1. MATRUSRI ENGINEERING COLLEGE
DEPARTMENT OF INORMATION TECHNOLOGY
SUBJECT NAME: BASIC ELECTRONICS
FACULTY NAME: K SHRAVAN KUMAR
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2. BASIC ELECTRONICS
COURSE OBJECTIVES:
1.To understand the characteristics of diodes and transistor
configurations.
2.To understand the design concepts of biasing of BJT and FET.
3.To understand the design concepts of feedback amplifiers and
oscillators.
4. To study the design concepts of OP Amp and data converters
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3. BASIC ELECTRONICS
COURSE OUTCOMES:
On successful completion of this course, the students will be able to:
1.Study and analyse the rectifier and regulator circuits.
2.Study and analyse the performance of BJTs, FETs on the basis of their
operation and working.
3.Ability to analyse and design oscillator circuits.
4.Ability to analyse different logic gates and multivibrator circuits.
5.Ability to analyse different data acquisition systems.
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4. UNIT-I
OUTCOMES:
To understand the semiconductor materials that are
suitable for electronic devices
To study the properties of materials for electronic devices
To understand the biasing of diode
Able to design Diode circuit
Able to apply some types of diodes
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Characteristics of PN junction: half wave rectifier, full wave
rectifiers, filters, ripple, regulation, TUF and efficiency , Zener
diode and Zener diode regulators.CRT construction and CRO
applications..

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OUTCOMES:
To understand the biasing of diode
Able to design Diode circuit
Able to apply some types of diodes
MODULE-I
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CONTENTS:
Characteristics of PN junction: half wave
rectifier, full wave rectifiers
Filters, ripple, regulation, TUF and
efficiency

6. Boron and gallium each have only three outer electrons. When mixed into the
silicon lattice, they form "holes" in the lattice where a silicon electron has
nothing to bond to. The absence of an electron creates the effect of a positive
charge, hence the name P-type. Holes can conduct current. A hole happily
accepts an electron from a neighbor, moving the hole over a space. P-type
silicon is a good conductor.
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PN Junction formation
P-Type Doping

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Phosphorus and arsenic each have five outer electrons, so they're out of place
when they get into the silicon lattice. The fifth electron has nothing to bond
to, so it's free to move around. It takes only a very small quantity of the
impurity to create enough free electrons to allow an electric current to flow
through the silicon. N-type silicon is a good conductor. Electrons have a
negative charge, hence the name N-type.
PN Junction formation
N-Type Doping

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PN Junction
We create a p-n junction by joining together two pieces of semiconductor,
one doped n-type, the other p-type. In the n-type region there are extra
electrons and in the p-type region, there are holes from the acceptor
impurities .

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PN Junction
In the p-type region there are holes from the acceptor impurities and in the n-
type region there are extra electrons. When a p-n junction is formed, some of
the electrons from the n-region which have reached the conduction band are
free to diffuse across the junction and combine with holes. Filling a hole makes a
negative ion and leaves behind a positive ion on the n-side. A space charge
builds up, creating a depletion region.

10. This causes a depletion zone to form around the junction (the join)
between the two materials. This zone controls the behavior of the diode.
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PN Junction

11. Forward biasing the p-n junction drives holes to the junction from the p-
type material and electrons to the junction from the n-type material. At
the junction the electrons and holes combine so that a continuous current
can be maintained.
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Forward Biasing

12. The application of a reverse voltage to the p-n junction will cause a
transient current to flow as both electrons and holes are pulled away
from the junction. When the potential formed by the widened depletion
layer equals the applied voltage, the current will cease except for the
small thermal current.
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Reverse Biasing

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V-I Characteristics
When forward-biased, there is a small amount of voltage necessary to get the
diode going. In silicon, this voltage is about 0.7 volts. This voltage is needed to
start the hole-electron combination process at the junction. When reverse-biased,
an ideal diode would block all current. A real diode lets perhaps 10 microamps
through -- not a lot, but still not perfect. Usually, the breakdown voltage is a lot
more voltage than the circuit will ever see, so it is irrelevant.

14. where
IS = reverse saturation current (A)
vD = voltage applied to diode (V)
q = electronic charge (1.60 x 10-19 C)
k = Boltzmann’s constant (1.38 x 10-23 J/K)
T = absolute temperature
n = non ideality factor (dimensionless)
VT = kT/q = thermal voltage (V) (25 mV at room temp.)
IS is typically between 10-18 and 10-9 A, and is strongly temperature
dependent due to its dependence on ni
2. The non ideality factor is
typically close to 1, but approaches 2 for devices with high current
densities. It is assumed to be 1 in this text.
Diode current equation
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15. IS, the reverse bias saturation current for an ideal p-n diode is given as
where
• IS is the reverse bias saturation current,
• e is elementary charge
• A is the cross-sectional area
• Dp,n are the diffusion coefficients of holes and electrons, respectively,
• ND,A are the donor and acceptor concentrations at the n side and p
side,respectively,
• ni is the intrinsic carrier concentration in the semiconductor material,
• τp,n are the carrier lifetimes of holes and electrons, respectively.
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Diode current equation(contd…)

16. Diode Current for Reverse, Zero, and Forward Bias
• Reverse bias:
• Zero bias:
• Forward bias:

i
D

I
Sexp
v
D
nV
T







1







I
S0

1
 


I
S

i
D

I
Sexp
v
D
nV
T







1







I
S1

1
 

0
i
D

I
Sexp
v
D
nV
T







1







I
S
exp
v
D
nV
T






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36. MATRUSRI
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Capacitor Filter

37. MATRUSRI
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Capacitor Filter

38. MATRUSRI
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Capacitor Filter

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40. MATRUSRI
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41. MATRUSRI
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42. MATRUSRI
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43. MODULE-II
CONTENTS:
Zener diode and Zener diode regulators.CRT construction and
CRO applications.
OUTCOMES:
To understand the operation, performance of Zener diode
To understand the operation and applications of CRO

44. Breakdown in diodes
• Avalanche Breakdown
Si diodes with VZ greater than about 5.6 volts breakdown according to
an avalanche mechanism. As the electric field increases, accelerated
carriers begin to collide with fixed atoms. As the reverse bias
increases, the energy of the accelerated carriers increases, eventually
leading to ionization of the impacted ions. The new carriers also
accelerate and ionize other atoms. This process feeds on itself and
leads to avalanche breakdown.
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45. Breakdown in diodes (cont.)
• Zener Breakdown
Zener breakdown occurs in heavily doped diodes. The heavy doping
results in a very narrow depletion region at the diode junction.
Reverse bias leads to carriers with sufficient energy to tunnel directly
between conduction and valence bands moving across the junction.
Once the tunneling threshold is reached, additional reverse bias leads
to a rapidly increasing reverse current.
• Breakdown Voltage Temperature Coefficient
Temperature coefficient is a quick way to distinguish breakdown
mechanisms. Avalanche breakdown voltage increases with
temperature, whereas Zener breakdown decreases with temperature.
For silicon diodes, zero temperature coefficient is achieved at
approximately 5.6 V.
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46. 46
A typical V-I curve for a Zener diode.
A Zener diode reference circuit.
The Zener diode is made to operate under reverse bias once a sufficiently high
voltage has been reached. The V-I curve of a Zener diode is shown in Figure.
Notice that under reverse bias and low voltage the current assumes a low
negative value, just as in a normal pn-junction diode. But when a sufficiently
large reverse bias voltage is reached, the current increases at a very high rate.
Zener Diode
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47. Voltage Regulator Using the Zener Diode
The Zener diode keeps the
voltage across load resistor
RL constant. For Zener
breakdown operation, IZ > 0.

I
S

V
S

V
Z
R

(
20

5
)
V
5
k


3
mA
I
L

V
Z
R
L

5
V
5
k


1
mA
| I
Z

I
S

I
L

2
mA

I
Z

V
S
R

V
Z
1
R

1
R
L









0
|R
L

R
V
S
V
Z

1













R
min
For proper regulation, Zener current must be
positive. If the Zener current < 0, the Zener
diode no longer controls the voltage across
the load resistor and the voltage regulator is
said to have “dropped out of regulation”.
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48. Line and Load Regulation
Line regulation characterizes how sensitive the output voltage is to input
voltage changes. (Circuit on previous page, Figure 3.41)

Line Regulation

dV
L
dV
S
mV/V
For a fixed load curren
Line Regulation
=
R
Z
R
+
R
Z
Load regulation characterizes how sensitive the output voltage is to
changes in load current withdrawn from regulator.
Load regulation is the Thévenin equivalent resistance looking back into
the regulator from the load terminals.

Load Regulation

dV
L
dI
L

For changes in load
Load Regul


R
Z
R






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49. CRO
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50. CRO
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Cathode ray tube - it is the heart of the oscilloscope. When the electrons emitted by the electron gun
strikes the phosphor screen, a visual signal is displayed on the CRT.
Vertical amplifier - the input signals are amplified by the vertical amplifier. Usually, the vertical amplifier
is a wide band amplifier which passes the entire band of frequencies.
Delay line - as the name suggests, this circuit is used to delay the signal for a period of time in the vertical
section of CRT. The input signal is not applied directly to the vertical plates because the part of the signal
gets lost, when the delay time is not used. Therefore, the input signal is delayed by a period of time.
Time base (sweep) generator - time base circuit uses a uni-junction transistor, which is used to produce
the sweep. The saw tooth voltage produced by the time base circuit is required to deflect the beam in the
horizontal section. The spot is deflected by the saw tooth voltage at a constant time dependent rate.
Horizontal amplifier - the saw tooth voltage produced by the time base circuit is amplified by the
horizontal amplifier before it is applied to horizontal deflection plates.
Trigger circuit - the signals which are used to activate the trigger circuit are converted to trigger pulses for
the precision sweep operation whose amplitude is uniform. Hence input signal and the sweep frequency
can be synchronized.
Power supply - the voltages required by CRT, horizontal amplifier, and vertical amplifier are provided by
the power supply block. It is classified into two types -
(1) negative high voltage supply
(2) positive low voltage supply
The voltage of negative high voltage supply is from -1000V to -1500V. The range of positive voltage
supply is from 300V to 400V.

51. CRO
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Working of cathode ray oscilloscope:
when the electron is injected through the electron gun, it passes through the
control grid. The control grid controls the intensity of electron in the vacuum tube. If the
control grid has high negative potential, then it allows only a few electrons to pass through it.
Thus, the dim spot is produced on the lightning screen. If the negative potential on the control
grid is low, then the bright spot is produced. Hence the intensity of light depends on the
negative potential of the control grid.
After moving the control grid the electron beam passing through the focusing and
accelerating anodes. The accelerating anodes are at a high positive potential and hence they
converge the beam at a point on the screen.
After moving from the accelerating anode, the beam comes under the effect of the
deflecting plates. When the deflecting plate is at zero potential, the beam produces a spot at
the centre. If the voltage is applied to the vertical deflecting plate, the electron beam focuses
at the upward and when the voltage is applied horizontally the spot of light will be deflected
horizontally.
Applications:
used for display of waveforms, measurement of voltages, measurement of current,
measurement of frequency and measurement of phase difference.

52. CRO
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Applications:
used for display of waveforms, measurement of voltages, measurement of current,
measurement of frequency and measurement of phase difference.

53. Problems
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1.Determine the value of forward current in the case of a PN Junction diode, with Is=10µA
and Vf=0.8V at T =300K, assume silicon diode.

54. Problems
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2.The reverse saturation current of a silicon PN Junction diode is 10µA. Calculate the diode
current for the forward bias voltage of 0.6V at 250C

55. Problems
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3.A half-wave rectifier, having a resistive load of 1000Ω, rectifiers an alternating voltage of 325V peak
value and the diode has a forward resistance of 100Ω.
Calculate a) peak, average and rms value of current
b)dc output power
c) ac input power
d) efficiency of the rectifier

56. Problems
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4.A 230V, 60Hz voltage is applied to the primary of a 5:1, step-down, center – tap transformer used in
a full-wave rectifier having a load of 900Ω . If the diode resistance and secondary coil resistance
together has a resistance of 100Ω, determine
a) dc voltage across the load
b) dc current flowing through the load
c) dc power delivered to the load
d) PIV across each diode
e) ripple voltage and its frequency
f) rectification efficiency

57. 5. The diode current is 0.6mA when the applied voltage is 400mV, and 20mA when the
applied voltage is 500mV. Determine η
Assume KT/q=25mv
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Problems

58. Short Answer Questions
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59. Long Answer Questions
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1.Draw and explain the V-I characteristics of a PN junction diode under forward and
reverse bias? Derive the total current equation of PN diode?
2.Draw the circuit diagram of a centre tapped full wave rectifier and explain its working.
Also define and derive the expression for (i) ripple factor (ii) conversion efficiency (iii)
TUF (iv) regulation.
3.Explain how does a zener diode behaves as voltage regulator?
4.A half wave rectifier circuit, having a resistive load of 1000Ω, rectifies an alternating
voltage of 325V peak value and the diode has forward resistance of 100Ω. Calculate (i)
peak (ii) average (iii) RMS value of current (iv) AC input power (v) DC output power
(vi) efficiency.
5.What is hall effect? Derive an expression for hall coefficient.
6.By drawing energy band diagrams show how fermi levels are indicated in intrinsic and
extrinsic materials in both n-type and p-type materials.
7.Explain capacitor filter with FWR with required waveforms and derive ripple factor
equation.
8.A 230V, 50hz voltage is applied to the primary of a 5:1 step down centre tapped
transformer in FWR having a load of 900Ω. Determine (i) DC voltage across the load (ii)
idc (iii) DC power delivered to the load (iv) PIV across each diode.

60. Long Answer Questions
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6.By drawing energy band diagrams show how fermi levels are indicated in
intrinsic and extrinsic materials in both n-type and p-type materials.
7.Explain capacitor filter with FWR with required waveforms and derive ripple
factor equation.
8.A 230V, 50hz voltage is applied to the primary of a 5:1 step down centre
tapped transformer in FWR having a load of 900Ω. Determine (i) DC voltage
across the load (ii) idc (iii) DC power delivered to the load (iv) PIV across each
diode.
9.What is an ideal diode? How can it be represented as switch? Draw the
equivalent circuit and its characteristics.

61. Questions & Answers
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