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Chapter 3: BIPOLAR JUNCTION TRANSISTOR

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Title: Chapter 3: BIPOLAR JUNCTION TRANSISTOR


1
Chapter 3 BIPOLAR JUNCTION TRANSISTOR
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Learning Outcomes
  • At the end of this chapter, the students should
    be able to
  • Understand the basic transistor construction,
    operation configuration
  • Discuss transistor parameters and
    characteristics
  • Develop working knowledge of a transistor
    through use of specification sheet

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Transistor Construction
There are two types of transistors (a) pnp and
(b) npn-type.
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Energy band of a NPN transistor
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  • Analysis of Common-base configuration for pnp
  • Current base, IB (?A) is small compare to current
  • emitter, IE (mA) and current collector,IC (mA).
  • The relationship among these current can be
    analyse
  • with KCL IE IB IC
  • Current collector is produce from the total sum
    of
  • current emitter and leakage current.
  • Current emitter that flow through collector known
    as
  • ??DC IE . The value is big compare to leakage
    current.

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The analysis can be understand by expression
below
IC IC(majority) IC(minority) IC ???IE
ICBO IC ???IE (ignore ICBO due to small
value) ???? Ideally ? 1, but in reality it
is between 0.9 and 0.998. thus, ????is
a common base current gain factor that shows
the efficiency
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Input characteristics for a common-base pnp
transistor
Output characteristics for a common-base pnp
transistor
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Transistor as an amplifier
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  • Common-emitter configuration (CE)
  • It is called common-emitter configuration since
  • emitter is common or reference to both i/p and
    o/p
  • terminals.
  • emitter is usually the terminal closest to or at
    ground
  • potential.
  • Almost amplifier design is using connection of CE
  • due to the high gain for current and voltage.
  • Two set of characteristics are necessary to
    describe the
  • behavior for CE input (base terminal) and output
    (collector
  • terminal) parameters.

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Characteristics of Common-Emitter npn transistor
(a) - Collector characteristics output
characteristics. (b) - Base characteristics
input characteristics.
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Simulation of transistor as an amplifier
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Solution (a)
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Solution (b)
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  • Beta (?)
  • The ratio of dc collector current (IC) to the dc
    base current (IB) is dc beta (?dc ) which is dc
    current gain where IC and IB are determined at a
    particular operating point, Q-point (quiescent
    point).
  • Its define by the following equation
  • 30 lt ?dc lt 300 ? 2N3904
  • On data sheet, ?dchFE with h is derived from ac
    hybrid equivalent cct. FE are derived from
    forward-current amplification and common-emitter
    configuration respectivley.

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  • For ac conditions an ac beta has been defined as
    the changes of collector current (IC) compared to
    the changes of base current (IB) where IC and IB
    are determined at operating point.
  • On data sheet, ?achfe
  • It can defined by the following equation

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Example From o/p characteristics of CE
configuration find ??ac and ??dc with an
operating point at IB25 ?A and VCE 7.5V.
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Solution
Q-point
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Relationship analysis
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  • Common-collector configuration (CC)
  • Also called emitter-follower (EF).
  • It is called common-emitter configuration since
    both the
  • signal source and the load share the collector
    terminal as a
  • common connection point.
  • The o/p voltage is obtained at emitter terminal.
  • The input characteristic of CC configuration is
    similar
  • with CE configuration.
  • All the current relationship for CE
    configuration are true for CC configuration.

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Example Calculate the emitter current for the
common-collector configuration below
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  • Limits of operation for transistor
  • Many BJT transistor used as an amplifier. Thus it
    is
  • important to notice the limits of operations.
  • At least 3 maximum values is mentioned in data
    sheet.
  • There are
  • a) Maximum power dissipation at collector PCmax
    or PD
  • b) Maximum collector-emitter voltage VCEmax
    sometimes named as VBR(CEO) or VCEO.
  • c) Maximum collector current ICmax
  • There are few rules that need to be followed for
    BJT
  • transistor used as an amplifier. The rules are
  • i) transistor need to be operate in active
    region!
  • ii) IC lt ICmax
  • ii) PC lt PCmax

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Transistor limits of operation
Note VCE is at maximum and IC is at minimum
(ICmaxICEO) in the cutoff region. IC
is at maximum and VCE is at minimum (VCEsat ) in
the saturation region. The transistor operates
in the active region between saturation and
cutoff.
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Example
Refer to the fig. Step1 The maximum collector
power dissipation, PDICmax x VCEmax (1)
18m x 20 360 mW Step 2 At any point on the
characteristics the product of and must be equal
to 360 mW. Ex. 1. If choose ICmax 5 mA,
subtitute into the (1), we get VCEmaxICmax 360
mW VCEmax(5 m)360/57.2 V Ex.2. If choose
VCEmax18 V, subtitute into (1), we
get VCEmaxICmax 360 mW (10) ICmax360m/1820 mA
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Derating PDmax
  • PDmax is usually specified at 25C.
  • The higher temperature goes, the less is PDmax
  • Example
  • A derating factor of 2mW/C indicates the power
    dissipation is reduced 2mW each degree centigrade
    increase of temperature.

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More Example Transistor 2N3904 used in the
circuit with VCE 20 V.This circuit used at
temperature 1250C. Calculate the new maximum IC.
Transistor 2N3904 have maximum power
dissipation is 625 mW. Derating factor is
5 mW/0C.
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Transistor 2N3904 used in the circuit with VCE
20 V.This circuit used at temperature 1250C.
Calculate the new maximum IC. Transistor
2N3904 have maximum power dissipation is 625 mW.
Derating factor is 5 mW/0C.
Solution Step 1 Temperature increase 1250C
250C 1000C Step 2 Derate transistor 5 mW/0C
x 1000C 500 mW Step 3 Maximum power
dissipation at 1250C 625 mW500 mW125
mW. Step 4 Thus ICmax PCmax / VCE125m/20
6.25 mA. Step 5 Draw the new line of power
dissipation at 1250C .
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Example The parameters of transistor 2N3055 as
follows - Maximum power dissipation _at_ 250C115
W - Derate factor0.66 mW/0C. This transistor
used at temperature 780C. a) Find the new
maximum value of power dissipation. b) Find the
set of new maximum of IC if VCE10V, 20 V and 40
V.
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Solution Step 1 Temperature increase 780C
250C 530C Step 2 Derate transistor 0.66mW/0C
x 530C 35 mW Step 3 Maximum power dissipation
at 780C 115W 35W80 mW. Step 4 ICmax PCmax
/ VCE80m/10 8 mA (point C) ICmax PCmax /
VCE80m/20 4 mA. (point B) ICmax PCmax /
VCE80m/40 2 mA (point A)
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Step 5 Draw the new line of power dissipation at
780C .
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Transistor Specification Sheet
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Transistor Testing
1. Curve Tracer Provides a graph of the
characteristic curves. 2. DMM Some DMMs will
measure ?DC or HFE. 3. Ohmmeter
1. Curve Tracer Provides a graph of the
characteristic curves. 2. DMM Some DMMs will
measure ?DC or HFE. 3. Ohmmeter
46
References 1. Thomas L. Floyd, Electronic
Devices, Eighth Edition, Prentice Hall, 2002.
2. Robert Boylestad, Electronic Devices and
Circuit Theory, Seventh edition, Prentice Hall,
2002. 3. Puspa Inayat Khalid, Rubita Sudirman,
Siti Hawa Ruslan, ModulPengajaran
Elektronik 1, UTM, 2002.
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