Transcript An Introduction to BJT

Recommended Books
• Robert Boylestad and Louis Nashelsky,
“Electronic Devices and Circuit Theory”,
Prentice Hall, 7th Edition or Latest.
• Thomas L. Floyd, “Electronic Devices”, Prentice
Hall, 7th Edition or Latest, ISBN: 0-13-127827-4
1
This Lecture
Current and Voltage Analysis of BJT – A Review
2
Types of Bipolar Junction Transistors
npn
pnp
n
p
C
n
C
E
p
n
p
C
Cross Section
B
B
B
B
Schematic
Symbol
Schematic
Symbol
•
•
•
E
E
Collector doping is usually ~ 106
Base doping is slightly higher ~ 107 – 108
Emitter doping is much higher ~ 1015
3
BJT Equations
IE
-
E
VCE
IC
+
IE
C
-
VBE
VBC
IB
+
+
B
npn
IE = I B + I C
VCE = -VBC + VBE
E
+
VEC
IC
-
C
+
+
VEB
VCB
IB
-
-
B
pnp
IE = I B + I C
VEC = VEB - VCB
4
DC Beta and DC Alpha
• DC Beta (dc) : The ratio of the dc collector current (Ic) to the
dc base current (IB) is the dc beta. It is also called the dc
current gain of a transistor.
Ic
 dc 
IB
– Typical values of dc range from less than 20 to 200 or
higher.
– If temperature goes up, dc goes up and vice versa.
• DC Alpha (dc): It is the ratio of dc collector current (Ic) to the
dc emitter current (IE).
Ic
dc 
IE
– Typically values of dc range from 0.95 to 0.99, but it is
always less than unity.
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Relationship between dc and dc
For an NPN transistor
IE  IB  Ic
Dividing each term by IC we get
IC
IE
IB


IC
IC
IC
or
IE
IB

1
IC
IC
1
1

1
dc dc
1   dc
1

dc
 dc
 dc
 dc 
1   dc
Similarly, we can prove that
 dc
 dc 
1   dc
6
Problems on dc and dc
1. Determine dc and IE for IB = 50A and IC =
3.65 mA.
Solution:
I C 3.65  103
 dc  
 73
6
I B 50  10
6
I E  I B  I C  50  10  3.65  10
3
 3.70  10 3 A  3.70mA
7
Problems on dc and dc
2. What is the dc when IC = 8.23mA and IE =
8.69 mA.
I C 8.23  103
 0.947
Solution: dc  
3
8.69  10
IE
3. A certain transistor exhibits an dc of 0.96.
Determine IC when IE = 9.35 mA.
Solution:
IC
dc 
IE
I
C
 dc  I E  0.96  9.35  8.976mA
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Current and Voltage Analysis
IB: dc base current
IE: dc emitter current
IC: dc collector current
VBE: dc voltage across
base-emitter junction
VCB: dc voltage across
collector-base junction
VCE: dc voltage from
collector to emitter
Transistor bias circuit.
9
Current and Voltage Analysis
• When the BE junction is forward-biased, it is like
a forward-biased diode. Thus; (Si = 0.7, Ge = 0.3)
VBE  0.7V
• From KVL, the voltage across RB is
• By Ohm’s law;
VR B  VBB  VBE
VR B  I B R B
• Solving for IB
IB 
VBB  VBE
RB
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Current and Voltage Analysis
• The voltage at the collector is;
VCE  VCC  VR C
• The voltage drop across RC is
VR C  I C R C
• VCE can be rewritten as
VCE  VCC  IC R C
• The voltage across the reverse-biased CB junction
is
VCB  VCE  VBE
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Problems
Determine IB, IC, IE, VBE, VCB
and VCE in the circuit. The
transistor has a dc = 150.
Solution:
VBB  VBE 5  0.7
IB 

 430A
RB
10,000
I C   dc I B  150  430  10 6
 64.5mA
VCB  VCE  VBE  3.55  0.7  2.85V
3
VCE  VCC  I C R C  10  (64.5  10 )(100)  3.55V
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Problems
A base current of 50A is
applied to the transistor in
the adjacent Fig, and a
voltage of 5V is dropped
across RC. Determine the dc
and dc of the transistor.
Solution:
VR C
5
IC 

 5  10 3  5mA
R C 1000
I C 5  103
 dc  
 100
6
I B 50  10
 dc
100
dc 

 0.99
 dc  1 100  1
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Problems
Find VCE, VBE and VCB in the given circuit.
Solution:
VBE  0.7V
VBB  VBE
5  0.7
IB 

 1.1mA
3
RB
3.9  10
I C   dc I B  50  1.1  10 3  55mA
VCE  VCC  I C R C
 15  55  10  3  180  5.10V
VBC  VBE  VCE  0.7  5.10  4.40V
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Problems: Homework
1. Find IB, IE and IC in
Fig.1. dc = 0.98.
Ans: IE = 1.3 mA, IB = 30,
IC = 1.27 mA.
2. Determine the terminal
voltages
of
each
transistor with respect to
ground for circuit in Fig.
2. Also determine VCE, VBE
and VBC.
Ans. VB = 10 V, VC = 20 V, VE =
Fig. 1
9.3 V, VCE = 10.7, VBE = 0.7
V, VBC = -10 V.
Fig. 2
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Modes of Operation
BJTs have three regions of operation:
1. Active: BJT acts like an amplifier (most common use)
2. Saturation - BJT acts like a short circuit BJT is used as a switch
3. Cutoff - BJT acts like an open circuit
By switching
between these
two regions.
IC(mA)
Saturation Region
IB = 200 A
30
Active Region
IB = 150 A
22.5
IB = 100 A
15
IB = 50 A
7.5
Cutoff Region
IB = 0
0
VCE (V)
0
5
10
15
20
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More about Transistor Regions
Cutoff: In this region,
IB = 0 and VCE = VCC.
That is, both the baseEmitter and the basecollector junctions are
reversed biased.
Under this condition, there is a very small amount of
collector leakage current ICE0 due mainly to
thermally produced carriers. It is usually neglected
in circuit analysis.
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More about Transistor Regions
Saturation: When the
Base-emitter junction is
forward biased and the
base current is increased,
The collector current also
Increases (IC = dcIB) and VCE
Decreases (VCE = VCC – ICRC). When VCE reaches its
saturation, there is no further change in IC.
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DC Load Line
The bottom of the load
Line is at ideal cutoff
where
IC = 0 and VCE = VCC.
The top of the load line
is at saturation where
IC = IC(sat) and VCE = VCE (sat).
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Quiescent-Point (Q-Point)
 Operating point of an amplifier to state the values of
collector current (ICQ) and collector-emitter voltage (VCEQ).
 Determined by using transistor output characteristic and
DC load line.
 Quiescent means quiet, still or inactive.
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Example
The transistor shown in Figure (a) is biased with
variable voltages VCC and VBB to obtain certain values
of IB, IC, IE and VCE. The collector characteristic curves
are shown in Figure (b). Find Q-point when:
(a) IB = 200A (b) 300A (c) 400A.
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Solution:
(a) IC = dcIB = 100200  10-6
= 20 mA
VCE = VCC – ICRC =
10 – 2010-3220 = 5.6 V
This Q-Point is shown as Q1.
(b) IC = dcIB = 100300  10-3
= 30 mA
VCE = VCC – ICRC =
10 – 3010-3220 = 3.4 V
This Q-Point is shown as Q2.
(c) IC = dcIB = 100400  10-6
= 20 mA
VCE = VCC – ICRC
= 10 – 4010-3220 = 1.2 V
This Q-Point is shown as Q3.
22
Problem
(a) Determine the intercept
points of the dc load line on
The vertical and horizontal
Axes of the
collector
characteristic curves in the
Fig.
(b) Assume that you wish to bias the transistor
with IB = 20A. To what voltage must you
change the VBB supply. What are IC and VCE at
the Q-point , given that dc = 50. VBE =0.7
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Problem
Solution:
(a) Horizontal intercept
VCE = VCC = 20 V
Vertical intercept
I C( sat) 
VCC
20

 2mA
RC
10000
(b) VBB = IBRB + VBE
= 2010-6 1 106 + 0.7
= 2.7 V
IC = dcIB = 502010-6 = 1 mA
VCE = VCC – ICRC = 20 - (110-3101000) = 10 V
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