Electronics and Communication Engineering

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Electronics - II Transistor Amplifier1
Transistor Amplifiers
 ANALYSIS OFANAMPLIFIER:
(i) Transistor biasing connections must be correct. Base-Emitter should be in forward
bias and Base-Collector should be in reverse bias.
(ii) Operating point „Q‟ should be adjusted at the center of the load line.
(iii)Operating point should be fixed, it should not shift or change with temperature.
 Circuit Description: In the amplifier circuit
1) Transistor is an active device, with amplifies the signal basically by amplifying the
current.
2) Transistor amplifies the signal by drawing current from DC supply (+ ccV ) i.e. energy
is supplied by + ccV .
3) Resistor 1R and 2R provide voltage divider bias to fix the operating point „Q‟ as
shown in the graph
4) Resistor RL is collector load resistor.
5) Resistor RE is connected to stabilize the operating point as explained earlier.
6) Capacitor EC is bypass capacitor, it will bypass AC signal across ER otherwise the
gain of amplifier is reduced.
7) Capacitor C1C and 2CC are coupling capacitors they will couple AC signal and block
DC flow.
 MULTISTAGE AMPLIFIER:
When a large gain is required then single stage transistor amplifier is unable to
produce large gain due to limited power dissipation capacity, and distortion may occur
if transistor is driven beyond saturation or below cut off. Therefore multistage amplifier

is connected; it may be two stage, three-stage amplifier. Fig. shows a three-stage
transistor amplifier and the phase shift at each stage.
The total gain of three-stage amplifier is calculated by a formula.
TotalA = 1 2 3A A A 
If Gain (A) of each stage is 100 and suppose input voltage ( inV ) is u1 V then
TotalA = 100 x 100 x 100
= 2
10 x 2
10 x 2
10  TotalA = 6
10
out Total in
6
6 6 6 0
V A V
= 10 1μV
= 10 1 10 1 10 Volt.
 

    
Refer fig. each transistor produces 0
180 phase shift then total phase shift of there stage
amplifier is
= 180 + 180 + 180
= 0 0 0 0 0
540 (360 180 )OR (0 180 )  
 Output signal is 0
180 out of phase with input signal
 COUPLING METHODS:
Basically coupling circuit is necessary avoid the shift of operating point, to block
DC reverse current flow, and to couple AC signal. Coupling circuit is designed
according to
(i)Frequency response (ii) Impedance matching (iii) Type of input signal.
There are four methods of amplifier coupling

(1) R-C coupling
(2) Impedance Coupling
(3) Transformer Coupling
(4) Direct Coupling
Let us see each type with their advantage, disadvantage and typical application.
(1) R-C Coupling:
The Collector load resistor ( CR ) and the Coupling Capacitor ( CC ) form a coupling
circuit as shown in fig (a). This type of coupling is more popular in audio amplifiers;
it couples AC and blocks DC from CCV to base of the transistor as shown in fig (b)
Capacitor coupling you can observe in voltage amplifier of radio receiver i.e. from
volume control to A.F. amplifier.
 Frequency Response Of R-C Coupling:
The frequency response of RC coupled amplifier is shown by fig. It also shows the
experimental set up for this purpose. Frequency response tells us about the
performance of a circuit against frequency variation from low to high. Frequency
response is essential because it gives an idea about the range of frequency for which
this coupling is suitable.
The graph of output or the gain of amplifier verses frequency is known as frequency
response. A signal generator is connected to apply various frequencies with small
and constant input voltage to a RC coupled amplifier. The output is measured either
on CRO or VTVM and graph is plotted. The curve shows 3 db curve it falls below 1F
and 2F .

The frequency response is flat from Z50H to Z20KHL but it is poor below Z50H
above Z20 KH . The gain of amplifier is falling below Z50H and above Z20KHL . This
curve can be explained as follows.
(i) At lower frequencies below Z50H reactance of coupling capacitor ( CC ) is high
because it is given by
C
1
X
2 fC

Therefore, more voltage drop takes place across it, which reduces the input thus
output is poor. Similarly emitter bypass capacitor EC is less effective at these
frequencies.
(ii) Above Z,20 KH the coupling capacitor behaves as a short circuit because its
reactance is negligible which increases the loading effect on the next stage.
Similarly inter-electrode capacitance from collector to base becomes effective its
equivalent capacitive reactance is low which gives negative feedback and it
affects the gain of amplifier.
 Advantages:
(i) RC coupled amplifier is suitable only for audio frequency signal Z20H to Z20KH .
(ii) Soldering of RC is convenient and cost of the coupling network is least expensive.
(iii) It is useful in record players, sound section of amplifiers etc.
 Disadvantages:
(i) It is not useful for RF signal.
(ii) Impedance matching is poor.
(2) Impedance Coupling:
In RC coupling a power loss ( 2
I R ) takes place across load resistance CR . An inductor
can replace this resistance. Inductor has got high impedance for high frequency.
Therefore large output AC Voltage can be developed across it and since its DC
resistance is very low, DC loss is negligible. This type of coupling is not suitable for
low frequencies it is useful for the high frequencies (RF) above Z20 KH . Impedance
coupling is used in oscillators circuits.
Disadvantages:-
Inductors or RF chokes are expensive and their impedance decreases at low
frequencies. The size of the inductor is also another drawback.

(3) Transformer Coupling:
Transformer coupling is also popular line RC coupling. In radio receivers
transformer coupling is always used in IF amplifiers connecting one IF amplifier to
the second IF amplifier of radio receiver. Transformer coupling serves three
important purposes.
(i) Output impedance of one stage matching with input impedance of the next stage.
(ii) It provides isolation from one stage to next stage because there is no direct
connection between primary and secondary winding of the transformer.
(iii) Output voltage of one stage can be stepped up with a step up transformer.
Transformer coupling also commonly used in RF amplifiers of radio and TV circuits.
Fig. shows that load resistance CR is replaced by primary of transformer there is no
capacitance from secondary to input of the next stage. DC isolation is achieved with
secondary winding itself because DC cannot be transformed from primary to
secondary.
Transformer coupling with shunt capacitor across it is another way of coupling in
amplifier. Such type of amplifier is known as “tuned-RF/(TRF)” amplifier. Here L-C
parallel resonance principle is used which is tuned to desired frequency so the
impedance of L-C circuit becomes maximum only at resonant frequency.
 Advantages:
1) This type of coupling is suitable for RF frequencies, as in radio circuits to amplify
station signals from Z550 KH to Z1600 KH medium wave band.
2) Transformer coupling provides good impedance matching and better gain.
3) Power loss across collector load is minimized up to negligible value.
 Disadvantages:
1) It has poor frequency response i.e. the gain varies with frequency.
2) The size of the transformer and its high cost.
3) Amount of „hum‟ or noise is comparatively high.

(4) Direct Coupling:
In this method there is no coupling component, the collector of one transistor is
directly connected to the base of the next transistor. These types of amplifiers are
used in DC amplification or amplification of very low frequencies below 10Hz. For
example, to drive an active load like DC relay, DC motor, headphones, etc. the main
drawback of such amplifier is, temperature effect It is called as temperature drift it
affects the biasing of transistor.
Fig. shows an example of direct coupling. When light falls on LDR is resistance
decreases therefore more DC voltage is developed across the base resistance BR .
This change in base voltage is amplified in the collector, which drives the next
transistor the collector goes low. The low collector voltage of the first transistor is
applied to the base of second transistor. The low base voltage will make the second
transistor OFF. Thus relay is operated in OFF state. When light is made cut-off the
base voltage decreases which is amplified by the first transistor. Its collector goes
high which is coupled to the second transistor. It will make the second transistor
ON and thus relay is operated into ON state. In this way it couples and amplifies
DC signal.
 Advantages:
i) The coupling circuit is simple because it requires minimum resistors.
ii)It is a low cost method because no coupling component is required.
 Disadvantages:
i) It is not useful for AC high frequency signal.
ii)Operating point is shifted due to temperature.
 POWER AMPLIFIERS:
Power amplification is necessary because the output stage of any instrument required
to be more powerful. It should supply maximum current and maximum voltage (I x V =
Power). Fig. shows the block diagram of an amplifier stage.

The output of voltage amplifier is not capable of driving the loudspeaker because of
heavy load current. Similarly in radio transmitter, TV transmitter circuit, the output of
amplifier is not capable of driving the antenna. Therefore a power amplifier is required
to deliver maximum current with maximum voltage to drive the loudspeaker, antenna
etc.
They are classified into four different types
i) Class A ii) Class AB iii) Class B iv) Class C
i) Class A power Amplifier:
The Class A power amplifier is operated at the center of the load line, as it is
illustrated in fig. the operating point „Q‟ is located at the center of load line.
 Advantages:
The output of Class A amplifier is undistorted and full cycle of input signal is
amplified. Transistor conducts for 360⁰ i.e., it amplifies complete input signal. The
transistor is operated in active region.
 Disadvantage:
Refer the fig. at zero signal condition, when input signal is zero or when input signal
is absent collector current of 100 mA flows through transistor and CEV (on X-axis) it is
8V therefore power loss across the transistor is lossP =8V×100mA=800mW The efficiency
of class A amplifier is only 25%, it is poor efficiency.

ii)Class B power Amplifier:
In class B power amplifier transistor is biased at cut off as illustrated in fig. The
operating point is adjusted at cut off. The result of this, under zero signal condition
transistors is in cut off therefore less wastage of power takes place under zero signal
condition.
 Advantage:-
Since, only half cycle is amplified as shown, during the negative half cycle of input
signal, transistor remains cut off but the efficiency increases about 78.5% because
when signal is zero transistor is OFF. Current through transistor is zero.
 Disadvantage:
Since, the transistor is biased at cut off it conducts only for positive half cycle of input
signal. It remains cut off for negative half cycle as shown. The transistor amplifies
only half cycle and thus extremely high distortion is produced. There is a crossover
distortion.
 Class B Push Pull Amplifier:-
The fig. shows a typical Push Pull Class B amplifier. In Positive half cycle transistor 1T
is ON for 180⁰ and current flows through output transformer. Similarly in negative half
cycle transistor 2T is ON and 1T is cut off it also flows through output transformer and
thus in the secondary side a amplified output is developed.
Push Pull arrangement eliminates much of the distortion but at the midpoint of two
half cycles a distortion occur known as Crossover distortion.
This crossover distortion is produced because of nonlinear characteristics of internal
diode of the transistor at and below 0.6 volt.

iii) Class AB Power Amplifier:
It is the solution to avoid crossover distortion of class B amplifier. It‟s
characteristics lies between class A and class B amplifier. It‟s operating point is
located slightly above cut off as shown in fig., it is in between class A and Class B.
Transistor in class AB conducts more than 180⁰ angle and less than 360⁰. The
efficiency of class AB is more than class A and less than Class B push full amplifier,
commonly used in radio circuit to drive the load speaker.
iv) Class C Power Amplifier:
In this class, transistor is conducting less than 180⁰. It‟s operating point lies
beyond cut off.
 Advantages:
Since transistor is operated for less than 180⁰ efficiency is very high about 100%. It is
used in tuned amplifiers and in oscillators.
 Disadvantage:
Extremely high distortion hence it is not useful for audio amplification. It is used in
oscillator circuits.
 COMPARISON:-
Class A Class B Class AB Class C
Operating
point Q located
At the center At cutoff Before Cutoff Beyond cutoff
Transistor
conducts for
360⁰ 180⁰ More than 180⁰ Less than 180⁰
Distortion Absent Crossover Less than 50% Highest
Efficiency Lowest Better Better Excellent
Application Voltage amplifier Push-pull
amplifier
Push-pull
amplifier
Oscillators

 DIFFERENTIAL AMPLIFIER:-
It is a special type of amplifier capable of amplifying the difference hence called as
difference amplifier or differential amplifier.
It plays very important role in “operational amplifier commonly known as “OPAMP”. It
is also useful in other industrial circuits.
The fig. shows the circuit diagram of differential amplifier, it is a symmetrical structure
in which two identical transistors are connected with equal collector resistors.
The differential amplifier has two inputs:-
(i) Inverting input and
(ii) Non-inverting input.
The basic difference between these two is inverting input produces output 180⁰ out of
phase with input while non-inverting does not produce phase shift between input and
output.
It has two inputs 1V and 2V the output is taken either between two collectors or from
one collector with respect to ground. The fig. (3.14) shows most practical and widely
used configuration, output is taken from collector of 1T with respect to ground.
Differential amplifier it can be operated in different modes such as:-
1) Differential input mode
2) Common input mode
3) Single ended input mode
1) Differential input mode:
In this mode signal is applied between terminal 1V and 2V and output is taken from
the collector. It is called as differential input single ended output mode.
As shown in figure (3.14) output is taken from either of collectors. Its output voltage
is given by equation –
 0 1 2V =A V -V
Since. 1V is 180⁰ output of phase with 2V but equal in amplitude.
  0 1 2 1V =A V -V =2AV
Thus, in differential mode maximum gain is obtained, twice input voltage multiplied
by gain (A).

2) Common input mode:
In this mode same signal is applied to both input terminals that is
 
 
1 2
0 1 2
1 1
V =V then
V =A V -V
=A V -V
=0 it means that for common signals like noise its gain is zero it rejects unwanted noise signals
DM
CM
A Differentialmodegain
CMRR = =
A Common modegain

3) Single ended input mode:
In this mode, input is applied to one of the inputs and other input is grounded,
output can be differential or it can be single ended. When output is single ended it is
called as an OP-AMP. In OP-AMP the mode is single ended input single ended output
or differential input, single ended output.
 TRANSISTOR AS A SWITCH:-
Normally transistor is used as an amplifier where it is operated below saturation and
above cut off; it is operated in active region. Another important application of transistor
in digital circuits, it can be operated as a switch. In switching application the transistor
is operated either is saturation or in cut off only and non in active region. It has got
following advantages over normal mechanical switch.
i) Low cost ii) Small size iii) High speed
Let us see how transistor is used as a switch.
Fig. shows a NPN transistor with emitter grounded. Input voltage is applied to the base
terminal and output is taken from collector ( CEV ). Here transistor is operated either in
saturation or in cut off instead of active region.

SOLVED PROBLEMS
1) What is the value of collector load resistance required to set CV as 4 volts and CI as
1mA for the collector supply voltage of 6 volts in Transistor Amplifier.
CC C C C
CE CC C C
C C CC C
CC C
C
C
-3
3
C
Solution: Given V =6 V, I =1mA R = ? V = 4V
Formula for collector voltageV = V -I R
I R = V - V
V - V
R =
I
6-4
=
1×10
R = 2×10 Ω= 2KΩCollector load should be 2KΩ
2) Find the total voltage gain of a three – stage amplifier in db if first stage has gain of
100 second has 300 and third has 400.
o
i
o
i
o
i
V
Solution: i) First Stage Voltage Gain in db = 20log db
V
= 20log100= 20×2= 40db
V
ii)Second Stage Voltage Gain in db = 20log db
V
= 20log300= 20×2.48= 49.6db
V
iii)ThirdStageVoltageGainin db= 20log db
V
= 20log 400= 20×2.60=52db
Total gain = 40+49.6+52=141db

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