Foundations of Electrical and Electronics Engineering

Unit -V
Semiconductor Devices and Circuits

By:
Dr. Tarkeshwar Mahto
EEE Department
SRM University AP
Semiconductor Devices
& Circuits

2
Materials
 Any materials Property is to control the flow of an
electrical current
 It includes
 Conductors (low resistivity)
 Low Resistance → Allows electrical current flow
 Insulators (high resistivity)
 High Resistance → Suppresses electrical current flow
 Semi Conductors (medium resistivity)
 Can allow or suppress electrical current flow

3
Conductors
• Good conductors have low resistance so
electrons flow through them with ease.
• Best element conductors include:
• Copper, silver, gold, aluminum, &
nickel
• Good conductors can also be liquid:
• Salt water

4
Conductor Atomic Structure
 Atomic Structure of a good
conductor includes only one
electron in their outer most
shell
 It is called Valance Electron
 It can be easily removed from
the atom contributing to
current flow

5
Insulators
 Insulators have a high resistance so current does not
flow in them.
 Good insulators include:
– Glass, ceramic, plastics, & wood
 Most insulators are compounds of several elements.
 The atoms are tightly bound to one another so
electrons are difficult to strip away for current flow.

6
Semiconductors
 A material whose properties are such that it is not
quite a conductor, not quite an insulator.
 Elemental
 Carbon, Silicon and Germanium
 Compound
 GaAs - Gallium arsenide
 AlAs - Aluminum arsenide

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Semiconductors – Atomic Structure
 The main characteristic of a semiconductor
element is that it has four electrons in its outer or
valence orbit

8
Semiconductors – Crystal Lattice

 Semiconductor atoms
can link together to form
Crystal Lattice
 The links are formed by
means of Covalent Bonds

9
Energy band

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Semiconductor – Material Types
 Pure Semiconductor
 Good Insulator
 Intrinsic Semiconductor

𝐷𝑜𝑝𝑖𝑛𝑔
 Intrinsic Semiconductor Extrinsic Semiconductor

 Doping → Addition of Impurities

 Extrinsic Semiconductor can conduct current based on
doping concentration

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N-Type
 Doping Pentavalent Impurity
 Arsenic, Antimony, Bismuth
 Additional Valence Electron
left over from bonding with
neighboring atoms
 Additional Electron →
Current flow
 Majority Charge Carrier →
Electrons

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P-Type
 Doping Trivalent Impurity
 Boron, Aluminium, Indium
 One of the covalent bond shared with the
neighboring atoms is missing one electron → Hole
 Hole → current carrier similar to electron to support
current flow
 Majority Charge Carrier → Holes

13
 Diodes
 Used in almost all the electronic circuits
 Allows current flow in only one direction
(unidirectional device)
 The usage of semiconductor materials to build the
electronic components was started with diodes
 Before the invention of diode there were with vacuum
tubes, where the applications of both these devices are
similar but the size occupied by the vacuum tube will
be much greater than the diodes

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PN Junction Diode
 Join the two types of semi-conductors P-type and N-
type together → P-N junction diode

 The arrow indicates the flow of current through it

when the diode is in forward biased mode, the dash or
the block at the tip of the arrow indicates the blockage
of current from the opposite direction
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16
Junction Formation
 When P-type and N-type are joined
 Excess electrons (from N) combines with excess holes (from
P) → Immobile Ions
 Immobile ions resists the flow of electrons or holes through it
which now acts as a barrier in between the two materials
 Barrier → Depletion Region
 Width of Depletion region depends on doping
concentration
 When a heavily doped and lightly doped semiconductors are
combined, depletion region will be more in lightly doped side and
less in heavily doped side

17
Forward Bias
 Battery ‘+’ terminal → P side
 Battery ‘-’ terminal → N side
 Due to forward bias, majority charge carriers in both
regions get repelled and enter into the depletion
region
 Diode conducts when the barrier formed in the path is
broken
 Applied voltage > 0.7 V (for silicon), 0.3 V (for germanium)
 Width of the depletion region decreases gradually

 0.7 V → Cut-in voltage; Offset voltage; Break-point voltage;

firing voltage; Threshold voltage

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Reverse Bias
 Battery ‘-’ terminal → P side
 Battery ‘+’ terminal → N side
 Majority charge carriers in both regions get attracted towards
source → large numbers of immobile ions
 Width of the depletion region increases gradually → difficult for
the electrons and holes to cross the junction → open circuit
forms and current flow stops
 When Vapplied increased → depletion region cannot withhold the
external force and the junction breaks down → diode
permanently fails
 The reverse voltage at which the diode conducts is called as
Break down voltage
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Diode Characteristics

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Diode Current
 PN junction current is given approximately by
𝑒𝑉
(
𝐼 = 𝐼𝑠 𝑒 η𝑘𝑇−1)
 where I is the current, e is the electronic charge, V is
the applied voltage, k is Boltzmann’s constant, T is the
absolute temperature and η (Greek letter eta) is a
constant in the range 1 to 2 determined by the junction
material
 for most purposes we can assume η = 1

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Ideal diode
 The diode conducts well in the forward direction and
poorly in the reverse direction. Ideally, a diode acts like a
perfect conductor (zero resistance) when forward Biased
and like a perfect insulator (infinite resistance) when
reverse biased
 An ordinary switch has zero resistance when closed and
infinite resistance when open. Therefore, an ideal diode
acts like a switch that closes when forward biased and
opens when reverse biased.
 The Ideal diode model treats a forward-biased diode like a
closed switch with a voltage drop of zero volts

22
Constant voltage drop model
Figure a shows a current versus voltage for Constant voltage drop
model.
No current - until 0.7 V appear across the diode. When the voltage
reaches 0.7 v the diode turns on, 0.7 V can appear across the diode.
Figure b shows the equivalent circuit for the Constant voltage drop
model a silicon diode. The diode act as a switch in series with a barrier
potential of 0.7 V.
If the voltage across the diode is greater than 0.7 V, the switch will
close. On the other hand, if the voltage is less than 0.7 V, the switch
will open. In this case, there is no current through the diode.

23
Problem
Calculate the forward bias current of a Si diode when forward
bias voltage of 0.4V is applied, the reverse saturation current is
1.17×10-9A and the thermal voltage is 25.2mV.
Solution
Equation for diode current
I=Is×(e(V/ηVT)-1)
where Is = reverse saturation current
η = ideality factor
V T = thermal voltage
V = applied voltage
Since in this question ideality factor is not mentioned it can be
taken as one.
I0 = 1.17 x 10-9A, V T = 0.0252V, η = 1, V = 0.4V
Therefore, I= 1.17×10-9x e 0.4/0.025 -1 = 9.156mA
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Diode parameters
Maximum power rating.
It is the maximum power that can be dissipated at the
junction without damaging it..
Maximum forward current.
It is the highest instantaneous forward current that a pn
junction can conduct without damage to the junction.
Peak inverse voltage.
It is the maximum reverse voltage that a diode can
withstand without destroying the junction.

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Diode parameters
Knee Voltage
In the forward region, the voltage at which the current starts
to increase rapidly is called the knee voltage or Cut in
voltage of the diode. The knee voltage equals the barrier
potential
Vk= 0.7 (Si)
Vk= 0.3 (Ge)
Reverse current or leakage current.
It is the current that flows through a reverse biased diode.

This current is due to the minority carriers.

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Problem
Use the ideal diode model to calculate the load voltage
and load current in the circuit.

Since the diode is forward biased, it is equivalent to a

closed switch. :
Vs= 10 V
With Ohm’s law, the load current is:

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Problem
Use Constant voltage drop model to calculate the load
voltage and load current,
Solution
Since the diode is forward biased, it is equivalent to a
battery of 0.7 V.
VL =10 V - 0.7 V =9.3 V
With Ohm’s law, the load current is:

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Problem
For the series diode configuration given in the Figure,
determine VD, VR, and ID.(Constant voltage drop
model)
The diode is in the “on” state,

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More Examples

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More Examples

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More Examples

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More Examples

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More Examples

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More Examples

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Example:

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Rectifiers
Introduction
 The application of rectifier circuits is in the conversion
of AC to DC power. A circuit that accomplishes this
conversion is usually called a DC power supply.

 Many familiar electrical and electronic appliances

(e.g., radios, personal computers, TVs) require DC
power to operate. For most applications, it is desirable
that the DC supply to be as steady and ripple-free as
possible.

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DC power supply

Fig.1 Schematic diagram of a DC power supply

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Classification of Rectifiers
 Half – Wave Rectifier
 Full Wave Rectifier
Center-Tapped Full Wave Rectifier
Bridge Rectifier

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Half Wave Rectifier
 A half wave rectifier allows one half-cycle of an
AC voltage waveform to pass, blocking the other half-cycle.
Half-wave rectifiers are used to convert AC voltage to
pulsating voltage, and require a single diode to construct

41
Half-wave rectifier

Fig.3 Conduction region ( T/2 to T)

Fig.4 Half wave rectified signal 42

Center tapped Full Wave Rectifier
A full-wave rectifier converts an ac voltage into a pulsating dc
voltage using both half cycles of the applied ac voltage.
A full-wave rectifier appears in Fig. 5 with two diodes but
requiring a center-tapped (CT) transformer to establish the
input signal across each section of the secondary of the
transformer. The diodes feed a common load R with the help of
a center-tap transformer.

Fig.5 Center-tapped transformer full-wave rectifier. 43

Center tapped Full Wave Rectifier
During the positive portion of Vi applied to the primary of the
transformer, the network will appear as shown in Fig. 6
D1 - Forward biased- short-circuit
D2 – Reverse biased- open-circuit
The output voltage appears as shown in Fig.6

Fig 6 Network conditions for the positive region of vi.

44
Center tapped Full Wave Rectifier
During the negative portion of the input the network appears as
shown in Fig.7
D1 – Reverse biased- open-circuit
D2 - Forward biased- short-circuit

Fig 7 Network conditions for the negative region of vi.

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Center tapped Full Wave Rectifier

Fig. 8 Input and output wave forms

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Bridge Rectifier
This type of full wave rectifier uses four diodes connected in a
bridge configuration to produce the desired output.
The main advantage of this bridge circuit is that it does not require
a special centre tapped transformer, thereby reducing its size and
cost. Four diodes labeled D1 to D4 are arranged as shown in Fig 9
with only two diodes conducting current during each half cycle

Fig 9. Full Wave Bridge Rectifier 47

Bridge Rectifier
During the positive half cycle of the supply, diodes D2 and
D3 conduct in series while diodes D1 and D4 are reverse
biased and the current flows through the load as shown
below

Fig.10 Conduction path for the positive region of vi

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Bridge Rectifier
During the negative half cycle of the supply, diodes D1 and
D4 conduct in series but diodes D2 and D 3 switch “OFF” as
they are reverse biased.
The current flowing through the load is the same direction as
before.

Fig. 11 Conduction path for the negative region of vi

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Bridge Rectifier
Over one full cycle the input and output voltages will appear
as shown in Fig.12

Fig.12 Input and output waveforms for a full-wave rectifier

50
Analysis of Full-Wave Rectifier
Peak Current
The value of peak current (Imax) can be derived with the help of
instantaneous value of applied voltage and the resistance of the
diodes. The value of instantaneous voltage applied to the rectifier
circuit can be given as:-

Let’s assume the forward resistance - Rf , load resistor RL then the

current flowing through the load resistor can be given as:-

The total current i can be obtained by the sum of i1 and i2 for the
whole cycle.

51
Analysis of Full-Wave Rectifier
Output Current
The current through the load is the same for both the cycles of
the ac signal thus, the dc output current can be given as

DC output voltage
The average dc voltage is given as

52
Analysis of Full-Wave Rectifier
RMS Current
The rms current through the load RL is given as

53
Analysis of Full-Wave Rectifier
RMS Voltage
The rms value of a voltage across the load is given as

Form factor
The form factor is the ratio of rms value to the dc output value
of current. It is given as

54
Analysis of Full-Wave Rectifier
Peak factor
It is the ratio of the peak value of current to the rms value of
current

Ripple factor

55
Analysis of Full-Wave Rectifier
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 .The peak inverse voltage of diode in center tapped full
wave rectifier is 2 Vsmax and Bridge rectifier is Vsmax.

Rectification Efficiency: The rectification efficiency of full

wave rectifier can be obtained by the ratio of dc power delivered
to load and ac power present in the output

For bridge rectifier,

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Advantages of Full Wave Rectifiers
Advantages of Full Wave Rectifiers
 The rectification efficiency of full wave rectifier is much
higher than that of half wave rectifier. It is approximately
double to that of half wave rectifier i.e. it is about 81%.
 The filtering circuit required in full wave rectifier is simple
because ripple factor in the case of full wave rectifier is very
low as compared to that of half wave rectifier. The value of
ripple factor in full wave rectifier is 0.482 while in half wave
rectifier it is about 1.21.
 The output voltage and output power obtained in full wave
rectifiers are much more than that of full wave rectifiers.
Disadvantages of Full Wave Rectifiers
 The full wave rectifiers need more circuit elements than half
wave rectifier which makes it costlier
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Examples

58
Examples

59
Examples

60
Examples

61
Examples

62
Clipping Circuits

63
 Clippers, also known as 'diode limiters' are usually a diode and resistor network
that “clips” away a portion of an input signal without distorting the remaining
part of the applied input signal. One of the simplest examples of a clipper circuit
is the half-wave rectifier circuit.
Series Clipper
The series configured clipper circuit is where the diode is placed in series with the load.

This is a negative series clipper and is also popularly known as the half-wave rectifier.
64
During the positive interval of the input signal, the diode is forward biased and is in the
on-state. Therefore, the diode conducts and the positive interval of the input signal
passes through the output. On ideal analysis, the voltage drop across the diode is 0V
and is represented by a short circuit across its pins when it is forward biased. During
this interval, the output is exactly the same as the input, therefore the output voltage is
just equal to the positive region of the input signal.

65
During the negative interval of the input signal, the diode is reversed biased. Therefore
the diode is in off-state creating an open circuit. This is represented in the schematic
below as an open circuit across the diode. Therefore, there is no output during this
interval. The circuit “clipped” off the negative region of the applied input signal. That is
why this circuit is called a negative series clipper.

If we reverse the diode compared to the circuit above, a positive series clipper
is created.

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Parallel Clipper
Parallel clipper is configured in a way where the diode is placed in a branch parallel to
the load.

During the positive interval of the applied signal, the diode is forward biased and is in
‘on-state’. The diode will be represented with a short across its terminals. Because of
this, the node where the resistor and the diode meet is shorted to the ground. And
hence, the output is also shorted to the ground. Therefore, there is no output during this
interval.

67
During the negative interval of the input signal, the diode is reversed biased and is
‘open’ as shown in the figure below. This makes the resistor in the circuit a floating
element. Also, making the output directly parallel to the input. Therefore, the output
during the negative interval is equal to the negative region of the input signal. Observing
the final waveform of the output diagram, the positive interval of the input signal was
“clipped off”, therefore this circuit is called a positive parallel clipper.

Connecting the diode in reverse on the circuit, will make it a negative parallel
clipper.

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69
 Biased Clippers
Biased clippers are formed by adding a DC supply to the network. The addition of a DC
supply can have a pronounced effect on the analysis of the clipper configuration. The
DC supply may aid or work against the input signal depending on how it is positioned. It
can be placed in the leg between the input and the output or in the branch parallel to
the output.
Biased - Series Clipper

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As the input signal transitions from 0V - 5V, the diode is ‘off’, the input signal must be
greater than the DC supply presented in this circuit for the diode to be in ‘on-state’.

Using Kirchhoff's Voltage Law, or KVL, we can

determine the peak voltage of the output.

V0=Vi-Vdc
Therefore, peak voltage of the output during
the positive interval is:
V0=20 V-5 V=15 V

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During the negative interval, the diode will remain open. Therefore, the output in this
interval is 0V.

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 Biased - Parallel Clipper
Here, a DC supply is added to the line which is parallel to the output. In this example,
the DC supply is placed where its negative terminal is connected to the ground of the
circuit, while its positive terminal is connected to the anode of the diode.

Looking at the circuit , during the positive interval, the diode is reverse biased by the
input signal. But also take note that it is being forward biased by the DC supply on a
certain time when the input is still less than the DC supply. Therefore, as the input
signal is increasing from 0V to 4V, the diode is ‘on’ through the DC supply. During this
time, the output is directly parallel to the 4Vdc supply. So the output voltage, Vo, during
this portion of the positive interval is equal to 4V.
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During the negative interval, the diode is
forward biased by the 4V DC supply.
Therefore, the diode is in the ‘on-state’,
making the output directly parallel to the 4V
DC supply for the whole duration of this
interval.

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Bipolar Junction Transistors

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History

 Invented in 1947 by Shockley, Bardeen, and Brattain at

Bell Laboratories
 Awarded with the Nobel Prize
 Transistors were originally manufactured using
Germanium
 Today’s Silicon-based transistors were adopted
because Germanium breaks down at 180 degrees F
 It’s a three terminal device.
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Intro-Transistor
 The name transistor is derived from the term transfer
resistance. Transistor – Trans-Resistor
 The current (voltage) through two of the terminals is
controlled by the current (voltage) through another
pair of terminals.
 Transistor can act as a switch.
 Transistor has the ability to amplify a signal between
one pair of terminals by using an input signal at
another pair of terminals. Thus, a transistor can also
act as an amplifier
77
 symbol

Construction
 The BJT is a three terminal device and has two junctions
 The N and P regions are different both geometrically and in
terms of the doping concentration of the regions
 Emitter region - this is usually a heavily doped region (P/N).
The emitter ‘emits’ the carriers into the base.
 Base region - this is a lightly doped (N/P) region. The base
region is also physically thin so that carriers can pass through
with minimal recombination.
 Collector region - this is a (P/N) type region. The collector
region has a larger width that the other two regions since
charge is accumulated here from the base.

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 Construction
 The doping concentrations in the collector, base and
emitter are not the same
• Collector doping is usually ~ 106
• Base doping is slightly higher ~ 107 – 108
• Emitter doping is much higher ~ 1015
 Therefore the behavior of the device is not electrically
and geometrically symmetric and the terminals cannot
be cannot be interchanged

79
 Types – PNP and NPN Transistors

Symbols

Note: NPN type is most commonly used Transistor. The majority charge carriers in
an NPN transistor are electrons and the majority carriers in a PNP transistor are holes.
80
The electrons have better mobility than holes and helps in better conduction.
Working Principle
 Emitter Base Junction → Forward Biased
 Collector Base Junction → Reverse Biased
 Forward bias at emitter base junction causes electrons
to move towards base leading to emitter current

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Working Principle
 As electrons flow towards p-type base, recombination
will happen with the holes
 Since the base is lightly doped, only few recombination
would happen within the base
 Small recombination results in small base current
 Remainder electrons cross the base and constitute the
collector current because the electrons are attracted
towards the collector due to high reverse bias

82
Operation of Transistor

83
BJT Configurations
A Transistor has 3 terminals, the emitter, the base and the collector.
Using these 3 terminals the transistor can be connected in a circuit with
one terminal common to both input and output in a 3 different possible configurations

 Common-Base (CB) :
 Input = VEB & IE
 Output = VCB & IC
 Common-Emitter (CE):
 Input = VBE & IB
 Output = VCE & IC
 Common-Collector (CC):
 Input = VBC & IB
 Output = VEC & IE

84
BJT Configurations-Comparison
COMMON
AMPLIFIER TYPE COMMON BASE COMMON EMITTER COLLECTOR

INPUT/OUTPUT
PHASE RELATIONSHIP 0° 180° 0°

VOLTAGE GAIN HIGH MEDIUM LOW

CURRENT GAIN LOW MEDIUM HIGH

POWER GAIN LOW HIGH MEDIUM

INPUT RESISTANCE LOW MEDIUM HIGH

OUTPUT RESISTANCE HIGH MEDIUM LOW

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Modes of Operation
 Active
 Most important mode of operation
 Central to amplifier operation
 The region where current curves are practically flat
 Saturation
 Barrier potential of the junctions cancel each other out
causing a virtual short- closed switch
 Cut-Off
 Current reduced to zero
 Ideal transistor behaves like an open switch

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Common Emitter Configuration
 In CE Configuration, the Emitter terminal of the transistor will be
connected common between the output and the input terminals.
 Input voltage VBE is applied between base and emitter terminals
and output voltage VCE is taken across emitter and collector.
 The output current IC is taken across the emitter and collector
terminals.
 Input side is forward biased and the output side is reverse biased.
 Input current IB is measured in µA because the base region is very
lightly doped
The CE configuration is the most widely used
configuration.
Common emitter transistors provides Moderate
current gain, Moderate voltage gain and High
power gain.
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CE Configuration-Input Characteristics.

 Input characteristics are the

relationship between the input
current and the input voltage
keeping output voltage constant.
 Input current is the base current
IB, input voltage is base emitter
voltage VBE and the output voltage
is collector emitter voltage VCE.
 when the input voltage VBE is
increased initially there is no
current produced, further when it is
increased beyond 0.7V the input
current IB increases steeply
 If the output voltage VCE is further
increased the curve shifts right
side.

88
CE Configuration-Output Characteristics.

 Output characteristics is the relationship

between the output current and the
output voltage keeping input current
constant.
 Output current IC and the output voltage
VCE is noted keeping input current IB
constant
 Active region when the output voltage
is increased there is very slight change
in the input current
 Cut off region is the region where the
input current is below zero.
 When both the junctions are forward
biased, it is in saturation region.
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CE Configuration

 Cutoff region
 both BE and BC reverse biased
 Active region
 BE Forward biased
 BC Reverse biased
 Saturation region
both BE and BC forward biased

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 Amplification factors
 The ratio of change in collector current with respect to base
current is known as the base amplification factor ꞵ
 The ratio of change in collector current with respect to emitter
current is known as the current amplification factor α

𝛽
𝛼=
𝛽+1

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Problem
Given: IB = 50  A , IC = 1 mA
Find: IE , ꞵ , and 
Solution:

IE = IB + IC = 0.05 mA + 1 mA = 1.05 mA

ꞵ = IC / IB = 1 mA / 0.05 mA = 20

 = IC / IE = 1 mA / 1.05 mA = 0.95238

 could also be calculated using the

value of ꞵ

= ꞵ = 20 = 0.95238
ꞵ+1 21 92
 Transistor as Switch
 Solid state switches are one of the
main applications for the use of
transistor to switch a DC output “ON”
or “OFF”.
 When transistor can be operated
between Saturation Region and the
Cut-off Region, it can be used as a
Switch
 Transistor used as a switch is driven
back and forth between its “fully-
OFF” (cut-off) and “fully-ON”
(saturation) by changing the biasing
suitably
 When transistor is operated in active
region it works as an Amplifier.
93
Thank you

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