01.pptx

1. Structure of the Lesson
Intro
Class
Class end
Study
Assessment
Review
1. intro – Overview of the lesson
2. Learning objective – present learning objective of the lesson
3. Table of Content – structure of the topics and subtopics in the
lesson
4. Lecture (75-90 minutes)
– present the lecture in detailed topics that covers all the
learning objectives of the lesson.
- each topics should be divided into subtopics
(5-15 min in length is recommended)
- if a subtopic goes over 15 minutes divide the subtopic into
series of subtopics.

Course
Circuit theory and
Laboratory
Lesson # Lesson 1
Title
Introduction-Diode
applications
SME Dr. Nguyen Vu Thang

Learning Objectives Table of Content
At the end of this lecture, you shou
ld be able to:
• Understand the configuration,
operation and measurement of
different applications of diode.
• The applications are: rectifier,
clippers, clampers, Zener diodes
and voltage multiplication
• Introduction
• Diode overview
• Rectifier
• Clipper
• Clamper
• Zener diode
• Voltage multiplication

• Introduction
• Diode overview
• Rectifier
• Clipper
• Clamper
• Zener diode
• Voltage multiplication
Content

 On completion of this course, the student will
understand
◦ Able to explain, describe, and use physics-based device and
circuit models for semiconductor devices
◦ Able to choose appropriate BJT and FET configuration
◦ Able to choose and calculate appropriate biasing
◦ Understand effect of source, load resistance; power,
frequency limitation
◦ Understand the advantages and method of analysis of
feedback
◦ Able to analyze and design electronic circuits
◦ Able to Identify the design issues, and develop solutions
Objectives

Grading
Activities Percentage
Homework 30%
Midterm exam 30%
Final exam 40%
Total 100%

 Text book:
◦ Robert Boylestad, Louis Nashelsky, Electronic Devices and
Circuit Theory, 2002
 Reference books:
◦ Richard C. Jaeger, Microelectronic Circuits Design, 2003
◦ Microelectronic Circuits; Fifth Edition by Sedra/Smith
◦ Microelectronic circuits, 5th edition, Behzad Razavi
 Websites:
◦ http://www.discovercircuits.com/list.htm
◦ http://www.epanorama.net/links/basics.html
◦ http://www.datasheetcatalog.com/
Reference books

Diode overview
It is a 2-terminal device
9

Diode overview
Ideally it conducts current in only one direction
and acts like an open in the opposite direction
10
10

Characteristics of an ideal diode:
Conduction Region
In the conduction region (the vertical blue line), ideally
• the voltage across the diode is 0V,
• the current is ,
• the forward resistance (RF) is defined as RF = VF/IF= 0
• the diode acts like a short.
11
11

Characteristics of an ideal diode:
Non-Conduction Region
In the non-conduction region (the horizontal blue line), ideally
• all of the voltage is across the diode,
• the current is 0A,
• the reverse resistance (RR) is defined as RR = VR/IR, = ∞
• the diode acts like open.
12
12

Actual Diode Characteristics
Note the regions for No Bias, Reverse Bias, and Forward
Bias conditions.
13
13

Practical Diode
Narrow temperature
range (lower than 1000C)
Wider temperature
range (up to 2000C)
Lower current rating
Higher current rating
Lower PIV ( 400V)
Higher *PIV ( 1000V)
Lower forward-bias
voltage (0.3V)
Higher forward-bias
voltage (0.7V)
Germanium
Silicon
* PIV = peak inverse voltage
14
14

Comparison of Si and Ge diodes
15
15

Diode specification sheets
16
16

• Types
–Half-wave
–Full-wave
–Full-wave bridge
–With capacitor
Rectifier circuits

• Vi(t)>0 => D on
• Vi(t)<0 => D off
Half-wave rectifier

• Effect of VT
Half-wave rectifier

• Example:
– (a) Sketch the output vo and determine the dc level of the
output for the network of figure above
– (b) Repeat part (a) if the ideal diode is replaced by a silicon
diode.
– (c) Repeat parts (a) and (b) if Vm is increased to 200 V and
compare solutions
Half-wave rectifier

• A. For ideal diode:
– Vdc = -0.318Vm = -0.318(20 V) = -6.36 V
• B. For Si diode:
– Vdc = -0.318(Vm - 0.7 V) = -0.318(19.3 V) = -6.14 V
• C. for ideal diode:
– Vdc =-0.318Vm = -0.318(200 V) = -63.6 V
• For Si diode:
– Vdc =-0.318(Vm -VT) = -0.318(200 V-0.7 V) = -63.38 V
Half-wave rectifier

• Center-taped transformer
• Bridge network
Full-wave rectifier

Full-wave rectifier
center-taped transformer
Circuit and input
Positive region
Vi>0 => D1 on, D2 off
Negative region
Vi<0 => D1 off, D2 on

Full-wave rectifier
Bridge rectifier
Circuit and input

Full-wave rectifier
Bridge rectifier
 Positive half:
 Vi>0 => D2, D4 on; D1, D3 off

Full-wave rectifier
Bridge rectifier
 Negative half:
 Vi<0 => D2, D4 off; D1, D3 on

Full-wave rectifier
Bridge rectifier
 Waveform of Full wave
 Ideal diode: Vdc = 0.636Vm
 Silicon diode: Vdc = 0.636(Vm - 2VT)

Full-wave rectifier
Rectifier with capacitor

Clippers
- Is a diode network that have the ability to “clip”
off a portion on the input signal without distorting
the remaining part of the alternating waveform.
- Used to eliminate amplitude noise or to fabricate
new waveforms from an existing signal.

Clipper
Series:
•The series configuration is
defined as one where the diode is
in series with the load.
Parallel:
•The series configuration is
defined as one where the diode is
parallel with the load.

Clipper
• Series:
– Vi>V => D on => Vo=Vi-V
– Vi<V => D off => Vo=0

Clipper
• Example: determine output waveform for the
network above

Clipper
Example:
• Repeat the previous example using for the
square wave input

Clipper
• Parallel network
• The diode connection is in parallel
configuration with the output.

Clipper
• Parallel:
– Vi>0 => D on => Vo=0
– Vi<0 => D off => Vo=Vi

Clipper
• Example: determine Vo sketch the output
waveform for the above network

Clipper
Positive region of vi
Negative region of vi
Solution

Clipper
• Example: repeat the previous example using a
silicon diode

Clipper
• Solution:
• Vi>3.3V => D on => Vo = Vi
• Vi<3.3V => D off => Vo = 3.3V t
0 T/2
Vo
16
3.3
T
Output waveform

Clamper
• The clamping network is to “clamp” a signal to a different dc level.
• Often used in TV receivers as a dc restorer.
• The network consists of:
– a) Capacitor
– b) Diode
– c) Resistive element
– d) Independent dc supply (option)
• The magnitude of R and C must be chosen such that the time
constant ζ = RC is large enough to ensure that the voltage across
the capacitor does not discharge significantly during the interval the
diode is nonconducting.
• Assume in our analysis that all capacitor is fully charge and
discharge in 5 time constant.

Clamper
Operation:
• 0 → T/2: D on
=> RC time constant is small because of
the resistance of the diode
=> capacitor charge to V volts quickly
=> Vo = 0 V
• T/2 → T: D off
=> RC time constant > 5T >> T/2
=> can assume capacitor keep all charges
and voltage during this period => Vo = -2V

51
Total swing output signal = the
total swing input signal
Clamper

52
Example: Determine vo for the network above for
the input indicated
Clamper

53
Solution:
•F=1000 Hz => interval between levels =
0.5 ms
•0 → t1: D off => Vo = 10 V
•t1 →t2: D on => network will appear as
shown in Fig. 2
Vc = V + Vi = 25 V
Vo = 5V
•t2 →t3: D off => network will appear as
shown in Fig. 3
Vo = Vc+Vi = 25 + 10 = 35 V
Clamper
Fig. 2
Fig. 3

54
Solution:
•Time Constance
ζ = RC = (100kΩ)(1 µF) = 0.1 s =100ms
•The total discharge time = 5ζ = 500 ms
=> the capacitor can hold the voltage during the interval
of 0.5 ms
Clamper

Zener diode
• The zener diode is a special type of diodes that is
designed to work in the reverse breakdown region.
• Can operate in the forward bias region.
• Application: always reverse bias
– Reference voltage for DC power supply

Zener diode simple application
• Example: Fixed DC voltage is applied in the network
above. Analyze the operation of the network.

Solution:
• Determine the state of the Zener diode by removing it from
the network:
V = VL = RLVi/(R+RL)
• If V > Vz => D on => Zener diode works as a DC source
• If V < Vz => D off => open circuit for Zener diode

Solution (continue):
• We have:
• IR=(Vin-Vz)/R;
• IL=Vz/RL;
• Pz=Iz*Vz<Pzmax

• Case 1: fixed Vin, variable RL
RLmax> RL >RLmin
RLmax=Vz/(IR-Izmax)
RLmin=RVz/(Vi-Vz)

• Case 2: fixed RL, variable Vin
RLmax> RL >RLmin
RLmax=Vz/(IR-Izmax)
RLmin=RVz/(Vi-Vz)

Example:
• Given the Zener diode network above
• a) Determine VL, VR, IZ, and PZ.
• b) Repeat with RL = 3kΩ

Solution: part a
• VZ = RLVi/(R+RL) = 8.73 V < 10 V
• => Zener diode is off
• VRL = 8.73 V
• IZ = 0 A
• PZ = 0 W

Solution (continue): part b
• VZ = RLVi/(R+RL) = 12 V > 10 V
• => Zener diode is on
• VRL = VZ = 10 V => VR = 6V
• IRL = 3.33 mA; IR = 6 mA; IZ = 2.67 mA
• PZ = IZVZ = 26.7 mW

Example:
• Given the network above
• a) Determine the range of RL and IL that will result in VRL
being maintained at 10 V.
• b) Determine the maximum wattage rating of the diode

Zener diode application
AC regulator

Zener diode application
Simple square generator

Voltage multiplier
• Voltage-multiplier circuits are employed to
maintain a relatively low transformer peak voltage
while stepping up the peak output voltage to two,
three, four, or more times the peak rectified
voltage

Double voltage
• Positive phase: D1 on, D2 off, VC1=Vm
• Nagative phase: D1 off, D2 on, VC2=Vm+VC1=2Vm

Multiple voltage
• Positive phase: D1 on, D2 off, VC1=Vm
• Negative phase: D1 off, D2 on, VC2=Vm+VC1=2Vm

• Rectifier Circuits
– Conversions of AC to DC for DC operated circuits
– Battery Charging Circuits
• Simple Diode Circuits
– Protective Circuits against Overcurrent
– Polarity Reversal Currents caused by an inductive kick
in a relay circuit
• Zener Circuits
– Overvoltage Protection
– Setting Reference Voltages
Real Diode applications

01.pptx

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