Chap7part2 - Analysis of Bipolar Transistor Amplifiers *...

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Ch. 7 Frequency Response Part 2 1 ECES 352 Winter 2007 Analysis of Bipolar Transistor Amplifiers * Single stage amplifiers Common Emitter ( CE ) Common Base ( CB ) Emitter Follower ( EF ) (Common Collector) * DC biasing Calculate I C , I B , V CE Determine related small signal equivalent circuit parameters h Transconductance g m h Input resistance r π * Low frequency analysis Gray-Searle ( Short Circuit ) Technique h Determine the pole frequencies ω PL1 , ω PL2 , . .. ω PLn Determine the zero frequencies ω ZL1 , ω ZL2 , . .. ω ZLn * High frequency analysis
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Ch. 7 Frequency Response Part 2 2 ECES 352 Winter 2007 CE Amplifier Frequency Analysis - Long and Difficult Way * DC analysis: I C , I B , V CE ; g m , r π * Draw ac equivalent circuit * Substitute hybrid-pi model for transistor * Obtain KVL equations (at least one for each capacitor in circuit (5)) * Solve set of 5 simultaneous equations to obtain voltage gain A V = V o /V s * Put expression in standard form for gain A V (ω) = A Vo F H (ω) F L (ω) * Identify midband gain A Vo * Determine F H (ω) part and factor to Determine high frequency poles ω PH1 and ω PH2 Determine high frequency zeros ω ZH1 and ω ZH2 * L + + + + = 2 1 2 1 1 1 1 1 ) ( P P Z Z H s s s s s F ϖ ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 + + + + + + = + + + + + + = s s s s s s s s s s s s s F P P P Z Z Z P P P Z Z Z L 3 2 1 3 2 1 3 2 1 3 1 1 1 1 1 1 1 1 ) (
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Ch. 7 Frequency Response Part 2 3 ECES 352 Winter 2007 CE Amplifier - Starting Point is DC Analysis * Q is quiescent point (DC bias point) * Q needs to be in the active region I C = β I B * If Q is in saturation ( V CE < 0.3 V ) , then I C < β I B and there is little or no gain from the transistor amplifier * If the transistor is in the cutoff mode, there is virtually no I C so there is no gain, i.e. g m h 0. * Q depends on the choice of R 1 and R 2 since they determine the size of I B . * Q point determines the size of the small signal parameters Transconductance g m = I C / V T V T = k B T/q = 26 mV π m Saturation region Cutoff region Active region Q point
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Ch. 7 Frequency Response Part 2 4 ECES 352 Winter 2007 Example of CE Amplifier - DC Analysis * GIVEN: Transistor parameters: Current gain β = 200 Base resistance r x = 65 Ω Base-emitter voltage V BE,active = 0.7 V Resistors: R 1 =10K, R 2 =2.5K, R C =1.2K, R E =0.33K * Form Thevenin equivalent for base; given V CC = 12.5V R Th = R B = R 1 ||R 2 = 10K||2.5K = 2K V Th = V BB = V CC R 2 / [R 1 +R 2 ] = 2.5V DC Base Current (use KVL base loop) h I B = [V Th -V BE,active ] / [R Th +(β +1)R E ] h I B = 26 μA * DC collector current I C = β I B I C = 200(26 μ A) = 5.27 mA R 1 = 10K R 2 = 2.5K R C = 1.2K R E = 0.33K
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Ch. 7 Frequency Response Part 2 5 ECES 352 Winter 2007 * Construct amplifier’s small signal ac equivalent circuit ( set DC supply to ground ) * Substitute small signal equivalent circuit ( hybrid-pi model ) for transistor * Neglect all capacitances Coupling and emitter bypass capacitors become shorts
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This note was uploaded on 04/28/2010 for the course ECECS 352 taught by Professor Cahay during the Spring '10 term at University of Cincinnati.

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Chap7part2 - Analysis of Bipolar Transistor Amplifiers *...

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