true that the actual state of the diode is “on.”
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EXAMPLE 9.1 Determining the
Conduction State of an Ideal
Diode
• Determine whether the ideal diode of figure
9.15 is conducting.
Figure 9.15
Figure 9.16
Figure 9.17
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EXAMPLE 9.2 Determining the
Conduction State of an Ideal Diode
Figure 9.18
Figure 9.19
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CHECK YOUR
UNDERSTADING
• Repeat the analysis of example 9.2
assuming that the diode is conducting, and
show that this assumption leads to
inconsistent results.
• Determine which of the diodes conduct in
the circuit shown in the figure, for each of
the following voltages. Treat the diodes as
ideal.
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CHECK YOUR
UNDERSTADING
Page 418
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Rectification
Half-wave rectifier
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Half-wave rectifier
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Offset Diode Model
Offset diode as an extension of
ideal diode model
Offset diode model
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EXAMPLE 9.3 Using the Offset
Diode Model in a Half-Wave
Rectifier
• Compute and plot the rectified load voltage
νR in the circuit of figure 9.24.
Figure 9.24
Figure 9.25
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EXAMPLE 9.3 Using the Offset
Diode Model in a Half-Wave
Rectifier
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CHECK YOUR
UNDERSTANDING
• Compute the DC value of the rectified
waveform for the circuit of figure 9.20 for
νi = 52 cos ωt V.
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EXAMPLE 9.4 Using the Offset
Diode Model Problem
• Using the offset diode model to determine
the value of ν1 for which diode D1 first
conducts in the circuit of figure 9.27.
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CHECK YOUR
UNDERSTANDING
• Determine which of the diode conduct in
the circuit in the circuit shown below. Each
diode has an offset voltage of 0.6 V.
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Small-Signal Diode Models
Diode circuit for illustration of load-line analysis
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Graphical solution of equations
9.13 and 9.14
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FOCUS ON METHODOLOGY
DETERMING THE OPERATING
POINT OF A DIODE
• Reduce the circuit to a Thévenin or Norton
equivalent circuit with the diode as the load.
• Write the load-line equation.
• Solve numerically two simultaneous equations in
two unknowns (the load-line equations and the
diode equation) for the diode current and voltage.
or
• Solve graphically by finding the intersection of
the diode curve (e.g., from a data sheet) with the
load-line curve. The intersection of the two curves
is the diode operating point.
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EXAMPLE 9.5 Using Load-Line
Analysis and Diode Curves to
Determine the Operating Point of a
Diode
• Determine the operating point of the 1N914
diode in the circuit of Figure 9.31, and
compute the total power output of the 12-V
battery.
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CHECK YOUR
UNDERSTANDING
• Use load –line analysis to determine the
operating point (Q point) of the diode in the
circuit shown in figure. The diode has the
characteristic curve of figure 9.32.
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Piecewise Linear Diode Model
Piecewise linear diode model
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EXAMPLE 9.6 Computing the
Incremental (Small-Signal)
Resistance of a Diode
• Determine the incremental resistance of a
diode, using the diode equation.
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CHECK YOUR
UNDERSTANDING
• Computer the incremental resistance of the
diode of Example 9.6 if the current through
the diode is 250mA.
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EXAMPLE 9.7 Using the
Piecewise Linear Diode Model
• Determine the load voltage in the rectifier
of figure9.36, using a piecewise linear
approximation.
Figure 9.36
Figure 9.37
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CHECK YOUR
UNDERSTANDING
• Consider a half-wave rectifier similar to
that of Figure 9.20, with νi =18 cost V, and
a 4-Ώ load resistor. Sketch the output
waveform if the piecewise linear diode
modle is used to represent the diode, with
Vγ= 0.6V and γD=1Ώ.what is the peak
value of the rectifier output waveform?
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The Full-Wave Rectifier
Full-wave rectifier
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Full-wave rectifier current and
voltage waveforms (RL = 1 Ω)
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The Bridge Rectifier
Operation of bridge rectifier
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(a) Unrectified source voltage; (b) rectified load
voltage (ideal diodes); (c) rectified load voltage
(ideal and offset diodes)
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Bridge rectifier with filter circuit
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EXAMPLE 9.8 Half-Wave
Rectifiers
• A half-wave rectifier, similar to that in
figure 9.25, is used to provide a DC supply
to a 50-Ώ load. If the AC source voltage is
20V (rms), find the peak and average
current in the load.Auusme an ideal diode.
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CHECK YOUR
UNDERSTANDING
• What is the peak current if an offset diode
model is used with offset voltage equal to
0.6V ?
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EXAMPLE 9.9 Bridge Rectifier
Figure 9.41
• A bridge rectifier, similar to that
in figure9.41, is used to provide
a 50-V, 5-A DC supply. What is
the resistance of the load that
will draw exactly 5A? What is
the required rms source voltage
to achieve the desired DC
voltage? Assume an ideal diode.
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CHECK YOUR
UNDERSTANDING
• Show that the DC output voltage of the
full-wave rectifier of figure 9.39 is
2Nνspeak /π. Compute the peak voltage
output of the bridge rectifier of figure9.40,
assuming diode with 0.6-V offset voltage
and a 110-V rms AC supply.
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DC Power Supplies, Zener Diode,
and Voltage Regulation
DC power supply
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EXAMPLE 9.10 Determining the
Power Rating of a Zener Diode
• We wish to design a regulator similar to the
one depicted in figure9.49(a). Determine
the minimum acceptable power rating of
the Zener diode.
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CHECK YOUR
UNDERSTANDING
• How would the power rating change if the
load were reduced to 100Ώ?
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EXAMPLE 9.11 Calculation of
Allowed Load Resistances for a
Given Zener Regulator
• Calculate the allowable range of load
resistances for the Zener regulator of figure
9.50 such that the diode power rating is not
exceeded.
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CHECK YOUR
UNDERSTANDING
• What should the power rating of the Zener
diode be to withstand operation with opencircuit load?
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EXAMPLE 9.12 Effect of Nonzero
Zener Resistance in a Regulator
• Calculate the amplitude of the ripple
present in the output voltage of the
regulator of figure9.51. The unregulated
supply voltage is depicted in figure 9.52.
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CHECK YOUR
UNDERSTANDING
• Compute the actual DC load voltage and
the percent of ripple reaching the load
(relative to the initial 100-mV ripple) for
the circuit of example 9.12 if γz = 1Ώ
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Homework Problem
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