Chapter 3 – Transistor Amplifiers – Part 1 Bipolar Transistor Amplifiers

Chapter 3 – Transistor Amplifiers – Part 1Chapter 3 – Transistor Amplifiers – Part 1

Bipolar Transistor Amplifiers

1)1) AMPLIFICATION AMPLIFICATION : Amplification is the process of increasing the strength of a SIGNAL (current, voltage, or power signal in a circuit).

2) 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.

1) Transistors are frequently used as amplifiers.

2) 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.

Three configurations in the active region are shown in below.

Biasing Configurations in active region

Biasing & Circuit Design

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

cut-off region and saturation region are not useful regions

Active RegionConditions:Vbe > 0.7 VVce 0.2 V.

Vbe < 0.7 V, hence Ib = 0. Ib = (6.0 – 0.7) V/ RB

Vce < 0.2 VVcc – Ic RL < 0.2 V

Thumb Rule 1) In cut-off region (not useful region) , Vbe < 0.7 V, hence Ib

= 0.2) If Ic/Ib is less than , the transistor is in saturation region.3) We need to ensure that the BJT is not in these regions.

Active Region : BJT has = 50, Find Ib, IC, and RL

1) 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

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.

Operation Operation RegionRegion

IB or VCE

Char. BC and BE BC and BE JunctionsJunctions

ModeMode

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

Operation region summary

RE = 1 k, R1 = 22 k, R2 = 3.3 k, RL = 6 k

Biasing of Transitor

Simple (or Fixed) Bias:

1)The base current Ib = Vcc / (RB + Emitter junction resistance) or Ib = Vcc / RB as the forward emitter junction resistance is small.

2)The collector current Ic = Ib

3) This simple bias circuit is not suitable as the operating point shifts with temperature.

Emitter junction resistance

Voltage Divider Bias

1) The satisfactory transistor bias circuit is obtained by adding the resistor in the emitter circuit.

2) The voltage drop across the RE provide bias to the emitter junction.

3) Voltage divider of R1 and R2 provides voltage to the base.

4) Base-emitter potential is in forward direction.

Collector Characteristics & Operating Point DC Collector Characteristics

Input circuit

Output circuit

(i) 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.

Grounded-emitter collector Characteristics

Load Line & Operating Point

Grounded-emitter collector characteristics

Load line & Operating point

1) The load line cuts through the different Base current values on the DC characteristics curves.

2) Find the equally spaced points along the load line.

3) These values are marked as points N and M on the line.

4) 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.

Base Current for distortion free output signal:

1) 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 distortiondistortion to the output signal output signal in the form of "clippingclipping".

Distorted Output Signal

Distortion Free Output Signal

Bias Line & Input Characteristics

For different value of Vce, the input characteristics Ib as a function of Vbe can be obtained.

Two coordinates are (Vbe, 0) & (0, Veq/Req).Draw a line joining two points

30μA

Vsig

Bias line

AC signal Transistor Amplifier

Frequency response of a typical amplifier

ProcedureProcedure: 1)Apply AC signal voltage to the input (base)2)Measure the output voltage across the collector3)Voltage gain = Output signal voltage/ Input signal voltage4) Vary the frequency, and perform steps 1-3

5) Plot the graph between frequency versus the voltage gain.6) This plotplot is called the frequency response frequency response of the amplifier.7) Use flat region of frequency response of the amplifier for the application.

Type of Signal

Type ofConfiguration Classification Frequency of

OperationSmall Signal

CommonEmitter

Class A Amplifier

Direct Current(DC)

Large Signal

CommonBase

Class B Amplifier

Audio Frequencies (AF)

  Common Collector

Class AB Amplifier

Radio Frequencies (RF)

    Class CAmplifier

VHF, UHF andSHFFrequencies

Classification of Amplifiers

Class A B C ABConductionAngle 360o 180o Less than

90o 180 to 360o

Position ofthe Q-point

Centre Point of theLoad Line

Exactly on theX-axis

Below theX-axis

In betweenThe X-axis And theCentre Load Line

OverallEfficiency

Poor25 to 30%

Better70 to 80%

Higherthan 80%

Better than Abut less than B 50 to 70%

SignalDistortion

None if CorrectlyBiased

At the X-axis Crossover Point

Large Amounts

Small Amounts

Power Amplifier Classes