chapter 3 – transistor amplifiers – part 1 bipolar transistor amplifiers
DESCRIPTION
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.TRANSCRIPT
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