The document discusses bipolar junction transistors (BJTs). It covers: 1) The basic structure of a BJT, which consists of three doped semiconductor regions forming two back-to-back PN junctions. 2) The common NPN and PNP transistor configurations and their symbols. BJTs operate by forward biasing one PN junction and reverse biasing the other. 3) Key characteristics including current gain (beta) and the different regions of operation - cutoff, active, and saturation. BJTs are commonly used as linear amplifiers and switches.

1. BIPOLAR JUNCTION TRANSISTORS (BJT)

2. INTRODUCTION – BASIC BJT
 A Transistor is often called a Bipolar Junction (BJT) or
bipolar transistor. The name bipolar is adopted from
its structure whereby it has ‘two junction’ and consist o
three doped region either ‘n-p-n’ or ‘p-n-p’
combination.

3. BIPOLAR JUNCTION TRANSISTOR
 The bipolar junction transistor is a semiconductor
device constructed with three doped regions.
 These regions essentially form two ‘back- to-back’ p-n
junctionsin the same block of semiconductor material
(silicon).
 The most common use of the BJT is in linear amplifier
circuits (linear means that the output is proportional
to input). It can also be used as a switched (in, for
example, logic circuits).

4. STRUCTURE AND SYMBOLS FOR
BJT
a. 3 Layer semiconductor device consisting:
- 2 n – and 1 p-type layers of material >NPN
transistor
- 2 p- and 1 n-type layers of material > PNP transistor
b. The term bipolar reflects the fact that holes and
electrons participate in the injection process into the
oppositely polarized material.
c. A single PN junction has two different types of bias:
- forward bias
- reverse bias

5. STRUCTURE AND SYMBOLS FOR
BJT
N
P
N
P
N
P

6. PHYSICAL STRUCTURE FOR BJT
 Consist of 3 alternate layers of n- and p-type
semiconductor is called emitter (E) , base (B) and
collector (C).
 Majority of current enters collectors , crosses base
region and exits through emitter. A small current also
enters base terminal, crosses base emitter junction and
exits through emitter.
 Carrier transport in the active base region directly
beneath the heavily doped(n) emitter dominates i-v
characteristics of BJT.

7. TRANSISTOR OPERATION
 The basic operation will be described using the pnp
transistor. The operation of the npn transistor is exactly the
same if the roles played by the electron and hole are
interchanged.
 One p-n junction of a transistor is reverse- biased,whereas
the other is forward-biased.

8. TRNSISTOR OPERATION
 Majority carriers can cross the reverse-biased
junction because the injected majority carriers in
the n-type material.
 Applying KCL to the transistor :
IE = IC + IB
 The comprises of two component – the majority
carriers
IC = Icmajority + ICOminority
 ICO – IC current with emitter terminal open and is
called leakage.

9. EXAMPLE CONNECTION

10. TRANSISTOR OPERATION
 We will consider npn transistors
 Pnp devices are similar but with different polarity of voltage
and currents.
 When using npn transistors.
Collecter is normally more positive than the emitter
VCE might be a few volts
Devices resembles two back – to – back diodes – but has very different
characteristics
With the base open – circuit negligilbe current flows from the
collector to the emitter.

11. BJT OPERATES AN AMPLIFIER AND
SWITCH
 A simple transistor amplifier
- Rb is used to ‘bias’ the transistor by injecting an
appropriate base current
 C is a coupling capacitor and is used to couple the AC
signal while preventing external circuits from affecting
the bias

12.  AC-coupled amplifier
- Vb is set by the conduction voltage of the base-emitter
junction and so is about 0.7V
- voltage across Rb is thus Vcc- 0.7
- this voltage divided by by Rb gives the base current Ib
- the collector current is then given by Ic= hFE Ib
- the voltage drop across Rc is given by IcRc
- the quiescent output voltage is therefore
Vo= Voc- IcRc

13. BJT OPERATES AS AN AMPLIFIER AND SWITCH

14.  Base current is small, so
Vb= Vcc R2/R1+R2= 10 10kΩ/27kΩ+10kΩ =2.7V
Emitter voltage
Ve= Vb- Vbe =2.7-0.7 =2.0V
Since Ib is small,collector current Ic=Ie= 2mA
Output voltage =Vcc-IcRc = 10- 2mA x 2.2kΩ= 5.6V

15.  A common-collector amplifier
- unity gain
- high input resistance
- low output resistance
- a very good buffer amplifier

16.  Regions of BJT operations
- Cut-off region: The transistor is off. There is no conduction
between the collector and the emitter. (Ib= 0 therefore Ic=0)
- Active region: The transistor is on. The output current(Ic) is
relatively insentive to Vce. In this region the transistor can be an
amplifier.
- Saturation region: The transistor is on. The collector current
varies very little with a change in the base current in the
saturation region. The collector crrent is strongly dependt on Vce
unlike the active region. It is desirable to operate transistor
switches in or near the saturation region when in their state.

17.  A common –emitter configuration
- emitter is a common or reference to both input and
output 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
behaviour for CE: input(base terminal) and ouptut
(collector terminal) parameters.

18. Common-emitter configuration
Active region Saturation region Cut-off region
• B-E junction is forward •B-E and C-B junction is •Region below Ib= 0µA is
bias forward bias, thus the to be avoided if an
•C-B junction is reverse value of Ib and Ic is too undistorted o/p signal is
bias big. required
•Can be employed for •The value of Vce is so •B-E junction and C-B
voltage, current, and small junction is reverse bias
power amplification •Suitable region when the •Ib= Iceo where is this
transistor as a logic switch current flow when B-E is
•NOT and avoid this reverse bias.
region when the
transistor as an amplifier.

19. Common-base configuration
Active region Saturation region Cut-off region
•Ie increased,Ic increased •BE and CB junction is •Region below the line of
•BE junction forward bias forward bias Ie= 0A
and CB junction reverse •Small changes in Vcb •BE and CB is reverse bias
bias will cause big different to •No current flow at
•Ic is not depends on Vcb Ic collector,only leakage
•Suitable region for the •The allogation for this current
transistor working as region is to the left of
amplifier Vcb= 0V

20. Common-collector configuration
 also called emitter-follower (EF)
 Is called common-emitter configuration since both signal
source and the load share the collector terminal as a
common connection point
 Output voltage is obtained at emitter terminal
 Input characteristic is similar with common-emitter
configuration
 Provided with the load resistor connected from emitter to
ground
 Primarily for impendance-matching purpose since it has
high impedance an low output impedance

21. DEFINE – β DC AND β AC
 Ratio of dc collector current(IC) to the dc base current
(IB) is dc beta(βdc) which is dc current gain where IC
and IB are determined at a particular operating
point, Q-point(quiescent point).
Ic
 30< βdc<300 2N3904
βdc IB
 On data sheet, βDC = hFE with h is derived from ac
hybrid equivalent cct.FE are derived from forward-
current amplification and common-emitter
configuration respectively.

22.  For ac conditions an ac beta has been defined as the
changes of collector current(Ic) compared to the
changes of base current6(IB).
 On data sheet, βac=hFE

23. Beta(β)
In the dc mode the level of Ic and Ib are related by a
quantity called beta and defined by the following equation:
βdc =Ic/IB
For the ac situation an ac beta has been defined as follows:
βac=∆ Ic/∆IB
β indicates the amplification factor of transistor. (β is
sometimes refer as hfe,a term used in transistor modeling
calculations)

24. Relationship between β and α
Both indicate
an amplification
factor.
α= β/ β +1
β=α/α-1
β provides
a Relationship
between Currents
Ic= β*IB
IE=(β+1)IB

25. Voltage-divider bias
Because the base current is small, the approximatioon
V =[R2/R1+R2] Vcc is useful for calculating the base voltage.
B
After calculating VB, you can find VE by subtracting 0.7V for VBE.
Next, calculate IE by applying Ohm’s law to RE : IE=VE/RE
Then apply the approximation Ic=IE
Finally, you can find the collector voltage from Vc=Vcc-IcRc