The document discusses electronic devices and circuits focused on the classification of components into active, passive, and electromechanical types. It covers the functioning of pn junction diodes, the principles of doping in semiconductors, and the operation of various rectifier circuits including half wave and full wave rectifiers. Additionally, it includes concepts related to integrated circuits and light-emitting diodes (LEDs), illustrating their characteristics and applications.

1. EC6202 ELECTRONIC DEVICES AND CIRCUITS
Unit 1
Dr Gnanasekaran Thangavel
Professor and Head
Electronics and Instrumentation
Engineering
R M K Engineering College

2. Electronic components
7/19/2017Dr Gnanasekaran Thangavel2

3. Classification
7/19/2017Dr Gnanasekaran Thangavel3
Active components
Rely on a source of energy and can inject power into a circuit
Passive components
Can't introduce net energy into the circuit and can't rely on a
source of power
Electromechanical
can carry out electrical operations by using moving parts or by
using electrical connections

4. Active components
7/19/2017Dr Gnanasekaran Thangavel4
Semiconductors
 Diodes
 Transistors
 Integrated circuits
 Optoelectronic devices
Display technologies
Vacuum tubes (valves)
Discharge devices
Power sources

5. Passive components
7/19/2017Dr Gnanasekaran Thangavel5
Resistors
Capacitors
 Magnetic (inductive) devices
 Memristor
 Networks
 Transducers, sensors, detectors
Antennas
Assemblies, modules

6. Electromechanical
7/19/2017Dr Gnanasekaran Thangavel6
Passive components that use piezoelectric effect:
Components that use the effect to generate or filter high frequencies
 Crystal – a ceramic crystal used to generate precise frequencies (See
the Modules class below for complete oscillators)
 Ceramic resonator – Is a ceramic crystal used to generate semi-
precise frequencies
 Ceramic filter – Is a ceramic crystal used to filter a band of
frequencies such as in radio receivers
 Surface acoustic wave (SAW) filters
Components that use the effect as mechanical transducers.
Ultrasonic motor – Electric motor that uses the piezoelectric effects
For piezo buzzers and microphones,

7. UNIT I PN JUNCTION DEVICES
PN junction diode –structure, operation and V-I
characteristics, diffusion and transient capacitance -
Rectifiers – Half Wave and Full Wave Rectifier,– Display
devices- LED, Laser diodes- Zener diode characteristics-
Zener Reverse characteristics – Zener as regulator
7 Dr Gnanasekaran Thangavel 7/19/2017
1. https://www.youtube.com/watch?v=OyC02DWq3mI
2. https://www.youtube.com/watch?v=d4zO39K_ce8
3. https://www.youtube.com/watch?v=AspBbh_jOuk
4. https://www.youtube.com/watch?v=UMgOG4OqBT0
5. https://www.youtube.com/watch?v=Kl8IOESVWlM

8. PN junction diode
Definition
“A semiconductor device with two terminals, typically
allowing the flow of current in one direction only.
“A diode is a specialized electronic component with two
electrodes called the anode and the cathode. They are
made with semiconductor materials such as silicon,
germanium, or selenium. The fundamental property of a
diode is its tendency to conduct electric current in only
one direction.”
“A Diode is an electronic device that allows current to flow
in one direction only. It is a semiconductor that consists of
a p-n junction. They are used most commonly to convert8 Dr Gnanasekaran Thangavel 7/19/2017

9. Semiconductors and Physical Operation of
Diodes
 Semiconductors
 Doping
• n-type material
• p-type material
 pn-Junctions
• forward, reverse, breakdown
• solar cells, LEDs, capacitance

10. Periodic Table of Elements
Relevant Columns: III IV V

11. The Silicon Atom
Nucleus:
14 protons
14 neutrons
10 core electrons:
1s22s22p6
-
-
-
-
4 valence
electrons
The 4 valence electrons are responsible
for forming covalent bonds

12. Silicon Crystal
Each Si atom has four nearest neighbors — one for each valence electron
0.5 nm

13. Two-dimensional Picture of Si
note: each line ( —) represents a valence electron
covalent bond
At T=0 Kelvin, all of
the valence electrons
are participating in
covalent bonds
There are no “free”
electrons, therefore no
current can flow in the
silicon  INSULATOR
Si

14. Silicon at Room Temperature
For T>0 K, the silicon atoms
vibrate in the lattice. This is
what we humans sense as
“heat.”
Occasionally, the vibrations
cause a covalent bond to break
and a valence electron is free
to move about the silicon.

15. Silicon at Room Temperature
-
-
For T>0 K, the silicon atoms
vibrate in the lattice. This is
what we humans sense as
“heat.”
Occasionally, the vibrations
cause a covalent bond to break
and a valence electron is free
to move about the silicon.
= free electron

16. Silicon at Room Temperature
The broken covalent bond site
is now missing an electron.
This is called a “hole”
The hole is a missing negative
charge and has a charge of +1.
= a hole
-
+
hole

17. Current Flow in Silicon
*
+ -
+-
a bar of silicon
I
V
Bond breaking
due to:
-heat (phonons)
-light (photons)
Conductance is
proportional to
the number of
electrons and
holes:
Si resistance
depends on temp.
and light

18. Some important facts
 The number of electrons = the number of holes
 that is, n = p in pure silicon
 this is called intrinsic material
 High temp more electrons/holes  lower resistance
 Very few electrons/holes at room temperature
 n=1.5x1010 per cm3, but nSi = 5x1022 per cm3
 n/nSi = 3x10-13 (less than 1 in a trillion Si bonds are broken
 This is a SEMICONDUCTOR

19. Important Facts (cont.)
 Band Gap: energy required to break a covalent bond and free
an electron
 Eg = 0.66 eV (germanium)
 Eg = 1.12 eV (silicon)
 Eg = 3.36 eV (gallium nitride)
 Metals have Eg= 0
 very large number of free electrons  high conductance
 Insulators have Eg > 5 eV
 almost NO free electrons  zero conductance

20. Doping
 Intentionally adding impurities to a semiconductor to create
more free electrons OR more holes (extrinsic material)
 n-type material
 more electrons than holes (n>p)
 p-type material
 more holes than electrons (p>n)
 HOW???

21. Periodic Table of Elements
Relevant Columns: III IV V

22. n-type silicon
add atoms from column V of the periodic table
Si
P
-
Column V elements have 5 valence
electrons
Four of the electrons form covalent bonds
with Si, but the 5th electron is unpaired.
Because the 5th electron is weakly bound,
it almost always breaks away from the P
atom
This is now a free electron.

23. VERY IMPORTANT POINT
Si
P+
-
The phosphorus atom has donated an
electron to the semiconductor (Column V
atoms are called donors)
The phosphorus is missing one of its
electrons, so it has a positive charge (+1)
The phosphorus ion is bound to the
silicon, so this +1 charge can’t move!
The number of electrons is equal to
the number of phos. atoms: n = Nd

24. Periodic Table of Elements
Relevant Columns: III IV V

25. p-type silicon
add atoms from column III of the periodic table
Si
B
Column III elements have 3 valence electrons
that form covalent bonds with Si, but the 4th
electron is needed.
This 4th electron is taken from the nearby
Si=Si bond

26. p-type silicon
add atoms from column III of the periodic table
Si
B
Column III elements have 3 valence electrons
that form covalent bonds with Si, but the 4th
electron is needed.
This 4th electron is taken from the nearby
Si=Si bond
This “stolen” electron creates a free hole.
hole

27. VERY IMPORTANT POINT
Si
B-
+
The boron atom has accepted an electron
from the semiconductor (Column III
atoms are called acceptors)
The boron has one extra electron, so it
has a negative charge (-1)
The boron ion is bound to the silicon, so
this -1 charge can’t move!
The number of holes is equal to
the number of boron atoms: p = Na

28. The pn Junction
p-type n-type
anode cathode
integrated circuit diode
metal
silicon oxide
doped silicon
wafer (chip)

29. Dopant distribution inside a
pn junction
p>>n n>>p
excess electrons diffuse
to the p-type region
excess holes diffuse
to the n-type region

30. n~0, and donor ions are
exposed
Dopant distribution inside a
pn junction
excess electrons diffuse
to the p-type region
excess holes diffuse
to the n-type region
DEPLETION REGION:
+
p~0, and acceptor ions are
exposed
p>>n n>>p
+
+
+-
-
-
-

31. Voltage in a pn junction
p>>n n>>p
+
+
+-
-
-
-
x
charge, r(x)
x
x
electric field,
E(x)
voltage,
V(x)
+
~0.7 volts
(for Si)

x
dxxxE
0
)(
1
)( r


x
dxxExV
0
)()(

32. Zero Bias
p>>n n>>p
+
+
+-
-
-
x
voltage,
V(x)
~0.7 volts
(for Si)
At zero bias (vD=0), very few electrons or holes can overcome this built-in
voltage barrier of ~ 0.7 volts (and exactly balanced by diffusion)
 iD = 0

33. Forward Bias
p>>n n>>p
+
+
+-
-
-
x
voltage,
V(x)
0.65 volts
As the bias (vD), increases toward 0.7V, more electrons and holes can
overcome the built-in voltage barrier . iD > 0
vD
0.50 volts
0.0 volts

34. Reverse Bias
p>>n n>>p
+
+
+-
-
-
x
voltage,
V(x)
-5 volts
As the bias (vD) becomes negative, the barrier becomes larger. Only
electrons and holes due to broken bonds contribute to the diode
current.  iD = -Is
vD
0.0 volts
1/2Is
1/2Is
Is

35. Breakdown
p>>n n>>p
+
+
+-
-
-
x
voltage,
V(x)
-50 volts
As the bias (vD) becomes very negative, the barrier becomes larger.
Free electrons and holes due to broken bonds are accelerated to
high energy (>Eg) and break other covalent bonds – generating
more electrons and holes (avalanche).
vD
0.0 volts
|I| >> Is
large reverse current

36. Solar Cell (Photovoltaic)
p>>n n>>p
+
+
+-
-
-
x
voltage,
V(x)
~0.7 volts
(for Si)
Light hitting the depletion region causes a covalent bond to break.
The free electron and hole are pushed out of the depletion region
by the built-in potential (0.7v).
Rload
light
Iph

37. Light Emitting Diode (LED)
7/19/2017Dr Gnanasekaran Thangavel37
 A light-emitting diode (LED) is a two-lead semiconductor light
source. It is a p–n junction diode that emits light when activated.
When a suitable voltage is applied to the leads, electrons are able to
recombine with electron holes within the device, releasing energy in
the form of photons. This effect is called electroluminescence, and
the colour of the light (corresponding to the energy of the photon) is
determined by the energy band gap of the semiconductor.

38. Light Emitting Diode (LED)
p>>n n>>p
+
+
+-
-
-
x
voltage,
V(x)
2.0 volts
In forward bias, an electron and hole collide and self-annihilate in the
depletion region. A photon with the gap energy is emitted. Only occurs
in some materials (not silicon).
vD
1.5 volts
0.0 volts
photon

39. Junction Capacitance
p>>n n>>p
+
+
+-
-
-
W
n=p~0
=11.9
semiconductor-”insulator”-semiconductor
The parasitic (unwanted) junction capacitance is
Cj = eA/W, where W depends on the bias voltage
A

40. Junction Capacitance (Cj)
 The junction capacitance must be charged and discharged
every time the diode is turned on and off
 Transistors are made of pn junctions. The capacitance due
to these junctions limits the high frequency performance of
transistors remember, Zc = 1/jwC becomes a short circuit
at high frequencies (Zc  0) this means that a pn junction
looks like a short at high frequency
 This is a fundamental principle that limits the performance
of all electronic devices

41. HALF WAVE RECTIFIER
 The Half wave rectifier is a circuit, which
converts an ac voltage to dc voltage.
The primary of the transformer is
connected to ac supply. This induces an
ac voltage across the secondary of the
transformer.
 During the positive half cycle of the input
voltage the polarity of the voltage across
the secondary forward biases the diode.
As a result a current IL flows through the
load resistor, RL. The forward biased
diode offers a very low resistance and
hence the voltage drop across it is very
small. Thus the voltage appearing across
the load is practically the same as the

42. HALF WAVE RECTIFIER …….
7/19/2017Dr Gnanasekaran Thangavel42
 During the negative half cycle of the input voltage the polarity of
the secondary voltage gets reversed. As a result, the diode is
reverse biased.
 Practically no current flows through the circuit and almost no
voltage is developed across the resistor. All input voltage
appears across the diode itself.
 Hence we conclude that when the input voltage is going through
its positive half cycle, output voltage is almost the same as the
input voltage and during the negative half cycle no voltage is
available across the load. This explains the unidirectional
pulsating dc waveform obtained as output. The process of
removing one half the input signal to establish a dc level is aptly

43. The output waveform
7/19/2017Dr Gnanasekaran Thangavel43

44. FULL WAVE RECTIFIER
• A Full Wave Rectifier is a circuit, which converts an ac voltage into
a pulsating dc voltage using both half cycles of the applied ac
voltage. It uses two diodes of which one conducts during one half
cycle while the other conducts during the other half cycle of the
applied ac voltage.

45. The output waveform
7/19/2017Dr Gnanasekaran Thangavel45

46.  Positive cycle, D2 off, D1 conducts;
Vo – Vs + V = 0
Vo = Vs - V
Full-Wave Rectification – circuit with center-
tapped transformer
 Since a rectified output voltage occurs during both positive and
negative cycles of the input signal, this circuit is called a full-
wave rectifier.
 Also notice that the polarity of the output voltage for both
cycles is the same
 Negative cycle, D1 off, D2 conducts;
Vo – Vs + V = 0
Vo = Vs - V

47. Vs = Vpsin t
V
-V
Notice again that the peak voltage of Vo is lower since Vo = Vs -
V
Vp
• Vs < V, diode off, open circuit, no current flow,Vo = 0V

48.  Positive cycle, D1 and D2 conducts, D3 and D4 off;
+ V + Vo + V – Vs = 0
Vo = Vs - 2V
Full-Wave Rectification –Bridge Rectifier
 Negative cycle, D3 and D4 conducts, D1 and D2 off
+ V + Vo + V – Vs = 0
Vo = Vs - 2V
Also notice that the polarity of the output voltage for both cycles is the same

49.  A full-wave center-tapped rectifier circuit is shown in Fig. 3.1. Assume that for each diode,
the cut-in voltage, V = 0.6V and the diode forward resistance, rf is 15. The load
resistor, R = 95 . Determine:
 peak output voltage, Vo across the load, R
 Sketch the output voltage, Vo and label its peak value.
25: 1
125 V (peak
voltage)
( sine wave )

50.  SOLUTION
 peak output voltage, Vo
Vs (peak) = 125 / 25 = 5V
V +ID(15) + ID (95) - Vs(peak) = 0 ID = (5 – 0.6) / 110
= 0.04 A Vo (peak) = 95 x 0.04 = 3.8V
3.8V
Vo
t

51. Duty Cycle: The fraction of the wave cycle over which the
diode is conducting.

52. EXAMPLE 3.1 – Half Wave Rectifier
Determine the currents and voltages of the half-wave rectifier circuit. Consider the half-wave rectifier circuit
shown in Figure.
Assume and . Also assume that
Determine the peak diode current, maximum reverse-bias diode voltage, the fraction of the wave cycle over
which the diode is conducting.
A simple half-wave battery charger circuit
-VR + VB + 18.6 = 0
VR = 24.6 V
- VR +
+
-

54. The peak inverse voltage (PIV) of the diode is the
peak value of the voltage that a diode can withstand
when it is reversed biased

55. Type of
Rectifier
PIV
Half Wave Peak value of the input secondary voltage, Vs (peak)
Full Wave :
Center-Tapped
2Vs(peak) - V
Full Wave: Bridge Vs(peak)- V

56. Example: Half Wave Rectifier
Given a half wave rectifier with input primary voltage, Vp = 80 sin t and the
transformer turns ratio, N1/N2 = 6. If the diode is ideal diode, (V = 0V), determine the
value of the peak inverse voltage.
1. Get the input of the secondary voltage:
80 / 6 = 13.33 V
1. PIV for half-wave = Peak value of the input voltage = 13.33 V

57. EXAMPLE 3.2
Calculate the transformer turns ratio and the PIV voltages for each type of the full wave rectifier
a) center-tapped
b) bridge
Assume the input voltage of the transformer is 220 V (rms), 50 Hz from ac main line source. The desired peak
output voltage is 9 volt; also assume diodes cut-in voltage = 0.6 V.

58. Solution: For the centre-tapped transformer circuit the peak voltage of the transformer secondary
is required
The peak output voltage = 9V
Output voltage, Vo = Vs - V
Hence, Vs = 9 + 0.6 = 9.6V
Peak value = Vrms x 2
So, Vs (rms) = 9.6 / 2 = 6.79 V
The turns ratio of the primary to each secondary winding is
The PIV of each diode: 2Vs(peak) - V = 2(9.6) - 0.6 = 19.6 - 0.6 = 18.6 V

59. Solution: For the bridge transformer circuit the peak voltage of the transformer secondary is
required
The peak output voltage = 9V
Output voltage, Vo = Vs - 2V
Hence, Vs = 9 + 1.2 = 10.2 V
Peak value = Vrms x 2
So, Vs (rms) = 10.2 / 2 = 7.21 V
The turns ratio of the primary to each secondary winding is
The PIV of each diode: Vs(peak)- V = 10.2 - 0.6 = 9.6 V

60. Laser diodes
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 LASER — Light Amplification by Stimulated Emission of Radiation
 The Laser is a source of highly directional, monochromatic, coherent light.
 The Laser operates under a “stimulated emission” process.
 The semiconductor laser differs from other lasers (solid, gas, and liquid lasers):
 small size (typical on the order of 0.1 × 0.1 × 0.3 mm3)
 high efficiency
 the laser output is easily modulated at high frequency by controlling the junction current
 low or medium power (as compared with ruby or CO2 laser, but is comparable to the He-Ne
laser)
 particularly suitable for fiber optic communication
 Important applications of the semiconductor lasers:
 optical-fiber communication, video recording, optical reading, high-speed laser printing.
high-resolution gas spectroscopy, atmospheric pollution monitoring.

61. 61
Comparison between an LD and LED
 Laser Diode
 Stimulated radiation
 narrow line width
 coherent
 higher output power
 a threshold device
 strong temperature dependence
 higher coupling efficiency to a fiber
 LED
 Spontaneous radiation
 broad spectral
 incoherent
 lower output power
 no threshold current
 weak temperature dependence
 lower coupling efficiency

62. Laser Diode Construction
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 The figure shows a simplified
construction of a laser diode, which is
similar to a light emitting diode (LED).
 It uses gallium arsenide doped with
elements such as selenium, aluminum,
or silicon to produce P type and N type
semiconductor materials.
 While a laser diode has an additional
active layer of undoped (intrinsic)
gallium arsenide have the thickness
only a few nanometers, sandwiched
between the P and N layers, effectively
creating a PIN diode (P type-Intrinsic-N
type). It is in this layer that the laser
light is produced.

63. How Laser Diode Work?
7/19/2017Dr Gnanasekaran Thangavel63
 Every atom according to the quantum
theory, can energies only within a certain
discrete energy level. Normally, the atoms
are in the lowest energy state or ground
state.
 When an energy source given to the atoms
in the ground state can be excited to go to
one of the higher levels. This process is
called absorption.
 After staying at that level for a very short
duration, the atom returns to its initial
ground state, emitting a photon in the
process, This process is called
spontaneous emission.
 These two processes, absorption and

64. How Laser Diode Work?
7/19/2017Dr Gnanasekaran Thangavel64
 In case the atom, still in an excited state, is struck by an outside photon having
precisely the energy necessary for spontaneous emission, the outside photon is
increased by the one given up by the excited atom, Moreover, both the photons
are released from the same excited state in the same phase, This process, called
stimulated emission, is fundamental for laser action (shown in above figure).
 In this process, the key is the photon having exactly the same wavelength as that
of the light to be emitted.
Amplification and Population Inversion
 When favorable conditions are created for the stimulated emission, more and more
atoms are forced to emit photons thereby initiating a chain reaction and releasing
an enormous amount of energy.
 This results in a rapid build up of energy of emitting one particular wavelength
(monochromatic light), travelling coherently in a particular, fixed direction. This
process is called amplification by stimulated emission.

65. Laser Diode
 Laser diode is an improved LED, in the sense that uses stimulated emission in semiconductor from
optical transitions between distribution energy states of the valence and conduction bands with
optical resonator structure such as Fabry-Perot resonator with both optical and carrier
confinements.

66. Laser Diode Characteristics
 Nanosecond & even picoseconds response time (GHz BW)
 Spectral width of the order of nm or less
 High output power (tens of mW)
 Narrow beam (good coupling to single mode fibers)
 Laser diodes have three distinct radiation modes namely,
longitudinal, lateral and transverse modes.
 In laser diodes, end mirrors provide strong optical feedback in
longitudinal direction, so by roughening the edges and cleaving
the facets, the radiation can be achieved in longitudinal
direction rather than lateral direction.

67. Zener Diode
A Zener diode is a type of diode that permits current not only in
the forward direction like a normal diode, but also in the reverse
direction if the voltage is larger than the breakdown voltage
known as "Zener knee voltage" or "Zener voltage".

68. Zener Diode - Voltage Regulator (reverse
biased)
7/19/2017Dr Gnanasekaran Thangavel68

69. References
1. David A. Bell ,”Electronic Devices and Circuits”, Prentice Hall of India,.
2. staff.iium.edu.my/.../L6%20and%20L7%20full%20wave%20rectifier,%20PIV.ppt
3. www.ece.neu.edu/.../eceg201/.../Semiconductors_and_Physical_Operation_of_Diodes...
4. http://www.electronics-tutorials.ws/diode/diode_3.html.
5. http://www.electronicsandyou.com/electronics-basics/diode.html
6. https://en.wikipedia.org/wiki/Light-emitting_diode
7. http://www.circuitstoday.com/half-wave-rectifiers
8. http://www.visionics.a.se/html/curriculum/Experiments/HW%20Rectifier/Half%20Wave%20Rectifi
er1.html
9. eshare.stust.edu.tw/EshareFile/2010_5/2010_5_4fc2dc4c.ppt
10. https://ece.uwaterloo.ca/~ece477/Lectures/ece477_4_0.ppt
11. https://www.elprocus.com/laser-diode-construction-working-applications/
12. www.ohio.edu/people/starzykj/network/Class/.../Lecture5%20Diode%20Circuits.ppt
69 Dr Gnanasekaran Thangavel 7/19/2017

70. Other presentations
http://www.slideshare.net/drgst/presentations
70 Dr Gnanasekaran Thangavel 7/19/2017

71. 71
Thank You
Questions and Comments?
Dr Gnanasekaran Thangavel 7/19/2017