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

2. 1)AMPLIFICATION 1)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.

3. 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.

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

5. Biasing & Circuit Design Design of Circuit for Amplifier Design of Circuit for Amplifier : Selection of appropriate values of (i) resistances ( R L & R B ), and (ii) voltages (V CC and V BB )

6. cut-off region and saturation region are not useful regions Active Region Conditions: V be > 0.7 V V ce  0.2 V. V be < 0.7 V, hence I b = 0. I b = (6.0 – 0.7) V/ R B V ce < 0.2 V V cc – I c R L < 0.2 V

7. Thumb Rule 1)In cut-off region (not useful region), V be < 0.7 V, hence I b = 0. 2)If I c / I b 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 I b, I C, and R L 1)In active region, V be > 0.7 V. V bb = I b R B + 0.7 6.0V = I b R B + 0.7V or I b = (6.0 – 0.7) V/ R B I c =  I b For Example : Let R B = 1kΩ, then I b = 5.3 V / 1000 Ω = 5.3 mA BJT has  = 50, therefore I c = 5.3 mA x 50 = 265 mA

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

9. Operation Region I B or V CE Char. BC and BE Junctions Mode CutoffI B = Very small Reverse & Reverse Open Switch SaturationV CE = SmallForward & Forward Closed Switch Active Linear V CE = Moderate Reverse & Forward Linear Amplifier Break- down V CE = Large Beyond Limits Overload Operation region summary

10. R E = 1 k , R 1 = 22 k , R 2 = 3.3 k , R L = 6 k  Biasing of Transitor

11. Simple (or Fixed) Bias: 1)The base current I b = V cc / (R B + Emitter junction resistance) or I b = V cc / R B as the forward emitter junction resistance is small. 2)The collector current I c =  I b 3) This simple bias circuit is not suitable as the operating point shifts with temperature. Emitter junction resistance

12. 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 R E provide bias to the emitter junction. 3) Voltage divider of R 1 and R 2 provides voltage to the base. 4)Base-emitter potential is in forward direction.

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

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

16. Grounded-emitter collector Characteristics

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

19. Load line & Operating point

20. 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.

21. 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 distortionoutput signal clipping 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 Signal

23. Bias Line & Input Characteristics

25. For different value of V ce, the input characteristics I b as a function of V be can be obtained. Two coordinates are (V be, 0) & (0, V eq /R eq ). Draw a line joining two points 30μA V sig Bias line

26. AC signal Transistor Amplifier

27. Frequency response of a typical amplifier Procedure Procedure: 1)Apply AC signal voltage to the input (base) 2)Measure the output voltage across the collector 3)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. plotfrequency response 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. Type of Signal Type of Configuration Classification Frequency of Operation Small Signal Common Emitter Class A Amplifier Direct Current (DC) Large Signal Common Base Class B Amplifier Audio Frequencies (AF) Common Collector Class AB Amplifier Radio Frequencies (RF) Class C Amplifier VHF, UHF and SHF Frequencies Classification of Amplifiers

30. ClassABCAB Conduction Angle 360 o 180 o Less than 90 o 180 to 360 o 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 Crossov er Point Large Amounts Small Amounts Power Amplifier Classes