TS21_01 v0.1.ppt

Diodes and Transistors Theory Support 21/1
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P-N Junction Diode
Topics covered in this presentation:
 Semiconductor Basics
 The Junction Diode

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Atomic Structure
An atom is the smallest portion of an element that cannot be reduced
without changing the properties of the element.
It consists of a nucleus of neutrons and
positively charged protons, with negatively
charged electrons orbiting around it.
In any atom there are the same number of
electrons as protons, so that it is
electrically neutral.
An element is made up of atoms of the same type. Aluminium, carbon,
potassium, silicon, sulphur and copper are all examples of elements with
common atomic structures.

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Shells
Orbiting electrons have restricted levels around the nucleus that are called
shells. Each shell may be divided into sub-shells and can contain a
maximum number of electrons.
The electrons in a partially
completed outer shell are called
valence electrons and are used for
chemical combinations and
determine electrical properties.
Inert gases have a valence of 0.
The Group of an element is the
same as the number of valence
electrons it has. Group 1 elements
have only one outer electron,
Group 4 has four, and so on.
Valence
8
8 or 18
8, 18 or 32
2
K
L
M
N
Maximum
Electrons
Shell
K
L
M
N
O
P
Q
2
8
8
8 or 18
8 or 18
8 or 18
8, 18
or 32

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Semiconductors
Conductors are Group 1 elements and have one loosely-bound electron in
the outer shell, giving good conductivity and low resistance.
Insulators have many valence electrons and therefore high resistance.
Group 4 elements have middle range resistances (semi-conductors) and
have some other very special properties. Some Group 4 elements are
shown in the table below.
K
Atomic
Number
Element L N
Carbon
Silicon
Germanium
4
8
8 4
2
2 4
32
2
6
14
M
18

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Silicon Atom
Silicon is a readily available element and has good electrical properties.
All semiconductor
elements are Group 4
materials.
Materials such as
germanium, gallium
arsenide, cadmium
sulphide and many others
are used for specialist
devices.
The valence electrons are the most significant part within an atom’s structure.
A diagram showing only the nucleus and valence electrons shows the
structures more simply.

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Electron Sharing
Atoms close together are able to share their valence electrons.
By sharing, the hydrogen atoms
appear to have a completed K shell,
two outer electrons. Figure (a). This
is called covalent bonding.
Group 4 elements have four
valence electrons, so need four
adjacent atoms to share with.
Hydrogen atoms are the simplest atoms, having only one proton and one
electron.
In semiconductors, bonding with an electron from each of four adjacent
atoms gives the appearance of eight electrons in a completed outer shell.
Figure (b).
Electron Sharing
between Hydrogen Atoms
(a)
Electron Sharing
in a Silicon Crystal
(b)

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Crystal Lattice
Group 4 elements, such as
germanium and silicon, readily
form into regular crystals when
cooling from the molten state.
Each atom is surrounded by four
other bonding atoms.
However, even at normal room temperatures, heat energy causes some
covalent bonds to break, releasing electron charge carriers.
There would not appear to be
any free charge carriers to allow
current to flow.

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Impurities
The resistivity of pure silicon is
300kW/cm and germanium is
47W/cm @27°C.
Free charge carriers are present in
pure (intrinsic) semiconductor due
to broken covalent bonds and
unwanted impurities that have not
been removed during refining.
These are called minority carriers.
Impurities are also introduced
intentionally during processing to
give the semiconductor certain
desired properties.
This technique is called doping.

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N-Type Semiconductor
Pure semiconductor can be doped with a Group 5 impurity such as antimony
or arsenic. Four of the element’s five valence electrons fit into the crystal
lattice but the fifth is free.
This is called N-Type
semiconductor since it
contains free electrons
that are Negative charge
carriers.
There is a large increase
in conductivity.
Very little impurity is
needed to give a large
increase in current flow.
Impurity
Atom
Surplus
Electron

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P-Type Semiconductor
If pure semiconductor is doped with a Group 3 impurity such as aluminium or
indium, the crystal lattice will be an electron short for each impurity atom.
The deficiency of one electron per atom creates a hole in the crystal lattice.
This is called P-Type
semiconductor.
The hole is free to migrate
away from its parent atom
and behaves as a Positive
charge carrier.
This results in an
increase in conductivity.
Impurity
Atom
Missing
Electron
or ‘Hole’

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Charge Carriers in N- and P-Type
Semiconductor
The lattice structure is omitted in this diagram for simplicity. Only the free
charge carriers, electrons or holes, are shown and provide the majority
carriers.
Current will flow in either direction if an external EMF is applied.
N-Type P-Type
Minority
Carrier
Minority
Carrier
Majority
Carriers
Majority
Carriers
Free Electrons Holes

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Current Flow in N- and P-Type Semiconductor
In N-Type semiconductor, free electrons will be repelled by the negative
terminal of the EMF and attracted by the positive.
In P-Type semiconductor, there is an excess positive charge due to the holes.
Electrons enter the semiconductor from the negative terminal of the EMF. Inside
the material the electrons move from hole to hole towards the positive terminal.
This makes the holes appear to migrate in the opposite direction.
For each electron released at the positive terminal another hole is produced.
N-Type P-Type
Free Electrons Holes
+ +
Large
Electron
Flow
Large
Electron
Flow

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P-N Junction Diode
The semiconductor P-N junction diode consists of an N-type region joined to
a P-type region.
At the boundary there will be a drift of charge carriers.
Electrons migrate from the N-region toward the P-region and holes move the
other way.
N P
Cathode
(K)
Anode
(A)
K A
Circuit Symbol

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Depletion Layer
Where the charge carriers meet at the junction they combine and disappear
(the number of charge carriers is depleted).
A depletion layer is formed where there are no free charge carriers of either
type.
N P
Cathode
(K)
Anode
(A)
Depletion
Layer

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Barrier Potential
When the electrons leave the N-region they leave their protons behind, giving
a positive potential.
An excess of electrons in the P-region cause a negative potential.
The result is a potential barrier that prevents any further migration of charge
carriers.
Cathode
(K)
Anode
(A)
+

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Reverse Bias
If an external battery is connected across the diode in the same direction as
the potential barrier, then the barrier is strengthened.
Only minority carrier
(leakage) current
can flow.
The diode is said to
be reverse biased.
There is a limit to
the amount of
reverse bias voltage
that the diode can
stand before it
breaks down.
+
+
Small
Electron
Flow
N P
Flow of minority
carriers gives
very small current

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Forward Bias
If the external battery is reversed an unlimited current can flow.
Holes repelled by the
positive region and
attracted by the negative
move toward the junction.
Similarly, electrons
are attracted to the
positive region where
they meet the holes
and recombine.
The depletion layer
closes entirely and
the diode is forward
biased.
+
+
Large
Electron
Flow
N P
Flow of majority
carriers gives
large current

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Forward Bias
Reverse Bias
Diode Conduction
A diode is forward biased when a
positive EMF terminal is
connected to the Anode (P-
region) and a negative terminal
to the Cathode (N-region); P to P
and N to N.
The diode will allow conduction
when connected this way.
A diode will not allow conduction
when a positive EMF terminal is
connected to the Cathode (N-
region) and a negative terminal
to the Anode (P-region).
Anode
(A)
Cathode
(K)
N
P
K
A
ON
_
K
A
OFF
+
_
+

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Half-Wave Rectification and Zener Diodes
 Half-Wave Rectifier
 The Zener Diode
 Simple Power Supply

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Half-Wave Rectifier
Electronic circuits need a DC supply. The function of any rectifier is to turn
the AC supply voltage into DC voltage. This process is rectification.
A diode only allows current to
flow in one direction to give
direct(ional) current, DC.
A single diode rectifier is cheap and simple, but the average output voltage
is low and has a lot of variation known as ripple.
The diode will only allow
the positive half-cycles of
the AC input to flow in the
load. No current flows in
the load during the
negative half-cycles of the
AC input.

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Reservoir Capacitor
A reservoir capacitor is used to store charge to keep the current flowing
when no signal is passing through the diode. A typical reservoir capacitor
value might be 200mF (microfarad).
When the diode conducts, the charge on the capacitor is “topped-up”.
During the negative
half-cycle of the AC
input, when the diode
is not conducting, the
capacitor continues
to supply the load
with current.
The capacitor discharges slightly as it feeds the load with current, so the
voltage across it falls a little (as shown in the waveform diagram).
Load
Reservoir Capacitor

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Power Supply Failures
There are two main categories of equipment failures:
A partial failure could cause a low voltage and too much ripple across the
load, as in the waveform diagram above.
Complete Failure when the equipment fails totally and gives no output.
Partial Failure the equipment operates, but not to specification.
Load
Reservoir Capacitor

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Zener Diode
A Zener diode has a low reverse breakdown voltage (Vz) that can be used
to stabilise an output voltage.
A limiting resistor is required to prevent damage to the Zener diode. Current
through this resistor reduces the supply voltage down to Zener voltage Vz.
The Zener diode is connected in reverse bias so that it will break down at
its ‘Zener’ voltage (Vz).
The power supply voltage can be any value greater than Zener voltage Vz.
Power Supply
Zener
Diode
Limiting Resistor
Vz
Vz
V
I
Reverse
Breakdown

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Zener Diode Ratings
The Zener diode has a maximum power rating, for example, 500mW.
If the Zener voltage is 6V, then
the maximum current that can be
allowed to flow is limited to:
= 83mA.
For a 12V power supply, the
voltage dropped across the
limiting resistor is: 12 - 6 = 6V.
The minimum value of the limiting resistor is: = 75W.
This will keep the Zener current to a safe value when no load is connected.
6V
500mW
80mA
6V
Power Supply
Zener
Diode
Limiting Resistor
Vz

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Zener Voltage Stabiliser
The Zener diode current reduces
by the same value of current that
the load draws.
As the load resistance varies, the
load current varies and the Zener
diode current varies by the same
amount to keep the power supply
current constant.
The total current through the limiting resistor is the sum of the load current
and the Zener diode current. This is constant, so the voltage dropped
across it is constant.
The output voltage across the load is stabilised at Zener voltage Vz for all
values of load current up to the design value (for example, 80mA).
Vz
Power Supply Load
Zener
Diode
Limiting Resistor

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Simple Power Supply
1. The diode rectifies the AC input by removing the negative half-cycles.
3. The Zener diode slices off the ripple and leaves a clean DC output
voltage stabilized at Zener voltage Vz.
2. The reservoir capacitor smoothes the output by charging to the peak
of the AC input and then releasing current to the load when the diode is
not conducting, giving a higher value of average output DC voltage with
a small amount of ripple.
1
2
3
Ave DC

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Light Emitting Diodes - LEDs
Light-emitting principles
Display devices
Characteristics
Protection and limiting

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P
N
P
N
Anode
Cathode
Light Emitting Diode - Basic
Any PN junction diode consists of a P-type
anode and an N-type cathode.
When forward-bias is applied, the free
charge carriers move towards the junction.
When they meet they recombine and light
and heat energy are released as the
mobile electrons drop to a lower energy
level.
Gallium-arsenide (GaAs) base material
gives the greatest light output.
There is no current in reverse bias so
forward bias is necessary for light output.

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Indicator LEDs
The LED “chip” is mounted between the cathode and anode wires.
The cathode is located by a flat on the
cap base.
The “flag” on the cathode carries the
active chip.
The anode usually has a longer wire.
Cathode
Anode
Flat
Long
wire
Lens
Flag
The circuit symbol is similar to an ordinary diode.
Symbol
A plastic lens magnifies, filters and
concentrates the light output.

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Each element can be illuminated independently for the
numbers and the decimal point.
Some letters can also be represented in upper case or in lower case.
Seven-Segment Display
Seven long LED elements and one dot are mounted in a
common package.
The elements have a labelling convention a-g.
Special decoder integrated circuits are needed to drive the elements of
the 7-segment display.

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Bar Graph Indicator
A bar-graph device has many LED bars. It is used to display a varying
voltage signal.
As the input voltage rises more bars light up.
1.5V
Each bar has a reference comparator.
3.0V

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Light Emitting Diode - Characteristic
The characteristics of an LED show that a higher switch-on voltage
(about +2V) is needed than for ordinary silicon diodes (+0.6V).
Red LEDs are the most efficient
requiring the lowest switch-on
voltage.
The required forward voltage is
about 2V.
To give a good light output, around
10mA of current needs to flow
through the LED.
0.5 1.5
1.0 2.0 2.5
5
10
15
20
Voltage
Current (mA)
+2V
Green LEDs are the least efficient
requiring highest switch-on voltage.
Yellow LEDs have an efficiency
between the Green and Red with
a mid-range switch on voltage.

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Need for Series Resistor
If an LED is connected directly across a
power supply, there is nothing to limit the
current flowing.
An unlimited current flowing through the
delicate components within an LED would
cause a large amount of energy to be
released. In this case, the LED would
over-heat and could easily be destroyed.
To prevent the LED being destroyed a
limiting resistor must be placed in series
with the LED to control the current and
prevent damage.

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Design criteria - Supply voltage = +5V , LED current = 10mA.
The value required for the resistor is therefore 3 (V)  10 (mA) = 300W
Calculation of Series Resistor Value
The LED forward voltage can be taken
as approximately 2.0V.
The voltage dropped across the
series protection resistor is
therefore: 5 - 2 = 3V.
The current through the resistor is
10mA, the same as the LED current,
since they are in series.
+5V
10mA
+2V
+3V
10mA
300W

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Transistor Characteristics
 Construction
 Operation
 Current Gain
 Voltage Gain

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Basic Transistor
The bipolar junction transistor (BJT) is a two-junction,
three-layer semiconductor device that is capable of
current, voltage and power amplification.
The three layers provide three electrodes:
Emitter (e) - This region injects charge
carriers into the base region when the
base/emitter junction is forward biased.
Base (b) - Early transistors were built up
on the base material.
Collector (c) - The collector/base junction
has a reverse bias voltage that is the right
polarity to attract (collect) free charge carriers from
the base region. This voltage will be positive for an
NPN transistor or negative for PNP.
Base
(b)
N
N
P
Emitter
(e)
Collector
(c)

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Transistor Types
A three-layer transistor sandwich can have two formats, NPN or PNP.
Each type has its own circuit symbol.
The arrow on the emitter points from P to N, positive to negative and allows
you to identify which type of transistor appears on a schematic diagram.
Collector
(c)
Emitter
(e)
N
P P
Base
(b)
Collector
(c)
(e)
(c)
(b)
NPN (e)
(c)
(b)
PNP
Base
(b)
Emitter
(e)
N N
P

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V
+6V
0V
I =
0.995mA
C
I = 1mA
E
I =5 A
B m
N
P
N
+600mV
0V
c
e
b
BE
V
CE
Transistor Bias (NPN)
The base-emitter junction is forward
biased by +600mV.
Most of the emitter current
flows out via the collector.
Large numbers of negative charge
carriers (electrons) enter the base
region from the emitter.
The positive voltage on the
collector attracts the negative
charge carriers in the base.
There is a current gain ratio
of 995  5 = 199 (IC  IB).
Consider the NPN transistor.

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+6V
0V
I =
1.99mA
C
I = 2mA
E
I = 10 A
B m
N
P
N
+620mV
0V
c
e
b
VBE
VCE
Control of Currents
If the base voltage is increased from 600mV
to 620mV the base current is increased.
Many of the free electrons in
the base region cannot find
holes to re-combine with and
are captured by the higher
positive collector voltage.
More electrons enter the base region from
the emitter.
More electron flow means
that the collector current is
increased.
The ratio remains the same:
1990  10 = 199

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Switching Off
The large collector current
is entirely controlled by the
small base current.
The base voltage and
current are the input and
the collector current is the
output.
+6V
0V
I = 0
C
I = 0
I = 0
B
N
P
N
0V
0V
c
e
b
VBE
VCE
E
With no electrons entering the base region
there are none to be captured by the collector.
If the base - emitter voltage is below 600mV it
will not conduct.

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Voltage Amplifier
A varying input voltage applied to the base
causes the base current to vary.
Varying the base current causes the
collector (output) current to vary.
The varying output current
flows in the Load Resistor
producing a varying output
voltage.
The varying output voltage is much larger than the original input voltage.
The circuit provides a voltage gain.
Load
Resistor
Input
Signal
Voltage
Output
Signal
Voltage

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LDR, Thermistor and Varistor
 LDR
 Thermistor
 Varistor

43 of 10
Cadmium Sulphide disk
Gold Contact Fingers
Contact
Contact
Circuit Symbol
Light Dependent Resistor (LDR)
The resistance of all semiconductors will vary with exposure to light.
Cadmium Sulphide is one of the best variations to operate as a light-to-
electrical transducer.
Two sets of gold contact
fingers are deposited on
the surface of the
crystal.
Conduction between the
fingers increases as light
energy causes the
semiconductor to release
more charge carriers.
This is also known as a photo-conductive cell. The device senses the
ambient light level of an environment and is therefore a Sensor.

44 of 10
1MW
10k
0.1
10W
Resistance
( )
W
Luminance (lux)
LDR Characteristic
Cell resistance is plotted against ambient light level.
At low light levels, cell
resistance is very high
(in the region of several
megohms).
Resistance falls rapidly as
light energy is absorbed by
the semiconductor and
releases extra current
carriers.

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Voltage Divider
A potential divider circuit containing
an LDR is used to convert varying
ambient light into varying output
voltage.
With high light levels the LDR
resistance falls and the output
voltage increases.
At low light levels the LDR has
high resistance resulting in a
low output voltage.
100kW
10V
1V
10V
900kW
10W

46 of 10
Thermistors
Thermally sensitive resistors.
A metal oxide element has
contact electrodes on either side.
The device is encapsulated to
exclude light.
These devices are designed to vary
their resistance with temperature.
Disk (above), plate, rod or bead
types (left) are in production.
S y m b o l
N T C
Contacts
active
disc
element

47 of 10
Negative Temperature Coefficient - NTC
Typical characteristic of
an NTC thermistor.
Negative temperature
coefficient means that, for an
NTC thermistor, the resistance
reduces with temperature.
1kW
5kW
3kW
7kW
2kW
6kW
4kW
8kW
10 20 30 40 50 60 70 80
0
°C

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Positive Temperature Coefficient - PTC
Typical characteristic
of a PTC thermistor.
Positive temperature coefficient
means that, for a PTC thermistor,
the resistance increases with
temperature.
Note the improved linearity over
this temperature range compared
to the NTC thermistor.
W
W
W
W
W
W
W
W
10 20 30 40 50 60 70 80
0
°C

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Varistor Construction
The Varistor is also known as a Voltage Dependent Resistor (VDR) or a
Metal Oxide Varistor (MOV).
Zinc oxide is the
modern active material.
Terminal leads are
soldered to silver
electrodes on each
side of the disk.
Operating voltage varies with the thickness of the active material, since each
granule has a barrier potential in a similar way to all semiconductor materials.
The range of available Varistor voltages is from 10V to 1000V.
Zinc oxide
granules
Solder
contact
Silver
electrodes

50 of 10
Varistor Characteristics
Above the specified voltage the resistance of the Varistor falls off very rapidly,
in a similar way to the Zener diode.
Transient (short duration) currents of
large quantities can be passed
without damage.
Protection is given against excess
voltages of either polarity.
+V
-V
+I
-I
Normal voltage

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Varistor Applications
1. To protect a circuit against
large transients on the line.
Interference “spikes” in excess of 1kV
can often be found on domestic AC
supply lines.
2. To protect a switch or transistor
against inductive back EMF
when switching off.
Currents due to back EMF on switch
off can very easily destroy a transistor,
or break down the insulation and cause
arcing across switch contacts.
V
LOAD
Fuse Switch
Varistor
V V
Varistor Varistor
Switch
DC Supply DC Supply
Transistor

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Transistor Switching Circuits
 Transistor as a Switch
 Darlington Pair Switch
 Light Operated Switch

53 of 9
Transistor as a Switch (OFF)
To operate as a switch the transistor must be either fully ON or fully OFF.
When the base input voltage is 0V
there will be no base current.
With no base current there will be no
collector current.
With no collector current flowing the
lamp is OFF.
+Vcc = 12V
0V
Input
LP1
TR1
b
c
e
Rb

54 of 9
With base current, Ib, flowing, the
transistor will be switched ON and
collector current, Ic, will flow.
With collector current flowing, the
lamp will be ON.
Transistor as a Switch (ON)
If the base input is taken to +12V,
base current will flow.
To limit power dissipation in the transistor, collector current must be
enough to drop all of the supply voltage across the lamp so that the
transistor is fully saturated.
+Vcc = 12V
0V
Input
LP1
TR1
b
c
e
Rb
Ic
Ib

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Transistor Saturation
The rating of the lamp determines the collector current required for saturation:
Current gain hFE of the transistor will
determine the base current required:
This is a minimum value, so the maximum
value of base resistor required is:
The base resistor can be smaller but must not
be larger or the transistor will not saturate.
I
=
P
V
=
1
W
1
2
V
=
8
3
m
A
I
b
=
8
3
m
A
2
0
0
=
4
1
5
A
m
Rb=
V
Ib
=
12V-0.6V
415 A
=
m
27.47kW
12V
1W
27kW
hFE=200
+Vcc = 12V
0V
Input
LP1
TR1
b
c
e
Rb
Ic
Ib

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Base Resistor Too Big
If the base resistor is larger than the required value, then not enough base
current will flow for saturation:
Only half of the supply voltage is
dropped across the lamp and the
transistor does not saturate.
The collector current will be less than
the design value:
Too much (wasted) power is dissipated
in the transistor:
200mA x 200 = 40mA
(Design value of 83mA)
6V x 40mA = 240mW
I
b
=
1
2
V
-
0
.
6
V
5
6
0
0
0
=
2
0
0
A
m +Vcc = 12V
0V
Input
LP1
TR1
b
c
e
Rb
56kW
6V
6V

57 of 9
Darlington Pair
An improved circuit uses two transistors with the emitter current of TR1
being used as the input (base) current for TR2.
If hFE for each transistor is 200 then the total
current gain is:
200 x 200 = 40 000.
Both output (collector) currents are of the same
polarity, so they can be connected together.
For 80mA of output current the input current
needed is:
Darlington Pair transistors are supplied as a single device, needing
only three terminals.
Ib =
80mA
40000
= 2 A
m
b
c
e
TR1
TR2
e
b
Ib2
400 A
m
Ic1 + Ic2
80mA
Ic2
Ic1
Ib1
2 A
m

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Light-Operated Switch
An LDR senses the ambient light level.
The lamp lights when the transistors switch ON.
The diagram shows the variable resistor
VR1 adjusted for <1.2V with threshold
ambient lighting.
The transistors are not switched on,
so the lamp is not lit.
If the ambient light reduces, the LDR
resistance increases and the potential
divider output voltage rises above 1.2V,
switching the transistors ON.
To switch a load of 80mA, 2mA is required at the base. To do this an
input voltage of 1.22V is required to switch on both transistors and
allow for a drop of 20mV across the 10kW base resistor.
Vcc
20kW
10kW
VR1
0.6V
0.6V
1.1V

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Light Operated Switch Light Conditions
With ambient light the LDR resistance is low and the transistor base voltage
is below threshold (1.2V).
Neither transistor will have the
0.6V needed for conduction.
With no collector current flowing,
no current flows through the lamp
so there is no voltage dropped
across it.
The full supply voltage (Vcc) appears
at the transistor collector electrodes.
Vcc
20kW
10kW
VR1
0.55V
<1.1V <1.1V
0V
12V

60 of 9
Light Operated Switch Dark Conditions
With low ambient light the LDR resistance goes high and the transistor
base voltage rises above threshold (to about 1.25V).
Each transistor has about 0.6V
across the base-emitter junction.
The transistors saturate and the
lamp is switched ON.
When saturated, the collector
voltages fall to <0.1V.
Vcc = 12V
20kW
10kW
VR1
0.6V
>1.2V 1.2V
12V
<0.1V

Transistor Amplifiers
 Basic Amplifier
 DC Amplifier
 AC Amplifier

What is an Amplifier?
The function of an amplifier is to provide
an output that is greater than the input.
The main types are:
Current amplifier -
The output current is greater than the input current, but without
gain in voltage.
Voltage amplifier -
Intended to provide voltage gain. Current gain is not important.
Power amplifier -
May increase the amplitude of both voltage and current.
Inverting amplifier -
These give an amplified output 180o
out of phase (inverse) of
the input.

Current Amplifier
The resultant two currents are greater than the input current.
The output can be taken from either
the collector or emitter.
The Bipolar Junction Transistor (BJT) behaves as a current amplifier due
to transistor action between the base current and the collector current.
The emitter current is the sum of the other two.
The input signal is applied to the base.
Base
Collector
Emitter

Voltage Amplifier
The small base current causes a much larger
collector current (Ic) to flow.
The output voltage (Vout) is much bigger
than the input voltage.
The large collector current flowing in the
collector resistor causes a large voltage
(Vc) to be developed across it. Input
Output
The emitter electrode is common to both input and output circuits, so
this is called a Common Emitter Amplifier.
The input signal voltage (Vin) sets up a small base current (Ib).
Vin
Ib
Ic
Vc
Vout

DC Amplifier
The input DC voltage must be greater than 600mV (0.6V) for the transistor
to conduct at all (collector current is zero).
With 620mV input, the base current = 50mA
Collector current = 1mA
Voltage across R = 1.0V (1mA x 1kW)
Output Voltage = 11.0V (12V - 1V)
With 640mV input, the base current = 100mA
Collector current = 2mA
Voltage across R = 2.0V (2mA x 1kW)
Output Voltage = 10.0V (12V - 2V)
Input change = 640 - 600 = 40mV
Output change = 12.0 - 10.0 = 2V
Voltage gain = 2V  40mV = 50
Input
Output
600m
V
50m
A
1m
A
1.0
V
620
mV
100m
A
2m
A
2.0
V
640
mV
0
0
0
R1
1k
Ω
+12
V
12.0V
11.0
V
10.0
V

AC Amplifier - Principle
The input (AC) signal is shown below
the horizontal axis.
This is projected up through the
characteristic to the output.
The Transfer Characteristic shows the relationship
between the Input and the Output.
Note that there is no response at the
output when the input is negative
because the transistor does not conduct.
When the input is positive, the
transistor conducts and the collector
voltage falls due to current through the
load resistor.
Input
Vc
c
Outpu
t

AC Amplifier - Circuit
The output voltage is limited by the value of the
supply voltage Vcc.
The input voltage must be more than
0.6V for any conduction to occur.
For an NPN transistor the input has to
be positive for the transistor to conduct.
As the transistor conducts the output
voltage falls due to the current through
load resistor R. The signal is inverted.
Only the positive half of the input will
give an output. It appears as a
negative output, due to the inversion.
Vc
c
Outpu
t
Vc
c
Inp
ut
R

68 of 15
Transistor Amplifier Biasing and
Troubleshooting
 Current Bias
 Potential Divider Bias
 Troubleshooting

69 of 15
Need for Bias
A simple transistor amplifier will produce a distorted version of the input
signal. To prevent distortion the transistor amplifier output is biased.
If the transistor is not biased, then it will not
conduct during negative half-cycles of the
input signal.
Only the (inverted) positive half-cycles
of the input will appear at the output.
The output is a different shape to the
input and is said to be distorted.
Output
Vcc
R
Vcc
Input

70 of 15
Effect of Bias
Biasing is used to prevent distortion of the
output signal.
The transistor is biased by applying a small,
steady DC input to the base. This input causes
the transistor to conduct slightly. This is known
as the quiescent state.
The input is chosen so that the quiescent output
voltage is set to approximately half the power
supply voltage.
Doing this allows the maximum possible swing
of the transistor output during both positive and
negative half-cycles, without any distortion.
Input
Output
Bias

71 of 15
Effect of Bias
The transfer characteristic shows that a steady
DC bias, represented by the dashed lines, has
been added to the input.
The AC input signal is shown oscillating around
the bias point.
The output is produced by projecting the input
signal through the transistor characteristics.
The bias allows positive and negative half-
cycles to be projected through the
characteristics giving an undistorted signal at
the output of the amplifier.
Input
Output
Bias

72 of 15
Current Bias
The amplifier circuit has to be biased
to produce a quiescent output
voltage that is approximately half
the power supply voltage, Vcc. This
allows the output voltage to have
maximum swing in both positive and
negative half cycles.
With current biasing, resistor Rb is
connected between the supply
voltage, Vcc and the base of the
transistor. This value should be such
that the base current, Ib produces a
collector current, Ic (Ic = Ib x hFE) that
causes a voltage drop of Vcc/2
across Rc.
With Vcc/2 dropped across Rc the quiescent output voltage is:
Vcc - Vcc/2 = Vcc/2
Rc
Rb
Vcc
Input
Output
Ib
Ic
VRc

73 of 15
Current Bias
The biasing of this type of amplifier is
highly dependent on the current gain
(hFE) of the transistor.
If the transistor is replaced with
another transistor that has a higher
hFE, Ib will remain constant but Ic will
increase. This results in the voltage
across Rc increasing and the
quiescent output voltage decreasing.
To return the quiescent output voltage
to Vcc/2, the value of Rb should be
increased to reduce Ib. As a result of
doing this Ic will also decrease, increasing the quiescent output voltage. If
this is done correctly the voltage gain of this circuit will be approximately the
same for any transistor.
Rc
Rb
Vcc
Input
Output
Ib
Ic
VRc

74 of 15
Potential Divider Bias
The current bias circuit gives a high
voltage gain, but with the drawback
that the circuit needs to be re-biased
when the transistor is changed.
The potential divider bias circuit
removes the requirement to re-bias
the amplifier for a different transistor.
Using potential divider biasing results
in the circuit being much less
dependent on the hFE of the transistor.
This is due to the bias circuit being
dependent on the base voltage, Vb,
of the transistor instead of the base
current, Ib. The value of Vb is fixed for
any transistor.
Input
Output
Rb1
Rc
Rb2 Re
Vcc
Ib
Vb

75 of 15
As with the current bias circuit the
quiescent output voltage of the biased
amplifier is required to be approximately
Vcc/2. This is to enable the output
signal to have maximum swing.
From the quiescent output voltage,
the voltage across Rc can be
calculated. The collector current, Ic,
flowing through Rc can be calculated
from this voltage using Ohm’s Law.
Assuming that the collector current,
Ic, is approximately equal to the
emitter current, Ie, the voltage across
Re can be calculated.
Input
Output
Rb1
Rc
Rb2 Re
Vcc
Ic
Ve
Ie

76 of 15
There is approximately 0.6V dropped
across the junction of the base and
emitter. The voltage across Rb2 (Vb)
can be calculated by adding 0.6V to the
voltage across Re.
The base current, Ib, can be
calculated using the formula:
A rule of thumb is used stating that
the current, I2, flowing through Rb2,
should be 10 times Ib. This means
that the current I1, flowing through
Rb1 is given by: I1 = I2 + Ib.
FE
h
Ic
Ib  Vb
Input
Output
Rb1
Rc
Rb2 Re
Vcc
Ic
Ve
Ib
0.6V
I2
I1

77 of 15
Knowing I2 and Vb, the value of Rb2
can be calculated.
The potential divider bias amplifier is
a voltage dependent circuit and can be
used for a range of transistor current
gains without needing to be re-biased.
The value of Rb1 can also be obtained
using I1 and the voltage drop across
Rb1.
The disadvantage of this circuit is that
the voltage gain of the circuit is low.
It is approximately:
Re
Rc
Gain 
Vb
Input
Output
Rb1
Rc
Rb2 Re
Vcc
VRb1
I2
I1

78 of 15
Emitter Decoupling Capacitor
To overcome the problem of low gain
in a potential divider bias amplifier a
decoupling capacitor is used across
the emitter resistor.
The capacitor is specifically chosen
to provide a short circuit across Re at
the operating frequency of the circuit.
This decreases the value of Re in the
voltage gain ratio and therefore
increases the voltage gain of the
circuit.
Output
Rc
Re
Input
Rb1
Rb2
Vcc

79 of 15
Troubleshooting - Quiescent Conditions
Start by finding the base voltage
from the potential divider formula:
The emitter voltage is 0.6V less:
Calculate the emitter current:
To find faults you must know what voltages to expect from a working circuit.
The voltage dropped across the collector load is: VRc = Ic x Rc
So the quiescent collector voltage, Vc, is: Vc = Vcc - VRc
Rb2
+
Rb1
Rb2
Vcc
=
Vb 
Re
Ve
Ie 
0.6
Vb
Ve 

It is assumed that Ic is approximately equal to Ie (Ic  Ie)
Input
Output
Rb1
Re
Rb2
Rc
Vb
Ve
V
Vc
Vcc
Rc

80 of 15
Troubleshooting - Quiescent Conditions
Using the circuit below as an example:
Start by finding the base voltage
from the potential divider formula:
The emitter voltage is 0.6V less:
Calculate the emitter current:
The voltage dropped across collector load is: VRc = 1.15 x 4.7 = 5.4V
So the collector voltage, Vc, is: Vc = 10 - 5.4 = 4.6V
1.75V
10
47
10
10
=
Vb 


1.15mA
1000
1.15
Ie 

1.15V
0.6
1.75
Ve 


It is assumed that: Ic  Ie
Input
Output
Rb1
47kW
Re
1kW
Rb2
10kW
Rc
4.7kW
1.75V
1.15V
5.4V
4.6V
Vcc = 10V

81 of 15
Fault Diagnosis Rules
If a voltage is lower than it should be,
then there may be an open circuit above
(toward Vcc) or a short circuit below
(toward ground).
If two voltages are identical, then there
is probably a short circuit between
them.
If there is more than about 0.8V across a
base-emitter junction, then it is open circuit.
If a voltage is higher than it should be,
then there may be a short circuit above
(toward Vcc) or an open circuit below
(toward ground).
47kW
1kW
10kW
4.7kW
Vcc = 10V
4.6V
1.15V
1.75V

82 of 15
Dynamic Testing
Inject a suitable AC signal at the
input.
Use an oscilloscope to check
for a signal at the output.
If there is no signal, then work
back toward the input to find
where the signal appears.
Complete loss of signal across a
capacitor may be caused by a
fault in the capacitor.
Output
Input
Vcc

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