1. Chapter 3 – Transistor Amplifiers – Part 1
Bipolar Transistor Amplifiers

2. AMPLIFICATION : Amplification is the process of increasing the strength of a SIGNAL (current, voltage, or power signal in a circuit).
AMPLIFIER : An amplifier is the device that provides amplification (the increase in current, voltage, or power of a signal) without appreciably altering the original signal.
3) To understand the overall operation of the transistor amplifier, only consider the input current and output current of the transistor.

3. Transistors are frequently used as amplifiers.
Three Types of Amplifiers are :
(i) CURRENT amplifiers, with a small load
resistance,
(ii) VOLTAGE amplifiers have a high load
resistance, and
(iii) POWER amplifiers.

4. Biasing Configurations in active region
Three configurations in the active region are shown in below.

5. Biasing & Circuit Design
Design of Circuit for Amplifier :
Selection of appropriate values of
(i) resistances ( RL & RB ), and (ii) voltages (VCC and VBB)

6. Vce < 0.2 V
Vcc – Ic RL < 0.2 V
Active Region
Conditions:
Vbe > 0.7 V
Vce  0.2 V.
Vbe < 0.7 V, hence Ib = 0.
Ib = (6.0 – 0.7) V/ RB
cut-off region and saturation region are not useful regions

7. Active Region : BJT has  = 50, Find Ib, IC, and RL
Thumb Rule
In cut-off region (not useful region) , Vbe < 0.7 V, hence Ib = 0.
If Ic/Ib is less than , the transistor is in saturation region.
We need to ensure that the BJT is not in these regions.
Active Region : BJT has  = 50, Find Ib, IC, and RL
In active region, Vbe > 0.7 V.
Vbb = Ib RB + 0.7
6.0V = Ib RB + 0.7V
or Ib = (6.0 – 0.7) V/ RB
Ic =  Ib
For Example : Let RB = 1kΩ, then Ib = 5.3 V / 1000 Ω
= 5.3 mA
BJT has  = 50, therefore Ic = 5.3 mA x 50 = 265 mA

8. 2) BJT should not be in saturation region. For active region, Vce  0
2) BJT should not be in saturation region. For active region, Vce  0.2 V.
In the limiting case, Vce = 0.2 V.
If Vce < 0.2 V, BJT will go in saturation region.
Therefore, Vcc – 265mA RL  0.2 V
or RL  (10 –0.2) / 265 mA = 9.8V/265mA
RL  37 Ω.
For Example :
Suppose RL > 37 Ω, say 500 Ω.
Then Ic < 9.8 V /500 Ω = 19.6 mA.
Therefore, The current gain Ic/Ib = 19.6 mA/ 5.3 mA
= 3.7 < active = 50.

9. Operation region summary
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

10. Biasing of Transitor
RE = 1 k, R1 = 22 k, R2 = 3.3 k, RL = 6 k

11. Simple (or Fixed) Bias: The base current Ib = Vcc / (RB +
Emitter junction resistance)
or Ib = Vcc / RB
as the forward emitter junction
resistance is small.
The collector current Ic =  Ib
This simple bias circuit is not
suitable as the operating
point shifts with temperature.
Emitter
junction
resistance

12. Voltage Divider Bias
The satisfactory transistor bias circuit is obtained by adding the resistor in the emitter circuit.
The voltage drop across the RE provide bias to the emitter junction.
Voltage divider of R1 and R2 provides voltage to the base.
Base-emitter potential is in forward direction.

13. Collector Characteristics & Operating Point
DC Collector Characteristics
Input
circuit
Output

15. We have two independent variables here,
(a) base current Ib (input circuit), and
(b) collector to emitter voltage VCE (output
circuit).
(ii) We first study the out put characteristics (i.e.
collector characteristics) for a fixed base
current value.
(iii) Determine collector current, Ic for various
values of collector-emitter voltage, Vce.

16. Grounded-emitter collector Characteristics

17. Load Line & Operating Point
Grounded-emitter collector characteristics

19. Load line & Operating point

20. The load line cuts through the different Base current values on the DC characteristics curves.
Find the equally spaced points along the load line.
These values are marked as points N and M on the line.
A minimum Base current = 20μA and
A maximum Base current = 80μA .
5) These points, N and M can be anywhere along the load line.
6) The operating point Q is a point on the load line which is equally spaced from M & N.

21. Base Current for distortion free output signal:
A minimum Base current = 20μA and
A maximum Base current = 80μA.
2) Difference = (80 – 20) = 60μA
3) So the peak-to-peak Base current without producing any distortion to the output signal = 60/2 = 30μA.
4) Any input signal giving a Base current greater than this value will drive the transistor to go beyond point N and into its Cut-off region or
beyond point M and into its Saturation region thereby resulting in distortion to the output signal in the form of "clipping".

22. Distorted Output
Signal
Distortion Free Output

23. Bias Line & Input Characteristics

25. Draw a line joining two points
For different value of Vce, the input characteristics Ib as a function of Vbe can be obtained.
30μA
Vsig
Bias line
Two coordinates are
(Vbe, 0) &
(0, Veq/Req).
Draw a line joining two points

26. AC signal Transistor Amplifier

27. Frequency response of a typical amplifier
Procedure:
Apply AC signal voltage to the input (base)
Measure the output voltage across the collector
Voltage gain = Output signal voltage/ Input
signal voltage
4) Vary the frequency, and perform steps 1-3
5) Plot the graph between frequency versus
the voltage gain.
6) This plot is called the frequency response of
the amplifier.
7) Use flat region of frequency response of the
amplifier for the application.

29. Classification of Amplifiers
Type of Signal
Type of Configuration
Classification
Frequency of Operation
Small
Signal
Common
Emitter
Class A
Amplifier
Direct Current
(DC)
Large
Base
Class B
Audio Frequencies (AF)
Collector
Class AB
Radio Frequencies (RF)
Class C
VHF, UHF and
SHF Frequencies

30. Power Amplifier ClassesB C AB
Conduction Angle
360o
180o
Less than 90o
180 to 360o
Position of the Q-point
Centre Point of the
Load Line
Exactly on the X-axis
Below the X-axis
In between
The X-axis
And the
Centre Load Line
Overall Efficiency
Poor 25 to 30%
Better 70 to 80%
Higher than 80%
Better than A but less than B 50 to 70%
Signal Distortion
None if Correctly Biased
At the X-axis
Crossover Point
Large
Amounts
Small Amounts