Chapter 3 – Transistor Amplifiers – Part 1 Bipolar Transistor Amplifiers

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.

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.

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

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

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

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

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.

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

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

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

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.

Collector Characteristics & Operating Point DC Collector Characteristics Input circuit Output

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

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.

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".

Distorted Output Signal Distortion Free Output

Bias Line & Input Characteristics

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

AC signal Transistor Amplifier

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.

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

Power Amplifier Classes B 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