The document discusses bipolar junction transistors (BJTs). It describes the basic construction of an NPN and PNP transistor including the emitter, base, and collector regions. It explains that the base-emitter junction must be forward biased and the base-collector junction must be reverse biased for the transistor to operate properly. The document also discusses BJT biasing circuits, operating regions including cutoff, saturation, and active modes, and uses of BJTs as switches and amplifiers.

1. Chapter 4
Bipolar Junction Transistors (BJTs)

2. Contents
• Bipolar Junction transistors
• Construction
• Circuit diagrams
• BJT biasing circuits
• Q-point
• BJT as a switch
• BJT as an Amplifier

3. • Transistor is a combination of two words i.e.
transfer and resistor. It is because a transistor
is basically a resistor that amplifies electrical
impulses as they are transferred through it
from its input to output terminal.
Transistor

4. • A transistor has three doped regions.
• The bottom region is called the emitter
• The middle region is the base
• And the top region is the collector.
• In an actual transistor, the base region is much thinner as compared to the
collector and emitter regions.
• The transistor Shown in figure (b) is an npn device because there is a p region
between two n regions. Recall that the majority carriers are free electrons in n-
type material and holes in p-type material.
• Transistors shown in figure (c) is an pnp. A pnp transistor has an n region
between two p regions. To avoid confusion between the npn and the pnp
transistors, our early discussions will focus on the npn transistor.
Architecture of a Transistor

5. • There are two types of BJTs, the npn and pnp.
• The two junctions are termed the base-emitter junction and
the base-collector junction
• The term bipolar refers to the use of both holes and electrons
as charge carriers in the transistor structure
• In order for the transistor to operate properly, the two
junctions must have the correct dc bias voltages
– the base-emitter (BE) junction is forward biased(>=0.7V
for Si, >=0.3V for Ge)
– the base-collector (BC) junction is reverse biased
Architecture of a BJTs

6. • An integrated circuit (IC) consists of
transistors, resistors, diodes and capacitors
combined together in one wafer-thin chip of
silicon. This one wafer-thin chip is called a
microchip. The microchip is only a few
millimeters square with a thickness of 0.5 mm.
IC?

7. Emitter
It is the most heavily doped part of the transistor. Its
major function is to supply the majority charge carriers
to base.
Base
It is the smallest part of the transistor with 10.6mm
area and it is lightly doped.
Collector
It is physically the largest part of the transistor. Its
major function is to collect the charge carriers.
Parts of a Transistor

8. Diode equivalent of a Transistor

9. The minus signs represent free electrons.
The heavily doped emitter has the following job:
to emit or inject its free electrons into the base.
The lightly doped base also has a well-defined
purpose: to pass emitter-injected electrons on to
the collector.
The collector is so named because it collects or
gathers most of the electrons from the base.
The left source VBB of forward-biases the emitter
diode, and the right source VCC reverse-biases the
collector Diode.
VBB forward-biases the emitter diode, forcing the
free electrons in the emitter to enter the base.
The thin and lightly doped base gives almost all
these electrons enough time to diffuse into the
collector. These electrons flow through the
collector, through RC, and into the positive
terminal of the VCC voltage source.
Basic Circuit of a BJT

10. Symbolic Representation

11. • Recall Kirchhoff’s current law. It says that the sum of all
currents into a point or junction equals the sum of all
currents out of the point or junction. When applied to a
transistor, Kirchhoff’s current law gives us this important
relationship:
IE = IC + IB
IC >> IB
IE = IC
• alpha (DC)
IC = DCIE
• beta (DC)
IC = DCIB
– DC typically has a value between 20 and 200.
DC Analysis of BJTs (Transistor Currents)

12. Operation of BJTs

13. Examples

14. DC Analysis of a Transistor
As the base-emitter junction is forward bias so,
VBE ≈0.7 V
Since the emitter is at ground (0V), by Kirchoff’s Voltage law
VRB=VBB – VBE
Also by OHM’S law VRB = IBRB
Substituting for VRB yields IBRB = VBB – VBE
Solving for IB IB = VBB – VBE
RB
The voltage at the collector with respect to the grounded emitter is
VCE = VCC – VRC
Since the drop across RC VRC=ICRC
VCE = VCC – ICRC
Where IC = DCIB
The voltage across the reverse-biased collector-base junction is
VCB = VCE - VBE
DC Analysis of BJTs

15. • DC voltages for the biased transistor:
• Collector voltage
VC = VCC - ICRC
• Base voltage
VB = VE + VBE
– for silicon transistors, VBE = 0.7 V
– for germanium transistors, VBE = 0.3 V
DC Analysis of BJTs

16. Example

17. • The base current, IB, is established by the base bias.
• The point at which the base current curve intersects the
dc load line is the quiescent or Q-point for the circuit.
Q-Point

18. • The voltage divider
biasing is widely used
• Input resistance is:
RIN  DCRE
• The base voltage is
approximately:
VB  VCCR2/(R1+R2)
DC Analysis of BJTs

19. • When used as an electronic switch, a transistor
normally is operated alternately in cutoff and
saturation.
– A transistor is in cutoff when the base-emitter junction is
not forward-biased. VCE is approximately equal to VCC
– When the base-emitter junction is forward-biased and
there is enough base current to produce a maximum
collector current, the transistor is saturated.
BJTs as a Switch

20. BJTs as a Switch

21. BJT Operating Regions
Operation
Region
IB or VCE
Char.
BC and BE
Junctions
Mode
Cutoff IB = Very
small
Reverse &
Reverse
Open
Switch
Saturation VCE = Small Forward &
Forward
Closed
Switch
Active
Linear
VCE =
Moderate
Reverse &
Forward
Linear
Amplifier
Break-
down
VCE =
Large
Beyond
Limits
Overload

22. •Common emitter mode
•Linear Active Region
•Significant current Gain
Example:
•Let Gain,  = 100
•Assume to be in active
region -> VBE=0.7V
•Find if it’s in active region
BJT as Amplifier

23. V
VRIRIVV
mAII
mA
RR
VV
I
IIII
VV
BEEECCCCCB
BC
EB
BEBB
B
BCBE
BE
93.3
7.0)0107.0*101)(2()07.1)(3(10
**
07.10107.0*100*
0107.0
402
7.05
101*
)1(
7.0














VCB>0 so the BJT is in active
region
BJT as Amplifier