day1 intro diodeapplication

1. Structure of the Lesson

Intro

Class

Class end

Study

Assessment

Review

1. intro – Overview of the lesson2. Learning objective – present learning objective of the lesson3. 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 ContentAt the end of this lecture, you

should 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


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 PercentageHomework 30%Midterm exam 30%Final exam 40%Total 100%

Text book: Robert Boylestad, Louis Nashelsky, Electronic Devices and

Circuit Theory, 2002Reference 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

http://www.discovercircuits.com/list.htm

http://www.epanorama.net/links/basics.html

http://www.datasheetcatalog.com/


Content

Diode overview

It is a 2-terminal device

9

Diode overviewIdeally it conducts current in only one direction

and acts like an open in the opposite direction

1010

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.

1111

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.

1212

Actual Diode Characteristics

Note the regions for No Bias, Reverse Bias, and Forward Bias conditions.

1313

Practical Diode

Narrow temperature range (lower than 1000C)

Wider temperature range (up to 2000C)

Lower current ratingHigher current rating

Lower PIV ( 400V)Higher *PIV ( 1000V)

Lower forward-bias voltage (0.3V)

Higher forward-bias voltage (0.7V)

GermaniumSilicon

* PIV = peak inverse voltage

1414

Comparison of Si and Ge diodes

1515

Diode specification sheets

1616

Diode examples

1717

Diode examples

1818


Content

• 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 regionVi>0 => D1 on, D2 off

Negative regionVi<0 => D1 off, D2 on

Full-wave rectifier Bridge rectifier

Circuit and input


Positive half: Vi>0 => D2, D4 on; D1, D3 off


Negative half: Vi<0 => D2, D4 off; D1, D3 on


Waveform of Full wave Ideal diode: Vdc = 0.636Vm

Silicon diode: Vdc = 0.636(Vm - 2VT)

Full-wave rectifierRectifier with capacitor


Content

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

Solution

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 t0 T/2

Vo16

3.3T

Output waveform

45

Clipper - summary

46

Clipper - summary


Content

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

• Network

ClamperOperation:• 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

55

Clamper


Content

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


Content

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

day1 intro diodeapplication

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